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The Oxford Solid State Basics

Solutions to Exercises

Steven H. Simon

Oxford University

CLARENDON PRESS .OXFORD

2015

iii

These are the solutions to exercises from the Book The Oxford Solid

State Basics by Steven H. Simon, published by Oxford University Press,

2013 edition. Please do everyone a favor and do not circulate these

solutions. Do not post these solutions on your website. Do not put them

on Russian websites. Do not copy them and hand them out to students.

While there is no way for me to enforce these reasonable rules, be assured

that I, being a professor at Hogwarts, am in possession of powerful hexes

which I have used to protect the secrecy of these solutions. Those who

attempt to circulate these solutions unlawfully will activate the hex and

will suﬀer thirty years of bad luck, including spiders crawling into your

underwear.

Some of these solutions have been tested through use in several years

of courses. Other solutions have not been completely tested. Errors

or ambiguities that are discovered in the exercises will be listed on my

web page. If you think you have found errors in the problems or the

solutions please do let me know, and I will make sure to ﬁx them in the

next version. Doing so will undoubtedly improve your Karma. ,

Steven H Simon

Oxford, United Kingdom

January 2014

Contents

1 About Condensed Matter Physics 1

2 Speciﬁc Heat of Solids: Boltzmann, Einstein, and Debye 3

3 Electrons in Metals: Drude Theory 15

4 More Electrons in Metals: Sommerfeld (Free Electron)

Theory 21

5 The Periodic Table 35

6 What Holds Solids Together: Chemical Bonding 39

7 Types of Matter 47

8 One-Dimensional Model of Compressibility, Sound, and

Thermal Expansion 49

9 Vibrations of a One-Dimensional Monatomic Chain 55

10 Vibrations of a One-Dimensional Diatomic Chain 71

11 Tight Binding Chain (Interlude and Preview) 81

12 Crystal Structure 95

13 Reciprocal Lattice, Brillouin Zone, Waves in Crystals 99

14 Wave Scattering by Crystals 111

15 Electrons in a Periodic Potential 125

16 Insulator, Semiconductor, or Metal 135

17 Semiconductor Physics 139

18 Semiconductor Devices 149

19 Magnetic Properties of Atoms: Para- and

Dia-Magnetism 159

vi Contents

20 Spontaneous Magnetic Order: Ferro-, Antiferro-, and

Ferri-Magnetism 167

21 Domains and Hysteresis 175

22 Mean Field Theory 179

23 Magnetism from Interactions: The Hubbard Model 191

About Condensed Matter

Physics 1

There are no exercises for chapter 1.

Speciﬁc Heat of Solids:

Boltzmann, Einstein, and

Debye 2

(2.1) Einstein Solid

(a) Classical Einstein (or “Boltzmann”) Solid:

Consider a three dimensional simple harmonic oscilla-

tor with mass mand spring constant k(i.e., the mass

is attracted to the origin with the same spring constant

in all three directions). The Hamiltonian is given in the

usual way by

H=p2

2m+k

2x2

Calculate the classical partition function

Z=Zdp

(2π~)3Zdx e−βH(p,x)

Note: in this problem pand xare three dimensional vec-

tors.

Using the partition function, calculate the heat ca-

pacity 3kB.

Conclude that if you can consider a solid to consist

of Natoms all in harmonic wells, then the heat capac-

ity should be 3NkB= 3R, in agreement with the law of

Dulong and Petit.

(b) Quantum Einstein Solid:

Now consider the same Hamiltonian quantum mechan-

ically.

Calculate the quantum partition function

Z=X

e−βEj

where the sum over jis a sum over all eigenstates.

Explain the relationship with Bose statistics.

Find an expression for the heat capacity.

Show that the high temperature limit agrees with

the law of Dulong of Petit.

Sketch the heat capacity as a function of tempera-

ture.

(See also exercise 2.7 for more on the same topic)

(a)

H=p2

2m+k

2x2

Z=Zdp

(2π~)3Zdx e−βH(p,x)

Since, Z∞

−∞

dy e−ay2=pπ/a

in three dimensions, we get

Z=h1/(2π~)pπ/(β/2m)pπ/(βk/2))i3= (~ωβ)−3

with ω=pk/m. From the partition function

U=−(1/Z)∂Z/∂β = 3/β = 3kBT

4Speciﬁc Heat of Solids: Boltzmann, Einstein, and Debye

Thus the heat capacity ∂U/∂T is 3kB.

(b) Quantum mechanically, for a 1d harmonic oscillator, we have

eigenenergies

En=~ω(n+ 1/2)

with ω=pk/m. The partition function is then

Z1d=X

n≥0

e−β~ω(n+1/2)

=e−β~ω/21/(1 −e−β~ω)

= 1/[2 sinh(β~ω/2)]

The expectation of energy is then

01 2

0.25

0.5

0.75

Fig. 2.1 Heat capacity in the Einstein

model (per atom) in one dimension.

Units are kbon vertical axis and kbT/ω

on horizontal. In three dimensions, the

heat capacity per atom is three times

as large.

hEi=−(1/Z)∂Z/∂β = (~ω/2) coth(β~ω/2)

=~ω(nB(β~ω) + 1

where nBis the boson occupation factor

nB(x) = 1/(ex−1)

(hence again the relationship with free bosons). The high temperature

limit gives nB(x)→1/(x+x2/2) = 1/x−1/2 so that hEi → kBT. More

generally, we obtain

C=kB(β~ω)2eβ~ω

(eβ~ω−1)2

In 3D,

En1,n2,n3=~ω[(n1+ 1/2) + (n2+ 1/2) + (n3+ 1/2)]

and

Z3d=X

n1,n2,n3≥0

e−βEn1,n2,n3= [Z1d]3

and correspondingly

hEi= 3~ω(nB(β~ω) + 1

So the high temperature limit is hEi → 3kBTand the heat capacity

C=∂hEi/∂T = 3kB. More generally we obtain

C= 3kB(β~ω)2eβ~ω

(eβ~ω−1)2

Plotted this looks like Fig. 2.1.

(2.2) Debye Theory I

(a)‡State the assumptions of the Debye model of heat

capacity of a solid.

Derive the Debye heat capacity as a function of

temperature (you will have to leave the ﬁnal result in

terms of an integral that cannot be done analytically).

From the ﬁnal result, obtain the high and low tem-

perature limits of the heat capacity analytically.

You may ﬁnd the following integral to be useful

Z∞

dx x3

ex−1=

∞

n=1 Z∞

x3e−nx = 6

∞

n=1

n4=π4

By integrating by parts this can also be written as

Z∞

dx x4ex

(ex−1)2=4π4

15 .

(b) The following table gives the heat capacity Cfor

potassium iodide as a function of temperature.

T(K) C(J K−1mol−1)

0.1 8.5×10−7

1.0 8.6×10−4

5 .12

8 .59

10 1.1

15 2.8

20 6.3

Discuss, with reference to the Debye theory, and

make an estimate of the Debye temperature.

(a) The key assumption of Debye theory is that the dispersion curve

is linear (ω=vk) up to a cut-oﬀ frequency ωDebye determined by the

requirement that the total number of vibrational modes is correct.

For a crystal containing Natoms, the low temperature limiting form

C=12NkBπ4

5T

TD3

(2.1)

and the high temperature limit is 3N kB. Here, TD=~ωDebye/kB.

The full derivation goes as follows. For oscillators with frequency ω(k)

a system has a full energy

E=L3Zd3k(2π)3~ω(k)[nB(β~ω(k)) + 1/2]

One includes also a factor of 3 out front to account for the three diﬀerent

sound modes (two transverse and one longitudinal) and we cut oﬀ the

integral at some cutoﬀ frequency ωcutof f . We use the assumption that

ω=v|k|although it is not much harder to consider three diﬀerent

velocities for the three diﬀerent modes. We thus obtain

E=Zωcutoff

dωg(ω)[nB(β~ω) + 1/2]~ω

where

g(ω) = N12πω2

(2π)3nv3=N9ω2

ω3

and we have replaced nL3=Nwhere nis the density of atoms. Here

ω3

d= 6π2nv3is the Debye frequency, and ~ωd=kBTDebye deﬁnes the

Debye temperature. Note that there is no dependence of g(ω) on the

density n(it cancels). This shows that until the cutoﬀ is imposed, there

6Speciﬁc Heat of Solids: Boltzmann, Einstein, and Debye

is actually no knowledge of the underlying lattice — only the overall

volume and sound velocity.

We should choose the cutoﬀ frequency such that we have the right

number of modes in the system, thus we have

3N=Zωcutoff

dωg(ω)

performing this integral, we ﬁnd that the proper value of ωcutof f is

exactly the Debye frequency ωdthat we just deﬁned.

The general heat Debye theory heat capacity will then be

C=dhEi/dT =kB

(kBT)2Zωd

dωg(ω)(~ω)2eβ~ω

(eβ~ω−1)2

Deﬁning x=~ω/kBTwe obtain

C=dhEi/dT =NkBT

TDebye 3

9Z~ωd/kBT

dxx4ex

(ex−1)2

This integral is known as the Debye integral. In the low temperature

limit, we can extend the integral out to inﬁnity whereupon it just gives

the constant 4π4/15 recovering the above claimed result Eq. 2.1.

In the high temperature limit, the exponents can be expanded such

that the Debye integral becomes

Z~ωd/kBT

dxx4ex

(ex−1)2=Z~ωd/kBT

dxx2= (1/3)(~ωd/kBT)3

which then recovers the law of Dulong-Petit C= 3NkB

(b) Given the heat capacity and the temperature, in the low Tlimit

we should have (from Eq. 2.1)

TD=12Rπ4T3

5C1/3

The table of heat capacity looks like

T(K) 0.1 1.0 5 8 10 15 20

C(J K −1mol −1) 8.5×10−78.6×10−41.2×10−15.9×10−11.1 2.8 6.3

12Rπ4T3

5C1/3(K) 132 131 127 119 121 132 135

So TDebye is about 130K. The fact that the T3ﬁt is not perfect is a

reﬂection of (a) that Debye theory is just an approximation (in particular

that phonons have a nonlinear spectrum!) and (b) that one needs to be

in the low Tlimit to obtain perfect T3scaling. (Note that at low enough

T, the T3scaling does indeed work).

(2.3) Debye Theory II

Use the Debye approximation to determine the heat

capacity of a two dimensional solid as a function of tem-

perature.

State your assumptions.

You will need to leave your answer in terms of an inte-

gral that one cannot do analytically.

At high T, show the heat capacity goes to a con-

stant and ﬁnd that constant.

At low T, show that Cv=KT nFind n. Find Kin

terms of a deﬁnite integral.

If you are brave you can try to evaluate the integral,

but you will need to leave your result in terms of the

Riemann zeta function.

In 2d there should be 2N modes. So high Theat capacity should be

C= 2kbN(Law of Dulong-Petit).

Assume longitudinal and transverse sound velocities are equal.

2N= 2AZ|k|=kDebye

d2k

(2π)2=2(πk2

Debye

(2π)2

with Athe area. So

kDebye =√4πn

with n=N/A the density. So ΘDebye =~kDebyecwith cthe sound

velocity.

Since phonons obey bose statistics we have

E= 2AZ|k|=kDebye

d2k

(2π)2ǫknB(βǫk)

= 2AZ|k|=kDebye

d2k

(2π)2~ck 1

eβ~ck −1

= 2A2π

(2π)2Z|k|=kDebye

k dk ~ck 1

eβ~ck −1

πZΘDebye

dǫ

~cǫ1

eβǫ −1

Let z=βǫ =ǫ/(kbT) and we get

E=A(kbT)3

π~2c2ZΘDebye /(kbT)

z2dz

ez−1

For large T, Θ/T is small so zis small, so

z2dz

ez−1=z

so we get ZΘDebye /(kbT)

zdz = (ΘDebye/(kbT))2/2

so in this limit

E=A(kbT)Θ2

Debye

2π~2c2=Ak2

DebyekbT

2π=A(4πN/A)kbT

2π= 2NkbT

8Speciﬁc Heat of Solids: Boltzmann, Einstein, and Debye

which gives

C=dE/dT = 2Nkb

as expected.

For small T, the upper limit of the integral goes to inﬁnity and we

have

E=A(kbT)3

π~2c2Z∞

z2dz

ez−1

Cv=KT 2

where

K=3Ak3

π~2c2Z∞

z2dz

ez−1

To evaluate the integral we have

Z∞

z2dz

ez−1=Z∞

z2dz

−z∞

n=0

e−nz

∞

n=1 Z∞

dzz2e−nz =

∞

n=1

2/n3= 2ζ(3)

Thus we obtain

K=6Ak3

bζ(3)

π~2c2

(2.4) Debye Theory III

Physicists should be good at making educated guesses:

Guess the element with the highest Debye temperature.

The lowest? You might not guess the ones with the abso-

lutely highest or lowest temperatures, but you should be

able to get close.

Largest Debye temperature should be the one with the highest speed

of sound which is probably the hardest element (ie., highest spring con-

stant) and/or smallest mass. Diamond is the obvious guess (and indeed

it does have the highest Debye temperature). ΘDebye = 2230K. The

lowest is harder to guess. One presumably wants a soft material of some

sort – also possibly a heavy material.

Material ΘDebye

Neon 75 K

Argon 92 K

Krypton 64 K

Xenon 64 K

Radon 64 K

Mercury 69 K

Potassium 91 K

Rubidium 56 K

Cesium 32 K

Some Low Debye Temperatures

Soft and heavy metals like mercury are good guesses. (in fact mercury

is liquid at room temperature and one has to go to low T to measure

a Debye temperature). Also good guesses are Noble gases where the

spring constant is very low (weak interaction between the atoms). Also

heavy soft group 1 metals are good guesses. Many of these are gas or

liquid at room T and a Debye temeperature can only be measured at

low T.

(2.5) Debye Theory IV

From Fig. 2.3 (main text) estimate the Debye temper-

ature of diamond. Why does it not quite match the result

listed in Table 2.2 (main text)?

Extracting the slope from the ﬁgure gives C/T 3≈1.9×10−7J/(mol −K4)

Then using the formula

C=12NkBπ4

5T

TD3

We obtain

TD≈2200K

The reason that it does not match the Debye temperature given in

the ﬁgure caption has to do with the comment in the caption. Debye

theory predicts the heat capacity at all possible temperatures. The

Debye temperature quoted in the text is chosen so as to give a good ﬁt

over the full temperature range. The Debye temperature measured here

is chosen to give a good ﬁt at the lowest temperatures (where Debye

theory can actually be exact).

(2.6) Debye Theory V*

In the text we derived the low temperature Debye heat

capacity assuming that the longitudinal and transverse

sound velocities are the same and also that the sound ve-

locity is independent of the direction the sound wave is

propagating.

(a) Suppose the transverse velocity is vtand the lon-

gitudinal velocity is vl. How does this change the Debye

result? State any assumptions you make.

(b) Instead suppose the velocity is anisotropic. For ex-

ample, suppose in the ˆx, ˆyand ˆzdirection, the sound ve-

locity is vx, vyand vzrespectively. How does this change

the Debye result?

(a) This is actually quite simple. The derivation of the heat capacity

follows the text (or exercise 2.1). The only diﬀerence is in the density

of states. In the isotropic calculation we use

g(ω) = N12πω2

(2π)3nv3

Recall the origin of these factors. Really we had (See Eq. 2.3 of the

main text)

g(ω) = 3L34πω2

(2π)3v3

where the factor of 3 out front is for the three polarizations of the sound

waves. One could just as well have written it as

g(ω) = L34πω2

(2π)31

v3+1

v3

10 Speciﬁc Heat of Solids: Boltzmann, Einstein, and Debye

separating out the three diﬀerent polarizations. Now, if the three polar-

izations have three diﬀerent velocities, we have

g(ω) = L34πω2

(2π)31

3

this is true since the density of states of the three diﬀerent excitation

modes simply add. In an isotropic solid, the two transverse mode have

the same velocity vtand the one longitudinal mode has velocity vland

we would have

g(ω) = L34πω2

(2π)32

l

The remainder of the derivation is unchanged. Thus, deﬁning ¯vsuch

that 3

¯v3=2

we obtain the low temperature capacity in the usual form

C=12NkBπ4

5T

TD3

where now

(kBTD)3= 6π2n~3¯v3

Note that the high frequency cutoﬀ is diﬀerent for the two types of modes

(but the kcutoﬀ is the same for both modes).

(b) If instead we have three diﬀerent sound velocities in three diﬀerent

directions, the situation is more complicated (and here we neglect the

diﬀerences between longitudinal and transverse modes). Here we must

make some assumption about the sound velocity in some arbitrary di-

rection. A reasonable guess would be as follows. If you consider a sound

wave in direction ˆ

k(with ˆ

k=k/|k|a unit vector), we would have

v(ˆ

k) = qv2

xˆ

x+v2

yˆ

y+v2

zˆ

Now, following the usual derivation of Debye theory, we start with

hEi= 3 L3

(2π)3Zdkxdkydkz~ω(k)nB(β~ω(k)) + 1

2.

And now

ω(k) = v(ˆ

k)|k|=qv2

xk2

x+v2

yk2

y+v2

zk2

Since the system is now not isotropic, we cannot do the usual thing

and convert to spherical polar coordinates directly. Instead, we rescale

the axes ﬁrst writing (with j=x, y, z)

Kj=kjvj

So that

ω(K) = v(ˆ

k)|k|=qK2

x+K2

y+K2

and

hEi= 3 L3

(2π)3vxvyvzZdKxdKydKz~ω(K)nB(β~ω(K)) + 1

2.

We can now use spherical symmetry to obtain

hEi= 3 4πL3

(2π)3vxvyvzZ∞

ω2dω(~ω)nB(β~ω) + 1

2.

(2.2)

The rest of the derivation follows as usual to give the usual expression

for heat capacity

C=12NkBπ4

5T

TD3

where now

(kBTD)3= 6π2n~3vxvyvz.

(2.7) Diatomic Einstein Solid*

Having studied exercise 2.1, consider now a solid made

up of diatomic molecules. We can (very crudely) model

this as a two particles in three dimensions, connected to

each other with a spring, both in the bottom of a har-

monic well.

H=p12

2m1

+p22

2m2

2x12+k

2x22+K

2(x1−x2)2

Here kis the spring constant holding both particles in the

bottom of the well, and Kis the spring constant holding

the two particles together. Assume that the two particles

are distinguishable atoms.

(For this problem you may ﬁnd it useful to transform to

relative and center-of-mass coordinates. If you ﬁnd this

diﬃcult, for simplicity you may assume that m1=m2.)

(a) Analogous to exercise 2.1 above, calculate the clas-

sical partition function and show that the heat capacity

is again 3kBper particle (i.e., 6kBtotal).

(b) Analogous to exercise 2.1 above, calculate the quan-

tum partition function and ﬁnd an expression for the heat

capacity. Sketch the heat capacity as a function of tem-

perature if K≫k.

(c)** How does the result change if the atoms are in-

distinguishable?

(a) We can write the partition function as

Z=Zdp1

(2π~)3

dp2

(2π~)3Zdx1dx2e−βH

Considering the momentum integrals ﬁrst, we have

Zdpe−βp2/(2m)=2πm

β3/2

Then the spatial integrals are made simple by transforming

Y=x1−x2

y= (x1+x2)/2

So the spatial integrals are

ZdYdye−β(−ky2−(k/4+K/2)Y2)=π

kβ 3/2π

β(k/4 + K/2)3/2

12 Speciﬁc Heat of Solids: Boltzmann, Einstein, and Debye

Putting these together we get a partition function

Z=π

kβ 3/2π

β(k/4 + K/2)3/22πm1

β3/22πm2

β3/2

∼β−6

The energy is

hEi=−∂ln Z/∂β = 6/β

So the heat capacity for the two particles is

C=∂hEi/∂T = 6kB

(b) The case where the two masses are identical is fairly simple. Again,

we construct

Y=x1−x2

y= (x1+x2)/2

and correspondingly

Q= (p1−p2)/2

q= (p1+p2)

Note that these two variables are constructed so that [Qj, Yk] = i~δjk

and [qj, yk] = i~δjk and all other commutators are zero (in other words,

these are canonical conjugates). The Hamiltonian is now written as

H=Q2

2(m/2) +q2

2(2m)+kY2+k

4+K

2y2

which comprises two independent three-dimensional harmonic oscillators

with frequencies

ω1=p4k/m

ω2=p(k/2 + K)/(2m)

The heat capacity is then (analogous to 2,1)

C= 3kB(β~ω1)2eβ~ω1

(eβ~ω1−1)2+ 3kB(β~ω2)2eβ~ω2

(eβ~ω2−1)2(2.3)

The case of unequal masses is more tricky. The general method is

similar to the discussion outlined in problem *** below. First, rescale

ri=pi√miand xi=zi/√miso that the Hamiltonian reads

H=r12

2+r22

2+k

2m1

z12+k

2m2

z22+K

2(z1/√m1−z2/√m2)2(2.4)

Note that rand zare canonically conjugate just like pand x. The

potential can be viewed as a matrix which we can write as

(k+K)/m1−K/√m1m2

−K/√m1m2(k+K)/m2

We can diagonalize this matrix to deﬁne two new decoupled degrees of

freedom representing two independent harmonic oscillators. The fre-

quencies of these oscillaltors squared are the eigenvalues of the above

matrix.

ω1=(k+K)(m1+m2) + p(k+K)2(m1−m2)2+ 4K2m1m2

2m1m2

ω2=(k+K)(m1+m2)−p(k+K)2(m1−m2)2+ 4K2m1m2

2m1m2

And the heat capacity is given by formula 2.3 using these two oscillator

frequencies. A plot is given in Fig. 2.2.

246 8 10 T

Ck_B

Fig. 2.2 Heat capacity of two einstein

oscillators. Here ω1= 1 and ω2= 10

(d) If the two atoms are indistinguishable then they must obey either

Bose or Fermi statistics depending on the atom type. The center of

mass degree of freedom (yabove) has the same Einstein heat capacity as

calculated before. However, the relative degree of freedom does not. Due

to the statistics, the relative wavefunction must obey Ψ(Y) = ±Ψ(−Y)

with the ±depending on whether we have bosons or fermions. Since

the three dimesional harmonic motion wavefunction can be decomposed

into three one-dimensional wavefunctions Ψ(r) = ψnx(x)ψny(y)ψnz(z),

and ψnis symmetric or antisymmetric depending on whether nis even

or odd, we must have that

nx+ny+nz= even for bosons, odd for fermions

So when we write the partition function for this oscillator, instead of

nx,ny,nz≥0

e−β~ω(nx+ny+nz+3/2)

as usual, we instead only include the terms of sum respecting the even/odd

symmetry. This restriction can be handled by writing

nx+ny+nz=even/odd →1

nx,ny,nz≥0

(1 ±(−1)nx+ny+nz)

The sum can then be evaluated to give a partition function

Zbose/fermi =e−3

2β~ω2"1

1−e−β~ω2

±1

1 + e−β~ω3#

which then can be diﬀerentiated to get the heat capacity. I obtained

C=kb

24e2ω/kbT1 + 2e2ω/kbT+ 5e4ω/kbT

1 + 2e2ω/kbT−3e4ω/kbT2(kbT)2

for the fermi case and

C=kb

24e2ω/kbT5 + 2e2ω/kbT+e4ω/kbT

−3 + 2e2ω/kbT+e4/kbT2(kbT)2

for the bose case. Note that both of these have the correct Dulong-Petit

high temperature limit of 3kb.

14 Speciﬁc Heat of Solids: Boltzmann, Einstein, and Debye

(2.8) Einstein Versus Debye*

In both the Einstein model and the Debye model the

high temperature heat capacity is of the form

C=NkB(1 −κ/T 2+...)

For the Einstein model calculate κin terms of the

Einstein temperature.

For the Debye model calculate κin terms of the

Debye temperature.

From your results give an approximate ratio

TEinstein/TDebye. Compare your result to the values for

silver given in Fig. 2.4 (main text). (The ratio you cal-

culate should be close to the ratio stated in the caption

of the Figure. It is not exactly the same though. Why

might it not be?).

Expanding the heat capacity of a single Einstein oscillator

C=kb(β~ω)2eβ~ω

(eβ~ω−1)2

∼kb(β~ω2)1 + β~ω+ (β~ω)2/2

[β~ω+ (β~ω)2/2 + (β~ω)3/6]2

∼kb

1 + β~ω+ (β~ω)2/2

[1 + (β~ω)/2 + (β~ω)2/6]2

∼kb

1 + β~ω+ (β~ω)2/2

[1 + (β~ω)/2 + (7/12)(β~ω)2]

∼kb1 + β~ω+ (β~ω)2/2[1 −(β~ω)/2 + (5/12)(β~ω)2]

∼kb1−(β~ω)2/12 + ...

So κ=T2

Einstein/12.

We can handle the Debye case by realizing that the heat capacity is

just an integration over Einstein oscillators. So

C=ZωD

dωg(ω)kb(β~ω)2eβ~ω

(eβ~ω−1)2

∼kbZωD

dωg(ω)1−(β~ω)2/12 + ...

where g(ω) = N9ω2/ω3

D. Note that the integration is cut oﬀ so that the

integral over 1 gives precisely 3Nkbas it should. Thus we obtain

C∼3kBN1−(β~ωD)2/20 + ...

So κ=T2

Debye/20. Setting T2

Einstein/12 = T2

Debye/20 we would predict

TDebye =p5/3TEinstein ≈1.29TEinstein

In the data from Fig 2.4 the ratio is 215/151 ≈1.42. The reason this

does not match perfectly with our prediction is mainly because TDebye

and TEinstein are likely ﬁt over the full range of the heat capacities

measured, not just in the high temperature limit. If they were ﬁt pa-

rameters for only the high temperature limit, the ratio would come out

as we predicted here.

Electrons in Metals: Drude

Theory 3

(3.1) Drude Theory of Transport in Metals

(a)‡Assume a scattering time τand use Drude theory

to derive an expression for the conductivity of a metal.

(b) Deﬁne the resistivity matrix ρ

eas E=ρ

ej. Use

Drude theory to derive an expression for the matrix ρ

efor

a metal in a magnetic ﬁeld. (You may assume Bparallel

to the ˆzaxis. The under-tilde means that the quantity ρ

is a matrix.) Invert this matrix to obtain an expression

for the conductivity matrix σ

Estimate the magnitude of the Hall voltage for a

specimen of sodium in the form of a rod of rectangular

cross-section 5mm by 5mm carrying a current of 1A down

its long axis in a magnetic ﬁeld of 1T perpendicular to

the long axis. The density of sodium atoms is roughly

1 gram/cm3, and sodium has atomic mass of roughly 23.

You may assume that there is one free electron per sodium

atom (sodium has valence 1).

What practical diﬃculties would there be in mea-

suring the Hall voltage and resistivity of such a specimen.

How might these diﬃculties be addressed).

(d) What properties of metals does Drude theory not

explain well?

(e)* Consider now an applied AC ﬁeld E∼eiωt which

induces an AC current j∼eiωt. Modify the above calcu-

lation (in the presence of a magnetic ﬁeld) to obtain an

expression for the complex AC conductivity matrix σ

e(ω).

For simplicity in this case you may assume that the metal

is very clean, meaning that τ→ ∞, and you may assume

that E⊥B. You might again ﬁnd it convenient to as-

sume Bparallel to the ˆzaxis. (This exercise might look

hard, but if you think about it for a bit, it isn’t really

much harder than what you did above!)

At what frequency is there a divergence in the con-

ductivity? What does this divergence mean? (When τis

ﬁnite, the divergence is cut oﬀ.)

Explain how could one use this divergence (known

as the cyclotron resonance) to measure the mass of the

electron. (In fact, in real metals, the measured mass of

the electron is generally not equal to the well-known value

me= 9.1095×10−31 kg. This is a result of band structure

in metals, which we will explain in Part VI.)

(a) We consider an electron with momentum pat time tand we ask

what momentum it will have at time t+dt. There is a probability dt/τ

that it will scatter to momentum zero. If it does not scatter to momen-

tum zero (with probability 1 −dt/τ) it simply accelerates as dictated by

its usual equations of motion dp/dt =FThus

hp(t+dt)i=1−dt

τ(p(t) + Fdt)

or dp

dt =F−p

τ(3.1)

where here the force Fon the electron is just the Lorentz force

F=−e(E+v×B)

16 Drude Theory

In absence of magnetic ﬁeld

dt =−eE−p

In steady state, dp/dt = 0 so we have

mv=p=−eτE

with mthe mass of the electron and vits velocity. If there is a density

nof electrons in the metal, and they are all moving at velocity vthen

the electrical current is given by

j=−env=e2τn

or in other words, the conductivity of the metal is

σ=e2τn

m(3.2)

(b) In both an electric and a magnetic ﬁeld

dt =−e(E+v×B)−p/τ

Again setting this to zero in steady state, and using p=mvand j=

−nev, we obtain an equation for the steady state current

0 = −eE+j×B

n+m

neτ j

E=1

nej×B+m

ne2τj

We now deﬁne the 3 by 3 resistivity matrix ρ

ewhich relates the current

vector to the electric ﬁeld vector

E=ρ

We then obtain components of this matrix

ρxx =ρyy =ρzz =m

ne2τ

and if we imagine Boriented in the ˆzdirection, then

ρxy =−ρyx =B

Inverting this equation we obtain a conductivity matrix

σzz =ne2τ/m

σxx =σyy =ρxx/(ρ2

xx +ρ2

xy) = σzz /[1 + (eBτ/m)2]

σyx =−σxy =ρxy/(ρ2

xx +ρ2

xy) = σzz (eBτ/m)/[1 + (eBτ/m)2]

with all other entries in the σmatrix being zero.

theory. If sodium n= 1 gm /cm3with atomic mass M= 23, this is then

a density of atoms of

n=NA×n/M = 2.6×1028m−3

hence the same density of electrons assuming one free electron per atom.

The cross section of the rod is Lby Lwith L= 5mm, so the current

density is j=I/L2. The Hall resistivity is ρxy =B/(ne) so the Hall

voltage is jρxyL. So the total Hall voltage is

V=IB

Lne = 4.8×10−8Volts

Some of the problems with making this measurement might be:

•This is a very small voltage: One needs a sensitive voltmeter

•There may be contact resistance: Use a high impedance voltmeter

•Contacts may not be perfectly aligned: Try varying (reversing) the

magnetic ﬁeld to pick out only the Bdependent part (I.e., measure

(V(B)−V(−B))/2).

•Could have heating

Tutors might also use this problem as an opportunity to discuss how

useful lock-in ampliﬁers are (which most students do not appear to

know).

(d) Drude theory fails to explain why the electrons do not carry heat

capacity of 3/2kBper electron as a classical gas would. This results in

incorrect predictions of, ex, thermoelectric coeﬃcients. Drude theory

also fails to explain why the sign of the Hall eﬀect can be diﬀerent in

diﬀerent samples. Drude theory does not explain why we should only

count valence electrons.

(e) Drude theory at ﬁnite frequency. We start with the equation of

motion simpliﬁed by setting τ→ ∞

dt =−e(E+v×B)

setting E= ˆxE0

xeiωt and B=Bˆz,

m˙vx=−eE0

xeiωt −evyB

m˙vy=evxB

We can diﬀerentiate the ﬁrst equation to give

m¨vx=−iωeE0

xeiωt −e˙vyB

then plug in the second equation to give

¨vx=−iωe(E0

x/m)eiωt −(eB/m)2vx

18 Drude Theory

which is the equation of a driven harmonic oscillator. We use the ansatz

solution vx=v0

xeiωt so we obtain

−ω2v0

x=−iω(eE0

x/m)−(eB/m)2v0

which we solve

x=−iω(eEx/m)

(eB/m)2−ω2

and similarly

y=−(eB/m)(eEx/m)

(eB/m)2−ω2

with the current being j=−env we obtain

σxx =iω(ne2/m)/[(eB/m)2−ω2]

σyx = (eB/m)(ne2/m)/[(eB/m)2−ω2]

The cyclotron frequency eB/m is the natural oscillation frequency of a

particle of charge −eof mass min magnetic ﬁeld B. This divergent

response is easy to detect experimentally as a strong absorbtion of the

ac electric ﬁeld at a particular frequency. (Then this obviously can be

converted into a measurement of the mass).

The motion in the z-direction is unaﬀected by the magnetic ﬁeld in

the zdirection so that we have

σzz =ne2/(iωm)

and oﬀ-diagonal terms including zare zero.

Note, the calculation may look a bit nicer if you set vor equivalently

j, and solve for Eto obtain the ﬁnite frequency resistance matrix, and

then invert last. Lets try doing it that way also. Starting with

dt =−e(E+v×B)

Writing E=E0eiωt and j=j0eiωt and also j=n(−e)v=n(−e)p/m

we then have

iωmj0/(n(−e)) = −e[E0+j0×B/(n(−e))]

E0=iωm

ne2j0−B×j0/(n(−e))

So assuming Bin the ˆzdirection, we have a resistivity matrix

e=



iωm

ne2B/(−ne) 0

B/(ne)iωm

ne20

0 0 iωm

ne2





which we invert to get the same result as above.

(3.2) Scattering Times

The following table gives electrical resistivities ρ, den-

sities n, and atomic weights wfor the metals silver and

lithium:

ρ(Ωm) n(g/cm3)w

Ag 1.59 ×10810.5 107.8

Li 9.28 ×1080.53 6.94

Given that both Ag and Li are monovalent (i.e.,

have one free electron per atom), calculate the Drude

scattering times for electrons in these two metals.

In the kinetic theory of gas, one can estimate the scat-

tering time using the equation

τ=1

nhviσ

where nis the gas density, hviis the average velocity (see

Eq. 3.4 main text), and σis the cross-section of the gas

molecule—which is roughly πd2with dthe molecule di-

ameter. For a nitrogen molecule at room temperature,

we can use d=.37nm.

Calculate the scattering time for nitrogen gas at

room temperature and compare your result to the Drude

scattering times for electrons in Ag and Li metals.

Note: the table should read 10−8not 108!

We use σ=ρ−1=Ne2τ /m with mthe free electron mass and where

Nhere is the electron density which we calculate by

N=nAvagadro Number

mol-weight in grams/cm3106

Solving for τwe get

τAg = 3.8×10−14sec

τLi = 8.3×10−15sec

The second part should say room temperature and pressure. The

weight of a Nitrogen molecule is about 28 times that of a proton (two

nitrogen atoms of atomic weight 14). So the velocity at 300 K is

hvi=s8kBT

π28mp≈475m/sec

uncoincidentally being close to the speed of sound in air. The density

is given by n=P/RT with Rthe gas constant. At P= 105pascals

and T= 300 K, this gives .025 mol/m3. (This should be the usual 22.4

moles per liter that people remember, but we used 300 K instead of 273

and we approximated the pressure). Multiplying by Avagadro’s number

give the density that we should use in the equation

τ=1

nhviσ≈2×10−10sec

So electrons scatter much much much more often — this is not surprising

considering how much higher their density is than that of the nitrogen

gas.

20 Drude Theory

(3.3) Ionic Conduction and Two Carrier Types

In certain materials, particularly at higher tempera-

ture, positive ions can move throughout the sample in

response to applied electric ﬁelds, resulting in what is

known as ionic conduction. Since this conduction is typ-

ically poor, it is mainly observable in materials where

there are no free electrons that would transport cur-

rent. However, occasionally it can occur that a material

has both electrical conduction and ionic conduction of

roughly the same magnitude—such materials are known

as mixed ion–electron conductors.

Suppose free electrons have density neand scattering

time τe(and have the usual electron mass meand charge

−e). Suppose that the free ions have density ni, scatter-

ing time τi, mass miand charge +e. Using Drude theory,

(a) Calculate the electrical resistivity.

(b) Calculate the thermal conductivity.

(c)* Calculate the Hall resistivity.

If we ﬁx the electric ﬁeld, both species respond to the electric ﬁeld in-

dependently. So the total conductivity is the sum of the two independent

conductivities

σ=σe+σi=e2neτe

+niτi

mi

And thus

ρ=1

e2neτe

me+niτi

mi

The thermal conductivity is similar – both pieces add

κ=κe+κi=4k2

πneτe

+niτi

mi

Note that the Weidemann-Franz law continues to hold here in the ratio

of σto κ.

The Hall resistivity is more complicated. To simplify, if we apply

magentic ﬁeld in the zdirection, we need only keep track of conductivity

in the x, y plane (i.e., we can think of this as a two dimensional problem).

For a single species, we have (See exercise 3.1)

ρ=r BR

−BR r 

where r=m/(nq2τ) and R=q/n with qthe charge on the charge

carrier. We deﬁne tensors ρeand ρifor the two separate species in

terms of rj=mj/(njq2

jτj) and Rj=qj/njwith j=eor i. The

conductivity tensors are σj=ρ−1

jand then the total conductivity tensor

is σ=σe+σi. Finally this is inverted to give the tensor ρtotal =σ−1.

There is a lot of algebra involved in this. I obtained

ρxx =B2(reR2

i+riR2

e) + rire(re+ri)

B2(Re+Ri)2+ (re+ri)2

ρxy =BB2ReRi(Re+Ri) + Rir2

e+Rer2

i

B2(Re+Ri)2+ (re+ri)2

More Electrons in Metals:

Sommerfeld (Free

Electron) Theory 4

(4.1) Fermi Surface in the Free Electron (Som-

merfeld) Theory of Metals

(a)‡Explain what is meant by the Fermi energy, Fermi

temperature and the Fermi surface of a metal.

(b)‡Obtain an expression for the Fermi wavevector and

the Fermi energy for a gas of electrons (in 3D).

Show that the density of states at the Fermi sur-

face, dN/dEFcan be written as 3N/2EF.

of sodium atoms is roughly 1 gram/cm3, and sodium has

atomic mass of roughly 23. You may assume that there

is one free electron per sodium atom (sodium has valence

one)]

(d) Now consider a two-dimensional Fermi gas. Ob-

tain an expression for the density of states at the Fermi

surface.

(a.i) Fermi Energy EFis chemical potential at T= 0. Note, if there is

a ﬁlled band the chemical potential is mid-gap, and this diﬀers from the

conventional intuition that it is the highest ﬁlled state at zero tempera-

ture. Note that some books deﬁne fermi energy to be chemical potential

as a function of temperature. This is annoying — why deﬁne a new

quantity if it is just another name for the old quantity?!

(a.ii) Fermi temperature TF=EF/kbwith kbbeing Boltzmann’s con-

stant.

(a.iii) Fermi surface is the surface in momentum space separating the

ﬁlled and unﬁlled states at zero temperature. (This is ill-deﬁned for the

case of a ﬁlled band – but we don’t do band theory until later in the

course). Or the manifold of states having energy EF. Note that it need

not be a sphere, for example, if the eﬀective mass (deﬁned later!) is

anisotropic you get an ellipsoid instead.

(b)

N= 2VZk<kF

(2π)3=2V

(2π)3

4πk3

Note the factor of 2 out front is for two species of spins. This result

implies

kF= (3π2N/V )1/3

22 Sommerfeld Theory

which gives us the Fermi energy

EF=~2(3π2N/V )2/3

2m(4.1)

with mthe (eﬀective) electron mass.

Here is a short way to show that the density of states at the Fermi

surface is 3N/2EF. We write the density as

EF=C(N)2/3

where Cis a bunch of constants (given in Eq. 4.1, although we actually

don’t care about its actual value). We then have

dEF/dN = (2/3)CN −1/3= (2/3)EF/N

which immediately gives us

dN/dEF= (3/2)N/EF

(1gm/cm3)(102cm/m)3(mole/23gm)(6.02 ×1023atoms/mole)

= 2.6×1028atom/m3.

(See also part (c) of problem 3.1.) With 1 free electron per atom, this

gives us the density N/V and we can plug this into Eq. 4.1 yielding

EF= 5 ×10−19J= 3.2eV

dividing by Boltzmann’s constant this gives about 37,000 Kelvin.

(d) For a 2d Fermi gas we have

N= 2AZk<kF

(2π)2=2A

(2π)2πk2

where Ais the (2d) area of the system. Thus we obtain

kF= (2πN/A)1/2

The Fermi energy is then

EF=~2(2πN/A)

2m=~2πN/A

The density of states is then independent of energy and is given by

dN/dE =Am/(~2π) = N/EF

One can ask the same question in ddimensions and use a similar scheme

to do the calculation.

(4.2) Velocities in the Free Electron Theory

(a) Assuming that the free electron theory is applica-

ble: show that the speed vFof an electron at the Fermi

surface of a metal is vF=~

m(3π2n)1/3where nis the

density of electrons.

(b) Show that the mean drift speed vdof an electron

in an applied electric ﬁeld Eis vd=|σE/(ne)|, where σ

is the electrical conductivity, and show that σis given

in terms of the mean free path λof the electrons by

σ=ne2λ/(mvF).

to copper:

(i) calculate the values of both vdand vFfor

copper at 300K in an electric ﬁeld of 1 V m−1

and comment on their relative magnitudes.

(ii) estimate λfor copper at 300K and com-

ment upon its value compared to the mean

spacing between the copper atoms.

You will need the following information: copper is mono-

valent, meaning there is one free electron per atom. The

density of atoms in copper is n= 8.45 ×1028 m−3. The

conductivity of copper is σ= 5.9×107Ω−1m−1at 300K.

(a) As in the previous problem, in 3D, kF= (3π2n)1/3with n=N/V .

Thus pF=~kFand the fermi velocity is vF=~kF/m.

(b) If an electric ﬁeld Eis applied, a current density j=σEﬂows, with

σthe conductivity. The electrical current is then given by j=−envd

where −eis the electron charge, nis the density of the electrons, and

vdis the average drift velocity. Thus we obtain

|vd|=|σE/(ne)|

Deriving the conductivity from the mean free path is an exercise in

Drude theory. As in problem 3.1, we have the Drude theory expression

dt =F−p

With F=−eEwe then have the steady state momentum p=−eE/τ

corresponding to the steady state drift velocity vd=−eE/(τm). The

current density is then j=ne2/(τm)Eyielding the usual expression for

the Drude conductivity

σ=ne2τ

Now if the typical velocity of an electron is on the order of the Fermi

velocity vF, then in the scattering time τthe distance traveled, i.e., the

mean free path is λ=τvF. Thus we can rewrite the Drude conductivity

σ=ne2λ

mvF

(c.i) On the scale of the Fermi temperature, 300K is close to zero

kelvin so we can ignore this temperature.

vF=~(3π2n)1/3

we obtain vF≈1.6×106m/sec. This is about 0.5% of the speed of light.

Very fast.

24 Sommerfeld Theory

The drift velocity is

vd=σE/(ne)

which gives vd≈4.4×10−3m/sec. I.e., very slow.

(c.ii) Use here

λ=vFσm/(ne2)

plugging in numbers this gives λ≈3.9×10−8m, or about 400 angstoms

– roughly 100 lattice spacings.

(4.3) Physical Properties of the Free Electron

Gas

In both (a) and (b) you may always assume that the

temperature is much less than the Fermi temperature.

(a)‡Give a simple but approximate derivation of the

Fermi gas prediction for heat capacity of the conduction

electron in metals.

(b)‡Give a simple (not approximate) derivation of

the Fermi gas prediction for magnetic susceptibility of

the conduction electron in metals. Here susceptibility is

χ=dM/dH =µ0dM/dB at small Hand is meant to

consider the magnetization of the electron spins only.

that of a classical gas of electrons?

What other properties of metals may be diﬀerent

from the classical prediction?

(d) The experimental speciﬁc heat of potassium metal

at low temperatures has the form:

C=γ T +α T 3

where γ= 2.08 mJ mol−1K−2and α=

2.6 mJ mol−1K−4.

Explain the origin of each of the two terms in this

expression.

Make an estimate of the Fermi energy for potassium

metal.

(a) Let the density of states at the Fermi surface be given by D(EF)

(and assume this is a nonzero quantity). At temperature T, electrons

within an energy kbTof the Fermi surface can be excited above the

Fermi energy. Typically if an electron is excited, it will gain an energy

of order kbT. Thus, the electron energy (above the ground state) is

roughly (kbT)2D(EF). Thus the heat capacity is approximately

C=dE/dT ≈2kb(kbT)D(EF)

In a slightly more careful (but still not exact) treatment, one approxi-

mates the smooth fermi function as a simple function as shown here in

Fig. 4.1.

Fig. 4.1 The Fermi function (green)

and a simple approximation to the

fermi function (blue).

In this picture, the approximation to the fermi function is given by

n(E) = 





1 (E−µ)/kbT < −2

0 (E−µ)/kbT > 2

1/2−(E−µ)/(4kbT) otherwise

In this approximation, one can calculate that the energy is given by

E(T) = Constant + DZ2/kbT

−2/kbT

dEE(1/2−E/(4kbT))

= Constant + D(kbT)2(4/3)

which results in C= (8/3)kb(kbT)D(EF).

An exact calculation (See Ashcroft+Mermin, or exercise 4.9.b below)

of this result is

C= (π2/3)kb(kbT)D(EF) = N(π2/2)kb(kbT)/EF(4.2)

This calculation is an exercise given as an additional problem.

Note that as discussed above, for a free fermi gas D(EF) = 3N/2EF.

Thus up to constants of order one we have

C∼kbN(kbT/EF)

which is very small since T≪TF.

(b) There are several ways that the electrons can respond to the mag-

netic ﬁeld. First, we assume that the magnetic ﬁeld couples only to

the spins of the electron (we ignore orbital eﬀects). The Hamiltonian

(neglecting the Lorentz force of the magnetic ﬁeld) becomes

H=p2

2m+gµBB·σ

where g= 2 is the g-factor of the electron Bis the externally applied

magnetic ﬁeld and σis the spin of the electron which takes eigenvalues

±1/2. Here µB≈.67(K/T )/kBthe conventional Bohr magneton. Thus

in the magnetic ﬁeld the energy of an electron with spin up or down

26 Sommerfeld Theory

(with up meaning it points the same way as the applied ﬁeld)

ǫ(k,↑) = ~2|k|2

2m+µb|B|

ǫ(k,↓) = ~2|k|2

2m−µb|B|

The spin magnetization of the system in the direction of the applied

magnetic ﬁeld will then be

M=−1

dB =−([# up spins] −[# down spins]) µB/V (4.3)

So when the magnetic ﬁeld is applied, it is lower energy for the spins to

be pointing down, so more of them will point down.

Let us now calculate the Pauli paramagnetism of the free electron

gas at T= 0. With zero magnetic ﬁeld applied, both the spin up and

spin down states are ﬁlled up to the Fermi energy (i.e, to the Fermi

wavevector). Near the Fermi level the density of states per unit volume

for spin up electrons is g(EF)/2 and similarly the density of states per

unit volume for spin down electrons is g(EF)/2. When Bis applied, the

spin ups will be more costly by an energy µBB. Thus, (assuming that the

chemical potential does not change) we will have (g(EF)/2)µBBfewer

spin ups electrons per unit volume. Similarly, the spin downs will be less

costly by the same amount, so we will have (g(EF)/2)µBBmore spin

downs per unit volume. Note that the total number of electrons in the

system did not change, so our assumption that the chemical potential did

not change is correct. (Recall that chemical potential is always adjusted

so it gives the right total number of electrons in the system). Thus,

using Eq. 4.3 the moment per unit volume is given by

M=g(EF)µ2

and hence the magnetic susceptibility χ=∂M/∂H is given (at T= 0

by)

χP auli =µ0µ2

Bg(EF)

equipartition law C= 3kBNwhich is larger than the result above by

roughly a factor of EF/kbTwhich could be a factor of 100 or more.

Similarly, for a single isolated spin 1/2 we can calculate the partition

function (this calculation was done in stat mech class last year).

Z=eβgµBB/2+e−βgµBB/2

The expectation of the moment (per spin) is then

m=−dlog Z/d(Bβ) = (gµB/2) tanh(βgµBB/2)

For small Bthis is

m= (gµB/2)2(B/kbT)

thus the total susceptibility for Nspins (recall susceptibility is measured

per unit volume) is

dM/dH =µ0dM/dB =Nµ0(gµB/2)2(1/kbT)

which, for any T≪TFis much larger than the Pauli susceptibility

calculated above (by a factor of approximately TF/T ).

Other properties that diﬀer from the classical prediction include: Ther-

mopower, Peltier Coeﬃcinent, Average Electron Velocity, Compressibil-

ity, ...

(d) The T3term is clearly from Debye phonon speciﬁc heat. The T-

linear term is the speciﬁc heat of free electrons. Using the above formula

Eq. 4.2 yields

EF= (π2/2)R(T/Clinear) =≈2×104KkB≈1.7eV

The real value is roughly 2.1 eV.

Another method would be to use the density of potassium and assume

the valence is 1. If you do this, you get something much closer to the

right answer.

(4.4) Another Review of Free Electron Theory

What is the free electron model of a metal.

Deﬁne Fermi energy and Fermi temperature.

Why do metals held at room temperature feel cold

to the touch even though their Fermi temperatures are

much higher than room temperature?

(a) A d-dimensional sample with volume Ldcontains

Nelectrons and can be described as a free electron model.

Show that the Fermi energy is given by

EF=~2

2mL2(Nad)2/d

Find the numerical values of adfor d= 1,2, and 3.

(b) Show also that the density of states at the Fermi

energy is given by

g(EF) = Nd

2LdEF

Assuming the free electron model is applicable, es-

timate the Fermi energy and Fermi temperature of a

one-dimensional organic conductor which has unit cell of

length 0.8 nm, where each unit cell contributes one mobile

electron.

culate the Fermi energy as a function of the density for

electrons in d= 1,2,3 and calculate the density of states

at the Fermi energy in each case.

The free electron model of a metal describes electrons in a metal as a

noninteracting gas of fermions at some ﬁxed density (usually chosen to

be velectrons per unit cell of the metal where vis the valence).

Fermi Energy EFis chemical potential at T= 0. Note, if there is a

ﬁlled band the chemical potential is mid-gap, and this diﬀers from the

conventional intuition that it is the highest ﬁlled state at zero tempera-

ture. Note that some books deﬁne fermi energy to be chemical potential

as a function of temperature. This is annoying — why deﬁne a new

quantity if it is just another name for the old quantity?!

Fermi temperature TF=EF/kbwith kbbeing Boltzmann’s constant.

Due to Pauli exclusion, a metal can have a very high Fermi tempera-

28 Sommerfeld Theory

ture (high chemical potential) even if the material is at zero temperature

— i.e., if the material is in its ground state. When you touch a material

and it feels hot, this is because heat has ﬂowed from the material to you.

If the material is in its ground state (despite having a high fermi temper-

ature) it cannot lower its own energy and therefore cannot transfer heat

to you. Note: Having a high chemical potential DOES mean that the

material might have a tendency to transfer electrons to another body

with fewer electrons (although this might create a charge imbalance that

then prevents further ﬂow of electrons). We discuss physics of this type

in Chapter 18.

(a) In any number of dimensions we can write

N= 2LdZ|k|<kF

ddk

(2π)d

with the 2 accounting for spin. The integration is over a d-dimensional

ball. In 1,2, 3d we obtain

N/Ld=





2kF/π 1d

F/(2π) 2d

F/(3π2) 3d

So that

kF=





(N/L)(π/2) 1d

(N1/2/L)(2π)1/22d

(N1/3/L)(3π2)1/33d

And in any dimension EF=~2k2

F/(2m), so that a1=π/2 and a2= 2π

and a3= 3π2.

(b) We have

EF=Cd/L2N2/d

for some constant Cd. Then

dEF/dN = (2/d)(EF/N)

and

g(EF) = (1/V )(dN/dEF) = (N/Ld)(d/(2EF)).

For a one dimensional system, as noted above EF= (~2/(2m))(π/2)2(N/L)2

where here N/L = 1/(.8nm) This gives me EF= 2.4×10−20Jor

TF=EF/kb= 1700K.

EF=c~|kF|=c~(N1/d/L)ad

So here we have

dEF/dN = (1/d)(EF/N)

and

g(EF) = (1/V )(dN/dEF) = (N/V )(d/EF)

(4.5) Chemical Potential of 2D Electrons

Show that for free electron gas in two dimensions, the

chemical potential µis independent of the temperature so

long as T≪µ. Hint: ﬁrst examine the density of states

in two dimensions.

The key here is to realize that the density of states is independent

of energy. From the previous problem, in 2d, we have EF∼N, and

g(EF)∼N/EFso that the density of states is a constant independent

of temperature. Now, given a ﬁxed density of electrons, the chemical

potential is set by

n=Z∞

dEg(E)1

eβ(E−µ)+ 1

where ghere is now constant. The point here is that except for correc-

tions exponentially small in βµ the value of the integral is independent of

β. Therefore the dependence of non µis to a very good approximation

independent of temperature.

To see this in more detail rewrite as

n/g =Z∞

−µ

dx 1

eβx + 1 =Z∞

−µ

dx e−βx

e−βx + 1

βln(eβµ + 1)

Now for large βµ we can expand to get

n/g =µ+O(e−βµ)

so, as claimed, the relationship of nto µis to a good approximation

independent of Tso long as µ≫kbT.

(4.6) Chemical Potential at T= 0

Consider a system of Nnon-interacting electrons. At

T= 0 the Nlowest-energy eigenstates will be ﬁlled and

all the higher energy eigenstates will be empty. Show that

at T= 0 the energy of the chemical potential is precisely

half way between the highest energy ﬁlled eigenstate and

the lowest-energy unﬁlled eigenstate.

This is a bit more diﬃcult than it appears as most generally one

should consider the possibility that the highest ﬁlled state at energy E1

or lowest unﬁlled state at energy E2are degenerate – i.e, there is more

than one eigenstate at the given energy. Let us call the degeneracies of

these levels g1and g2respectively. Let us focus on these two energies

only and assume for now (to be justiﬁed in retrospect) that we can

ignore any other states in the system with the intuition that any state

with energy below E1is completely ﬁlled and can be ignored any any

state with energy above E2is completely empty and can be ignored.

30 Sommerfeld Theory

We take our system (E1with degeneracy g1and E2with degeneracy

g2) and will ﬁll it with exactly g1electrons. (Were we to choose fewer,

the lowest unﬁlled would move down to E1were we to choose more the

highest ﬁlled would move up to E2). Fermi occupation then gives us

N=g1=g1

eβ(E1−µ)+ 1 +g2

eβ(E2−µ)+ 1

Deﬁning x=eβ(E1−µ)and z=e−β(E2−E1)this then becomes

g1=g1

x+ 1 +g2

x/z + 1

Solving for x(and taking the positive root only) gives

x=(g2−g1)z+p(g2−g1)2z2+ 4g1g2z

2g1

Since temperature is small, zbecomes very small, so we can expand this

expression to give

x=√g1g2z

Or equivalently taking log of both sides

β(E1−µ) = 1

2(ln(g1g2)−β(E2−E1))

Taking the limit of small temperature or large β, this becomes

µ= (E1+E2)/2 + ...

(4.7) More Thermodynamics of Free Electrons

(a) Show that the kinetic energy of a free electron gas

in three dimensions is E=3

5EFN.

(b) Calculate the pressure P=−∂E/∂V , and then the

bulk modulus B=−V ∂P/∂V .

2.53×1022 cm−3and that of potassium is 1.33×1022 cm−3,

and given that both of these metals are monovalent (i.e.,

have one free electron per atom), calculate the bulk modu-

lus associated with the electrons in these materials. Com-

pare your results to the measured values of 6.3 GPa and

3.1 GPa respectively.

(a) We begin with the fact that the density of state g(ǫ)∼ǫ1/2. To

see this note that ǫ∼k2so dǫ ∼kdk and dN ∼dk ∼k2dk, so we have

dN/dǫ ∼k∼ǫ1/2. We can thus write g(ǫ) = Cǫ1/2for some constant C

and

N=VZEF

dǫg(ǫ) = CV (2/3)E3/2

E=VZEF

ǫdǫg(ǫ) = CV (2/5)E5/2

Thus dividing one by the other we get

E/N = (3/5)EF

(b) From exercise 4.4.a or on dimensional grounds EF∼V−2/3, so

E∼V−2/3so the pressure is

P=−∂E

∂VN

V=2

NEF

Since P∼V−5/3we then have the bulk modulus

B=−V∂P

∂V =5

3P=10

V=2EF

10−19Jand EF= 3.3×10−19 for sodium and potassium respectively.

Thus we obtain B= 8.5 GPa and B= 2.9 GPa respectively. Not too

shabby!!

(4.8) Heat Capacity of a Free Electron Gas*

In Exercise 4.3.a we approximated the heat capacity of

a free electron gas

(a*) Calculate an exact expression for the heat capacity

of a 2d metal at low temperature.

(b**) Calculate an exact expression for the heat capac-

ity of a 3d metal at low temperature.

The following integral may be useful for these calcula-

tions: Z∞

−∞

dx x2ex

(ex+ 1)2=π2

3=ζ(2)/2

Note that for the 3d case you have to worry about the

fact that the chemical potential will shift as a function of

temperature. Why does this not happen (at least for low

T) in the 2d case?

(a) As pointed out in exercise 4.6, in two dimensions µis indepdendent

of temperature — a result of the fact that the density of states is a

constant (the answer to the last part of this question). First we need to

evaluate the value of the density of states. Using the result of 4.4, we

obtain at the fermi energy (and therefore at all energies)

g=m

~2π

We thus have

µ=EF=n/g

We can then write the total energy as

E=VZ∞

dǫ ǫ g

eβ(ǫ−µ)+ 1 =V g Z∞

−µ

dx x

eβx + 1 +V g Z∞

−µ

dx µ

eβx + 1

where we have deﬁned x=ǫ−µ. We recognize the second term here

as simply being µN , and as discussed in 4.6, this is independent of

temperature. To evaluate the ﬁrst term, we integrate by parts to get

E−Nµ =gV x2/2

eβx + 1 

∞

−µ−gV Z∞

−µ

dx x2

2−βeβx

(eβx + 1)2

32 Sommerfeld Theory

Assuming βµ is large, to within corrections that are of order of e−βµ

we can throw away the ﬁrst term and we can extend the integral in

the second term all the way to −∞ (since the argument is peaked near

x= 0 wiht only exponentially small tails). Scaling out the factor of β

and using the integral given we obtain

E−Nµ =V g(kbT)2π2

6+O(e−βµ)

Diﬀerentiating to obtan the heat capacity we obtain

C/V =gkb(kbT)π2/3

which is precisely the result claimed above in exercise 4.3 (see Eq. 4.2).

However, here note that up to exponentially small corrections, this is

exact – in other words, there is no subleading Tnwith n > 1 term at

all!

(b) In the more general case where g(E) is not a constant, one must use

the so-called Sommerfeld expansion. This is quite a bit more complex.

First we will quote the key formula and use it to derive our result for

the heat capacity, then we will go back to derive Sommerfeld’s formula.

Deﬁning the Fermi function

nF(ǫ) = 1

eβ(ǫ−µ)+ 1

The Sommerfeld formula is

Z∞

∞

H(ǫ)nF(ǫ)dǫ =Zµ

∞

H(ǫ)dǫ +π

6(kbT)2H′(µ) + O(T4) (4.4)

The intuition behind this formula is that at low temperature the fermi

function is a step function (given by the ﬁrst term). The ﬁnite slope of

the fermi function where it is almost a step creates the small correction

term which is determined by H′. Note that as we determinmed in 4.6

and above, if the argument His constant, then up to exponentially small

terms, there is no correction term at any order in T. One could carry

this expansion to higher order and pick up terms related to H′′ etc as

well.

Expecting that the chemical potential will remain very close to the

Fermi energy at low temperatures, we can conclude

Z∞

∞

H(ǫ)nF(ǫ)dǫ =ZEF

∞

H(ǫ)dǫ + (µ−EF)H(EF) + π

6(kbT)2H′(EF) + ...

Using this equation to write an expression for the density

V=Z∞

g(ǫ)nF(ǫ)dǫ (4.5)

=ZEF

∞

dǫg(ǫ) + (µ−EF)g(EF) + π

6(kbT)2g′(µ) + ...

=N(T= 0)

V+n(µ−EF)g(EF) + π

6(kbT)2g′(EF)o(4.6)

Since we are ﬁxing the density N, the term in brackets must remain

zero for all temperatures (accurate to order T2). This conﬁrms that

µ=EF+O(T2) and conﬁrms our suspicion that it should be very close

to EFat low temperature. Note also that for g′= 0, such as in 2d,

µ=EFindependent of temperature.

We can then write a similar expression for the energy density

V=Z∞

ǫg(ǫ)nF(ǫ)dǫ

=ZEF

∞

ǫg(ǫ)dǫ +EF(µ−EF)g(EF) + π

6(kbT)2(ǫg(ǫ))′

ǫ=EF+...

=E(T= 0)

V+EFn(µ−EF)g(EF) + π

6(kbT)2g′(EF)o

+π2

6(kbT)2g(EF) + ... (4.7)

Note that in the ﬁnal equation the term in brackets is the same as the

term in brackets from Eq. 4.6 which we have set to zero in order to keep

the density constant as a function of temperature. Thus we obtain

V=E(T= 0)

V+π2

6(kbT)2g(EF) + ...

Which we diﬀerentiate to obtain the heat capacity

C/V =π2

3kb(kbT)g(EF)

as claimed.

Finally we return to prove the Sommerfeld formula 4.4. The quantity

we would like to evaluate is

I=Z∞

∞

H(ǫ)nF(ǫ)dǫ

Let us deﬁne a function

K(ǫ) = Zǫ

−∞

dǫ′H(ǫ′)

So that, integrating by parts, we have

I=Z∞

∞

K(ǫ)−∂nF(ǫ)

∂ǫ dǫ

there are no boundary terms (to accuracy of O(e−βµ) because ∂nF/∂ǫ

decays very rapidly away from the chemical potential. Note now that

the function ∂nF/∂ǫ is a symmetric (even) function around the chemical

potential. Thus let if expand Kin a taylor series around the chemical

potential only the even terms will have a nonzero contribution, thus we

have

I=Z∞

∞K(µ) + 1

2(ǫ−µ)2K′′(µ) + ...−∂nF(ǫ)

∂ǫ dǫ (4.8)

34 Sommerfeld Theory

Z∞

∞

−∂nF(ǫ)

∂ǫ = 1

the ﬁrst term in the expansion of Eq. 4.8 is just

K(µ) = Zµ

−∞

H(ǫ)dǫ

as required. The next term gives a prefactor of K′′(µ) = H′(µ) and

requires that we evaluate the integral

Z∞

∞

2(ǫ−µ)2−∂nF(ǫ)

∂ǫ dǫ =1

2Z∞

∞

x2eβx

(eβx + 1)2= (kbT)2π2/6

using the given integral. Thus we have obtained the ﬁrst two terms of

Eq. 4.8. By followin a similar procedure one can evaluate higher terms

in the expansion and in particular we will ﬁnd that the next term in the

expansion must be proportional to T4.

The Periodic Table 5

(5.1) Madelung’s Rule

Use Madelung’s rule to deduce the atomic shell ﬁll-

ing conﬁguration of the element tungsten (symbol W)

which has atomic number 74.

Element 118 has recently been discovered, and is

expected to be a noble gas, i.e., is in group VIII. (No

real chemistry tests have been performed on the ele-

ment yet, as the nucleus decays very quickly.) Assuming

that Madelung’s rule continues to hold, what should the

atomic number be for the next noble gas after this one?

Angular momentum lorbitals (l= 0 is called s, l= 1 is called p, etc)

contain up to 2(2l+1) electrons. Madulung’s rule ﬁlls orbitals according

to the diagram Fig. 5.1: from lowest n+lto highest, and for cases of

the same n+l, ﬁll the lower nﬁrst. So we have

Tungsten atomic number 74:

1s22s22p63s23p64s23d10 4p65s24d10 5p66s24f14 5d4.

Note that the “exponents” add to 74. Or equivalently we write

[Xe] 4f14 5d4.

Note that the noble gases occur whenever a p-shell has just ﬁlled.

Element 118 has a ﬁlled 7p shell. Madelung’s rule tells us that we then

have to ﬁll 8s, 5g, 6f, 7d and ﬁnally 8p. This brings us to element

168. For entertainment sake (and you can try to prove this) note that

the sequence of nobel gas element numbers 2, 10, 18, 36, 54, 86, 118,

168 has successive diﬀerences which are twice the perfect squares each

occurring twice. (10-2)/2 = 4 , (18-10)/2 = 4, (36-18)/2 = 9, (54-36)/2

= 9, (86-54)/2 = 16, (118-86)/2=16, (168-118)/2 = 25, and so forth.

Fig. 5.1 Ordering of ﬁlling orbitals in

atoms (Madelung’s rule).

(5.2) Eﬀective Nuclear Charge and Ionization

Energy

(a) Let us approximate an electron in the nth shell

(i.e., principal quantum number n) of an atom as being

like an electron in the nth shell of a hydrogen atom with

an eﬀective nuclear charge Z. Use your knowledge of

the hydrogen atom to calculate the ionization energy of

this electron (i.e., the energy required to pull the electron

away from the atom) as a function of Zand n.

(b) Consider the two approximations discussed in the

text for estimating the eﬀective nuclear charge:

•(Approximation a)

Z=Znuc −Ninside

•(Approximation b)

Z=Znuc −Ninside −(Nsame −1)/2

where Znuc is the actual nuclear charge (or atomic num-

ber), Ninside is the number of electrons in shells inside of n

(i.e., electrons with principal quantum numbers n′< n),

and Nsame is the total number of electrons in the nth

principal shell (including the electron we are trying to

remove from the atom, hence the −1).

36 The Periodic Table

Explain the reasoning behind these two approxima-

tions.

Use these approximations to calculate the ioniza-

tion energies for the atoms with atomic number 1 through

21. Make a plot of your results and compare them to the

actual ionization energies (you will have to look these up

on a table).

Your results should be qualitatively quite good. If you

try this for higher atomic numbers, the simple approxi-

mations begin to break down. Why is this?

Neglecting ﬁne structure, the energy of an electron in the nth shell of

hydrogen is

En=−Ry

where Ry=13.6 eV is the Rydberg constant. For a hydrogenic atom

with nuclear charge Z, the Coulomb interaction is Ztimes as strong

as in hydrogen, resulting in binding energy which is Z2as strong. To

see this in detail, one can solve the Schroedinger equation in detail.

However, without doing this one can get it by a scaling argument as

well. The bound state is a balancing of the kinetic with the potential

energy. So roughly one should be able to estimate the binding energy

by setting these equal to each other. Setting the the length scale to a,

(i.e, deﬁne ato be the eﬀective Bohr radius) we have

KE =~2

ma2=P E =Ze2

4πǫ0a

solving for aobtains a∼1/Z and plugging back into P E or KE we

determine that the kinetic energy should scale as Z2.

Fig. 5.2 Ionization Energy (eV) as a

function of atomic number. Exact com-

pared to the two proposed approxima-

tions 0

100

150

200

250

0 5 10 15 20 25

Ionization Energies

experiment

"approx1.txt"

"approx2.txt"

The two approximations are plotted here with the exact ionization

energies. Qualitatively they are OK, but quantitatively not so good I

guess.

One can do a bit better (See Fig. 5.3) by assuming that p-shells are

“outside” of s-shells. In other words, a single electron in a p-shell sees

a charge of Z= 1 since the entire s-shell is inside of it. In this case one

gets the following ﬁgure (in the two apperoximations discussed above).

In fact, this is getting to be pretty decent. Notice with this second

approximation one obtains a dip in the ionization energy for ﬁlled s-

shells (such as atomic numbers 4 and 12) which is seen in the experiment,

although is weaker in reality.

100

120

140

0 5 10 15 20 25

Ionization Energies

experiment

"approx3.txt"

"approx4.txt"

Fig. 5.3 Ionization Energy (eV) as

a function of atomic number. Exact

compared to the two proposed approx-

imations with the modiﬁcation that p-

shells are declared to be outside of s-

shells.

Once you get to the transition metals, the d-shells really are not very

easily described as being inside or outside of anything. And often when

transition metals ionize, they lose their s-electrons.

(5.3) Exceptions to Madelung’s Rule

Although Madelung’s rule for the ﬁlling of electronic

shells holds extremely well, there are a number of excep-

tions to the rule. Here are a few of them:

Cu = [Ar] 4s13d10

Pd = [Kr] 5s04d10

Ag = [Kr] 5s14d10

Au = [Xe] 6s14f145d10

What should the electron conﬁgurations be if these

elements followed Madelung’s rule and the Aufbau prin-

ciple?

Explain how the statement “3d is inside of 4s”

might help justify this exception in copper.

Madelung’s rule incorrectly predicts:

Cu = [Ar] 4s23d9

Pd = [Kr] 5s24d8

Ag = [Kr] 5s24d9

Au = [Xe] 6s24f145d9

For copper, the fact that 3d is inside 4s makes the 4s electron less well

bound than you might otherwise expect. Thus the d electrons can ﬁll

preferentially over the s in some cases.

(5.4) Mendeleev’s Nobel Prize

Imagine writing a letter to the Nobel committee nom-

inating Mendeleev, the creator of the periodic table, for

a Nobel Prize. Explain why the periodic table is so im-

portant. Remember that the periodic table (1869) was

devised many years before the structure of the hydrogen

atom was understood. (If you do not already have some

background in chemistry, you may want to read the next

chapter before attempting this exercise.)

38 The Periodic Table

Dear Nobel Committee,

Do I have to smack you upside the head? Do

the right thing and give the prize to Mendeleev for

God sake!

Sincerely,

Professor Steven H. Simon

What Holds Solids

Together: Chemical

Bonding 6

(6.1) Chemical Bonding

(a) Qualitatively describe ﬁve diﬀerent types of chem-

ical bonds and why they occur.

Describe which combinations of what types of

atoms are expected to form which types of bonds (make

reference to location on the periodic table).

Describe some of the qualitative properties of ma-

terials that have these types of bonds.

(Yes, you can just copy the table out of the chapter

summary, but the point of this exercise is to learn the

information in the table!)

(b) Describe qualitatively the phenomenon of van der

Waals forces. Explain why the force is attractive and pro-

portional to 1/R7where Ris the distance between two

atoms.

(a) Just look at the table in the Chapter Summary of chapter 6.

(b) van der Waals forces are from correlated dipole ﬂucuations. If

the electron is a given ﬁxed position, there is a dipole moment p=er

where ris the vector from the electron to the proton. With the electron

“orbiting” (i.e, in an eigenstate), the average dipole moment is zero.

However, if an electric ﬁeld is applied to the atom, the atom will develop

a polarization (i.e., it will be more likely for the electron to be found on

one side of the nucleus than on the other). We write

p=χE

with χthe polarizability. .

Now, let us suppose we have two such atoms, separated a distance r

in the ˆxdirection. Suppose one atom momentarily has a dipole moment

p1(for deﬁniteness, suppose this dipole moment is in the ˆzdirection).

Then the second atom will feel an electric ﬁeld

E=p1

4πǫ0r3

in the negative ˆzdirection. The second atom then, due to its polariz-

ability, develops a dipole moment p2=χE whi ch in turn is attracted

to the ﬁrst atom. The potential energy between these two dipoles is

U=−|p1||p2|

4πǫ0r3=−p1χE

(4πǫ0)r3=−|p1|2χ

(4πǫ0r3)2

40 Chemical Bonding

corresponding to a force which is attractive and proportional to 1/r7.

Note that while for a single isolated atom hpi= 0 the result is propor-

tional instead to h|p|2i ∼ h|r|2i ∼ with rthe position of an electron, is

nonzero. This calculation is done more carefully in problem 6.6 below.

(6.2) Covalent Bonding in Detail*

(a) Linear Combination of Atomic Orbitals:

In Section 6.2.2 we considered two atoms each with a

single atomic orbital. We called the orbital |1iaround nu-

cleus 1 and |2iaround nucleus 2. More generally we may

consider any set of wavefunctions |nifor n= 1,...,N.

For simplicity, let us assume this basis is orthonormal

hn|mi=δn,m (More generally, one cannot assume that

the basis set of orbitals is orthonormal. In Exercise 6.5

we properly consider a non-orthonormal basis.)

Let us write a trial wavefunction for our ground state

|Ψi=X

φn|ni.

This is known as a linear combination of atomic orbitals,

LCAO, or tight binding (it is used heavily in numerical

simulation of molecules).

We would like to ﬁnd the lowest-energy wavefunction

we can construct in this form, i.e., the best approxima-

tion to the actual ground-state wavefunction. (The more

states we use in our basis, generally, the more accurate

our results will be.) We claim that the ground state is

given by the solution of the eﬀective Schroedinger equa-

tion

Hφ=Eφ(6.1)

where φis the vector of Ncoeﬃcients φn, and His the

Nby Nmatrix

Hn,m =hn|H|mi

with Hthe Hamiltonian of the full system we are consid-

ering. To prove this, let us construct the energy

E=hψ|H|ψi

hψ|ψi

Show that minimizing this energy with respect to

each φngives the same eigenvalue equation, Eq. 6.1.

(Caution: φnis generally complex! If you are not com-

fortable with complex diﬀerentiation, write everything in

terms of real and imaginary parts of each φn.) Similarly,

the second eigenvalue of the eﬀective Schroedinger equa-

tion will be an approximation to the ﬁrst excited state of

the system.

(b) Two-orbital covalent bond

Let us return to the case where there are only two or-

bitals in our basis. This pertains to a case where we have

two identical nuclei and a single electron which will be

shared between them to form a covalent bond. We write

the full Hamiltonian as

H=p2

2m+V(r−R1) + V(r−R2) = K+V1+V2

where Vis the Coulomb interaction between the electron

and the nucleus, R1is the position of the ﬁrst nucleus

and R2is the position of the second nucleus. Let ǫbe the

energy of the atomic orbital around one nucleus in the

absence of the other. In other words

(K+V1)|1i=ǫ|1i

(K+V2)|2i=ǫ|2i

Deﬁne also the cross-energy element

Vcross =h1|V2|1i=h2|V1|2i

and the hopping matrix element

t=−h1|V2|2i=−h1|V1|2i

These are not typos!

Why can we write Vcross and tequivalently using

either one of the expressions given on the right-hand side?

Show that the eigenvalues of our Schroedinger

equation Eq. 6.1 are given by

E=ǫ+Vcross ± |t|

Argue (perhaps using Gauss’s law) that Vcross

should roughly cancel the repulsion between nuclei, so

that, in the lower eigenstate the total energy is indeed

lower when the atoms are closer together.

This approximation must fail when the atoms get

suﬃciently close. Why?

(a) writing

|ψi=X

φn|ni

E=hψ|H|ψi

hψ|ψi=Pn,m φ∗

nHnmφm

Pn|φn|2(6.2)

We can extremize by diﬀerentiating with respect to φ∗

n. Note: When

working with complex quantities we can simplify life by treating φnand

φ∗

nas independent variables. We thus have

0 = ∂E

∂φ∗

=PmHnmφm

Pp|φp|2−Pn,m φ∗

nHnmφm

Pn|φn|2φn

Pp|φp|2

0 = X

mHnmφm−Eφn

where we have used Eq. 6.2 to identify E.

(b) The 2 by 2 matrix given by Hin this basis is

ǫ+Vcross t

t∗ǫ+Vcross 

which has eigenvalues E=ǫ+Vcross ± |t|. Note that tcan always be

taken as real by simply making a gauge transform on the single particle

wavefunction (i.e., redeﬁning the phase of one of the wavefunctions to

absorb the phase of t).

Here Vcross is the potential felt by the electron on atom 2 due to the

nucleus of atom 1. Since the charge distribution of the electron on atom

2 is roughly spherical we can use Gauss’s law to calculate its interaction

energy with the nucleus of atom 1. If the nucleus of atom 1 is outside

of this spherical distribution of charge of the electron, Gauss’s law tells

us that we can treat the entire spherical distribution of charge as if it is

all at the center of the sphere. In this way, the charge of the electron on

atom 2 exactly cancels the charge of the nucleus of atom 2.

When the two nuclei get close, this argument no longer works, as the

nucleus is then inside much of the distribution of the electron charge. If

the electron charge distribution remains spherical, the nucleus of atom

2 will only see electron charge that is at smaller distances (inside) to

the center of this spherical distribution. When the nuclei are very close

together, the nucleus does not see the electron charge at all, and only

sees the other nucleus. Also, when the atoms get close together, the

assumption of orthogonality of grounds tate orbitals on diﬀerent sites

breaks down. (One has to choose whether you want orthogonal orbitals

or orbitals that are eigenstates of the single atom Hamiltonian.)

42 Chemical Bonding

(6.3) LCAO and the Ionic–Covalent Crossover

For Exercise 6.2.b consider now the case where the

atomic orbitals |1iand |2ihave unequal energies ǫ0,1and

ǫ0,2. As the diﬀerence in these two energies increases

show that the bonding orbital becomes more localized on

the lower-energy atom. For simplicity you may use the

orthogonality assumption h1|2i= 0. Explain how this

calculation can be used to describe a crossover between

covalent and ionic bonding.

Here we have instead, the two by two hamiltonian matrix

ǫ1t

t∗ǫ2

where we have now absorbed Vcross into the values of ǫi. The lower

energy eigenstate is

Eground =1

2n(ǫ1+ǫ2) + p(ǫ1−ǫ2)2+ 4t2o

with normalized eigenvector

ψ=(X, 2t)

√4t2+X2

with

X=E2−E1+p(E2−E1)2+ 4t2

-7.5 -5 -2.5 2.5 5 7.5 HE2 -E1Lt

0.2

0.4

0.6

0.8

Probability

Fig. 6.1 Probability (squared ampli-

tude) of ground state wavefunction be-

ing on site 1 (solid) or site 2 (dashed)

as a function of E2−E1.

When E2−E1≫tthen X≫tand all of the wavefunction ends

up on the ﬁrst atom (i.e., the one with the lower energy). Similarly, if

E1−E2≫tthen conversely all of the weight of the wavefunction ends

up on the second atom. In Fig. 6.1 it is shown how the weight of the

wavefunction moves from towards the lower energy atom as a function

of energy.

When the energies on the two sites are equal, one has an equal shar-

ing of the wavefunction in the ground state (as in the prior problem).

However, as the energy diﬀerence is increased, the ground state moves

more towards the lower energy site, until the bond is completely ”ionic“

meaning that the electron is completely transferred from one atom to

the other.

(6.4) Ionic Bond Energy Budget

The ionization energy of a sodium atom is about 5.14

eV. The electron aﬃnity of a chlorine atom is about 3.62

eV. When a single sodium atom bonds with a single chlo-

rine atom, the bond length is roughly 0.236 nm. As-

suming that the cohesive energy is purely Coulomb en-

ergy, calculate the total energy released when a sodium

atom and a chlorine atom come together to form a NaCl

molecule. Compare your result to the experimental value

of 4.26 eV. Qualitatively account for the sign of your er-

ror.

Ecoh =e2

4πǫ0d= 6.10eV

Thus the total bonding energy is

E=−5.14eV + 3.62eV + 6.10eV = 4.58eV

which is slightly larger than the experimentally measured bonding en-

ergy. The reason for the discrepancy is that there must be a repulsive

force in addition to the coulomb attractive force which reduces the mag-

nitude of the cohesive (binding) energy.

In the Fig. 6.2 the lower curve is the pure Coulomb energy. The upper

curve includes a short range repulsion. The repulsion must be there, or

there would be no minimum in the curve!

✻

✲

V(x)

Fig. 6.2 Probability (squared ampli-

tude) of ground state wavefunction be-

ing on site 1 (solid) or site 2 (dashed)

as a function of E2−E1.

(6.5) LCAO Done Right*

(a)* In Exercise 6.2 we introduced the method of lin-

ear combination of atomic orbitals. In that exercise we

assumed that our basis of orbitals is orthonormal. In this

exercise we will relax this assumption.

Consider now many orbitals on each atom (and poten-

tially many atoms). Let us write

|ψi=

i=1

φi|ii

for an arbitrary number Nof orbitals. Let us write the

Nby Noverlap matrix Swhose elements are

Si,j =hi|ji

In this case do not assume that Sis diagonal.

Using a similar method as in Exercise 6.2, derive the

new “Schroedinger equation”

Hφ=ESφ(6.3)

with the same notation for Hand φas in Exercise 6.2.

This equation is known as a “generalized eigenvalue prob-

lem” because of the Son the right-hand side.

(b)** Let us now return to the situation with only two

atoms and only one orbital on each atom but such that

h1|2i=S1,26= 0. Without loss of generality we may as-

sume hi|ii= 1 and S1,2is real. If the atomic orbitals

are s-orbitals then we may assume also that tis real and

positive (why?).

Use Eq. 6.3 to derive the eigenenergies of the system.

(a) This is very similar to 6.2.

E=hψ|H|ψi

hψ|ψi=Pn,m φ∗

nHnmφm

Pn,m φ∗

nSnmφm

(6.4)

We can extremize by diﬀerentiating with respect to φ∗

nto give

0 = ∂E

∂φ∗

=PmHnmφm

Pn,m φ∗

nSnmφm− Pn,m φ∗

nHnmφm

Pn,m φ∗

nSnmφm!Pn,m Snmφm

Pn,m φ∗

nSnmφm

0 = X

mHnmφm−EX

Snmφm

where we have used Eq. 6.4 to identify E.

(b) An s-orbital can be taken to be manifestly postive everywhere (no

nodes), so overlaps Sij must be real and positive. Here the easiest thing

44 Chemical Bonding

to do is to just apply S−1to both sides of the equation to give the

eigenvalue problem

SHφ=Eφ

Here, let us write the Hamiltonian as

H=ǫ t

t∗ǫ

where we have absorbed Vcross into ǫ, and

S=1S

S1

with S=S12. The matrix we want to diagonalize is then

S−1H=1

1−S2ǫ−St t −ǫS

t−ǫS ǫ −St 

The eigenvalues are easily seen to be

E=1

1−S2([ǫ−St]± |t−ǫS|)

(6.6) Van der Waals Bonding in Detail*

(a) Here we will do a much more precise calculation

of the van der Waals force between two hydrogen atoms.

First, let the positions of the two nuclei be separated by

a vector R, and let the vector from nucleus 1 to electron

1 be r1and let the vector from nucleus 2 to electron 2 be

r2as shown in the following ﬁgure.

■✒

r1r2

✲

Let us now write the Hamiltonian for both atoms

(assuming ﬁxed positions of nuclei, i.e., using Born–

Oppenheimer approximation) as

H=H0+H1

H0=p12

2m+p22

2m−e2

4πǫ0|r1|−e2

4πǫ0|r2|

H1=e2

4πǫ0|R|+e2

4πǫ0|R−r1+r2|

−e2

4πǫ0|R−r1|−e2

4πǫ0|R+r2|

Here H0is the Hamiltonian for two non-interacting hy-

drogen atoms, and H1is the interaction between the

atoms.

Without loss of generality, let us assume that Ris in

the ˆxdirection. Show that for large Rand small ri, the

interaction Hamiltonian can be written as

H1=e2

4πǫ0|R|3(z1z2+y1y2−2x1x2) + O(1/R4)

where xi, yi, ziare the components of ri. Show that this

is just the interaction between two dipoles.

(b) Perturbation Theory:

The eigenvalues of H0can be given as the eigen-

values of the two atoms separately. Recall that the

eigenstates of hydrogen are written in the usual nota-

tion as |n, l, miand have energies En=−Ry/n2with

Ry = me4/(32π2ǫ2

0~2) = e2/(8πǫ0a0) the Rydberg (here

l≥0, |m| ≤ land n≥l+ 1). Thus the eigenstates

of H0are written as |n1, ll, m1;n2, l2, m2iwith energies

En1,n2=−Ry(1/n2

1+ 1/n2

2). The ground state of H0is

|1,0,0; 1,0,0i.

Perturbing H0with the interaction H1, show that

to ﬁrst order in H1there is no change in the ground-state

energy. Thus conclude that the leading correction to the

ground-state energy is proportional to 1/R6(and hence

the force is proportional to 1/R7).

Recalling second-order perturbation theory show

that we have a correction to the total energy given by

δE =

n1, n2

l1, l2

m1, m2

|<1,0,0; 1,0,0|H1|n1, ll, m1;n2, l2, m2i|2

E0,0−En1,n2

Show that the force must be attractive.

(c)*Bounding the binding energy:

First, show that the numerator in this expression is

zero if either n1= 1 or n2= 1. Thus the smallest En1,n2

that appears in the denominator is E2,2. If we replace

En1,n2in the denominator with E2,2then the |δE|we

calculate will be greater than than the |δE|in the exact

calculation. On the other hand, if we replace En1,n2by

0, then the |δE|will always be less than the δE of the

exact calculation.

Make these replacements, and perform the remain-

ing sum by identifying a complete set. Derive the bound

6e2a5

4πǫ0R6≤ |δE| ≤ 8e2a5

4πǫ0R6

You will need the matrix element for a hydrogen atom

h1,0,0|x2|1,0,0i=a2

where a0= 4πǫ0~2/(me2) is the Bohr radius. (This last

identity is easy to derive if you remember that the ground-

state wavefunction of a hydrogen atom is proportional to

e−r/2a0.)

(a) In fact, this is more or less the deﬁnition of dipole interaction! Let

us start by deriving

|R−a|=1

Rq1 + 2a·R

R2+a2

R1−a·R

R2+−1

R2+3

(a·R)2

R4+...(6.5)

Applying this to all the terms in

H1=e2

4πǫ0|R|+e2

4πǫ0|R−r1+r2|

−e2

4πǫ0|R−r1|−e2

4πǫ0|R+r2|

We discover that the ﬁrst two leading orders completely cancel, thus

leaving us with only the contribution of the terms coming from the

square bracketed terms of Eq. 6.5. We then obtain

H1=e2

4πǫ0Rr2

1+r2

2− |r1−r2|2

2R2+3(R·(r2−r1))2−(R·r1)2−(R·r2)2)

R4

with Rin the ˆxdirection, this simpliﬁes to

H1=e2

4πǫ0R3r1·r2+ 3 (x1−x2)2−x2

1−x2

2

which then simpliﬁes to the desired answer.

(b) The expectation of xor yor zin the ground state of the hydrogen

atom |100iis zero due to the fact that the state is spherically symmetric.

As a result taking the expectation of H1in the ground state gives zero.

Thus the leading correction to the energy occurs at second order in H1

46 Chemical Bonding

and is proportional to 1/R6(assuming it does not vanish, which we will

show next).

Given then the form of the 2nd order perturbation theory, note that

every term has an overall negative sign (numerator is positive, denomi-

nator is negative). Thus the interaction is of the form −C/R6for some

positive constant C(to be calculated), and hence the force is attractive.

state, this means H1has zero matrix element unless n1>1 and n2>1

(for n= 1 we must have l=m= 0).

First let us consider the quantity

I=X

n1, n2

l1, l2

m1, m2

|<1,0,0; 1,0,0|H1|n1, ll, m1;n2, l2, m2i|2(6.6)

n1, n2

l1, l2

m1, m2

h1,0,0; 1,0,0|H1|n1, ll, m1;n2, l2, m2i(6.7)

× hn1, ll, m1;n2, l2, m2|H1|1,0,0; 1,0,0i(6.8)

We notice the complete set in the middle here, so we do the sum over

the set to obtain

I=h1,0,0; 1,0,0|H2

1|1,0,0; 1,0,0i

=e2

4πǫ0R2hy2

1ihy2

2i+hz2

1ihz2

2i+ 4hx2

1ihx2

2i

with all expectations being in the ground state of the respective hydrogen

atom (each bracket gives a2

0). Thus we obtain

I=e2

4πǫ0R32

6a4

Now, the upper bound is deﬁned by setting En1,n2=E2,2whereas

the lower bound is deﬁned by setting En1,n2= 0 so we obtain

E0,0

<|δE|<I

E0,0−E2,2

Here using the excitation spectrum of the hydrogen atom, we have

E0,0=-2Ry and E2,2=−Ry/2 so E0,0−E2,2=−(3/2)Ry, and us-

ing the fact that the Rydberg is e2/(8πǫ0a0) we have

W=I/Ry = 12e2a5

4πǫ0R6

so our inequality is W

2<|δE|<2W

which is the required result.

Types of Matter 7

There are no exercises for chapter 7.

One-Dimensional Model of

Compressibility, Sound,

and Thermal Expansion 8

(8.1) Potentials Between Atoms

As a model of thermal expansion, we study the distance

between two nearest-neighbor atoms in an anharmonic

potential that looks roughly like this

✻

✲

✻

kBT

V(x)

where xis the distance between the two neighboring

atoms. This potential can be expanded around its mini-

mum as

V(x) = κ

2(x−x0)2−κ3

3! (x−x0)3+. . . (8.1)

where the minimum is at position x0and κ3>0. For

small energies, we can truncate the series at the cubic

term. (Note that we are deﬁning the energy at the bot-

tom of the well to be zero here.)

A very accurate approximate form for interatomic po-

tentials (particularly for inert atoms such as helium or

argon) is given by the so-called Lennard-Jones potential

V(x) = 4ǫσ

x12 −σ

x6+ǫ(8.2)

where ǫand σare constants that depend on the particular

atoms we are considering.

What is the meaning of the exponent 6 in the sec-

ond term of this expression (i.e., why is the exponent

necessarily chosen to be 6).

By expanding Eq. 8.2 around its minimum, and

comparing to Eq. 8.1, calculate the values of the coef-

ﬁcients x0,κ, and κ3for the Lennard-Jones potential in

terms of the constants ǫand σ. We will need these results

in Exercise 8.3.

The exponent 6 determines the long range behavior of the potential

and is ﬁxed by the form of the Van der Waals interaction.

By setting dV/dx = 0 we ﬁnd the minimum at x0= 21/6σ.

Taking second and third derivatives at this position

d2V

dx2x=x0

= 36 ×22/3ǫ/σ2=κ≈57ǫ/σ2(8.3)

and d3V

dx3x=x0

=−756√2ǫ/σ3=−κ3≈1069ǫ/σ3(8.4)

50 Compressibility, Sound, and Thermal Expansion

(8.2) Classical Model of Thermal Expansion

(i) In classical statistical mechanics, we write the ex-

pectation of xas

hxiβ=Rdx x e−βV (x)

Rdx e−βV (x)

Although one cannot generally do such integrals for ar-

bitrary potential V(x) as in Eq. 8.1, one can expand the

exponentials as

e−βV (x)=e−βκ

2(x−x0)21 + βκ3

6(x−x0)3+...

and let limits of integration go to ±∞.

Why is this expansion of the exponent and the ex-

tension of the limits of integration allowed?

Use this expansion to derive hxiβto lowest order in

κ3, and hence show that the coeﬃcient of thermal expan-

sion is

α=1

dT ≈1

dhxiβ

dT =1

kBκ3

2κ2

with kBBoltzmann’s constant.

In what temperature range is the above expansion

valid?

While this model of thermal expansion in a solid is

valid if there are only two atoms, why is it invalid for the

case of a many-atom chain? (Although actually it is not

so bad as an approximation!)

hxiβ=Rdx x e−βV (x)

Rdx e−βV (x)

with

e−βV (x)=e−βκ

2(x−x0)21 + βκ3

6(x−x0)3+...

Redeﬁne y= (x−x0) so we have

hxiβ=Rdy (y+x0)e−βκ

2y2h1 + βκ3

6y3+...i

Rdye−βκ

2y2h1 + βκ3

6y3+...i

=x0+

βκ3

6Rdy y4e−βκ

2y2

Rdy e−βκ

2y2+...

Using Zdxe−ax2=pπ/a

as a generating function, we have

Zdxx4e−ax2= (d/da)2Zdxe−ax2= (d/da)2pπ/a = (3/4)pπ/a5

gives

hxiT=x0+κ3(kbT)

2κ2+...

Thus 1

dhxiT

dT =1

κ3kb

2κ2

In order for this calculation to be valid, since we have treated the cubic

term perturbatively, this term actually must be small compared to the

leading term.

Roughly, if the leading term is most important, we have κ(x−x0)2∼

kbTwhich means that the typical deviation is |x−x0| ∼ pkbT/κ. Then

in order to have the leading term be larger than the cubic term, we have

κ|x−x0|2≫κ3|x−x0|3

kbT≫κ3(kbT/κ)3/2

or equivalently

kbT≪κ3/κ2

For a many atom chain, one must solve for the normal modes of the

chain. Then at ﬁnite temperature one should “occupy” the phonons

thermally and then calculate the eﬀect of the nonlinear terms in this

state. What we have done in this problem is more or less the thermal

expansion of an Boltzmann model of a solid.

(8.3) Properties of Solid Argon

For argon, the Lennard-Jones constants ǫand σfrom

Eq. 8.2 are given by ǫ= 10meV and σ=.34nm. You will

need to use some of the results from Exercise 8.1.

(a) Sound

Given that the atomic weight of argon is 39.9, estimate

the sound wave velocity in solid argon. The actual value

of the longitudinal velocity is about 1600 m/sec.

(b) Thermal Expansion

Using the results of Exercise 8.2, estimate the thermal

expansion coeﬃcient αof argon. Note: You can do this

part even if you couldn’t completely ﬁgure out Exercise

8.2!

The actual thermal expansion coeﬃcient of argon is

approximately α= 2 ×10−3/K at about 80K. However,

at lower temperature αdrops quickly. In the next ex-

ercise will use a more sophisticated quantum model to

understand why this is so.

Sound: From the text in one dimension we have

v=qκx2

0/m

where x0is the neighbor distance. From our above problem on Lennard-

Jones, we have

κx2

0= 72ǫ

and the mass m=.0399kg/NA. Which gives

v= 1320m/s

Thermal expansion. Plugging in our above expression for x0,κand

κ3for the Lennard Jones case, we obtain

α=1

κ3kb

2κ2=7kb

48ǫ≈.0012/K

52 Compressibility, Sound, and Thermal Expansion

(8.4) Quantum Model of Thermal Expansion

(a) In quantum mechanics we write a Hamiltonian

H=H0+δV

where

H0=p2

2m+κ

2(x−x0)2(8.5)

is the Hamiltonian for the free Harmonic oscillator, and

δV is the perturbation (see Eq. 8.1)

δV =−κ3

6(x−x0)3+...

where we will throw out quartic and higher terms.

What value of mshould be used in Eq. 8.5?

Using perturbation theory it can be shown that, to

lowest order in κ3the following equation holds

hn|x|ni=x0+Enκ3/(2κ2) (8.6)

where |niis the eigenstate of the Harmonic oscillator

whose energy is

En=~ω(n+1

2) + O(κ3)n≥0

with ω=pκ/m. In (c) we will prove Eq. 8.6. For now,

take it as given.

Note that even when the oscillator is in its ground

state, the expectation of xdeviates from x0. Physically

why is this?

(b)* Use Eq. 8.6 to calculate the quantum expectation

of xat any temperature. We write

hxiβ=Pnhn|x|nie−βEn

Pne−βEn

Derive the coeﬃcient of thermal expansion.

Examine the high temperature limit and show that

it matches that of Exercise 8.2.

In what range of temperatures is our perturbation

expansion valid?

In light of the current quantum calculation, when

is the classical calculation from Exercise 8.2 valid?

Why does the thermal expansion coeﬃcient drop

at low temperature?

(c)** Prove Eq. 8.6 by using lowest-order perturbation

theory.

Hint: It is easiest to perform this calculation by using

raising and lowering (ladder) operators. Recall that one

can deﬁne operators aand a†such that [a, a†] = 1 and

a†|ni0=√n+ 1|n+ 1i0

a|ni0=√n|n−1i0.

Note that these are kets and operators for the unper-

turbed Hamiltonian H0. In terms of these operators, we

have the operator x−x0given by

x−x0=r~

2mω (a+a†).

(a) For a system of two atoms, one should use the ”reduced” mass

µ=m1m2/(m1+m2) which for two identical atoms is m/2.

Quantum mechanically the oscillator does not sit in its exact mini-

mum, rather it has “quantum ﬂuctuations” around this minimum . Since

the well is asymmetric around the minimum, the expectation value of

the position is slightly greater than x0even in the ground state.

(b)

hxiβ=Pnhn|x|nie−βEn

Pne−βEn=Pnhn|(x0+Enκ3/(2κ2))|nie−βEn

Pne−βEn

=x0+hEiβκ3

2κ2

where hEiβis the energy expectation of a harmonic oscillator of fre-

quency ωat temperature β= 1/(kbT). As derived above when we

discussed Einstein model, this expectation is

hEiβ= (nB(β~ω) + 1

2)~ω=~ω

2coth(β~ω/2)

with nBthe boson occupation factor. Thus we obtain

hxiβ=x0+ (κ3~ω/(4κ2)) coth(β~ω/2)

The coeﬃcient of thermal expansion is then

α=1

dhxiβ

dT =κ3

2x0κ2

dhEi

where C=dhEi/dT is exactly the speciﬁc heat of a harmonic oscillator

(recall einstein model). Thus we obtain

α=κ3C

2x0κ2=κ3

2x0κ2kb(β~ω)2eβ~ω

(eβ~ω−1)2

We can see that the thermal expansion drops as modes ”freeze” out,

entirely analogous to the speciﬁc heat dropping at low temperature.

In the high Tlimit, the speciﬁc heat Cgoes to kb, so we obtain

α= (κ3kb/(2x0κ2))

in agreement with the classical result.

For the classical calculation to be valid, we must have kbT≫~ωso

that quantum mechanics can be replaced by classical mechanics.

the raising and lowering operators

a=rmω

2~[(x−x0) + (i/mω)p]

a†=rmω

2~[(x−x0)−(i/mω)p]

(8.7)

So that the position is given by

x−x0=p~/(2mω)(a+a†)

Write the eigenstates |ni0to mean the eigenstates of the unperturbed

oscillator with energy En=~ω(n+ 1/2). In lowest order perturbation

theory, the perturbed eigenket is given by

|ni=|ni0+−κ3

6~

2mω 3/2X

m|mi00hm|(a+a†)3|ni0

En−Em

So that we have

hn|x−x0|ni=−2κ3

6~

2mω 2X

0hn|a+a†|mi0 0hm|(a+a†)3|ni0

En−Em

There are two nonzero terms of the sum, the one where n=m+1 which

gives us a denominator ~ωand a numerator

0hn|a†|n−1i0 0hn−1|a†aa +aa†a+aaa†|ni0

=√n√n(n−1) + n√n+ (n+ 1)√n= 3n2

54 Compressibility, Sound, and Thermal Expansion

and also the term with n=m−1 which gives a denominator −~ωand

a numerator

0hn|a|n+ 1i0 0hn+ 1|a†a†a+a†aa†+aa†a†|ni0

=√n+ 1 n√n+ 1 + (n+ 1)√n+ 1 + (n+ 2)√n+ 1= 3(n+ 1)2

Thus we get

hn|x−x0|ni=2κ3

6~

2mω 21

~ω(6n+ 3) = Enκ3/(2κ2)

Vibrations of a

One-Dimensional

Monatomic Chain 9

(9.1) Classical Normal Modes to Quantum

Eigenstates

In Section 9.3 we stated without proof that a classical

normal mode becomes a quantum eigenstate. Here we

prove this fact for a simple diatomic molecule in a poten-

tial well (see Exercise 2.7 for a more diﬃcult case, and see

also Exercise 9.7 where this principle is proven in more

generally).

Consider two particles, each of mass min one dimen-

sion, connected by a spring (K), at the bottom of a poten-

tial well (with spring constant k). We write the potential

energy as

U=k

2(x2

1+x2

2) + K

2(x1−x2)2

Write the classical equations of motion.

Transform into relative xrel = (x1−x2) and center

of mass xcm = (x1+x2)/2 coordinates.

(a) Show that in these transformed coordinates, the

system decouples, thus showing that the two normal

modes have frequencies

ωcm =pk/m

ωrel =p(k+ 2K)/m.

Note that since there are two initial degrees of freedom,

there are two normal modes.

Now consider the quantum-mechanical version of the

same problem. The Hamiltonian is

H=p2

2m+p2

2m+U(x1, x2)

Again transform into relative and center of mass co-

ordinates.

Deﬁne the corresponding momenta prel = (p1−p2)/2

and pcm = (p1+p2).

(b) Show that [pα, xγ] = −i~δα,γ where αand γtake

the values cm or rel.

Hamiltonian decouples into two independent harmonic

oscillators with the same eigenfrequencies ωcm and ωrel.

Conclude that the spectrum of this system is

Enrel,ncm =~ωrel(nrel +1

2) + ~ωcm(ncm +1

where ncm and nrel are non-negative integers.

(d) At temperature Twhat is the expectation of the

energy of this system?

(a) The equations of motion are

m¨x1=−kx1−K(x1−x2)

m¨x2=−kx2−K(x2−x1)

Taking the sum and diﬀerence of these two equations gives

m¨xcm =−kxcm

m¨xrel =−(k+ 2K)xrel

56 Vibrations of a One-Dimensional Monatomic Chain

which we identify as harmonic oscillators of frequencies

ωcm =pk/m

ωrel =p(k+ 2K)/m

(b) We have [pi, xj] = −i~δij . Deﬁning prel = (p1−p2)/2 and pcm =

p1+p2and xrel =x1−x2and xcm = (x1+x2)/2 we obtain

[prel, xrel] = 1

2([p1, x1]−[p1, x2]−[p2, x1] + [p2, x2])

2(−i~)(1 −0−0 + 1) = −i~

[prel, xcm] = 1

4([p1, x1] + [p1, x2]−[p2, x1]−[p2, x2])

4(−i~)(1 + 0 −0−1) = 0

[pcm, xrel] = ([p1, x1]−[p1, x2] + [p2, x1]−[p2, x2])

= (−i~)(1 + 0 −0−1) = 0

[pcm, xcm] = 1

2([p1, x1] + [p1, x2] + [p2, x1] + [p2, x2])

2(−i~)(1 + 0 + 0 + 1) = −i~

One could alternately dmonstrate that prel =−i~∂/∂xrel etc using ja-

cobians to conﬁrm these commutations.

p1=pcm/2 + prel

p2=pcm/2−prel

x1=xcm +xrel/2

x2=xcm −xrel/2

So the Hamiltonian becomes

H=1

2m(p2

1+p2

2) + k

2(x2

1+x2

2) + K

2(x1−x2)2

2m(p2

cm/2 + 2p2

rel) + k

2(2x2

cm +x2

rel/2) + K

2x2

rel

which decouples into (note the total mass is 2mand the reduced mass

is m/2)

H1=p2

2(2m)+2k

2x2

H2=p2

rel

2(m/2) +(k/2 + K)

2x2

rel

which are two independent harmonic oscilators with frequencies

ωcm =pk/m

ωrel =p(k+ 2K)/m

such that the spectrum can be written as

Enrel ,ncm =~ωrel(nrel +1

2) + ~ωcm(ncm +1

At temperature T, the expectation of the energy of this system will

correspondingly be

hEi=~ωrel(nB(β~ωrel) + 1/2) + ~ωcm(nB(β~ωcm) + 1/2)

where nB(x) = 1/(ex−1) is the usual Bose factor.

The purpose of this exercise is not just to do another quantum me-

chanics problem. It is here to point out that coupled deg rees of freedom

act just like a single simple harmonic oscillator once the degrees of free-

dom are “rediagonalized”. This is important motivation for treating

phonons (coupled modes of springs) as individual harmonic oscillators.

(9.2) Normal Modes of a One-Dimensional

Monatomic Chain

(a)‡Explain what is meant by “normal mode” and by

“phonon”.

Explain brieﬂy why phonons obey Bose statistics.

(b)‡Derive the dispersion relation for the longitudinal

oscillations of a one-dimensional mass-and-spring crystal

with Nidentical atoms of mass m, lattice spacing a, and

spring constant κ(motion of the masses is restricted to

be in one dimension).

(c)‡Show that the mode with wavevector khas the

same pattern of mass displacements as the mode with

wavevector k+ 2π/a. Hence show that the dispersion

relation is periodic in reciprocal space (k-space).

How many diﬀerent normal modes are there.

(d)‡Derive the phase and group velocities and sketch

them as a function of k.

What is the sound velocity?

Show that the sound velocity is also given by vs=

1/√βρ where ρis the chain density and βis the com-

pressibility.

(e) Find the expression for g(ω), the density of states

of modes per angular frequency.

Sketch g(ω).

(f) Write an expression for the heat capacity of this

one-dimensional chain. You will inevitably have an inte-

gral that you cannot do analytically.

(g)* However, you can expand exponentials for high

temperature to obtain a high-temperature approxima-

tion. It should be obvious that the high-temperature

limit should give heat capacity C/N =kB(the law of

Dulong–Petit in one dimension). By expanding to next

non-trivial order, show that

C/N =kB(1 −A/T 2+. . .)

where

A=~2κ

6mk2

(a) A normal mode is a periodic collective motion where all particles

move at the same frequency. A phonon is a quantum of vibration.

[I do not like the deﬁnition ”a quantum of vibrational energy”. The

vibration does carry energy, but it carries momentum as well, so why

specify energy only?]

Each classical normal mode of vibration corresponds to a quantum

mode of vibration which can be excited multiple times. A single mode

may be occupied by a single phonon, or it may be occupied with mul-

tiple phonons corresponding to a larger amplitude oscillation. The fact

58 Vibrations of a One-Dimensional Monatomic Chain

that the same state may be multiply occupied by phonons means that

phonons must be bosons.

(b) The equation of motion for the nth particle along the chain is given

m¨xn=κ(xn+1 −xn) + κ(xn−1−xn) = κ(xn+1 +xn−1−2xn)

note that na is the equilibrium position of the nth particle. Using the

ansatz

xn=Aeiωt−ikna

we obtain

−ω2meiωt−ikna =κeiωt(eik(n+1)a+eik(n−1)a−2eikn)

ω2m= = 2κ(cos(ka)−1)

ω=p(2κ/m)(cos(ka)−1) = 2pκ/m |sin(ka/2)|

Fig. 9.1 Dispersion relation for

vibrations of the one-dimensional

monatomic harmonic chain. The

dispersion is periodic in k→k+ 2π/a.

0 k=+π/a k=−π/a

ω = 2√ m

(c)

e−i(k+2π/a)na =e−i(k+2π/a)na =e−ikna

If you assume periodic boundary conditions, then k= 2πm/L but kis

identiﬁed with k+ 2π/a so that there are therefore exactly N=L/a

diﬀerent normal modes.

(d)

vphase =ω(k)/k = 2pκ/m |sin(ka/2)|/k

and

vgroup =dω(k)/dk =pκ/m a cos(|k|a/2) = (a/2)ω0q1−ω2/ω2

Fig. 9.2 The monatomic harmonic chain. Right: Phase velocity. Left: Group

velocity. Note velocities are signed quantities, to the left of the origin, the velocity

should have negative sign.

where ω0= 2pκ/m. Note that the phase velocity is not periodic in the

Brillouin zone! One can understand this if you think carefully about

aliasing of waves. The phase velocity is the velocity at which the peaks

of waves move. However, the waves are only deﬁned at the position of

the masses along the chain. We write cos(kna) for the positions of the

masses at some time, but this only deﬁnes the value of the wave for

integer n. For integer n, we have kis the same as k+ 2π/a. However,

the ”peak” of this function may be between the integer values of n.

However, when we make nnon-integer, then kis no longer the same as

k+ 2π/n.

For sketches see ﬁgure 9.2

The sound velocity is the velocity at small k. This is

v=apκ/m

. The density of the chain is ρ=m/a and the compressibility is β=

−(1/L)dL/dF = 1/(κa). Thus we obtain v−2=ρβ

(e) Note ﬁrst that

(ω(k)/2)2+ (vgroup(k)/a)2=κ/m (9.1)

Density of states is uniform in k. If there are Nsites in the system,

there are Nmodes total. The density of states in kis therefore dN/dk =

Na/(2π) = L/(2π) where Lis the length of the system.

Thus we have

g(ω) = dN/dω = (dN/dk)(dk/dω) = Na

2πvgroup

2πpκ/m cos(|k|a/2)

=2N

2πp(κ/m)−(ω(k)/2)2(9.2)

where we have used Eq. 9.1.

60 Vibrations of a One-Dimensional Monatomic Chain

Fig. 9.3 The one dimensional harmonic chain. Density of states g(ω). Note that the

DOS diverges at ω= 2pk/m where the group velocity goes to zero.

The additional factor of 2 that appears up top is to account for the

fact that for each value of ω > 0 there are actually two values of kwith

that ω. (Note if you integrate over frequency you correctly get back N

degrees of freedom).

(f) The energy stored in the chain is given by

U=Zdωg(ω)~ω(nB(ω) + 1/2)

so the heat capacity is C=∂U/∂T . Note that we can drop the +1/2

since it has no derivative.

(g) To recover the law of Dulong-Petit, one takes the high temperature

limit of nB(ω) = kBT/~ωso th at we have

C=∂

∂T Zdωg(ω)(kBT) = kBZdωg(ω) = kBN

To go further, we use the high temperature expansion (expanding

1/(ex−1) for small x)

nB(ω) + 1/2 = kBT

~ω+1

~ω

kBT+...

So that we now have

C=∂U

∂T =kBN−1

T2Zdω~ωg(ω)1

~ω

kB

So that the coeﬃcient Adeﬁned in the problem has the values

A=~2

12Nk2

BZdωω2g(ω)

Inserting our expression for g(ω) we obtain

A=~2

12πk2

BZωmax

dω ω2

p(κ/m)−(ω/2)2

Deﬁning x= (ω/2)pm/κ we obtain

A=~2

12πk2

8κ

mZ1

dx x2

√1−x2

The integral is evaluated to give π/4 (make the substitution x= sin θ).

Thus we obtain

A=~2

6k2

as required.

(9.3) More Vibrations

Consider a one-dimensional spring and mass model of

a crystal. Generalize this model to include springs not

only between neighbors but also between second nearest

neighbors. Let the spring constant between neighbors be

called κ1and the spring constant between second neigh-

bors be called κ2. Let the mass of each atom be m.

(a) Calculate the dispersion curve ω(k) for this model.

(b) Determine the sound wave velocity. Show the group

velocity vanishes at the Brillouin zone boundary.

(a) Use the same approach

m¨xn=κ1(xn+1 −xn) + κ1(xn−1−xn) + κ2(xn+2 −xn) + κ1(xn−2−xn)

=κ1(xn+1 +xn−1−2xn) + κ2(xn+2 +xn−2−2xn)

Using the same ansatz

xn=Aeiωt−ikna

we obtain

−mω2= 2κ1(cos(ka)−1) + 2κ2(cos(2ka)−1) (9.3)

ω=r2κ1

m(cos(ka)−1) + 2κ2

m(cos(2ka)−1)

(b) To obtain the sound velocity, expand for small kto obtain

ω=r2κ1

(ka)2

2+2κ2

(2ka)2

2= arκ1+ 4κ2

m!k

Thus the sound velocity is

vs=arκ1+ 4κ2

The easiest way to examine ∂ω/∂k at the zone boundary is to diﬀer-

entiate Eq. 9.3 to given

mω∂ω/∂k =−2aκ1sin(ka)−4aκ2sin(2ka)

At the zone boundary k=π/2 both terms on the right hand side are

zero, hence we have zero group velocity.

62 Vibrations of a One-Dimensional Monatomic Chain

(9.4) Decaying Waves

In the dispersion curve of the harmonic chain (Eq. 9.3),

there is a maximum possible frequency of oscillation

ωmax. If a vibration with frequency ω > ωmax is forced

upon the chain (say by a driving force) the “wave” will

not propagate along the chain, but rather will decay as

one moves away from the point where the oscillation is im-

posed (this is sometimes known as an “evanescent” wave).

With ω > ωmax solve Eq. 9.3 for a complex kto determine

the decay length of this evanescent wave. What happens

to this length as ω→ωmax ?

The usual form is

Ω = ω/ωmax =±sin(ka/2)

with ±chosen so Ω is positive. Thus

ka =±2 sin−1Ω

When Ω = 1 then ka =π. When Ω >1 then ka becomes complex

with a real part of π. The complex part gets smaller and vanishes as

Ω approaches 1 from above. The complex part of kgives the inverse

length scale of the wave’s decay (and it can decay left-going or right-

going depending on the ±sign). So as Ω approachs 1 from above, the

length scale of decay grows longer and longer until at 1 it extends over

the entire system and is a nondecaying wave.

See also the solution of 9.6 below, which is quite similar.

To check these properties of the arcsin, consider

y= sin(x) = z−z−1

where z=eix. Multiplying by zwe obtain the quadratic

z2−2izy −1 = 0

Solving this equation gives

z=iy ±p1−y2

Note that for y > 1 we have z=eix along the positive imaginary axis,

which means x=π/2 + real. For y= 1 we have x=π/2 whereas for

y < 1 we obtain a complex valued zwith unti magnitude, thus a real x.

If one is concerned that this solution might not work, one can always

go back to the equations of motion

m¨xn=κ(xn+1 −xn) + κ(xn−1−xn) = κ(xn+1 +xn−1−2xn)

Using the ansatz

xn=Aeiωt−ikna

we obtain

−ω2meiωt−ikna =κeiωt(eik(n+1)a+eik(n−1)a−2eikn)

ω2m= = 2κ(cos(ka)−1)

Note that nowhere did we specify that kis real (!) so we can happily

extend into the complex plane. Using double angle formula

ω2=4κ

msin2(ka/2)

The only thing we would need to be careful about is choosing the sign

of the square root.

(9.5) Reﬂection at an Interface*

Consider a harmonic chain of equally spaced identical

masses of mass mwhere left of the n= 0 mass the spring

constant is κLbut right of the n= 0 mass, the spring

constant is κR, as shown in this ﬁgure.

n= 0

mmmmmmm

κLκLκLκRκRκR

A wave with amplitude Iis incident on this interface

from the left, where it can be either transmitted with

amplitude Tor reﬂected with amplitude R. Using the

following ansatz form

δxn=T eiωt−ikLna n≥0

Ieiωt−ikRna +Reiωt+ikRna n < 0

derive T/I and R/I given ω,κL,κRand m.

Oops sorry about the typo, it should read

δxn=T eiωt−ikRna n≥0

Ieiωt−ikLna +Reiωt+ikLna n < 0

On both the left and the right, we must have the usual relations

between kL,R and ω

ω= 2qκL,R/m|sin(kL,Ra/2)|

kL,R =2

asin−1ω

2qm/κL,R

with these values of k, the ansatz form will then satisfy the equations of

motion except near the boundary.

Near the boundary, we have equations of motion for δx−1and δx0

which involve δx coordinates on both sides of the junction. We write

m¨

δx−1=κL(δx−2+δx0−2δx−1)

m¨

δx0=κL(δx−1−δx0) + κR(δx1−δx0)

64 Vibrations of a One-Dimensional Monatomic Chain

Plugging in the ansatz form of the wavefunction and doing a bit of

algebra gives us

(−mω2+ 2κL)IeikLa+Re−ikLa=κLIeikL2a+Re−ikL2a+T

(−mω2+κL+κR)T=κLIeikLa+Re−ikLa+κRT e−ikRa

(9.4)

The ﬁrst equation simpliﬁes quite a bit by rewriting the dispersion rela-

tion as

−mω2= 2κL(cos(kLa)−1)

So that

−mω2+ 2κL=κLeikLa+e−ikLa(9.5)

and the ﬁrst equation becomes simply

I+R=T

which is just extending the wavefunction from the left hand side until

it hits position zero and matching it to the value of the wave from the

right at this position.

Plugging this into the second equation of Eq. 9.4 and solving to obtain

T/I =κL2isin(kLa)

−mω2+κL(1 −e−ikLa) + κR(1 −e−ikRa)

In the denominator we can simplify further applying 9.5 to give us

−mω2/2 + κL(1 −e−ikLa) = iκLsin(kLa)

and similarly for the right hand wave so that we get

T/I =2

1 + κRsin(kRa)

κLsin(kLa)

(9.6) Impurity Phonon Mode*

Consider a harmonic chain where all spring constants

have the same value κand masses have value m, except

for the mass at position n= 0 which instead has value

M < m as shown in this ﬁgure:

n= 0

mmmMmmm

κκκκκκ

Along with traveling wave solutions, there can be a

standing wave normal mode localized near the impurity.

Use an ansatz of the form

δxn=Aeiωt−q|n|a

with qreal to solve for the frequency of this impurity

mode. Consider your result in the context of Exercise

9.4.

Another error here: qneeds to have both real and imaginary parts.

This problem is fairly easy once you have solved 9.4. For 0 < ω <

ωmax the only solution to our equations of motion (except at the impu-

rity site) is for the usual waves

−mω2= 2κ(cos(ka)−1) (9.6)

or equivalently

ω=ωmax|sin(ka/2)|

ka = 2 sin−1(ω/ωmax)

with ω > ωmax = 2pκ/m we have decaying solutions of the form given

by the usual where (as shown in Exercise 9.4)

ka =π±iqa

with qhere real. Then on the left we choose + and on the right we

choose −so that the wave decays away from the impurity. So we have

δxn=Aeiωt+iπn+qan n≤0

Aeiωt−iπn−qan n≥0

with q > 0 this solves all of the equations of motion except for the one

at n= 0

M¨

δx0=κ(δx−1+δx1−2δx0)

Plugging in our ansatz, we obtain the three equations

−M ω2=κ(−e−qa −e−qa −2)

Taking this equation with to Eq. 9.6 and eliminating ω2we obtain

M(cosh(qa) + 1) = m(e−qa + 1)

Now, if M=mthe only place the two sides are equal is at q= 0, which

means there is no bound state. Further, since the left hand side grows

with increasing qand the right shrinks with qit is clear that there can

be no solution for M > m. However, for M < m there is one crossing

of the two curves, which ﬁnds the qappropriate for the bound state. In

fact, one can go a bit further analytically, writing z=eqa we have

(M/m)((z+z−1)/2 + 1) = z−1+ 1

which is a quadratic equation in z, which we can solve to given

z=eqa =2m

M−1

66 Vibrations of a One-Dimensional Monatomic Chain

qa = log 2m

M−1

which, since we require q > 0 has a solution only for M < m. Finally

to obtain the frequency, one just has to plub back into our dispersion

relation to give

−mω2= 2κ(cos(π+iqa)−1)

which gives a real frequency greater than ωmax

(9.7) General Proof That Normal Modes Be-

come Quantum Eigenstates∗

This proof generalizes the argument given in Exercise

9.1. Consider a set of Nparticles a= 1,...N with masses

mainteracting via a potential

U=1

a,b

xaVa,b xb

where xais the deviation of the position of particle afrom

its equilibrium position and Vcan be taken (without loss

of generality) to be a symmetric matrix. (Here we con-

sider a situation in 1d, however, we will see that to go

to 3d we just need to keep track of three times as many

coordinates.)

(i) Deﬁning ya=√maxa, show that the classical equa-

tions of motion may be written as

¨ya=−X

Sa,b yb

where

Sa,b =1

√ma

Va,b

√mb

Thus show that the solutions are

y(m)

a=e−iωmts(m)

where ωmis the mth eigenvalue of the matrix Swith

corresponding eigenvector s(m)

a. These are the Nnormal

modes of the system.

(ii) Recall the orthogonality relations for eigenvectors

of hermitian matrices

[s(m)

a]∗[s(n)

a] = δm,n (9.7)

[s(m)

a]∗[s(m)

b] = δa,b.(9.8)

Since Sis symmetric as well as hermitian, the eigenvec-

tors can be taken to be real. Construct the transformed

coordinates

Y(m)=X

s(m)

axa√ma(9.9)

P(m)=X

s(m)

apa/√ma(9.10)

show that these coordinates have canonical commutations

[P(m), Y (n)] = −i~δn,m (9.11)

and show that in terms of these new coordinates the

Hamiltonian is rewritten as

H=X

m1

2[P(m)]2+1

2ω2

m[Y(m)]2.

(9.12)

Conclude that the quantum eigenfrequencies of the sys-

tem are also ωm. (Can you derive this result from the

prior two equations?)

Yet another typo, in part (i) it should say −ω2

mis the mth eigenvaluer.

(i) First given the expression for U, diﬀerentiate with resepect to xa

to obtain the force on particle a

Fa=−∂U

∂xa

=−X

Va,bxb

The equation of motion is then

ma¨xa=−X

Va,bxb

Now deﬁning xa=ya/√mawe obtain

¨ya=−X

√ma

Va,b

√ma

yb=−X

Sa,byb

as required. Using the ansatz ya=e−iωtsawe have

ω2sa=X

Sa,bsa(9.13)

which is solved by sabeing the eigenvector of Sand ω2its eigenvalue.

(ii) Given

Y(m)=X

s(m)

axa√ma

P(n)=X

s(n)

bpb/√mb

we calculate

[P(n), Y (m)] = X

a,b

s(n)

bs(m)

a[pb, xa]pma/mb

Using the canonical commutations

[pb, xa] = −i~δab

we have

[P(n), Y (m)] = −i~X

s(m)

as(n)

a=−i~δnm

where we have used the orthogonality of eigenstates.

Next let us look at the terms of the proposed Hamiltonian

2[P(m)]2=X

a,b

2s(m)

as(m)

bpapb/√mamb

2ma

where we have used the orthogonality of sto collapse the msum and

generate δab.

Now examining the second term of the proposed Hamiltonian

2ω2

m[Y(m)]2=X

a,b

2s(m)

aω2

ms(m)

bxaxb√mamb

But here we can use Eq. 9.13 to replace ω2with the matrix Sgiving us

2ω2

m[Y(m)]2=X

a,b,c

2s(m)

aSb,cs(m)

cxaxb√mamb

a,b

2Sb,axaxb√mamb=X

a,b

2xaVa,bxb

68 Vibrations of a One-Dimensional Monatomic Chain

where we have used the orthogonality of sand taken the sum over mto

give a δac.

Thus the given Hamiltonian is equivalent to the original Hamitlonian

H=X

2ma

a+X

a,b

2xaVa,bxb

The new form of the Hamiltonian in terms of Pand Yhas each m

coordinate completely decoupled. Further the Pand Ysatisfy canonical

commutations for momenta and positon so each mis simply a harmonic

oscillator with frequency ωm.

If one wanted to derive the spectrum from the Hamiltonian, one could

follow the usual procedure of writing

a†=1

√2~P/√ω+i√ωY 

to rewrite the Hamiltonian for each decoupled mode as

H=~ω(a†a+ 1/2)

(9.8) Phonons in 2d* Consider a mass and spring model of a two-dimensional

triangular lattice as shown in the ﬁgure (assume the lat-

tice is extended inﬁnitely in all directions). Assume that

identical masses mare attached to each of their six neigh-

bors by equal springs of equal length and spring con-

stant κ. Calculate the dispersion curve ω(k). The two-

dimensional structure is more diﬃcult to handle than the

one-dimensional examples given in this chapter. In Chap-

ters 12 and 13 we study crystals in two and three dimen-

sions, and it might be useful to read those chapters ﬁrst

and then return to try this exercise again.

Ok, this one is pretty hard if you have never seen such a thing before.

Probably it should have two stars.

First of all, one assumes that the crystal starts in equilibrium with

the springs unstretched. Forces on the springs will be proportional to

the amount of stretching. Note however, that if one of the masses is

displaced in a direction perpendicular to one of its attaching springs, to

linear order, the spring is not stretched at all (it is only rotated)! So in

other words, we need only keep track of the stretching in the direction

parallel to the spring.

Let us let the edge have unit length for simplicity of notation. If

we let the position of a sites in equilibrium be called rnand let the

displacements from this equilibrium be called urn. Now let us deﬁne

three unit vectors aipointing in the three directions of the lattice edges

a1= (1,0),a2= (1/2,√3/2),a2= (−1/2,√3/2). The total potential

energy of the system can then be written as

-6 -4 -2 0 2 46

-6

-4

-2

Fig. 9.4 Contour plot of the transverse

mode frequency of the triangular lattice

phonons in kx, kyspace.

U=κ

j=1 urn−urn+aj·aj2

where here we have included each spring once in the sum and we have

dotted the displacement with its direction so as to only count stretching

of the spring and not rotation. The force on a mass, and hence using

newton’s law we have

-6 -4 -2 0 2 46

-6

-4

-2

Fig. 9.5 Contour plot of the longitu-

dinal mode frequency of the triangular

lattice phonons in kx, kyspace.

m¨uα

rn=Fα

rn=−∂U

∂uα

where αis a basis direction (say, xor y). Carefully taking the derivative

we get

m¨uα

rn=−κ



j=1

aα

j

X

[uβ

rn+aj+uβ

rn−ap−2uβ

rn]aβ

j





Using a wave ansatz

uα

rn=Uαeiωt−ik·rn

We obtain the two by two eigenvalue problem

ω2Uα=X

Dαβ (k)Uβ

where the so-called dynamical matrix is given by

Dαβ(k) = κ

j=1

aα

jaβ

j(eik·aj+e−ik·aj−2)

Note that this matrix is symmetric in αand β. Plugging in the values

of ajand using some trig identities we gets

D=κ

m 4 0

0 0 sin2(k·a1

2) + 1√3

√3 3 sin2(k·a2

2) + 1−√3

−√3 3 sin2(k·a3

2)

For each kthere will be two eigenvalues which correspond to the longi-

tudinal and transverse phonon modes. The eigenvalues of this becomes

ω2=2κ

mS1+S2+S3±qS2

1+S2

2+S2

3−S1S2−S1S3−S2S3

where

Si= sin2(k·ai

-3 -2 -1 1 2 3

0.5

1.5

Fig. 9.6 Cut along k= (kx, 0) of

the spectrum of the triangular lattice

phonons.

These spectra are shown in ﬁgures 9.4 - 9.6. Note that the spectrum

is periodic in the Brillouin zone – which has the shape of a hexagon.

Vibrations of a

One-Dimensional Diatomic

Chain 10

(10.1) Normal modes of a One-Dimensional Di-

atomic Chain

(a) What is the diﬀerence between an acoustic mode

and an optical mode.

Describe how particles move in each case.

(b) Derive the dispersion relation for the longitudi-

nal oscillations of a one-dimensional diatomic mass-and-

spring crystal where the unit cell is of length aand each

unit cell contains one atom of mass m1and one atom of

mass m2connected together by springs with spring con-

stant κ, as shown in the ﬁgure (all springs are the same,

and motion of particles is in one dimension only).

m1m2

κ κ

tical modes at k= 0 as well as at the Brillouin zone

boundary.

Describe the motion of the masses in each case (see

margin note 4 of this chapter!).

Determine the sound velocity and show that the

group velocity is zero at the zone boundary.

Show that the sound velocity is also given by vs=

pβ−1/ρ where ρis the chain density and βis the com-

pressibility.

(d) Sketch the dispersion in both reduced and extended

zone scheme.

If there are Nunit cells, how many diﬀerent normal

modes are there?

How many branches of excitations are there? I.e.,

in reduced zone scheme, how many modes are there there

at each k?

(e) What happens when m1=m2?

The following ﬁgure depicts a long wavelength acoustic wave: All

atoms in the unit cell move in-phase with a slow spatial modulation.

Acoustic waves ω∼kfor small k.

m1m2

κ κ

acoustic

72 Vibrations of a One-Dimensional Diatomic Chain

The following depicts a long wavelength optical wave: The two diﬀer-

ent types of atoms move out of phase, with a slow spatial modulation.

(In general a long wavelength optical mode is any long wavelength mode

where not all atoms in the unit cell are moving in phase). Note that the

amplitude of motion of the diﬀerent atoms in the cells is generally not

the same. Optical modes have ωnonzero as k→0.

m1m2

κ κ

optical

(b) Let xnbe the position of the nth particle of mass m1and ynbe

the position of the nth particle of mass m2. We can assume that the

equilibrium position of xnis given by na and the equilibrium position

of ynis given by na +d.

We write the equations of motion for the deviations from these equi-

librium positions δxnand δyn.

m1¨

δxn=−κ(δxn−δyn−1)−κ(δxn−δyn)

m2¨

δyn=−κ(δyn−δxn)−κ(δyn−δxn+1)

Writing the ans¨atze

δxn=Axeikan−iωt

δyn=Ayeikan−iωt

we obtain the equations

−m1ω2Axeikna =−2κAxeikna +κAy(eikna +eik(n−1)a)

−m2ω2Ayeikna =−2κAyeikna +κAx(eikna +eik(n+1)a)

ω2Ax= 2(κ/m1)Ax−(κ/m1)(1 + e−ika)Ay(10.1)

ω2Ay= 2(κ/m2)Ay−(κ/m2)(1 + eika)Ax(10.2)

which is an eigenvalue problem from ω2. Thus we need to ﬁnd the roots

of the determinant

2(κ/m1)−ω2−(κ/m1)(1 + e−ika)

−(κ/m2)(1 + eika) 2(κ/m2)−ω2

which gives the equation

0 = ω4−ω2(2κ(1/m1+ 1/m2)) + κ2

m1m24−(1 + eika)(1 + e−ika)

0 = ω4−ω22(m1+m2)κ

m1m2+κ2

m1m2

(2 −2 cos(ka)))

with the solution (skipping a few steps)

ω2=κ

m1m2m1+m2±qm2

1+m2

2+ 2m1m2cos(ka)

=κ

m1m2m1+m2±q(m1+m2)2−4m1m2sin2(ka/2)

the optical mode has energy

ω=s2κ(m1+m2)

m1m2

At the zone boundary cos(ka) = −1, so the two modes have energy

ω=r2κm1

m1m2

and r2κm2

m1m2

the greater of which is the optical mode, the lesser being the acoustic

mode.

To ﬁnd the motions corresponding to these modes we need to plug

our frequencies back into Eqs. 10.1 and 10.2 to ﬁnd the relation between

Axand Ay. For the acoustic mode at k= 0 we obtain Ax=Aywhich

means the two masses move in phae with the same amplitude. For the

optical mode at k= 0 we have Ax=−(m2/m1)Aymeaning that the

two diﬀerent masses move in opposite directions with the heavier mass

moving with lower amplitude. At the zone boundary the two modes cor-

respond to one of the masses staying still and the other mass moving.

For example, for the lower frequency mode, the higher mass particles

move and the lower mass particle stays ﬁxed. Since we are at the zone

boundary, every other higher mass particle moves in the opposite di-

rection (thus compressing symmetrically around the ﬁxed particle. An

example of a zone boundary mode is shown in the following ﬁgure

m1m2

κ κ

zone boundary

74 Vibrations of a One-Dimensional Diatomic Chain

To ﬁnd the sound velocity, expand the cos around k= 0, one obtains

the acoustic mode velocity ω=vk with

v=arκ

2(m1+m2)

We check that v−2=ρβ. The density of the chain is ρ= (m1+m2)/a,

the compressibility of two springs in series is κ/2 so the compressibility

of the chain is β=−(1/L)dL/dF = 2/(κa).

Near the zone boundary, since the group velocity is dω/dk and since

dω/d cos(ka) is nonsingular, the group velocity must be zero by using

the chain rule since dcos(ka)/dk =asin(ka) = 0 at the zone boundary

(k=π/a).

Fig. 10.1 Diatomic Chain. Top Reduced Zone Scheme. Bottom Extended Zone

Scheme. Both pictures use m1/m2=.4.

If there are Nunit cells, therefore 2Natoms, there are 2N modes.

There are 2 modes per kin the reduced zone scheme, therefore two

branches.

(e) When m1=m2the unit cell is now of size a/2 so the Brillouin

zone is doubled in size. In this limit, the gap at the Brillouin zone

boundary vanishes and the two branches become the single branch of

the monatomic chain (this is most easily described in extended zone

scheme).

(10.2) Decaying Waves

Consider the alternating diatomic chain dispersion as

discussed in the text Eq. 10.6 and shown in Fig. 10.6.

For frequencies above ω+(k= 0) there are no propagat-

ing wave modes, and similarly for frequencies between

ω−(k=π/a) and ω+(k=π/a) there are no propagating

wave modes. As in Exercise 9.4, if this chain is driven at

a frequency ωfor which there are no propagating wave

modes, then there will be a decaying, or evanescent, wave

instead. By solving 10.6 for a complex k, ﬁnd the length

scale of this decaying wave.

As in problem 9.4, we simply want to analytically extend kto complex

numbers. From the text Eq. 10.6 we can rearrange to obtain

ka = cos−11

2κ1κ2(mω2−κ1−κ2)2−κ2

1−κ2

2

The argument on the right hand size is greater than 1 for ωlarger than

the q= 0 optical mode freuquency, whereas for ωbetween the zone

boundary acoustic and optical model frequencies, the right hand side is

less than −1. The arccos of a number greater than one is pure imaginary

and grows from 0 as the argument increases from unity. Whereas the

arcos of a number less than -1 is π+ imaginary with the imaginary part

growing from zero as the argument decreases from -1. The length scale

of decay is always given by L=a/q with qthe imaginary part of k.

(10.3) General Diatomic Chain*

Consider a general diatomic chain as shown in Fig. 10.1

with two diﬀerent masses m1and m2as well as two dif-

ferent spring constants κ1and κ2and lattice constant a.

(a) Calculate the dispersion relation for this system.

(b) Calculate the acoustic mode velocity and compare

it to vs=pβ−1/ρ where ρis the chain density and βis

the compressibility.

(a) This is the same approach as the prior cases, just a bit more

algebra to keep track of

m1¨

δxn=−κ1(δxn−δyn−1)−κ2(δxn−δyn)

m2¨

δyn=−κ2(δyn−δxn)−κ1(δyn−δxn+1)

Writing the ans¨atze

δxn=Axeikan−iωt

δyn=Ayeikan−iωt

76 Vibrations of a One-Dimensional Diatomic Chain

we obtain the equations

−m1ω2Axeikna =−(κ1+κ2)eiknaAx+ (κ2eikna +κ1eik(n−1)a)Ay

−m2ω2Ayeikna =−(κ1+κ2)eiknaAy+ (κ2eikna +κ1eik(n+1)a)Ax

ω2Ax=κ1+κ2

m1Ax+κ2

+e−ika κ1

m1Ay

ω2Ay=κ1+κ2

m2Ay+κ2

+eika κ1

m2Ax

which is an eigenvalue equation for ω2. The secular equation is then

0 = (κ1+κ2−m1ω2)(κ1+κ2−m2ω2)−κ2

1−κ2

2−2κ1κ2cos(ka)

ω4−(κ1+κ2)(m1+m2)

m1m2

ω2−4κ1κ2

m1m2

sin2(ka/2) = 0

with the solution (skipping a few steps)

ω2=(κ1+κ2)(m1+m2)

2m1m2

±1

2s(κ1+κ2)2(m1+m2)2

(m1m2)2−16κ1κ2

m1m2

sin2(ka/2)

(b) Simply taylor expanding the dispersion gives us

ω=rκ1κ2

(κ1+κ2)(m1+m2)ak

vs=sa2κ1κ2

(κ1+κ2)(m1+m2)

Now the hydrodynamic approach. The density of the chain is ρ=

(m1+m2)/a, the spring constant of the two springs in series is κ1κ2/(κ1κ2)

so the compressibility of the chain is β=−(1/L)dL/dF = (κ1κ2)/(κ1+

κ2)/. Plugging this into vs=pβ−1/ρ gives the same result.

(10.4) Second Neighbor Diatomic Chain* Con-

sider the diatomic chain from Exercise 10.1. In addition

to the spring constant κbetween neighboring masses, sup-

pose that there is also a next nearest-neighbor coupling

with spring constant κ′connecting equivalent masses in

adjacent unit cells. Determine the dispersion relation for

this system. What happens if κ′≫κ?

We write the equations of motion

m1¨

δxn=−κ(δxn−δyn−1)−κ(δxn−δyn)−κ′(δxn−δxn−1)−κ′(δxn−δxn+1)

m2¨

δyn=−κ(δyn−δxn)−κ(δyn−δxn+1)−κ′(δyn−δyn−1)−κ′(δyn−δyn+1)

Writing the usual ans¨atze

δxn=Axeikan−iωt

δyn=Ayeikan−iωt

we obtain the equations

−m1ω2Axeikna =−2(κ+κ′)Axeikna +κAy(eikna +eik(n−1)a) + κ′Ax(eik(n+1)a+eik(n−1)a)

−m2ω2Ayeikna =−2(κ+κ′)Ayeikna +κAx(eikna +eik(n+1)a) + κ′Ay(eik(n+1)a+eik(n−1)a)

m1ω2Ax= 2 [(κ+κ′)−κ′cos(ka)] Ax−κ(1 + e−ika)Ay

m2ω2Ay= 2 [(κ+κ′)−κ′cos(ka)] Ay−κ(1 + eika)Ax

which is an eigenvalue problem for ω2. Using some trig identies and

some algebra the secular equation can be reduced to

0 = m1m2ω4−(m1+m2)2κ+ 4κ′sin2(ka/2)ω2+[4κ2+ 16κκ′] sin2(ka/2) + 16κ′2sin4(ka/2)

with the solution

ω2=1

2m1m2(m1+m2)(2κ+ 4κ′sin2(ka/2))

±q(m1+m2)2(2κ+ 4κ′sin2(ka/2))2−4m1m2([4κ2+ 16κκ′] sin2(ka/2) + 16κ′2sin4(ka/2))

m1m2(m1+m2)(κ+ 2κ′sin2(ka/2))

±q4m1m2κ2cos2(ka/2) + 4(m1−m2)2κ′2sin4(ka/2) + κκ′sin2(ka/2)

Note that when κ2= 0 we recover the diatomic chain from Exercise

10.1. When κ′≫κwe get two decoupled monatomic chains (as should

be expected!).

78 Vibrations of a One-Dimensional Diatomic Chain

(10.5) Triatomic Chain*

Consider a mass-and-spring model with three diﬀerent

masses and three diﬀerent springs per unit cell as shown

in this diagram.

m3m1m2m3m1m2

κ3κ1κ2κ3κ1

As usual, assume that the masses move only in one

dimension.

(a) At k= 0 how many optical modes are there? Cal-

culate the energies of these modes. Hint: You will get a

cubic equation. However, you already know one of the

roots since it is the energy of the acoustic mode at k= 0

(b)* If all the masses are the same and κ1=κ2de-

termine the frequencies of all three modes at the zone

boundary k=π/a. You will have a cubic equation, but

you should be able to guess one root which corresponds

to a particularly simple normal mode.

(c)* If all three spring constants are the same, and

m1=m2determine the frequencies of all three modes at

the zone boundary k=π/a. Again you should be able to

guess one of the roots.

With three particles in the unit cell, in one dimesion there is one

acoustic mode and two optical modes.

Declaring the unit cell to be one set of m1, m2, m3with positions

x, y, z our equations of motion are

m1¨

δxn=−κ3(δxn−δzn−1)−κ1(δxn−δyn)

m2¨

δyn=−κ1(δyn−δxn)−κ2(δyn−δzn)

m3¨

δzn=−κ2(δzn−δyn)−κ3(δzn−δxn+1)

Plugging in the usual wave ansatz gives us the secular determinant

0 = 

κ3+κ1−m1ω2κ1κ3e−ika

κ1κ1+κ2−m2ω2κ2

κ3eika κ2κ2+κ3−m3ω2

which multiplied out gives the secular polynomial

0 = m1m2m3ω6

−[κ1(m1+m2)m3+κ3m2(m1+m3) + κ2m1(m2+m3)] ω4

+ [(κ2κ3+κ1κ2+κ3κ1)(m1+m2+m3)] ω2

−[κ1κ2κ3(2 −2 cos(ka))] (10.3)

At k= 0, the ﬁnal term vanishes and the secular polynomial clearly

has ω= 0 a root. This we could have guessed in advance since we know

the acoustic mode should come down to zero frequency at k= 0. Once

we remove the obvious factor of ω2the remaining polynomial is just

quadratic and we can obtain the two roots in the obvious way.

(b) Setting all the masses the same and κ1=κ2, and setting ka =π

we obtain

0 = m3ω6−2m2(2κ1+κ3)ω4

+(2κ1κ3+κ2

1)(3m)ω2−κ2

1κ3(4)(10.4)

Here we have a diﬃcult cubic. However, with some intuition we realize

that there should be a mode at the zone boundary where the κ3spring

does not compress at all – the two masses connected to κ3moving in

unison, but since we are at the zone boundary every other κ3spring

moves in the opposite direction. The two masses moving in unison have

mass 2mand have a restoring force of 2κ1from the two springs on the

two sides. Thus we expect a frequency of ω2=κ1/m. Indeed, this

frequncy solves Eq. 10.4. We can thus factor out this root to obtain a

quadratic which we then solve to get the two other roots at the zone

boundary

ω2=3κ1+ 2κ3±p9κ2

1−4κ1κ3+ 4κ2

setting ka =πwe obtain

0 = m2

1m3ω6−[2κm1(2m3+m1)] ω4

+(2m1+m3)(3κ2)ω2−4κ3(10.5)

Very similar reasoning suggests a mode at this wavevector with frequency

ω2=κ/m1(which is indeed a solution) which we can then factor out to

give the two additional zone boundary modes

ω2=κ 3m3+ 2m1±p9m2

3−4m1m3+ 4m2

2m1m3!

Tight Binding Chain

(Interlude and Preview) 11

(11.1) Monatomic Tight Binding Chain

Consider a one-dimensional tight binding model of elec-

trons hopping between atoms. Let the distance between

atoms be called a, and here let us label the atomic or-

bital on atom nas |nifor n= 1 . . . N (you may assume

periodic boundary conditions, and you may assume or-

thonormality of orbitals, i.e., hn|mi=δnm). Suppose

there is an on-site energy ǫand a hopping matrix element

−t. In other words, suppose hn|H|mi=ǫfor n=mand

hn|H|mi=−tfor n=m±1.

Derive and sketch the dispersion curve for electrons.

(Hint: Use the eﬀective Schroedinger equations of Exer-

cise 6.2a. The resulting equation should look very similar

to that of Exercise 9.2.)

How many diﬀerent eigenstates are there in this

system?

What is the eﬀective mass of the electron near the

bottom of this band?

What is the density of states?

If each atom is monovalent (it donates a single elec-

tron) what is the density of states at the Fermi surface?

Give an approximation of the heat capacity of the

system (see Exercise 4.3).

What is the heat capacity if each atom is divalent?

Write a general wavefunction as

Ψ = X

φn|ni

The schroedinger equation (See problem 6.2a) is

Eφn=ǫφn−t(φn+1 +φn−1)

write the ansatz

φn=eikna

and we obtain

E(k) = ǫ−2tcos(ka)

(I won’t sketch this because it is rather trivial). Note however, for t > 0

the minimum occurs at k= 0 whereas for t < 0 the minimum is at the

zone boundary.

If there are Nsites in the system, the total length of the system (as-

sumed periodic) is Na. Thus kmust be quantized k= 2πm/(N a). Thus

there are exactly Ndiﬀerent values of kwithin the Brillouin zone. This

is to be expected. If we started with a Hilbert space of Ndimensions

(described by states |ni), then we expect that there should be exactly

82 Tight Binding Chain (Interlude and Preview)

Neigenstates. (And one gets twice as many eigenstates if you include

spin).

Expanding around the minimum one obtains

E= const + |t|a2k2

which we set equal to a free particle dispersion

~2k2/(2m∗) = |t|a2k2

yielding an eﬀective mass

m∗=~2/(2|t|a2)

The density of states in kspace is uniform as usual dN/dk =L/(2π) =

Na/(2π) (I have not included spin here yet) so the density of states in

energy

dN/dE = (dN/dk)(dk/dE)

Now using

dE/dk = 2ta sin(ka) = 2|t|aq1−((E−ǫ)/(2t))2

we obtain

dN/dE =N

2π

2∗2

2|t|q1−((E−ǫ)/(2t))2

and, as in exercise 9.2 there is an extra factor of 2 being that one has

two possible kstates for each energy Eand still another factor of 2

to include the 2 spin states. (Note if you integrate over frequency you

correctly get back 2Ndegrees of freedom, now with the factor of 2 for

spin included).

If each atom is monovalent, the band is exactly half ﬁlled (since two

electrons can go into each orbital). Thus the density of states is

dN/dE =N

|t|

The heat capacity is given in terms of the density of states as

CV= ˜γk2

Bg(EF)V T

with ˜γ=π2/6 (the exact number is from a more complicated calculation

that the students are not responsible for). Here g(EF)V=N

|t|.

If there were two electrons per atom, then the band would be com-

pletely full, and there would be no freedom in the system at all. The

heat capacity would be zero. The spin susceptibility would be zero too.

(11.2) Diatomic Tight Binding Chain

We now generalize the calculation of the previous exer-

cise to a one-dimensional diatomic solid which might look

as follows:

−A−B−A−B−A−B−

Suppose that the onsite energy of type Ais diﬀerent from

the onsite energy of type B. I.e, hn|H|niis ǫAfor nbeing

on a site of type Aand is ǫBfor nbeing on a site of type

B. (All hopping matrix elements −tare still identical to

each other.)

Calculate the new dispersion relation. (This is ex-

tremely similar to Exercise 10.1. If you are stuck, try

studying that exercise again.)

Sketch this dispersion relation in both the reduced

and extended zone schemes.

What happens if ǫA=ǫB?

What happens in the “atomic” limit when tbe-

comes very small.

What is the eﬀective mass of an electron near the

bottom of the lower band?

If each atom (of either type) is monovalent, is the

system a metal or an insulator?

*Given the results of this exercise, explain why LiF

(which has very ionic bonds) is an extremely good insu-

lator.

The unit cell ais now the distance from an A atom to the next A

atom. Let φA

nbe the amplitude of the wavefunction on the nth site of

type Aand φB

nbe the amplitude on the nth site of type B.

The Schroedinger equation becomes

EφA

n=ǫAφA

n−t(φB

n+φB

n−1)

EφB

n=ǫBφB

n−t(φA

n+φA

n+1)

Using the ans¨atz

φA

n=Aeikna

φB

n=Beikna

gives

EA =ǫAA−t(1 + e−ika)B

EB =ǫBB−t(1 + eika)A

again giving a two by two eigenvalue problem, where we solve for the

roots of the determinant

ǫA−E−t(1 + e−ika)

−t(1 + eika)ǫB−E

yielding the secular equation

0 = E2−E(ǫA+ǫB) + (ǫAǫB−t2(2 + 2 cos(ka)))

with the solutions

E±(k) = 1

2ǫA+ǫB±p(ǫA−ǫB)2+ 4t2(2 + 2 cos(ka))

Note that in the limit that ǫA=ǫBwe recover

E=ǫ±2|t|cos(ka/2)

84 Tight Binding Chain (Interlude and Preview)

Fig. 11.1 Diatomic Tight Binding Problem in 1D. Left: Reduced Zone scheme.

Right: Extended Zone scheme. In both pictures we have chosen ǫA= 2.56 and

ǫB= 1.56

which matches the solution of part (a) above, but for the change in the

deﬁnition of the size of the unit cell. (See the ﬁgure for sketches of the

dispersion)

In the limit of small tthe bands become very ﬂat at energies ǫAand

ǫB.

Expanding around the minimum gives

E= constant + 2t2(ka)2

p(ǫA−ǫB)2+ 16t2

which we set equal to ~2k2/(2m∗) to yield

m∗=~2p(ǫA−ǫB)2+ 16t2

4t2a2

If each atom is monovalent, there are now two electrons per unit cell,

and this ﬁlls exactly the lower band (leaving the upper band empty).

The system is therefore an insulator.

If ǫA=ǫBthis becomes a monotonic chain, the gap closes and it

becomes a metal.

For LiF we can expect a much lower energy for electrons on F than

on Li (F high high electron aﬃnity, Li has low ionization energy). So

we can set ǫA≪ǫB. What happens in this limit is that the band

are extremely far apart – thus a very good insulator. Further, if we

look a the eigenvectors in the lower band, we will ﬁnd that they are

almost completely on the lower energy atoms. Thus the free electron is

transferred almost completely from the higher to the lower energy atom.

(11.3) Tight Binding Chain Done Right

Let us reconsider the one-dimensional tight binding

model as in Exercise 11.1. Again we assume an on-site en-

ergy ǫand a hopping matrix element −t. In other words,

suppose hn|H|mi=ǫfor n=mand hn|H|mi=−tfor

n=m±1. However, now, let us no longer assume that or-

bitals are orthonormal. Instead, let us assume hn|mi=A

for n=mand hn|mi=Bfor n=m+ 1 with hn|mi= 0

for |n−m|>1.

Why is this last assumption (the |n−m|>1 case)

reasonable?

Treating the possible non-orthogonality of orbitals here

is very similar to what we did in Exercise 6.5. Go back

and look at that exercise.

Use the eﬀective Schroedinger equation from Ex-

ercise 6.5 to derive the dispersion relation for this one-

dimensional tight binding chain.

One can assume that orbitals far apart have no overlap since the tails

of wavefunctions decay exponentially.

The generalized Schoedinger equation is

Hnmφm=EX

Snmφm

where here

Hnm =ǫ0δnm −t(δn,m+1 +δn,m−1)

Snm =Aδnm +Bδn,m+1 +B∗δn,m−1

(note we did not promise that Bis real!). Plugging in our wave ansatz

φm=eimka into the Schroedinger equation, we obtain

E(A+Beik +B∗e−ik) = ǫ0−2tcos(ka)

E=ǫ0−2tcos(ka)

A+Beik +B∗e−ik

(11.4) Two Orbitals per Atom

(a) Consider an atom with two orbitals, Aand Bhav-

ing eigenenergies ǫA

atomic and ǫB

atomic. Now suppose we

make a one-dimensional chain of such atoms and let us

assume that these orbitals remain orthogonal. We imag-

ine hopping amplitudes tAA which allows an electron on

orbital Aof a given atom to hop to orbital Aon the neigh-

boring atom. Similarly we imagine a hopping amplitude

tBB that allows an electron on orbital Bof a given atom

to hop to orbital Bon the neighboring atom. (We as-

sume that V0, the energy shift of the atomic orbital due

to neighboring atoms, is zero).

Calculate and sketch the dispersion of the two re-

sulting bands.

If the atom is diatomic, derive a condition on the

quantities ǫA

atomic −ǫB

atomic, as well as tAA and tBB which

86 Tight Binding Chain (Interlude and Preview)

determines whether the system is a metal or an insulator.

(b)* Now suppose that there is in addition a hopping

term tAB which allows an electron on one atom in orbital

Ato hop to orbital Bon the neighboring atom (and vice

versa). What is the dispersion relation now?

(a) The two orbitals are completely decoupled from each other. So we

get two independent dispersions (we assume t’s real here)

E=ǫA−2tAA cos(ka) (11.1)

and

E=ǫB−2tBB cos(ka) (11.2)

A sketch is given in Fig. 11.2.

-3 -2 -1 1 2 3

-2

-1

Fig. 11.2 Dispersion in the two orib-

tal tight binding model (Eqs. 11.1 and

11.2). Here, tAA = 1, tBB = 2, ǫA= 0,

ǫB= 4. Horizontal axis is ka.

It should say ”’if the atom is divalent” not diatomic (doh!). If the

atom is divalent, then it will be a metal if the bands overlap and an

insulator (or semiconductor) if the bands do not overlap. A band with

the dispersion as in Eqs. 11.1 has a band which extends from ǫA−2|tAA|

to ǫA+ 2|tAA|(and similar for the other band – this holds even if t’s are

complex). In order for the bands to not overlap we must have either

ǫA+ 2|tAA|< ǫB−2|tBB |

ǫB+ 2|tBB |< ǫA−2|tAA|

If neither of these is satisﬁed the bands must overlap and we have a

metal. Note if we are tricky we can summarize the two above conditions

as one condition which, if satisﬁed, tells us we have an insulator

|ǫA−ǫB|>2 (|tAA| − |tBB |)

(b) When there is hopping tAB between an Aorbital on one site and a

Borbital on a neighboring site, we have a matrix schroedinger equation

EφA

n=ǫAφA

n−tAAφA

n+1 −t∗

AAφA

n−1−tABφB

n+1 −t∗

ABφB

n−1

EφB

n=ǫBφB

n−tBB φB

n+1 −t∗

BB φB

n−1−tABφA

n+1 −t∗

ABφA

n−1

Using the usual ans¨atz

φA

n=Aeikna

φB

n=Beikna

we obtain

EA =ǫA−tAAeika −t∗

AAe−ikaA+−tABeika −t∗

ABe−ikaB

EB =ǫB−tBB eika −t∗

BB e−ikaB+−tAB eika −t∗

ABe−ikaA

which is an eigenvalue equation for E. Deﬁning

TAA =tAAeika +t∗

AAe−ika = 2Re[tAAeika](11.3)

TBB =tBB eika +t∗

BB e−ika = 2Re[tBBeika] (11.4)

TAB =tABeika +t∗

ABe−ika = 2Re[tAB eika] (11.5)

We can write the secular equation as

0 = E2+E(TAA +TBB −ǫA−ǫB)

+ (ǫA−TAA) (ǫB−TBB)−T2

AB (11.6)

which we solve as a quadratic equation to give

E=1

2ǫA+ǫB−TAA −TBB ±q(ǫA−ǫB−TAA +TBB)2−4T2

AB

(11.5) Electronic Impurity State*

Consider the one-dimensional tight binding Hamilto-

nian given in Eq. 11.4. Now consider the situation where

one of the atoms in the chain (atom n= 0) is an impurity

such that it has an atomic orbital energy which diﬀers by

∆ from all the other atomic orbital energies. In this case

the Hamiltonian becomes

Hn,m =ǫ0δn,m −t(δn+1,m +δn−1,m) + ∆δn,mδn,0.

(a) Using an ansatz

φn=Ae−qa|n|

with qreal, and athe lattice constant, show that there

is a localized eigenstate for any negative ∆, and ﬁnd the

eigenstate energy. This exercise is very similar to Exercise

9.6.

(b) Consider instead a continuum one-dimensional

Hamiltonian with a delta-function potential

H=−~2

2m∗∂2

x+ (a∆)δ(x).

Similarly show that there is a localized eigenstate for any

negative ∆ and ﬁnd its energy. Compare your result to

that of part (a).

(a) Consider ﬁrst without the impurity. Propose exponentially decay-

ing or growing solutions

φn=Ae±qa|n|

plugging these in we obtain

E=ǫ0−2tcosh(qa) (11.7)

these give solutions with qreal for (ǫ0−E)/2t > 1, or in other words,

for energies below the bottom of the band. These are analogous to the

evanescent waves discussed in Exercises 9.4 and 10.2.

Now, we can patch together two of these evanescent waves at the

impurity. Examining the schroedinger equation at position zero, we

have

Eφ0= (ǫ0+ ∆)φ0−t(φ1+φ−1)

which gives us

E−ǫ0−∆ = 2te−qa

Then plugging in the value of Ein terms of qfrom Eq. 11.7, we obtain

−∆ = 2 sinh(qa)

recalling that ∆ was negative we have a solution for any value of ∆ <0

with

qa = sinh−1(|∆|/(2t))

88 Tight Binding Chain (Interlude and Preview)

Plugging back into Eq. 11.7 we obtain the bound state energy

E=ǫ0−2tp[∆/(2t)]2+ 1

and note that for ∆ = 0 this gives us the energy at the bottom of the

band.

(b) For the continuum problem we have similarly a decaying waves of

the form e−qx with energies

E=−~2q2

2m∗

Then we need to patch together solutions at zero, we have

∂ψ−−∂ψ+= 2q= (a|∆|)(2m∗)/~2

q= (a|∆|m∗)/~2

with energy

E=−a2m∗∆2/(2~2)

To compare to the above tight binding problem, we use the value of

the eﬀective mass m∗=~2/(2ta2) to obtain

q=|∆|/(2t)

which matches our above result for small ∆. Correspondingly the energy

E=−∆2/(4t)

which also matches the above result for small ∆ so long as we choose

the bottom of the band to be called zero energy.

(11.6) Reﬂection from an Impurity*

Consider the tight binding Hamiltonian from the pre-

vious exercise representing a single impurity in a chain.

Here the intent is to see how this impurity scatters a

plane wave incoming from the left with unit amplitude

(this is somewhat similar to Exercise 9.5). Use an ansatz

wavefunction

φn=T e−ikna n≥0

e−ikna +Re+ikna n < 0

to determine the transmission Tand reﬂection Ras a

function of k.

Use the schroedinger equation at position n=−1

Eφ−1=ǫ0φ−1−t(φ0+φ−2)

and at position zero

Eφ0= (ǫ0+ ∆)φ0−t(φ1+φ−1)

and we plug in the ansatz form. The former equation simply tells us

that we must have 1 + R=T(To see this recall that the relationship

between Eand kis ﬁxed by the plane wave away from positon zero.

So this ﬁrst equation tells us that position φ0must just be the usual

continuation of this wave-front with no changes.) The second equation

then gives us

E T = (ǫ0+ ∆)T−t(T eika +eika +Re−ika)

Using 1 + R=Tas well as using E=ǫ0+ 2tcos(ka) this becomes

T=1

∆

2it sin(ka)+ 1

Note that when ∆ = 0, we have complete transmission T= 1 and

R= 0. With R=T−1, we have |R|2+|T|2= 1 which indicates current

conservation.

(11.7) Transport in One Dimension*

(a) Consider the one-dimensional tight binding chain

discussed in this chapter at (or near) zero temperature.

Suppose the right end of this chain is attached to a reser-

voir at chemical potential µRand the left end of the

chain is attached to a reservoir at chemical potential µL

and let us assume µL> µR. The particles moving to-

wards the left will be ﬁlled up to chemical potential µR,

whereas the particles moving towards the right will be

ﬁlled up to chemical potential µL, as shown in the bot-

tom of Fig. 11.4, and also diagrammed schematically in

the following ﬁgure

µLµR

µR

µL

(i) Argue that the total current of all the particles mov-

ing to the right is

jR=Z∞

πv(k)nF(β(E(k)−µL))

with v(k) = (1/~)dǫ(k)/dk the group velocity and nFthe

Fermi occupation factor; and an analogous equation holds

for left moving current.

(ii) Calculate the conductance Gof this wire, deﬁned

Jtotal =GV

where Jtotal =jL−jRand eV =µL−µR, and show

G= 2e2/h with hPlanck’s constant. This “quantum” of

conductance is routinely measured in disorder free one-

dimensional electronic systems.

(iii) In the context of Exercise 11.2, imagine that an im-

purity is placed in this chain between the two reservoirs

to create some backscattering. Argue that the conduc-

tance is reduced to G= 2e2|T|2/h. This is known as the

Landauer formula and is a pillar of nano-scale electron-

ics.

(b) Now suppose that the chemical potentials at both

reservoirs are the same, but the temperatures are TLand

TRrespectively.

(i) Argue that the heat current jqof all the particles

moving to the right is

R=Z∞

πv(k) (E(k)−µ)nF(βL(E(k)−µ))

and a similar equation holds for left-moving heat current.

(ii) Deﬁne the thermal conductance Kto be

Jq=K(TL−TR)

where Jq=jq

L−jq

Rand TL−TRis assumed to be small.

Derive that the thermal conductance can be rewritten as

K=−2

hT Z∞

−∞

dE(E−µ)2∂

∂E nF(β(E−µ)).

Evaluating this expression, conﬁrm the Wiedemann–

Franz ratio for clean one-dimensional systems

T G =π2k2

3e2

(Note that this is a relationship between conductance and

thermal conductance rather than between conductivity

90 Tight Binding Chain (Interlude and Preview)

and thermal conductivity.) In evaluating the above inte-

gral you will want to use

Z∞

−∞

dx x2∂

∂x

ex+ 1 =−π2

If you are very adventurous, you can prove this nasty

identity using the techniques analogous to those men-

tioned in footnote 20 of Chapter 2, as well as the fact that

the Riemann zeta function takes the value ζ(2) = π2/6

which you can prove analogous to the appendix of that

chapter.

(a)(i) Genereally the velocity of a wavepacket is the group velocity

v=dω/dk. Here, ω=Ek/~To ﬁnd the total right going current, one

integrates over all possible k, the occupancy of that ktimes the group

velocity, to obtain the current. There should be an additional factor of

−eout front to turn this into an electrical current. The integral over

kusually has 2πdownstairs, but the 2 is cancelled by a factor of 2 out

front for two spins. The integral over kalso usually comes with a factor

of volume (or length in this case) out front. To see why this is missing we

have to think carefully about the deﬁnition of current – which counts the

number of particles crossing a given point in some unit of time. If one

has a particle in a keigenstate, it is delocalized over the entire system.

Thus the ”probability“ that with velocity vit crosses that given point

in a unit of time is given by v/L.

(ii) We can write the total current as

J=−e

πZ∞

dE(k)

dk nF(β(Ek−µR))dk −Z−∞

dE(k)

dk nF(β(Ek−µL))dk

=−e

~πZ∞

dEnF(β(E−µR)) −Z∞

dEnF(β(E−µL))

=−e

~π[µL−µR]

where in the last step we have assumed that µ≫kBTso that the

occupancy at E= 0 is unity — in which case the integral of the Fermi

function simply gives the chemical potential (one can see this by starting

at T= 0 where the Fermi function is just a step function. As we raise

the temperature, the change in the fermi function is symmetric around

the chemical potential, so even though the fermi function gets smaller

below µand larger above µthe integral is unchanged).

Plugging in the expression for voltage

J=−e2

π~V

or G= 2e2/h.

(iii) If the amplitude of transmission through the impurity is Tthen

the probability of transmission is |T|2. The resulting current is then

reduced by the probability that each particle is transmitted. Thus we

obtain

G= 2e2|T|2/h

Note that as we discovered in exercise 11.6, Tis a function of wavevector

and hence energy. Once should use knear kFto calculate T. The

reason for this is that all of the net current is coming from near the

Fermi surface. Well below the fermi surface the currents of left and

right movers exactly cancel.

(b) (i) The argument here is quite similar. Here each electron is

moving with velocity v, but is carrying energy E−µ. Here energy is

measured with respect to the chemical potential – this is an appropriate

deﬁnition since even at T= 0 the addition of an electron carries the

chemical potential worth of energy. The diﬀerence from this T= 0

energy is the excess heat added.

(ii) We take the diﬀerence of right moving and left moving energy

currents (the sign gives current moving from left to right)

Jq=−1

πZ∞

dE(k)

dk (Ek−µ)nF(βR(Ek−µ))dk −Z−∞

dE(k)

dk (Ek−µ)nF(βL(Ek−µL))dk

=−1

~πZ∞

dE(E−µ)nF(βR(E−µ)) −Z∞

dE(E−µ)nF(βL(E−µ))

=−1

~π(βR−βL)Z∞

dE(E−µ)dnF(β(E−µ))

dβ

(11.8)

where we have made the approximation that βR−βLis small to

replace the ﬁnite diﬀerence with a derivative with respect to β. This

then becomes.

Jq=−1

~πβ (βR−βL)Z∞

dE(E−µ)2dnF(β(E−µ))

then using

βR−βL≈TL−TR

kBT2

gives the desired result. Note that the formula also extends the lower

limit of integration to −∞. This is allowed since dnF/dE is strongly

peaked around µ.

Finally scaling out some factors of βwe obtain

K=−2T

hZ∞

−∞

dxx2d

ex+ 1 =2π2k2

Thus dividing by the expresion for Gwe obtain

GT =π2k2

3he2

which is the Weideman-Franz law.

For completeness, we evaluate the integral. First, note that the inte-

92 Tight Binding Chain (Interlude and Preview)

grand is symmetric around zero so we can write

I=Z∞

−∞

dxx2d

ex+ 1 = 2 Z∞

dxx2d

ex+ 1 =−4Z∞

dxx e−x

1 + e−x

= 4 Z∞

dxx

∞

n=1

(−ex)n= 4

∞

n=1

(−1)n

=−4"∞

n=1

n2−2

∞

n=1,evens

n2#= 4(1 −2

4)ζ(2) = 2ζ(2)

The method of calculating ζ(2) proceeds similar to the appendix of

Chapter 1. First, consider the function xin the range [−π, π] as a fourier

series. It can be writen as

∞

n=1

ansin(nx)

with

an=1

πZπ

−π

xsin(nx) = −2(−1)n/n

The calculate Zπ

−π

dx x2= 2π3/3

but also caculate the same quantity in terms of its fourier transforms

(using percival’s theorem)

Zπ

−π

dx x2=Zπ

−π

dx =

∞

n=1 |an|2sin2(nx) =

∞

n=1

n2π

setting these two expressions equal to each other gives

∞

n=1

n2=π2

(11.8) Peierls Distortion*

Consider a chain made up of all the same type of atom,

but in such a way that the spacing between atoms alter-

nates as long-short-long-short as follows

−A=A−A=A−A=A−

In a tight binding model, the shorter bonds (marked with

=) will have hopping matrix element tshort =t(1 + ǫ)

whereas the longer bonds (marked with −) have hopping

matrix element tlong =t(1 −ǫ). Calculate the tight-

binding energy spectrum of this chain. (The onsite en-

ergy ǫis the same on every atom). Expand your result to

linear order in ǫ. Suppose the lower band is ﬁlled and the

upper band is empty (what is the valence of each atom

in this case?). Calculate the total ground-state energy of

the ﬁlled lower band, and show it decreases linearly with

increasing ǫ.

Now consider a chain of equally spaced identical A

atoms connected together with identical springs with

spring constant κ. Show that making a distortion

whereby alternating springs are shorter/longer by δx

costs energy proportional to (δx)2. Conclude that for a

chain with the valence as in the ﬁrst part of this problem,

a distortion of this sort will occur spontaneously. This is

known as a Peierls distortion.

OK, I messed up this problem a bit, so bear with me.

First we need to evalulate the spectrum. Consider the unit cell to be

a single unit like this A=A−. Call the left side the A site and the

right site the B site. Set the onsite energy epsilon0to zero for simplicity

and assume tis real (without loss of generality). We then have the

tight-binding schroedinger equation

EφA

n=−t(1 + ǫ)φB

n−t(1 −ǫ)φB

n−1

EφB

n=−t(1 + ǫ)φA

n−t(1 −ǫ)φA

n+1

Using the usual ansatz this gives us

EA =−t(1 + ǫ)B−t(1 −ǫ)Be−ika

EB =−t(1 + ǫ)A−t(1 −ǫ)Aeika

which is an eigenvalue problem with solutions

E(k) = ±|2t[cos(ka/2) + iǫ sin(ka/2)]|=±|2t|pǫ2+ (1 −ǫ2) cos2(ka/2)

To ﬁnd the total energy of a ﬁlled lower band, we need

Etot = 2LZdk

2πE(k)

which is some horrid ellipic integral. Ideally, we just evaluate this horrid

integral and we are done, but this is very diﬃcult analytically (we could

do it numerically though!).

However, all we need to do is to make an estimate of how this changes

to lowest order in ǫ. Near the zone boundary, since the cosine term is

small the eﬀect of epsilon is most pronounced where the energy reduction

in the lower band is −|2tǫ|. As we move away from the zone boundary

the cosine term becomes more important. Deﬁne κa =π−ka to be the

distance to the zone boundary (focusing for now on the zone boundary

at ka =π).

For small κwe see that the two term in the square root become roughly

equal when

κa ≈ǫ

And when κa ≫ǫthe cosine term dominates. Let us then split the

integration roughly at this intermediate point κa ≈ǫ. For smaller κ, the

energy reduction can be approximated as −|2tǫ|for each value of κand

integrating then a range of κthat is ǫlarge we get an energy reduction

of ∼L|t|ǫ2, which we are not interested in. For the region further from

the zone boundary, we have roughly

δE ≈ |2tL|Zcutoﬀ

κa=ǫ

dκ hpǫ2+ (κa)2/4−κa/2i

≈ |2tL|Zcutoﬀ

κa=ǫ

dκǫ2

κ≈ |2tL|ǫ2log(ǫ/cutoﬀ)

94 Tight Binding Chain (Interlude and Preview)

here the cutoﬀ is some arbitrary momentum much further from the zone

boundary (it will not matter where we choose the cutoﬀ!). Thus we see

that the energy saving is proportional to ǫ2log ǫ.

(ii) The energy of stretching springs is always quadratic in the stretch-

ing (hookes’ law), hence proportional to ǫ2. For small ǫthe electronic

energy saving ǫ2log ǫalways wins, so the system always distorts sponta-

neously!

(11.9) Tight Binding in 2d*

Consider a rectangular lattice in two dimensions as

shown in the ﬁgure. Now imagine a tight binding model

where there is one orbital at each lattice site, and where

the hopping matrix element is hn|H|mi=t1if sites nand

mare neighbors in the horizontal direction and is = t2if

nand mare neighbors in the vertical direction. Calculate

the dispersion relation for this tight binding model. What

does the dispersion relation look like near the bottom of

the band? (The two-dimensional structure is more diﬃ-

cult to handle than the one-dimensional examples given

in this chapter. In Chapters 12 and 13 we study crys-

tals in two and three dimensions, and it might be useful

to read those chapters ﬁrst and then return to try this

exercise again.)

This is a lot easier than it looks! Let us label the sites (m,n). Let

the wavefunctions on the sites be φn,m accordingly. The Schroedinger

equation is then

Eφn,m =ǫ0φn,m −t1(φn+1,m +φn−1,m)−t2(φn,m+1 +φn,m−1)

Using an ansatz

φn,m =eikxnax+kymay

where axand ayare the lattice distances in the two directions.

and assuming both t1and t2real

E=ǫ0−2t1cos(kxax)−2t2cos(kyay)

Assuming both t1and t2positive, the minimum occurs at kx=ky= 0.

Near this minimum, we can expand to gets

E≈(ǫ0−2t1−2t2) + 1

2t1(kxax)2+t2(kyay)2

which has equal-Econtours which are ellipses. Thus near the bottom of

the band we have an ellipsoidal bowl.

Crystal Structure 12

(12.1) Crystal Structure of NaCl Consider the

NaCl crystal structure shown in Fig. 12.21. If the lattice

constant is a= 0.563 nm, what is the distance from a

sodium atom to the nearest chlorine? What is the dis-

tance from a sodium atom to the nearest other sodium

atom?

Super easy: The Na-Cl distance is a/2 = .2815 nm whereas the Na-Na

distance is a√2/2 = .398 nm.

(12.2) Neighbors in the Face-Centered Lattice.

(a) Show that each lattice point in an fcc lattice has

twelve nearest neighbors, each the same distance from the

initial point. What is this distance if the conventional

unit cell has lattice constant a?

(b)∗Now stretch the side lengths of the fcc lattice

such that you obtain a face-centered orthorhombic lattice

where the conventional unit cell has sides of length a,b,

and cwhich are all diﬀerent. What are the distances to

these twelve neighboring points now? How many nearest

neighbors are there?

(a) Given the primitive lattice vectors, one can deﬁne the fcc lattice

as being vectors of the form a

2(n, m, l) where n, m, l are either all even

or two odd and one even. Among this sets of possible lattice points, the

closest ones to [0,0,0] are of the form a

2[1,1,0]. Here the two entries that

are 1 could be either ±1 which gives four possibilities. Further the 0 can

be in one of three places. Thus we have 12 possibilities. The distance

from [0,0,0] to any of these points is a√2/2 (analogous to previous

problem).

(b) In units of the three diﬀerent (unequal) conventional unit cell

lattice constants, the 12 points are still of the form 1

2[1,1,0] and permu-

tations. The distances to these 12 points are then

[±1,±1,0] →d=1

2pa2+b2

[±1,0,±1] →d=1

2pa2+c2

[0,±1,±1] →d=1

2pb2+c2

There are four nearest neighbors corresponding to the smaller two of

a, b, c.

96 Crystal Structure

(12.3) Crystal Structure

The diagram of Fig. 12.22 shows a plan view of a

structure of cubic ZnS (zincblende) looking down the z

axis. The numbers attached to some atoms represent the

heights of the atoms above the z= 0 plane expressed as a

fraction of the cube edge a. Unlabeled atoms are at z= 0

and z=a.

(a) What is the Bravais lattice type?

(b) Describe the basis.

neighbor Zn–Zn, Zn–S, and S–S distances.

Zn= S =

Fig. 12.23 Plan view of conventional unit cell of

zincblende.

(a) The lattice is FCC (otherwise known as ”cubic F”)

(b) The basis can be described as Zn at position [0,0,0] and S at

position [ 1

4,1

4,3

4] (or equivalently, but less standard notation [1

4,1

4,−1

all in units of the lattice constant.

nm; nearest neighbor Zn-S is ap1/42+ 1/42+ 1/42= 0.234 nm; nearest

neighbor S-S is a/√2 = 0.383 nm;

(12.4) Packing Fractions

Consider a lattice with a sphere at each lattice point.

Choose the radius of the spheres to be such that neighbor-

ing spheres just touch (see for example, Fig. 12.8. The

packing fraction is the fraction of the volume of all of

space which is enclosed by the union of all the spheres

(i.e., the ratio of the volume of the spheres to the total

volume).

(a) Calculate the packing fraction for a simple cubic

lattice.

(b) Calculate the packing fraction for a bcc lattice.

(a) The volume of a conventional unit cell is Vcell =a3. Each cell cor-

responds to a single sphere. The radius of this sphere is a/2 so its volume

is Vsphere = 4π/3(a/2)3. Thus the packing fraction is Vsphere/Vcell =

π/6≈.52.

(b) The volume of a conventional unit cell is Vcell =a3. Each conven-

tional unit cell contains two lattice points which are a distance a√3/2

apart from each other. Thus the radius of the each sphere is a√3/4 so

has volume Vsphere = 4π(a√3/4)3/3. Thus we have a packing fraction

2Vsphere/Vcell =π√3/8≈.68

tional unit cell contains four lattice points which are a distance a√2/2

apart from each other. Thus the radius of the each sphere is a√2/4 so

has volume Vsphere = 4π(a√2/4)3/3. Thus we have a packing fraction

4Vsphere/Vcell =π/(3√2) ≈.74

(12.5) Fluorine Beta Phase

Fluorine can crystalize into a so-called beta-phase at

temperatures between 45 and 55 Kelvin. Fig. 12.24 shows

the cubic conventional unit cell for beta phase ﬂuorine in

three-dimensional form along with a plan view.

How many atoms are in this conventional unit cell?

What is the lattice and the basis for this crystal?

4and 3

Fig. 12.23 A conventional unit cell for ﬂuorine beta

phase. All atoms in the picture are ﬂuorine. Lines are

drawn for clarity Top: Three-dimensional view. Bottom:

Plan view. Unlabeled atoms are at height 0 and 1 in units

of the lattice constant.

There are 8 atoms in the conventional unit cell. 1 in the corner (8

times 1/8). Each atom on the face counts 1/2. Then there is one in the

center.

The lattice is simple cubic. The basis is

[0,0,0],[0,1/2,1/4],[0,1/2,3/4],[1/4,0,1/2],[3/4,0,1/2],[1/2,1/4,0],[1/2,3/4,0],[1/2,1/2,1/2]

Reciprocal Lattice,

Brillouin Zone, Waves in

Crystals 13

(13.1) Reciprocal Lattice

Show that the reciprocal lattice of a fcc (face-centered

cubic) lattice is a bcc (body-centered cubic) lattice. Cor-

respondingly, show that the reciprocal lattice of a bcc

lattice is an fcc lattice. If an fcc lattice has conventional

unit cell with lattice constant a, what is the lattice con-

stant for the conventional unit cell of the reciprocal bcc

lattice?

Consider now an orthorhombic face-centered lattice

with conventional lattice constants a1, a2, a3. What it

the reciprocal lattice now?

Brief Solution

Say we have an fcc lattice in real space. This can be written as cubic

with a basis R1= [0,0,0], and R2= [1/2,1/2,0] and R3= [1/2,0,1/2]

and R4= [0,1/2,1/2] in units of the lattice constant a. In reciprocal

space, we propose a basis for the reciprocal lattice which is S1= (0,0,0)

and S2= (1/2,1/2,1/2) in units of the reciprocal cubic lattice constant

4π/a. For these to be reciprocal, we must therefore have

ei2πRi·Sj= 1

for all iand jwhich is easy to check is true. Further, we can show that

given the fcc basis there is no third point that can be added to the bcc

basis which would still have the same property. Similarly, given the bcc

basis there is no ﬁfth point we can add to the fcc basis.

For the orthorhombic face centered system one would similarly obtain

a reciprocal lattice which is an orthorhombic body centered systems with

basis vectors 4π/ai.

Detailed Solution

The more straightforward way to ﬁnd the reciprocal lattice of a direct

lattice is by construction. Let us start with the direct lattice of the BCC

100 Reciprocal Lattice, Brillouin Zone, Waves in Crystals

lattice which has primitive lattice vectors

a1= [1,0,0]a

a2= [0,1,0]a

a3= [1/2,1/2,1/2]a

we then construct the primitive lattice vectors of the reciprocal lattice

via

bi=2πbfaj×ak

a1·a2×a3

with i, j, k cyclic. This gives

b1= (1/2,0,−1/2) 4π

b2= (0,1/2,−1/2) 4π

b3= (0,0,1) 4π

we can then reassemble these vectors to give the standard primitive

lattice vectors for FCC

b′

1=b1+b3= (1/2,0,1/2) 4π

b′

2=b2+b3= (0,1/2,1/2) 4π

b′

3=b1+b2+b3= (1/2,1/2,0) 4π

which shows us tht the fcc lattice has lattice constant 4π/a. NOTE

about making this transformation from binto b′. We are guaranteed

that that bare primitive lattice vectors. We assemble them together

with integer coeﬃcients to make b′. However we must also be able to

reassemble the b′with integer coeﬃcients to get back the bin order for

this to be an allowed transformation from one set of primitive lattice

vectors to another.

Slick Solution

Once one has read chapter 14 on scattering and selection rules, we

simply note that the selection rules on miller indices tell us the form

of the reciprocal lattice. For example, if the direct lattice is BCC, the

selection rule is that h+k+lis even. This means that either all h, k, l

are even or two are odd and one is even. Thus if we write

G(hkl) = 4π

ah

2ˆx+k

2ˆy+l

2ˆz

we must have either all h/2, k/2, l/2 integer or two half-odd integer and

one integer. This is precisely the condition we would use to deﬁne an

FCC lattice. The prefactor 4π/a then gives us the lattice constant.

101

Similarly in reverse, if the direct space lattice is FCC, the selection

rule is h, k, l all even or all odd. This means (h/2, k/2, l/2) all integer or

all half-odd integer. This is precisely the condition for deﬁning a BCC

lattice.

The moral of this story is that scattering occurs when (h, k, l) repre-

sents a reciprocal lattice vector. The selection rules tell you when this

is so.

(13.2) Lattice Planes

Consider the crystal shown in Exercise 12.3. Copy this

ﬁgure and indicate the [210] direction and the (210) fam-

ily of lattice planes.

(210) planes

[210]

NOTE: As drawn here this is a family of planes (it is a family of

lattice planes for the corresponding simple cubic). For the FCC to be

a family of lattice planes, it must have half the plane spacing so that it

is called (420) and it cuts though every lattice point. So the question is

not correctly written as it stands.

(13.3) Directions and Spacings of Crystal

Planes

‡Explain brieﬂy what is meant by the terms “crys-

tal planes” and “Miller indices”.

Show that the general direction [hkl] in a cubic

crystal is normal to the planes with Miller indices (hkl).

Is the same true in general for an orthorhombic

crystal?

Show that the spacing dof the (hkl) set of planes

in a cubic crystal with lattice parameter ais

d=a

√h2+k2+l2

What is the generalization of this formula for an

orthorhombic crystal?

102 Reciprocal Lattice, Brillouin Zone, Waves in Crystals

Acrystal plane is a plane which intersects at least three non-colinear

(and therefore an inﬁnite number of) points of the lattice.

Miller indices are a set of three integers which specify a set of parallel

planes (or equivalently specify a vector in reciprocal space). If the axes

of a lattice are mutually orthogonal , the Miller indices (hkl) speciﬁes

the reciprocal lattice vector (2π)(hˆ

x/ax+kˆ

y/ay+lˆ

z/azwhere ax,ayand

azare the length of the three basis vectors. In terms of lattice planes,

one can determine the lattice vectors by picking any given plane from

the family, and ﬁnding its intersection points with the three coordinate

axes. These intersection points (kx,0,0) and (0, ky,0) and (0,0, kz) have

the property that they are in the ratios

h:k:l=1

The actual values of h:k:lare then the smallest integer values with

these ratios. By convention note that negative numbers are denoted as

an integer with a bar on top, ex (1,1,¯

1) to denote (1,1,−1).

Let the basis vectors of the lattice be a,band c, assumed to be orthog-

onal. Let the lengths of these three lattice vectors be a, b, c respectively.

The plane (hkl) can be deﬁned as the plane connecting the points a/h,

b/k and c/l. To construct a vector normal to this plane, take any two

(noncolinear) vectors in this plane and take their cross product

n= (a/h −b/k)×(a/h −c/l)

=abc

hkl h

a2a+k

b2b+l

c2c

This is only parallel to the vector [hkl] in the case of the cubic crystal

a=b=c.

Much more succinctly one could also note that for orthogonal axes,

bi= 2π/ai, and the family of lattice planes (hkl) is normal to the cor-

responding reciprocal lattice vector.

The unit normal in this direction is

n=1

p(h/a)2+ (k/b)2+ (l/c)2h

a2a+k

b2b+l

c2c

To ﬁnd the inter-planar spacing, take any vector that connects two ad-

jacent planes and ﬁnd the component in the direction of the normal to

the plane. For example, the vector a/h connects two adjacent planes,

and we can resolve it parallel to nby taking the dot product with the

unit normal vector ˆ

n. Thus we obtain

dhkl =ˆ

n·a/h =1

p(h/a)2+ (k/b)2+ (l/c)2(13.1)

with the case a=b=cappropriate for cubic crystals.

Note: In cases where the axes are not orthogonal, this formula does

not work.

103

(13.4) ‡Reciprocal Lattice

(a) Deﬁne the term Reciprocal Lattice.

(b) Show that if a lattice in 3d has primitive lattice

vectors a1,a2and a3then primitive lattice vectors for

the reciprocal lattice can be taken as

b1= 2πa2×a3

a1·(a2×a3)(13.2)

b2= 2πa3×a1

a1·(a2×a3)(13.3)

b3= 2πa1×a2

a1·(a2×a3)(13.4)

What is the proper formula in 2d?

orthorhombic lattice, show that |bj|= 2π/|aj|. Hence,

show that the length of the reciprocal lattice vector

G=hb1+kb2+lb3is equal to 2π/d, where dis the

spacing of the (hkl) planes (see question 13)

(a) Given a lattice of points (in 3d it would look like R[uvw]=ua1+

va2+wa3with [uvw] integers), the reciprocal lattice is deﬁned by the

set of points in kspace such that

eik·R= 1

for all points Rin the lattice . Note that this set of points forms a lattice

of values of k.

(b) If we take

b1= 2πa2×a3

a1·(a2×a3)

b2= 2πa3×a1

a1·(a2×a3)(13.5)

b3= 2πa1×a2

a1·(a2×a3)

Then we can show the key formula

bi·aj= 2πδij .

If these are our basis vectors for the reciprocal lattice, we then have a

general reciprocal lattice point given by

Ghkl =hb1+kb2+lb3(13.6)

with h, k, l integers. It is trivial to then see that

eiK·R= 1 (13.7)

for any lattice vector Rand any reciprocal lattice vector K. However,

this does not quite prove the desired statement. What it proves is that

vector Kof the form of Eq. 13.6 are in the reciprocal lattice. We need

to show that there are no other vectors in the reciprocal lattice as well

so that b1,b2, and b3can indeed be taken as the basis.

104 Reciprocal Lattice, Brillouin Zone, Waves in Crystals

To show this, consider an arbitrary vector Kof the form of Eq. 13.6

but do not require that h, k, l are integers. Given an arbitrary Ruvw of

the real space lattice, in order that Eq. 13.7 is satisﬁed, we must have

uk +vh +wl = integer

for this to be true for arbitrary u, v, w which are integers, we can conclude

that k, h, l are integers.

We can derive the 2d analogous formula by setting a3=ˆ

zthe unit

vector normal to the plane, more conveniently written as

b1= 2πa2×ˆz

z·(a1×a2)

b2= 2πˆ

z×a1

z·(a1×a2)

are normal to each other, and two of them are the same length but the

third is a diﬀerent length. An orthorhombic lattice is a lattice (in 3d)

where all three basis vectors are normal to each other and all three have

diﬀerent lengths. (Note: crystals may have orthorhombic or tetragonal

symmetry even if lattice constants in the three directions are all equal.

The symmetry of a crystal has to do with whether it looks the same

under various types of rotations. An orthorhombic crystal does not look

the same under the three 90 degree rotations – this may be the case even

if all three lattice constants are the same).

For an orthorhombic crystal, without loss of generality, let us write

a1=a1ˆ

xand a2=a2ˆ

yand a3=a3ˆ

z. We then just use the above

formula Eq. 13.5 to obtain

b1= 2πa2a3

a1a2a3

y×ˆ

z= 2πˆ

x/a1

b2= 2πa2a3

a1a2a3

z×ˆ

x= 2πˆ

y/a2

b3= 2πa2a3

a1a2a3

x×ˆ

y= 2πˆ

z/a3

So |bi|=2π/|ai|. Thus the length of G=hb1+kb2+lb3is given by

|Ghkl|=ph2|b1|2+k2|b2|2+l2|b3|2=p(2πh/a1)2+ (2πk/a2)2+ (2πl/a3)2= 2π/dhk

as given in Eq. 13.1.

(13.5) More Reciprocal Lattice

A two-dimensional rectangular crystal has a unit cell

with sides a1= 0.468 nm and a2= 0.342 nm.

(a) Draw to scale a diagram of the reciprocal lattice.

Label the reciprocal lattice points for indices in the

range 0 ≤h≤3 and 0 ≤k≤3.

(b) Draw the ﬁrst and second Brillouin zones using the

Wigner–Seitz construction.

The reciprocal lattice vectors are b1= 2πˆ

x/|a1|, and b2= 2πˆ

y/|a2|.

Their magnitudes are

|b1|= 13.4nm−1|b2|= 18.4nm−1

105

(a) A diagram of the reciprocal lattice is given in the ﬁgure.

Fig. 13.1 Figures for problem 13. First Two Brillouin Zones. Red is the 1st Zone.

Blue is the 2nd zone.

To ﬁnd the Brillouin zones, one ﬁrst constructs perpendicular bisectors

between the origin and any given lattice point (shown as dotted lines on

the plot). Then starting at the origin, the region one can get to without

crossing a dotted line is the ﬁrst zone. Crossing only one dotted line

gets one to the second zone. etc.

(13.6) Brillouin Zones

(a) Consider a cubic lattice with lattice constant a.

Describe the ﬁrst Brillouin zone. Given an arbitrary

wavevector k, write an expression for an equivalent

wavevector within the ﬁrst Brillouin zone (there are sev-

eral possible expressions you can write).

(b) Consider a triangular lattice in two dimensions

(primitive lattice vectors given by Eqs. 12.3). Find the

ﬁrst Brillouin zone. Given an arbitrary wavevector k(in

two dimensions), write an expression for an equivalent

wavevector within the ﬁrst Brillouin zone (again there

are several possible expressions you can write).

(a) The reciprocal lattice of a cubic lattice with lattice constant a

is a cubic lattice with lattice constant 2π/a. The ﬁrst Brillouin zone

is a cube centered around the origin in reciprocal space (0,0,0), with

side length 2π/a. It spans kx, ky, kz∈[−π/a, π/a]. Given an arbitrary

k= (kx, ky, kz) we can write the equivalent wavevector as (k′

x, k′

y, kz)′

where

k′

j=kj−[[(kj+π/a)/(2π/a)]] ∗2π/a (13.8)

with j=x, y, z and [[ ]] is the ﬂoor function – meaning the greatest

integer less than its argument. The point of Eq. 13.8 is that it takes kin

units of the reciprocal lattice constant (2π/a) and returns a number be-

tween -1/2 and +1/2 of the reciprocal lattice constant by appropriately

106 Reciprocal Lattice, Brillouin Zone, Waves in Crystals

adding or subtracting integer units of 2π/a.

(b) Start with the primitive lattice vectors

a1=aˆx

a2= (a/2)ˆx + (a√3/2)ˆy

We need to ﬁnd vectors b1,b2such that ai·bj= 2πδi,j . One way to

do this is to use the usual 3d formula

b1= 2πa2×a3

a1·(a2×a3)

b2= 2πa3×a1

a1·(a2×a3)

b3= 2πa1×a2

a1·(a2×a3)

then assigning a3=ˆz. The denominator is a2√3/2. Thus we have

b1= 2π[(a√3/2)ˆx −(a/2)ˆy]/(a2√3/2) = (2π/a)[ˆx −(√3/3)ˆy]

b2= 2πˆya/(a2√3/2) = (2π/a)ˆy(2√3/3)

These form the primitive lattice vectors of a triangular lattice in k-space.

The ﬁrst Brillouin zone is the Wigner-Seitz cell of the reciprocal lattice

— which is a hexagon centered around zero wavevector (see ﬁg 12.6 of

the book).

Two write a formula that translates any kinto the ﬁrst Brillouin zone,

we ﬁrst write any kpoint in terms of the primitive lattice vectors

k=α1b1+α2b2

We are going to need to solve for αiin terms of k. To do this we write

k·b1=α1|b1|2+α2b1·b2

k·b2=α1b1·b2+α2|b2|2

and we will want to solve for the αs. In this particular case, things are

quite easy since |b1|2=|b2|2=−2b1·b2= (2π/a)2(4/3). So this

becomes

βk·b1=α1−α2/2

βk·b2=−α1/2 + α2

where we have deﬁned

β=3

(2π)2

Solving this system of equations we obtain

α1= (β/3)(2k·b1+k·b2)

α2= (β/3)(k·b1+ 2k·b2)

107

Now to obtain αwithin the ﬁrst brillouin zone we need to shift both

α’s by integers until they lie in the range [−1/2,1/2]. To do this we

write

α′

j=αj−[[αj+ 1/2]]

again with [[ ]] being the ﬂoor function. Once we have α′’s we construct

k′=α′

1b1+α′

2b2

which is now within the ﬁrst Brillouin zone.

(13.7) Number of States in the Brillouin Zone

A specimen in the form of a cube of side Lhas a primi-

tive cubic lattice whose mutually orthogonal fundamental

translation vectors (primitive lattice vectors) have length

a. Show that the number of diﬀerent allowed k-states

within the ﬁrst Brillouin zone equals the number of prim-

itive unit cells forming the specimen. (One may assume

periodic boundary conditions, although it is worth think-

ing about whether this still holds for hard-wall boundary

conditions as well.)

We are given a cubic lattice with lattice constant aand an overall size

Lon a size. The number of atoms in this sample are N= (L/a)3. It

is simplest to assume periodic boundary conditions. Since eikx(x+L)=

eikxxwe must have kx= 2πn/L for nan integer. Thus (considering all

three directions) there is one eigenstate per volume (2π/L)3in k-space.

The ﬁrst Brillouin zone extends from −π/a ≤kx, ky, kz≤π/a. This

has volume in k-space of (2π/a)3. Dividing this by the volume occupied

by one eigenstate gives a total of (L/a)3states in the ﬁrst Brillouin zone.

This agreement is not coincidental. If one thinks in a tight-binding or

atomic orbital picture, there should be exactly one orbital per atom per

band when hopping is turned oﬀ. Once hopping is turned back on, each

atomic orbital spreads into a band that ﬁlls the Brillouin zone, but the

total number of states stays constant.

One can also consider the same problem with hard wall boundary

conditions, although it is less convenient. In this case the eigenstates

are not of the form eikx but are rather sin(kx) where k > 0, so one is

not really talking about pure plane waves. In this case the analogue of

the Brillouin zone goes from 0 ≤kx, ky, kz≤π/a, but the density of

kstates is doubled, k=π/L in each direction, so the total number of

states remains the same as using periodic boundary conditions.

108 Reciprocal Lattice, Brillouin Zone, Waves in Crystals

(13.8) Calculating Dispersions in d > 1*

(a) In Exercises 9.8 and 11.9 we discussed dispersion

relations of systems in two dimensions (if you have not

already solved those exercises, you should do so now).

In Exercise 11.9 describe the Brillouin zone (you

may assume perpendicular lattice vectors with length a1

and a2). Show that the tight-binding dispersion is peri-

odic in the Brillouin zone. Show that the dispersion curve

is always ﬂat crossing a zone boundary.

In Exercise 9.8, describe the Brillouin zone. Show

that the phonon dispersion is periodic in the Brillouin

zone. Show that the dispersion curve is always ﬂat cross-

ing a zone boundary.

(b) Consider a tight binding model on a three-

dimensional fcc lattice where there are hopping matrix

elements −tfrom each site to each of the nearest-neighbor

sites. Determine the energy spectrum E(k) of this model.

Show that near k=0the dispersion is parabolic.

(a) [First see solutions for 9.8 and 11.9]

In Exercise 11.9, for a rectangular lattice with lattice vectors of length

a1and a2the reciprocal lattice is rectangular with lattice vectors of

length b1= (2π)/a1and b2= (2π)/a2. The Brillouin zone is hence a

rectangle extending between kx∈[−π/a1, π/a1] and ky∈[−π/a2, π/a2].

Given the derived dispersion

E=ǫ0−2t1cos(kxa1)−2t2cos(kya2)

at the zone boundary, the cos is at its maximum, and its derivative is

zero. Thus the dispersion is always ﬂat approaching the zone boundary.

In Exercise 9.8 we have a triangular lattice, whose reciprocal lattice is

also triangular and has a Wigner-Seitz cell which is hexagonal (see exer-

cise 13.6.b also). Note also the ﬁgures shown in the solution of 9.8 which

clearly show the hexagonal periodicity of the dispersion curve. Note that

the dispersion must be symmetric as kapproaches the zone boundary.

Let the zone boundary wavevector be bi/2 + k⊥where k⊥·bi= 0.

Imagine approaching the zone boundary via k=αbi/2 + k⊥and taking

αto one. Note that the dispersion must be symmetric around α= 1

since (by reﬂection symmetry of the problem) the frequency at kis the

same as the frequency at k′=−αbi/2 + k⊥. However, one can translate

this by a reciprocal lattice vector to get (2 −α)bi/2 + k⊥which is the

same point reﬂected around the zone boundary. Thus, so long as the

dispersion does not have a cusp at the zone boundary, it must have zero

derivative.

Recall from Eq. 9.8 that the derived freuquency spectrum is

ω2=2κ

mS1+S2+S3±qS2

1+S2

2+S2

3−S1S2−S1S3−S2S3

where

Si= sin2(k·ai

where aiare three independent vectors along the lattice. In terms of

primitive lattice vectors here we can take a1and a2to be primitive

lattice vector, and then a3=a2−a1.

109

The frequency is an analytic function in kso long as the argument of

the square-root is nonzero. By rewriting the argument as

2(S1−S2)2+ (S2−S3)2+ (S3−S1)2

it is clear the argument must be nonzero except at k=0. Thus we can

conclude that the dispersion is analytic and symmetric around the zone

boundary and therefore must have zero slope. This may seem like a bit

of a cheat, but it is perfectly rigorous.

For those who would prefer to see this statment proven a bit more

directly, we take the following approach. Let us start with the primitive

lattice vectors

a1=aˆx

a2= (a/2)ˆx + (a√3/2)ˆy

Recall that the reciprocal lattice vectors (see 13.6.b) are given by

b1= 2π[(a√3/2)ˆx −(a/2)ˆy]/(a2√3/2) = (2π/a)[ˆx −(√3/3)ˆy]

b2= 2πˆya/(a2√3/2) = (2π/a)ˆy(2√3/3)

For simplicity let us imagine approachig the zone boundary

k=αb2/2 + k⊥

(the other zone boundaries will be the same by symmetry). Here the

direction orthogonal to b2is b1+b2/2. So we can write k⊥=β(b1+

b2/2) for some value of β. We then have (using ai·bj= 2πδij for

i, j ∈1,2)

k·a1= (β+α)π

k·a2=β2π

k·a3= (β−α)π

Now let us examine the terms in the dispersion. In particular we are

concerned with the dependence of the dispersion on αnear α= 1. Again

all we need to do is to show that the dispersion is symmetric and analytic

around α= 1. Here S2is independent of αso we are not concerned with

that piece. Using sin2(x/2) = (1 −cos(x))/2 we have that

S1= [1 −cos((β+α)π)]/2

S2= [1 −cos((β−α)π)]/2

The cos is periodic in α→α±2, and reﬂection of αaround 1 (i.e.,

α→2−α) turns S1into S2, hence leaves the freqeuncy unchanged.

Thus we have the dispersion symmetric and analytic around the zone

boundary, thus having zero slope.

(b) For an fcc lattice, if the conventional unit cell has lattice constant

a, the vectors to the 12 nearest neighbors of a lattice point are given by

110 Reciprocal Lattice, Brillouin Zone, Waves in Crystals

(see exercise 12.2)

[±1,±1,0]a/2

[±1,0,±1]a/2

[0,±1,±1]a/2

Thus we have the schroedinger equation (setting the onsite energy ǫ0to

zero for simplicity)

Eψ(r) = −tX

ψ(r+u)

where the sum is over ubeing these 12 vectors. Using the usual ansatz

ψ(r) = Aeik·rwe obtain

E=−tX

α=±1,β=±1heik·[α,β,0]a/2+eik·[α,0,β]a/2+eik·[0,α,β]a/2i

E=−4t[cos(kxa/2) cos(kya/2) + cos(kxa/2) cos(kza/2) + cos(kya/2) cos(kza/2)]

Expanding around k= 0 to second order we obtain

E=−12t+tk2

x+k2

y+k2

z(a2) + ...

which is parabolic as claimed.

Wave Scattering by

Crystals 14

(14.1) Reciprocal Lattice and X-ray Scattering

Consider the lattice described in Exercise 13.5 (a two-

dimensional rectangular crystal having a unit cell with

sides a1= 0.468 nm and a2= 0.342 nm). A collimated

beam of monochromatic X-rays with wavelength 0.166 nm

is used to examine the crystal.

(a) Draw to scale a diagram of the reciprocal lattice.

(b) Calculate the magnitude of the wavevectors kand

k′of the incident and reﬂected X-ray beams, and hence

construct on your drawing the “scattering triangle” cor-

responding to the Laue condition ∆k=Gfor diﬀraction

from the (210) planes (the scattering triangle includes k,

k′and ∆k).

A diagram of the reciprocal lattice is given in the ﬁgure. The scatter-

Fig. 14.1 A picture of the reciprocal Lattice with the scattering triangle for the

(210) reciprocal wavevector.

ing triangle is the triangle such that |k|=|k′|and k+G=|k′|(Note

for ﬁxed Gthere are two such possible triangles).

112 Wave Scattering by Crystals

(14.2) ‡X-ray scattering II

BaTiO3has a primitive cubic lattice and a basis with

atoms having fractional coordinates

Ba [0,0,0]

Ti [ 1

2,1

O [ 1

2,1

2,0], [ 1

2,0,1

2], [0,1

2,1

Sketch the unit cell.

Show that the X-ray structure factor for the (00l)

Bragg reﬂections is given by

S(hkl)=fBa + (−1)lfT i +h1 + 2(−1)lifO

where fBa is the atomic form factor for Ba, etc.

Calculate the ratio I(002)/I(001), where I(hkl)is the

intensity of the X-ray diﬀraction from the (hkl) planes.

You may assume that the atomic form factor is propor-

tional to atomic number (Z), and neglect its dependence

on the scattering vector. (ZBa = 56, ZTi = 22, ZO= 8.)

Fig. 14.2 Unit cell of BaTiO3

The X-ray structure factor is given by

S(hkl)=X

fdeik(hkl)·Rd=X

fde2πi(hxd+kyd+lzd)

where Rd= (xd, yd, zd) are the positions of atom din the unit cell, and

fdit the corresponding form factor (which we take to be proportional

to Zd). If we are interested in (00l) we set h=k= 0 and obtain

S(00l)=fBa + (−1)lfT i + [1 + 2(−1)l]fO

The Bragg peak intensity is proportional to the square of the structure

factor (times a multiplicity factor, but the multiplicity for all (00l) are

the same!), thus we obtain

I002

I001

=(fBa +fT i + 3fO)2

(fBa−fT i −fO)2≈15.4

In reality the form factor depends on the scattering vector, and the

variation is diﬀerent from each atom, so this is just an approximation.

113

(14.3) ‡X-ray scattering and Systematic Ab-

sences

(a) Explain what is meant by “Lattice Constant” for a

cubic crystal structure.

(b) Explain why X-ray diﬀraction may be observed in

ﬁrst order from the (110) planes of a crystal with a body-

centered cubic lattice, but not from the (110) planes of a

crystal with a face-centered cubic lattice.

Derive the general selection rules for which planes

are observed in bcc and fcc lattices.

what atoms are in the primitive unit cell, so long as the

lattice is bcc or fcc respectively.

(d) A collimated beam of monochromatic X-rays of

wavelength 0.162 nm is incident upon a powdered sample

of the cubic metal palladium. Peaks in the scattered X-

ray pattern are observed at angles of 42.3◦, 49.2◦, 72.2◦,

87.4◦, and 92.3◦from the direction of the incident beam.

Identify the lattice type.

Calculate the lattice constant and the nearest-

neighbor distance.

If you assume there is only a single atom in the ba-

sis does this distance agree with the known data that the

density of palladium is 12023 kg m−3? (Atomic mass of

palladium = 106.4.)

(e) How could you improve the precision with which

the lattice constant is determined. (For one suggestion,

see Exercise 14.10.)

(a) For a cubic lattice, the lattice constant is the distance between

one lattice point and the nearest neighbor lattice point.

(b) The (110) planes of a body-centred cubic lattice contain all the

lattice points, whereas the (110) planes of a crystal with a face-centred

cubic lattice contain only half the lattice points. The remaining fcc

lattice points lie on a set of planes half-way in between the (110) planes,

and so X-rays reﬂected from these planes interfere destructively with

X-rays reﬂected from the (110) planes.

Let us see this more analytically now:

We view both the bcc and fcc lattices as being a cubic lattice with a

basis. For a cubic lattice, we must have khkl = (2π/a)(hˆ

x+kˆ

y+lˆ

z) (or

in other words we consider reciprocal lattice vectors (h, k, l).)

Generally we sum over lattice points Riin a unit cell to get the struc-

ture factor

S(hkl)∼X

eiRi·k(hkl)

The bcc lattice is a cubic lattice with a basis (0,0,0) and (1/2,1/2,1/2).

So we obtain

Sbcc

(hkl)∼1 + eik(hkl)·[(1/2)ˆ

x+(1/2)ˆ

y+(1/2)ˆ

z]= 1 + (−1)h+k+l

This is only nonzero when h+k+l= even.

The fcc lattice is a cubic lattice with with basis (0,0,0) and (1/2,1/2,0)

and (1/2,0,1/2) and (0,1/2,1/2). Analogously we obtain

Sfcc

(hkl)∼1 + (−1)h+k+ (−1)h+l+ (−1)k+l

This is only nonzero when h, k, l are either all even or all odd.

Note that these selection rules hold for an orthorhombic lattice as

well as for cubic (nowhere did we use the lattice constants in the three

114 Wave Scattering by Crystals

directions!).

Suppose we have a Bravais lattice (either fcc or bcc) and then we have

a basis of matoms which are at positions rj(j= 1,...m) with respect

to the lattice points. These atoms may have scattering form factors fj

as well.

As above, we treat the Bravais lattice as a cubic lattice with a basis

Ri(i=1,2 for bcc and i= 1 ...4 for fcc). So positions of all of the atoms

(2mor 4mof them) in the full cubic unit cell can be written as

Rij =Ri+rj

Writing the structure factor

S(hkl)=X

i,j

fjeik(hkl)Rij =X

fj"X

eik(hkl)Ri#eik(hkl)rj(14.1)

The factor inside the brackets is exactly the structure factor for the Bra-

vais lattice as calculated above. So if the fcc structure factor vanishes,

then the structure will also vanish for any lattice having an fcc lattice

with any basis.

(d) Note that the angle given is 2θ(the full deﬂection angle)! We are

given 2θthen we use Bragg’s law λ= 2dsin θto get the distance between

lattice planes. We then examine the ratio’s between these distances.

2θd=λ/(2 sin θ) (dmax/d)2aestimate =d√h2+k2+l2

42.3◦.224 nm 1 .389 nm

49.2◦.195 nm 1.33 .389 nm

72.2◦.137 nm 2.66 .389 nm

87.4◦.117 nm 3.67 .389 nm

92.3◦.112 nm 4.00 .389 nm

we see that the values of 1/d2are in ratios 3:4:8:11:12. Recall (see

problem 13.3) that for a cubic lattice (a2/d2) = h2+k2+l2. So the

observed peaks are of the types (1,1,1), (2,0,0), (2,2,0), (1,1,3), and

(2,2,2). These are all even or all odd, which is characteristic of the fcc

lattice. All measured scattering points give the same estimate of the

lattice constant ato three digits (which is the same number of digits

that the measurement contains... so no higher accuracy estimate can be

made from this data).

12023 kg

.1064 kg

6.022 ×1023 atoms = 1.4682 ×10−29m3/atom

Then for an fcc there are 4 atoms per unit cell, so the volume of the unit

cell is 5.872 ×10−29m3. The cube root of this gives 3.887 nm which is

in good agreement. An equivalent calculation is to calculate the density

115

based on the lattice constant

4∗.1064 kg

6.022 ×1023 atoms 1

.389 ×10−9m3

= 12006 kg/m3

in good agreement.

(e) to improve the precision of the measurement one would need to ob-

tain more “digits” of resolution in the measurement of the angle. There

are several things that will help this. From Bragg’s law d∼1/sin θso

∂d

∂θ = cot θ

So if there is an error in measurement of θone ends up with a fractional

error in dgiven by

δlog d= (cot θ)(δθ)

As a result, error is minimized if one can work as close as possible to

scattering angle of 90 degrees (See exercise 14.10).

In addition some experimental issues can be listed. A higher bright-

ness source will frequently reduce the noise and make the signal easier to

analyze. Better columnation will make measurements of angle more pre-

cise. At least as important is the calibration of the device that measures

the angles! Deﬁning the wavelength λmore precisely becomes necessary

at some point. Temperature control is necessary since lattice constants

do change as a function of temperature due to thermal expansion.

(14.4) ‡Neutron Scattering

(a) X-ray diﬀraction from sodium hydride (NaH) estab-

lished that the Na atoms are arranged on a face-centered

cubic lattice.

Why is it diﬃcult to locate the positions of the H

atoms using X-rays?

The H atoms were thought to be displaced from the Na

atoms either by [ 1

4,1

4] or by [ 1

2,1

2], to form the ZnS

(zincblende) structure or NaCl (sodium chloride) struc-

ture, respectively. To distinguish these models a neutron

powder diﬀraction measurement was performed. The in-

tensity of the Bragg peak indexed as (111) was found to

be much larger than the intensity of the peak indexed as

(200).

Write down expressions for the structure factors

S(hkl)for neutron diﬀraction assuming NaH has

(i) the sodium chloride (NaCl) structure

(ii) the zinc blende (ZnS) structure.

Hence, deduce which of the two structure mod-

els is correct for NaH. (Nuclear scattering length of Na

= 0.363 ×105nm; nuclear scattering length of H =

−0.374 ×105nm.)

(b) How does one produce monochromatic neutrons for

use in neutron diﬀraction experiments?

What are the main diﬀerences between neutrons

and X-rays?

Explain why (inelastic) neutron scattering is well

suited for observing phonons, but X-rays are not.

[This problem is based on a classic experiment by Shull et al, who

later won a Nobel prize for his work on neutron scattering]

Note, the scattering length should be in units of 10−5nm not 105nm.

It doesn’t change the problem much though.

It is diﬃcult to see H atoms with X-rays since the form factor (am-

plitude) of scattering is proportional to the charge of the nucleus (the

atomic number), and the charge of H is very small.

116 Wave Scattering by Crystals

From Eq. 14.1 above we have

S(hkl)= X

eik(hkl)Ri!

X

bjeik(hkl)rj



where jare the elements of the basis and iare the 4-elements of the fcc

lattice basis when fcc is viewed as a cubic lattice with a basis. Here I have

replaced fjby bjwhich is the conventional notation for the scattering

length (which is analogous to the form factor in X-rays) in neutron

scattering. Note that the ﬁrst factor in brackets is the scattering from

the fcc lattice which always gives (See problem 14)

Sfcc

hkl = (1 + eiπ(h+k)+eiπ(k+l)+eiπ(h+l))

For the case of the NaCl structure, we have a basis of Na at [0,0,0]

and H at [1/2,1/2,1/2]. So we obtain

S(hkl)= (1 + eiπ(h+k)+eiπ(k+l)+eiπ(h+l))(bNa +bHeiπ(h+k+l))

For the case of the ZnS structure we have a basis of Na at [0,0,0] and

H at [1/4,1/4,1/4] so we obtain

S(hkl)= (1 + eiπ(h+k)+eiπ(k+l)+eiπ(h+l))(bNa +bHei(π/2)(h+k+l))

The powder scattering intensity is proportional to the structure factor

squared times a multiplicity factor that counts how many symmetry re-

lated vectors give the same scattering pattern. The multiplicity of (111)

is 8 (since each factor of 1 could have been ¯

1), whereas the multiplicity

of (200) is 6 since the 2 could be of either sign, and any one of the 3

entries could contain the 2.

Thus we obtain in the NaCl case a ratio of scattering intensities

I(111)

I(200)

=(bNa −bH)2

(bNa +bH)2

6= 6000

whereas in the ZnS case we have

I(111)

I(200)

=(b2

Na +b2

(bNa −bH)2

6= 0.67

Thus we conclude that NaH has NaCl structure.

(b) Two typical schemes for producing monochromatic neutrons. (1)

time of ﬂight (a ”chopper”) can select a particular velocity (2) Bragg

scattering of a given angle oﬀ of a crystal can select a particular wave-

length.

The main things to know about the diﬀerences between X-rays and

neutrons are:

X-rays are scattered by electrons. The electromagnetic interaction is

relatively strong, and so X-rays typically penetrate only a few microns

into a sample. The intensity of scattering varies with the atomic form

117

factor f(∆k), which approximately scales with the number of electrons

in the atom.

Neutrons, on the other hand, are electrically neutral, and so interact

weakly with matter. They penetrate typically a few cm into materi-

als. Neutrons may scatter from nuclei (via the strong nuclear force)

or unpaired electrons (via the magnetic dipole-dipole interaction). The

diﬀraction theory is the same as for X-rays, except that for scattering

from nuclei the atomic form factor f(∆k) is replaced by the nuclear scat-

tering length b, which is independent of ∆k. (Note, we typically make

the approximation that fis independent of ∆k, to make calculations

easier, but it is not really such a good approximation. In the neutron

case, it really is a good approximation because the scattering is essen-

tially oﬀ of a point nucleus). The scattering length bvaries irregularly

from nucleus to nucleus so that the scattering from light elements is sim-

ilar in strength to that from heavy elements. Neutrons can thus easily

see small atoms, and can easily distinguish atoms with similar atomic

number.

Magnetic neutron diﬀraction can also be used to determine magnetic

structures (in which case there is a form factor similar to that for X-

rays).

The reason inelastic scattering of neutrons is more useful than X-rays

for studying phonons is because the X-ray velocity (velocity of light)

is huge. This makes it very diﬃcult to conserve both momentum and

energy in any inelastic process involving light since a very small mo-

mentum change corresponds to a huge energy change. Although with

modern tools measurement of phonons using X-rays is now possible.

(14.5) And More X-ray Scattering

A sample of aluminum powder is put in an Debye–

Scherrer X-ray diﬀraction device. The incident X-ray ra-

diation is from Cu–Ka X-ray transition (this just means

that the wavelength is λ= .154 nm). The following scat-

tering angles were observed:

19.48◦22.64◦33.00◦39.68◦41.83◦50.35◦57.05◦59.42◦

Given also that the atomic weight of Al is 27, and the

density is 2.7 g/cm3, use this information to calculate

Avagadro′s number. How far oﬀ are you? What causes

the error?

Note: The angles listed are θnot 2θhere. If you use 2θby mistake

118 Wave Scattering by Crystals

you will discover that the ratios of dvalues do not make sense!

θ d =λ/(2 sin θ) (dmax/d)2aestimate =d√h2+k2+l2

19.48◦.2309 nm 1 .3999 nm

22.64◦.2000 nm 1.33 .4001 nm

33.00◦.1414 nm 2.67 .3999 nm

39.68◦.1206 nm 3.66 .4000 nm

41.83◦.1155 nm 4.00 .4000 nm

50.35◦.1000 nm 5.33 .4000 nm

57.05◦.0918 nm 6.33 .4000 nm

59.42◦.0894 nm 6.66 .4000 nm

From the third column we see that we have an fcc lattice. From the

ﬁnal column we conclude the conventional lattice constant is about .4000

nm. There are four atoms per unit cell, so we would predict an atomic

density of 4/(4nm)3= 6.28 ×1028m−3. On the other hand, with atomic

weight of 27 and density 2.7 g/cm3we predict an atomic density of

NA×105m−3with NA= 6.022 ×1023 being Avagadro’s number. (Or

in other words, our slighly inaccurate prediction of Avagadro’s number

is 6.28 ×1023). So we are oﬀ by abour 4%. Where does the error come

from? There are a few possible places. First, the atomic mass is not

quite 27, it is actually about 26.982 (depending a bit on isotopic abun-

dances, but this is typical). But this is an error of much less than an

percent. Probably the biggest possible error is that the density changes

as a function of temperature. If the density is measured at one tem-

perature, but the lattice constant is measured at another temperature,

then there can be a substantial disagreement. In fact the linear thermal

expansion coeﬃcient of aluminum is about (22 ×10(−6))K−1which

would give a density diﬀerent between T= 0 and room temperature of

a few percent (and indeed, the measurements could even be done above

room temperature!). In the literature X-ray measurements of the lattice

constant of alumium give numbers of about .404 nm at room tempera-

ture (which would match the stated density much more closely). Thus I

suspect the data given above was measured at low temperature, hence

the disagreement.

(14.6) More Neutron Scattering

The conventional unit cell dimension for a particular

bcc solid is .24nm. Two orders of diﬀraction are observed.

What is the minimum energy of the neutrons? At what

temperature would such neutrons be dominant if the dis-

tribution is Maxwell–Boltzmann.

For bcc, the lowest two order diﬀraction peaks are the (110) and the

(200). The corresponding plane spacings are

d=a/ph2+k2+l2

119

Or d=a/√2 and d=a/2. Using Bragg’s law,

d=λ/(2 sin θ)

the wavelength must be at least as small as 2din order to see a particular

peak. Or in other words λ≤a=.24nm. The corresponding energy of

neutrons is

E=~2k2

2mn

=~2(2π)2

2mnλ2= 2.28 ×10−21J

and

E/kB= 165K

(14.7) Lattice and Basis

Prove that the structure factor for any crystal (de-

scribed with a lattice and a basis) is the product of the

structure factor for the lattice times the structure factor

for the basis (i.e., prove Eq. 14.14).

We have the deﬁnition of the structure factor

S(hkl)=X

atoms jin (conventional) unit cell

fje2πi(hxj+kyj+lzj)

When we have a lattice and a basis, all of the positons rjof atoms jin

the unit cell can be written as the sum of a lattice point Raand a basis

vector uα. Thus we can write

rj=Ra+uα

and the sum over jbecomes a sum over aand α. Thus we have

S(hkl)=X

α,a

fαe2πi(h(Rx,a+ux,α )+k(Ry,a +uy,α)+l(Rz,a+uz,α ))

Note that the atom type depends only on the basis vector not on the

lattice point. We split this sum into two

S(hkl)=X

fαe2πi(hux,α)+kuy,α +luz,α)X

e2πi(hRx,a+kRy,a +lRz,a)

which is simply the product of basis and lattice structure factors

S(hkl)=Sbasis

(hkl)Slattice

(hlk)

120 Wave Scattering by Crystals

(14.8) Cuprous Oxide and Fluorine Beta

(a) The compound Cu2O has a cubic conventional unit

cell with the basis:

O [000] ; [ 1

2,1

Cu [ 1

4,1

4] ; [ 1

4,3

4] ; [ 3

4,1

4,3

4] ; [ 3

4,3

4,1

Sketch the conventional unit cell. What is the lattice

type? Show that certain diﬀraction peaks depend only

on the Cu form factor fCu and other reﬂections depend

only on the O form factor fO.

(b) Consider ﬂuorine beta phase as described in exer-

cise 12.5. Calculate the structure factor for this crystal.

What are the selection rules?

(a) the conventional unit cell of Cu2O is shown in Figure 14.3.

The structure factor is

S(hkl)=fO(1 + eiπ(h+k+l))

+fCu heiπ

2(h+k+l)+eiπ

2(h+3k+3l)+eiπ

2(3h+k+3l)+eiπ

2(3h+3k+l)i

Note that sqaure bracketed expression can be simpliﬁed to

eiπ

2(h+k+l)h1 + eiπ(k+l)+eiπ(h+l)+eiπ(k+l)i

Note that the coeﬃcient of f0vanishes unless h+k+lis even. Indeed,

Fig. 14.3 Conventional unit cell of

Cu2O. The darker atoms at the corners

and the center are O and the lighter col-

ored atoms are Cu.

this tells us that the oxygens alone form a bcc lattice. The coeﬃcient of

the fCu vanish unless h, k, l are all even or all odd, which tells us that

the Cu alone form an fcc lattice (perhaps less obvious!). Thus peaks

such as (110) depend only on fOwheras peaks such as (111) depend

only on fCu.

(b) From 12.5 the basis of Flourine beta is

[0,0,0],[0,1/2,1/4],[0,1/2,3/4],[1/4,0,1/2],

[3/4,0,1/2],[1/2,1/4,0],[1/2,3/4,0],[1/2,1/2,1/2]

The structure factor is then

1 + eiπk+iπl/2+eiπk+iπ3l/2+eiπl+iπh/2+eiπl+iπ3h/2

+eiπh+iπk/2+eiπh+iπ3k/2+eiπ(h+k+l)

which can be simpliﬁed to

1 + (−1)h+k+l+ (−1)k[il+i−l] + (−1)l[ih+i−h] + (−1)h[ik+i−k]

Note that the ﬁrst two terms cancel if h+k+lis odd and otherwise

give 2. Let us examine the last three terms. Note that

ik+i−k=





0l= odd

2l= 0 mod 4

−2l= 2 mod 4

121

Thus we obtain the following rules for (hkl) where the structure factor

does not vanish

one even two odd if the even one is 2 mod 4

one odd two even if the two even indices are not the same mod 4

all even with either one or all three indices being 2 mod 4

(14.9) Form Factors

(a) Assume that the scattering potential can be written

as the sum over the contributions of the scattering from

each of the atoms in the system. Write the positions of

the atoms in terms of a lattice plus a basis so that

V(x) = X

R,α

Vα(x−R−yα)

where Rare lattice points, αindexes the particles in the

basis and yαis the position of atom αin the basis. Now

use the deﬁnition of the structure factor Eq. 14.5 and de-

rive an expression of the form of Eq. 14.8 and hence derive

expression 14.9 for the form factor. (Hint: Use the fact

that an integral over all space can be decomposed into a

sum over integrals of individual unit cells.)

(b) Given the equation for the form factor you just de-

rived (Eq. 14.9), assume the scattering potential from an

atom is constant inside a radius aand is zero outside that

radius. Derive Eq. 14.10.

(c)* Use your knowledge of the wavefunction of an elec-

tron in a hydrogen atom to calculate the X-ray form factor

of hydrogen.

(a) Start with 14.5

S(G) = Zunitcell

dx eiG·xV(x) = X

R,α Zunitcell

dx eiG·xVα(x−R−yα)

we then using eiG·R= 1 we have

S(G) = X

R,α Zunitcell

dx eiG·(x−R)Vα(x−R−yα)

Now we note that the integral dx over a unit cell summed over all

possible unit cells indexed by Ris equivalent to a single integral over all

of space, so we have

S(G) = X

αZdx eiG·xVα(x−yα) = X

αZdx eiG·(x+yα)Vα(x)

eiG·yαfα(G)

with

fα(G) = Zdx eiG·xVα(x)

as required.

(b) Assuming the potential is V0inside a radius aand zero outside we

have

fα(G) = V0Zradius=a

dx eiG·x= 2πV0Za

r2dr Zsin θdθeirGcosθ

122 Wave Scattering by Crystals

where θis the angle between xand G. So we have

fα(G) = 2πV0Za

r2dr Z1

−1

dzeirGz = 2πV0Za

r2dr 2 sin(rG)

= 4πV0a3sin(x)−xcos(x)

where x=Ga. Now comparing this to 14.10, it looks identical except for

the prefactor. Note however, that the charge of atom is spread out over

a volume 4πa3/3. So setting V04πa3/3 = Zmakes the two equations

match.

of a hydrogen atom is

ψ=1

√4π

a3/2

e−r/a0

with a0the Bohr radius. The scattering potential is proportional to the

electron density |ψ|2. Let us call the constant of proportionality K. We

then have (using similar calculation as above)

f(G) = KZdx eiG·x|ψ(r)|2=K2πZ∞

r2dr 2 sin(rG)

rG |ψ(r)|2

So we want

f(G) = 2K

0Z∞

r2dr 2 sin(rG)

rG e−2r/a0

The integration is not too diﬃcult and gives the result

f(G) = 16K

(a2

0G2+ 4)2

(14.10) Error Analysis

Imagine you are trying to measure the lattice constant

aof some crystal using X-rays. Suppose a diﬀraction peak

is observed at a scattering angle of 2θ. However, suppose

that the value of θis measured only within some uncer-

tainty δθ. What is the fractional error δa/a in the result-

ing measurement of the lattice constant? How might this

error be reduced? Why could it not be reduced to zero?

See 14.3.e.

(e) From Bragg’s law d∼1/sin θso

∂d

∂θ = cot θ

So if there is an error in measurement of θone ends up with a fractional

error in dgiven by

δlog d= (cot θ)(δθ)

As a result, error is minimized if one can work as close as possible to

scattering angle of 90 degrees. Although at 90 degrees one could have

123

this expression be exactly zero. However, this does not mean the error

is actually zero. One has only made the error in measurement zero to

lowest order. One also has to worry about higher derivative term

∂2d

∂θ2= 1 6= 0

at 90 degrees.

Electrons in a Periodic

Potential 15

(15.1) ‡Nearly Free Electron Model

Consider an electron in a weak periodic potential in one

dimension V(x) = V(x+a). Write the periodic potential

V(x) = X

eiGxVG

where the sum is over the reciprocal lattice G= 2πn/a,

and V∗

G=V−Gassures that the potential V(x) is real.

(a) Explain why for knear to a Brillouin zone bound-

ary (such as knear π/a) the electron wavefunction should

be taken to be

ψ=Aeikx +Bei(k+G)x(15.1)

where Gis a reciprocal lattice vector such that |k|is close

to |k+G|.

(b) For an electron of mass mwith kexactly at a zone

boundary, use the above form of the wavefunction to show

that the eigenenergies at this wavevector are

E=~2k2

2m+V0± |VG|

where Gis chosen so |k|=|k+G|.

Give a qualitative explanation of why these two

states are separated in energy by 2|VG|.

Give a sketch (don’t do a full calculation) of the

energy as a function of kin both the extended and the

reduced zone schemes.

zone boundary. Give an expression for the energy E(k)

correct to order (δk)2where δk is the wavevector diﬀer-

ence from kto the zone boundary wavevector.

Calculate the eﬀective mass of an electron at this

wavevector.

(a) A periodic lattice can only scatter a wave by a reciprocal lattice

vector (Bragg diﬀraction). In the nearly free electron picture, the scat-

tering perturbation is weak, so that we can treat the scattered wave in

perturbation theory. In this case, there is an energy denominator which

suppresses mixing of k-vectors which have greatly diﬀerent unperturbed

energies. Thus, the only mixing that can occur is between two states

with similar energies that are separated by a reciprocal lattice vector.

Degenerate perturbation theory tells us that we should ﬁrst diagonalize

within the degenerate space spanned by only these two eigenstates.

(b)We have our (variational) trial wavefunction given by

|ψi=A|ki+B|k+Gi(15.2)

or equivalently

ψ= (Aeikx +Bei(k+G)x)/√L

To maintain normalization we can insist that |A|2+|B|2= 1. Taking k

and k+Gboth on a Brillouin zone boundary we have k=nπ/a and k+

126 Electrons in a Periodic Potential

G=−nπ/a, where here we have chosen the nth zone boundary, and we

must have G=−2nπ/a the reciprocal lattice vector. The Hamiltonian

Hin question is the usual Kinetic term plus V(x).

Approach 1: Diagonalize Hwithin the 2d degenerate space

hk|H|ki=~2(nπ/a)2/(2m) + V0

hk+G|H|k+Gi=~2(nπ/a)2/(2m) + V0

hk|H|k+Gi=V2nπ/a

hk+G|H|ki=V−2nπ/a

Diagonalizing this two by two matrix

~2(π/a)2/(2m) + V0V2nπ/a

V−2nπ/a ~2(π/a)2/(2m) + V0

gives eigenstates

E=~2(nπ/a)2

2m+V0± |V2nπ/a|

where we have used VG=V∗

−G. The eigenstates are correspondingly

|ψ±i=|ki+|k+Gi

which are (proportional to) the functions sin(2nπx/a) and cos(2nπx/a).

Interpretation: If we have considered only the V2nπ/a and V−2nπ/a

Fourier modes of the potential then we have V= 2V2nπ/a cos(2nπr/a).

Assuming V2nπ/a >0, then the higher energy state is the ψ= cos(2nπr/a)

which puts the maximum amplitude of the wavefunction exactly at the

maxima of the potential. Similarly, the lower energy wavefunction is the

sin(2nπr/a) which has the minimum amplitude of the wavefunction at

the maximum of the potential. In the case of V2nπ/a <0 the sin is the

higher energy wavefunction.

Approach 2: Variational.

If we simply calculate the expectation value of Hin the trial state

given by Eq. 15.2 we obtain

hψ|H|ψi=~2(nπ/a)2/(2m) + V0+A∗B V ∗

2nπ/a +B∗A V2nπ/a

Using the variational principle, the eigenstate is the trial wavefunction

which minimizes the total energy while preserving the normalization.

One way to do this is to write A= cos(θ) and B=eiχ sin(θ) which is

the most general form we can write while still preserving |A|2+|B|2= 1

(we can arbitrarily choose Ato be real, since that only introduces an

irrelevant overall phase). In terms of these parameters we have

hψ|H|ψi=~2(nπ/a)2/2m+V0+ 2Re[V2nπ/aeiχ sin(θ) cos(θ)]

for V2nπ/a >0 this is minimized for χ=πand θ=π/4 (or equiv-

alently χ= 0 and θ= 3π/4). It gives minimum energy states for

ψ= sin(2nπr/a) as above.

127

See ﬁgure 15.1 for a sketch of the bands in the extended zone scheme.

0π/a

π/a

−π/a

2π/a

−2π/a

Reduced

Zone

Scheme

Zone

Extended

1st Brillouin Zone

2nd Zone 2nd Zone

Fig. 15.1 Diagram of the dispersion in a nearly free electron model. Top: extended

zone scheme. Bottom: reduced zone scheme. Note that gaps open up at all zone

boundaries

need to consider knot on the zone boundary. Letting k=nπ/a +δk

and k+G=−nπ/a +δk we have

hk|H|ki=~2(δk +nπ/a)2/(2m) + V0

hk+G|H|k+Gi=~2(δk −nπ/a)2/(2m) + V0

hk|H|k+Gi=V2nπ/a

hk+G|H|ki=V−2nπ/a

which we now need to diagonalize. We obtain

E±=~2[(δk)2+ (nπ/a)2]

2m+V0±s~22(δk)nπ/a

2m2

+|V2nπ/a|2

expanding the square-root we obtain

E±=~2(nπ/a)2

2m+V0± |V2nπ/a|+~2(δk)2

2m1±~2(nπ/a)2

m|V2nπ/a|

which is a quadratic correction as we move away from the Brillouin

zone. Note that for this expansion to remain valid we must have the

bracketed term in the square root two equations up small compared to

the |V2nπ/a|2term.

The eﬀective mass is then obtained by setting

2m∗=1

2m1±~2(nπ/a)2

m|V2nπ/a|

128 Electrons in a Periodic Potential

or equivalently

m∗=

1±~2(nπ/a)2

m|V2nπ/a |

with the + being for the upper band.

(15.2) Periodic Functions

Consider a lattice of points {R}and a function ρ(x)

which has the periodicity of the lattice ρ(x) = ρ(x+R).

Show that ρcan be written as

ρ(x) = X

ρGeiG·x

where the sum is over points Gin the reciprocal lattice.

Generally we can always write ρ(x) in terms of its Fourier transform

ρ(x) = VZdk

(2π)3ρkeik·x

Now by periodicity we know that ρ(x) = ρ(x+R) for any lattice vector

R, so let us take a sum over all Nlattice vectors in the system

ρ(x) = 1

ρ(x+R) = V

RZdk

(2π)3ρkeik·(x+R)

NZdk

(2π)3ρkeik·xX

eik·R

The sum gives X

eik·R=(2π)3

δ3(k−G)

where the sum is over all Gwhich are reciprocal lattice vectors and vis

the volume of the unit cell. Letting the delta function act and cancelling

some factors, we then directly obtain

ρ(x) = X

eiG·xρG

(15.3) Tight Binding Bloch Wavefunctions

Analogous to the wavefunction introduced in Chapter

11, consider a tight-binding wave ansatz of the form

|ψi=X

eik·R|Ri

where the sum is over the points Rof a lattice, and |Ri

is the ground-state wavefunction of an electron bound to

a nucleus on site R. In real space this ansatz can be

expressed as

ψ(r) = X

eik·Rϕ(r−R).

Show that this wavefunction is of the form required by

Bloch’s theorem (i.e., show it is a modiﬁed plane wave).

129

Start by writing the function ϕin its fourier representation

ϕ(r) = VZdq

(2π)3eiq·rϕq

so that

ψ(r) = X

eik·RVZdq

(2π)3eiq·(r−R)ϕq

The sum over Rgives

ei(k−q)·R=(2π)3

δ3(k−q−G)

and we allow the delta function to act, giving

ψ(r) = eik·rV

e−iG·rϕk−G

The sum over Gis a function periodic in r→r+R(since eiG·R= 1)

hence this is in Bloch form.

(15.4) *Nearly Free Electrons in Two Dimen-

sions Consider the nearly free electron model for a square

lattice with lattice constant a. Suppose the periodic po-

tential is given by

V(x, y) = 2V10[cos(2πx/a) + cos(2πy/a)]

+ 4V11[cos(2πx/a) cos(2πy/a)]

(a) Use the nearly free electron model to ﬁnd the energies

of states at wavevector G= (π/a, 0).

(b) Calculate the energies of the states at wavevec-

tor G= (π/a, π/a). (Hint: You should write down a 4

by 4 secular determinant, which looks diﬃcult, but actu-

ally factors nicely. Make use of adding together rows or

columns of the determinant before trying to evaluate it!)

Note, we should not call the points (π/a, 0) and (π/a, π/a) as Gsince

they are not reciprocal lattice vectors!

(a) The ﬁrst part of this problem is no diﬀerent from the one di-

mensional problem posed in 15.1! The fourier components V10 couple

(π/a, 0) and (−π/a, 0). The energies of the states are

E=~2(π/a)2/(2m)± |V10|

(b) The second part of the problem is more complicated. Here, all four

points (±π/a, ±π/a) all have the same energy and are coupled together

by the scattering potential. Therefore we must treat all of these states

in degenerate perturbation theory. Let us label the points

|1i= (+π/a, +π/a)

|2i= (+π/a, −π/a)

|3i= (−π/a, −π/a)

|4i= (−π/a, +π/a)

130 Electrons in a Periodic Potential

Writing a general wavefunction within this space as Piφi|iithe Hamil-

tonian matrix within this reduced Hilbert space is







ǫ V10 V11 V10

V10 ǫ V10 V11

V11 V10 ǫ V10

V10 V11 V10 ǫ







where ǫ=~2(π/a)2/m. To ﬁnd the eigenenergies we would thus like to

solve the determinant equation

0 = 

ǫ−E V10 V11 V10

V10 ǫ−E V10 V11

V11 V10 ǫ−E V10

V10 V11 V10 ǫ−E



Adding and subtracting rows and columns leaves the determinant un-

changed. So we can subtract row 3 from row 1 and subtract row 4 from

row 2. Then add column 3 to column 1 and add column 4 to column 2.

The result is

0 = 

0 0 V11 −ǫ+E0

0 0 0 V11 −ǫ+E

V11 +ǫ−E2V10 ǫ−E V10

2V10 V11 +ǫ−E V10 ǫ−E



Because of all the zeros, method of minors can then evaluate the deter-

minant easily to given

0 = (V11 −ǫ+E)2(V11 +ǫ−E)2−4V2

10

Thus giving solutions

E=ǫ+V11 (two solutions)

E=ǫ−V11 ±2|V10|

(15.5) Decaying Waves

As we saw in this chapter, in one dimension, a peri-

odic potential opens a band gap such that there are no

plane-wave eigenstates between energies ǫ0(G/2) − |VG|

and ǫ0(G/2) + |VG|with Ga reciprocal lattice vector.

However, at these forbidden energies, decaying (evanes-

cent) waves still exist. Assume the form

ψ(x) = eikx−κx

with 0 < κ ≪kand κreal. Find κas a function of energy

for k=G/2. For what range of VGand Eis your result

valid?

Here, we can think of the wavevector kas taking an imaginary part

(i.e., absorb κinto k)

k=−G/2 + iκ

131

so that

ǫ0(k) = ~2k2

2m=~2((G/2)2−iGκ −κ2)

2m=ǫR−iǫI

ǫ0(k+G) = ~2(k+G)2

2m=~2((G/2)2+iGκ −κ2)

2m=ǫR−iǫI

where we have deﬁned

ǫR=~2((G/2)2−κ2)

ǫI=~2(Gκ)

As worked out in the text (Eq. 15.8), the characteristic equation is

0 = (ǫ0(k)−E) (ǫ0(k+G)−E)− |VG|2

= (ǫR−iǫI−E)(ǫR+iǫI−E)− |VG|2

= (ǫR−E)2+ǫ2

I− |VG|2(15.3)

Thus, we have

E=ǫR±q|VG|2−ǫ2

Note here that for |ǫI|>|VG|the energy becomes imaginary and the

solution is not valid. Thus we must have

|VG| ≥ |~2Gκ/(2m)|

In Eq. 15.3 we have a quadratic equation for κ2(there are no lone factors

of κ), which can be solved to then give only one (possibly) positive

solution

~2κ2

2m=−~2(G/2)2

2m−E+r4E~2(G/2)2

2m+|VG|2

In order for this solution to be valid, we must have the right hand side

be positive. We can write this condition as

~2(G/2)2

2m+E2

≤4E~2(G/2)2

2m+|VG|2

Or equivalently 

~2(G/2)2

2m−E≤ |VG|

or in other words, that the energy is inside the gap!

(15.6) Kronig–Penney Model*

Consider electrons of mass min a so-called “delta-

function comb” potential in one dimension

V(x) = aU X

δ(x−na)

132 Electrons in a Periodic Potential

(a) Argue using the Schroedinger equation that in-

between delta functions, an eigenstate of energy Eis al-

ways of a plane wave form eiqExwith

qE=√2mE/~.

Using Bloch’s theorem conclude that we can write an

eigenstate with energy Eas

ψ(x) = eikx uE(x)

where uE(x) is a periodic function deﬁned as

uE(x) = Asin(qEx) + Bcos(qEx) 0 < x < a

and uE(x) = uE(x+a) deﬁnes uoutside of this interval.

(b) Using continuity of the wavefunction at x= 0 de-

rive

B=e−ika[Asin(qEa) + Bcos(qEa)],

and using the Schroedinger equation to ﬁx the disconti-

nuity in slope at x= 0 derive

qEA−eikak[Acos(qEa)−Bsin(qEa)] = 2maUB/~2

Solve these two equations to obtain

cos(ka) = cos(qEa) + mUa

~2qE

sin(qEa)

The left-hand side of this equation is always between −1

and 1, but the right-hand side is not. Conclude that there

must be values of Efor which there are no solutions of the

Schroedinger equation—hence concluding there are gaps

in the spectrum.

result agrees with the predictions of the nearly free elec-

tron model (i.e., determine the size of the gap at the zone

boundary).

Oops another error in this problem. Actually this one is really bad.

The ﬁrst part of the problem should ask you to show that

ψ(x+a) = eikaψ(x)

and it should be ψthat has the form Asin(qx) + Bcos(qx) not u. Then

in the second to last equation there is an extra random factor of kin

front of the brackets, it should be qEnot k. AND the exponent is e−ika.

. Probably too many martinis.

(a) Bloch’s theorem tells us

ψ(x) = eikxu(x)

with uperiodic in the unit cell. This then implies ψ(x+a) = eikaψ(x)

as required.

Between delta functions, there is no potential so the solution must be

plane waves e±iqEx. Since the Hamiltonian is time reversal invariant we

can choose the wavefunction to be real. As a result, we must have ψof

the form Asin(qEx) + Bcos(qEx).

(b) Using ψ(a+) = eikaψ(0) = Beika and ψ(a−) = Asin(qEa) +

Bcos(qEa) gives the required result immediately. The Schroedinger

equation evaluated near x=a

2m∂2

xψ(x) = aUδ(x−a)ψ(x)

is equivalent to

2m∂xψ(a+)−∂xψ(a−)=aUψ(a)

133

Using

ψ(a) = Beika

∂xψ(a+) = eikaqEA

∂xψ(a−) = qE[Acos(qEa)−Bsin(qEa)]

gives the desired result which now reads (after removing errors!)

A−e−ika[Acos(qEa)−Bsin(qEa)] = 2maU B/(qE~2)

The two equations (with errors removed) have the required solution.

VG=1

LZdxV (x)eiGx =1

Ua =U

Note this is true also at V0=U, so we expect an overall energy shift

of U(from V0) and then a gap of magnitude 2|U|opening at the zone

boundary as well.

Setting k=π/a and qE=k+δwe have

−1 = cos(±π+δa) + mU a

~2(π/a +δ)sin(π+δ)

Expanding for small δand solving to lowest order we get two solutions

δ= 0 or 2mUa

~2π

Substituting back into E=~2(π/a +δ)2/(2m) we obtain

E=E0or E0+ 2U

giving the gap of 2|U|at the zone boundary as expected (and the mid-

point of the gap shifted up by Udue to V0as well).

Insulator, Semiconductor,

or Metal 16

(16.1) Metals and Insulators

Explain the following:

(a) sodium, which has two atoms in a bcc (conventional

cubic) unit cell, is a metal;

(b) calcium, which has four atoms in a fcc (conventional

cubic) unit cell, is a metal;

tional cubic) unit cell with a basis, is an electrical insula-

tor, whereas silicon and germanium, which have similar

structures, are semiconductors. (Try to think up several

possible reasons!)

Why is diamond transparent?

(a) The conventional bcc unit cell contain 2 atoms, but the primitive

unit cell contains only a single atom. Thus sodium, as a monovalent

atom, results in a half-ﬁlled 1st BZ – and thus gives a metal.

(b) The conventional fcc unit cell contains 4 atoms, but the primitive

unit cell contains only one atom. So calcium, which is divalent, could

be either a metal or an insulator — depending on the strength of the

periodic potential. The fact that it is a metal tells us that the potential

is not strong enough to make it an insulator.

The fcc unit cell has 4 atoms, but the primitive unit cell has 1 atom,

and therefore 4 electrons. This could therefore constitute 2 completely

ﬁlled bands. In the case of C (diamond) this is indeed an insulator with

a large band gap (5.5 eV). For Si and Ge, the band gap is smaller (1.1

eV and .67 eV respectively), hence they are semiconductors.

There are various explanations for this eﬀect. One argument is that

for carbon, the periodic potential is extremely strong (no inner shell

electrons to screen it, and atoms much closer together) therefore it is

an insulator with a large band gap. For Si and Ge, the potential is less

strong, so the gap is smaller. Hence they are semiconductors.

Another argument is given by thinking in the tight binding picture.

Before you bring the atoms together to make a solid, one needs to con-

sider single-electron orbitals. The higher orbitals in an atom are typically

closer together in energy (as they are for hydrogen). For C the valence

electrons are in the 2p shell whereas for Si and Ge, the valence electrons

are 3p and 4p respectively. Hence we expect smaller gaps for Si and Ge.

In a more chemical language, 4 electrons per atom can therefore form

4 covalent bonds with each of the 4 neighbors in the diamond lattice

136 Insulator, Semiconductor, or Metal

structure. The electrons are all therefore tied up in covalent bonds,

making the material an insulator.

Diamond is transparent because it is an insulator with band gap that

is larger than any visible light frequency. Thus light cannot be absorbed

(or reﬂected) by any electronic transition.

(16.2) Fermi Surface Shapes

(a) Consider a tight binding model of atoms on a (two-

dimensional) square lattice where each atom has a single

atomic orbital. If these atoms are monovalent, describe

the shape of the Fermi surface.

(b) Now suppose the lattice is not square, but is rect-

angular instead with primitive lattice vectors of length

axand ayin the xand ydirections respectively, where

ax> ay. In this case, imagine that the hoppings have

a value −txin the x-direction and a value −tyin the y-

direction, with ty> tx. (Why does this inequality match

ax> ay?)

Write an expression for the dispersion of the elec-

tronic states ǫ(k).

Suppose again that the atoms are monovalent, what

is the shape of the Fermi surface now?

(a) The dispersion of the tight binding model is given by

ǫ(kx, ky) = −2t(cos(kxa) + cos(kya))

A contour plot of this energy is given in the left of Fig. 16.1. If we are

considering a monovalent unit cell, then the Brillouin zone is half ﬁlled.

This then gives a Fermi surface in the shape of a diamond, as shown in

the right of Fig. 16.1.

-3 -2 -1 012 3

-3

-2

-1

-3 -2 -1 012 3

-3

-2

-1

Fig. 16.1 Left: Dispersion of a 2D tight binding model on a square lattice. Right:

the Fermi surface for one electron per unit cell.

(b) If ax> ayone expects the hopping magnitude to be smaller in

the xdirection since the atoms are further apart (although this is not

holy, as the orbitals, such as pxorbitals, may not be isotropic). We then

expect a dispersion of the form

ǫ(kx, ky) = −2txcos(kxax)−2tycos(kyay)

137

As an example, let us choose ty= 2txbut ax=ay=afor simplicity.

A contour plot of the energy is given in the left of Fig. 16.2. If we are

considering a monovalent unit cell, then the Brillouin zone is half ﬁlled.

This then gives a Fermi surface in the shape of ... well, i’m not sure

what to call it. But it is shown in the right of Fig. 16.2.

-3 -2 -1 012 3

-3

-2

-1

-3 -2 -1 012 3

-3

-2

-1

Fig. 16.2 Left: Dispersion of a 2D tight binding model on a square lattice with

anisotropic hoppings. Right: the corresponding Fermi surface for one electron per

unit cell.

(16.3) More Fermi Surface Shapes*

Consider a divalent atom, such as Ca or Sr, that forms

an fcc lattice (with a single atom basis). In the absence

of a periodic potential, would the Fermi surface touch the

Brillouin zone boundary? What fraction of the states in

the ﬁrst Brillouin zone remain empty?

This is some nasty geometry. First, recall that the reciprocal lattice

of an FCC lattice with lattice constant ais a BCC lattice with lattice

constant 4π/a (See excercise 13.1). The BCC lattice has two lattice

points per conventional unit cell, so the primitive unit cell (or the Bril-

louin zone in this case) has volume 1

2(4π/a)3in k-space. Now since we

have a divalent unit cell in real space, we should have enough electrons

to ﬁll exactly the volume of the Brillouin zone. Thus for electrons in the

absence of a periodic potential we have

4πk3

3=1

24π

a3

≈992/a3

kF= (24π2)1/3/a ≈6.187/a

We would like to know if this hits the Brillouin zone boundary or not.

For a BCC lattice, the nearest neighbor of the point [0,0,0] is the point

138 Insulator, Semiconductor, or Metal

[1/2,1/2,1/2] in units of the lattice constant. The perpendicular bisector

to this point is thus a distance √3/4. For our reciprocal lattice with

lattice constant 4π/a, the distance to this perpendicular bisector (i.e.,

to the Brillouin zone boundary) is then

dk =π√3/a ≈5.441/a

which is less than the Fermi wavevector, thus telling us that the Fermi

surface hits the Brillouin zone boundary (it is obvious that it must hit

the Brillouin zone boundary since the volume of the fermi surface must

equal the volume of the Brillouin zone and they are not the same shape!).

The Fermi surface thus goes into the 2nd Brillouin zone as a spherical

cap of radius 6.187/a where the Brillouin zone boundary is of radius

5.441/a. Note that the center of this spherical cap is in the Ldirection

(in the language of Brillouin zones, see ﬁg 13.6 of the book). We should

check that the spherical cap remains on the L-face. We can check this

by noting that the angle subtended by this spherical cap is only θ=

cos−1(5.441/6.187) ≈28.425 degrees, which is much smaller than the

angle to say the Kpoint.

The height of the spherical cap is h= 6.187/a −5.441/a = 0.745/a.

A well known geometric formula gives us that the volume of a cap is

V=πh2

3(3r−h) = 10.4/a3

However, note that we have 8 such spherical caps in all of the 8 equivalent

Ldirections, thus giving a total volume of

V= 83/a3

in the 2nd Brillouin zone, compared to the total volume of the fermi

surface which is 992/a3. Thus the caps in the second Brillouin zone

account for roughly 8% of the ﬁlled states.

Semiconductor Physics 17

(17.1) Holes

(a) In semiconductor physics, what is meant by a hole

and why is it useful?

(b) An electron near the top of the valence band in a

semiconductor has energy

E=−10−37|k|2

where Eis in Joules and kis in m−1. An electron is re-

moved from a state k= 2 ×108m−1ˆx, where ˆxis the unit

vector in the x-direction. For a hole, calculate (and give

the sign of!)

(i) the eﬀective mass

(ii) the energy

(iii) the momentum

(iv) the velocity.

If there is a density p= 105m−3of such holes all

having almost exactly this same momentum, calculate the

current density and its sign.

(a) A hole is the absence of an electron in an otherwise ﬁlled valence

band. This is useful since instead of describing the dynamics of all the

(many) electrons in the band, it is equivalent to describe the dynamics

of just the (few) holes.

(b) Eﬀective mass ~2k2/(2m∗) = (10−37Joule ·meter2)k2. So m∗=

5×10−32 kg or .05 the mass of the electron. This mass is positive in the

usual convention. The energy is E= (10−37Joule·meter2))k2= 4×10−21

J, or about 0.025 eV. This energy is positive (it takes energy to ”push”

the hole down into the fermi sea, like pushing a balloon under water).

Getting the momentum and velocity right are tricky. First, note that

the velocity of an eigenstate is the same whether or not the state is ﬁlled

with an electron. It is always true that the velocity of an electron in a

state is ∇kEk/~where Ekis the electron energy. Thus the hole velocity

here is negative v=−~k/m∗=−3.8×105m/s (i.e the velocity is in

the negative ˆx) direction.

For momentum, since a ﬁlled band carries no (crystal) momentum,

and for electrons crystal momentum is always ~k, the removal of an

electron leaves the band with net momentum −~kwhich we assign as

the momentum of the hole. Thus we obtain hole momentum −~k=

−2.1×10−26 kg-m/s which is also in the negative ˆxdirection. (this

matches well to the intuition that p=mv with a positive eﬀective mass

for holes). With pthe density of such holes, the total current density is

pev =−6×10−9Amp/m2also in the negative ˆxdirection (noting that

the charge of the hole is positive).

Note that it is typical to deﬁne the wavevector of a hole to be negative

140 Semiconductor Physics

of the wavevector of the missing electron.

(17.2) Law of Mass Action and Doping of Semi-

conductors

(a) Assume that the band-gap energy Egis much

greater than the temperature kBT. Show that in a pure

semiconductor at a ﬁxed T, the product of the number of

electrons (n) and the number of holes (p) depends only on

the density of states in the conduction band and the den-

sity of states in the valence band (through their eﬀective

masses), and on the band-gap energy.

Derive expressions for nfor pand for the product

np. You may need to use the integral R∞

0dx x1/2e−x=

√π/2.

(b) The band gaps of silicon and germanium are 1.1

eV and 0.75 eV respectively. You may assume the ef-

fective masses for silicon and germanium are isotropic,

roughly the same, and are roughly .5 of the bare electron

mass for both electrons and holes. (Actually the eﬀective

masses are not quite the same, and furthermore the eﬀec-

tive masses are both rather anisotropic, but we are just

making a rough estimates here.)

Estimate the conduction electron concentration for

intrinsic (undoped) silicon at room temperature.

Make a rough estimate of the maximum concen-

tration of ionized impurities that will still allow for this

“intrinsic” behavior.

Estimate the conduction electron concentration for

germanium at room temperature.

tween charge-carrier concentration for a certain n-doped

semiconductor.

Estimate the band gap for the semiconductor and

the concentration of donor ions.

Describe in detail an experimental method by

which these data could have been measured, and suggest

possible sources of experimental error.

Fig. 17.1 Figure for Exercise 17.

(a) The density of states per unit volume of free electron with disper-

sion E=~2k2/(2m) is given by (including spin)

g(E) = √2Em3/2

~3π2

So if the dispersion near the valence band edge and conduction band

edges are

Ee(k) = Ec+~2k2/(2me)

Eh=Ev−~2k2/(2mh)

we obtain density of states for conduction electrons and valence holes

given by

ge(E > Ec) = p2(E−Ec)m3/2

~3π2

gh(E < Ev) = p2(Ev−E)m3/2

~3π2

141

At ﬁxed chemical potential µand temperature βthe number density of

electrons in the conduction band is

n=Z∞

ge(E)nF(β(E−µ))dE

where nF(x) = 1/(ex+ 1) is the Fermi occupation factor. Assuming

that µis well below the conduction band (by at least energy kbT), then

x=β(E−µ) is very positive and it is acceptable to replace nF(x) by

the Boltzmann factor e−x, thus we obtain.

n=Z∞

ge(E)e−β(E−µ)dE

Similarly, the number of holes in valence band is given by

p=ZEv

−∞

gh(E)(1 −nF(β(E−µ))dE

Assuming that µis well above the valence band (by at least energy kbT)

then x=β(E−µ) is very negative and we can replace the fermi factor

1−nF(x) by exresulting in

p=ZEv

−∞

gh(E)eβ(E−µ)dE

It is then clear immediately, that when we multiply np the variable µ

completely vanishes.

A bit more manipulation obtains

n=eβµ√2m3/2

~3π2Z∞

Ecp(E−Ec)e−βEdE

redeﬁning variables y=E−Ecand performing the integral, one obtains

n=1

42mekbT

π~23/2

e−β(Ec−µ)(17.1)

and similarly

p=1

42mhkbT

π~23/2

e−β(µ−Ev)(17.2)

Obtaining

np =1

16 2√mhmekbT

π~23

e−β(Ec−Ev)

which depends only on the band gap, Tand the eﬀective masses.

(b) In the undoped case n=p, and we are assuming mh=me=m/2

as well so we have

n=1

42(m/2)kbT

π~23/2

e−β(Eg/2)

142 Semiconductor Physics

with Egthe gap value. Plugging in numbers gives

n= 5.26 ×1015m−3

for Silicon and

n= 4.54 ×1018m−3

for Germanium. When the doping level gets to on the order of the

expected intrinsic level, then you no longer have intrinsic behavior.

tration at low T(which I estimate from the ﬁgure to be about 2 ×1019

m−3). Note that at very low temperature one could get carrier freeze-out

where the density drops again.

To estimate the gap, we need to measure the slope of the curve at

high temperature.

Extracting a slope

log n/(m3) = −1500K/T +Constant.

Or equivalently

n∼e−1500K/T

We then set 1500K=Eg/(2kb) and obtain a band gap of about .26 eV.

It might be useful to also mention to the students that Si and Ge

both have valley degeneracies (i.e, multiple equi-energy minima in the

conduction band). These may add an additional factor to the law of

mass action.

The concentration is most likely measured by a Hall eﬀect measure-

ment. Several possible sources of error can occur here. First, when there

are both electrons and holes present, then you measure some (nontriv-

ial) combination of the Hall resistivities weighted by their concentrations

and by their mobilities (in a very nontrivial way).

RH=Reρ2

e+Rhρ2

(ρe+ρh)2

(See exercise 17.9). One only gets an accurate absolute measurement of

the electron concentration to the extent that the electron resistivity is

much lower than the hole resistivity.

There are other more obvious sources of experimental error such as

heating when one runs current through a sample to measure it – thus it

requires measuring small voltages accurately. To measure hall resistivity,

without mixing in longitudinal resisitivity, one needs to align contacts

exactly parallel to each other in a hall bar. One has to also make sure

that the voltages being measured are due to the sample and not the

contacts/wires/ampliﬁers etc.

143

(17.3) Chemical Potential

(a) Show that the chemical potential in an intrinsic

semiconductor lies in the middle of the gap at low tem-

perature.

(b) Explain how the chemical potential varies with tem-

perature if the semiconductor is doped with (i) donors (ii)

acceptors.

density of 1023 electrons/m3. Calculate the hole density

at room temperature given that the gap is 1.0 eV, and

the eﬀective mass of carriers in the conduction and va-

lence band are 0.25 and 0.4 electron masses respectively.

Hint: use the result of Exercise 17.2.a.

(a) In an intrinsic semiconductor n=pso we can set

p= 1

Referrring back to the previous problem we can insert the expressions

Eq. 17.1 and Eq. 17.2 for nand prespectively. Almost all of the nasty

prefactors cancel and we obtains

m3/2

ee−β(Ec−µ)

m3/2

he−β(µ−Ev)= 1

We can solve this trivially to obtain

µ=Ec+Ev

2+3

4(kBT) log(mh/me)

So at low temperature the chemical potential lies mid-gap. (Incidentally,

this is why it is never a good idea to say that the fermi energy is the

energy of the highest ﬁlled state. There may be a very large diﬀerence

between the highest ﬁlled state and the chemical potential!).

(b) Let us assume for a moment we are well above the freezeout tem-

perature, so the doping can be thought of as going directly into the

conduction band. For simplicity let us assume me=mh, so the intrinsic

behavior is then simply that the chemical potential is ﬁxed as a func-

tion of temperature. If the doner density is much higher than nint then

n≈ndopant (i.e., the thermally excited electrons are irrelevant compared

to those that are there from doping). Then looking at Eq. 17.1 ﬁxing n

to be ndopant and solving for µwe have

µ≈Ec−kBTlog 

ndopant

42mekbT

π~23/2



Similarly for acceptor impurities

µ≈Ev+kBTlog 

pacceptor

42mhkbT

π~23/2



So at low tempearture, the chemical potential is essentially right at the

conduction (donor) or valence (acceptor) band and moves towards mid-

gap as the temperature is increased. When the intrinsic density exceeds

144 Semiconductor Physics

the dopant density, then one expects to have µgiven by the intrinsic

expression from part a, which is roughly to have the chemical potential

mid-gap with a small slope dependent on the ratio of masses.

Now the story is a bit more complicated if one wants to think about

the very low temperature regime where there is carrier freezeout. At

zero temperature all of the electrons are bound to their dopant nuclei

and one can think of this as being a ﬁlled “impurity band” playing the

role of a ﬁlled valence band, and the nearby conduction band is empty.

As is usually the case, at zero temperature the chemical potential is

midway between the top of the impurity band and the bottom of the

conduction band. As the temperature is increased, since the real valence

band is very far away compared to the temperature so we can ignore it.

It is then only a matter of ﬁguring out how the chemical potential moves

between the ﬁlled impurity band and the empty conduction band. Since

the density of states in the conduction band is larger than the those

in the sparse impurity band, as the temperature is raised, the chemical

potential moves up towards the conduction band.

the law of mass action with n=p=nintrinsic. At T=293 Kelvin, I

obtained

nintrinsic = 1016m−3

Then since ndopant ≫nintrinsic we can set p=n2

intrinsic/n (from the

law of mass action) to obtain

p= 109m−3

(17.4) Energy Density

Show that the energy density of electrons in the valence

band of a semiconductor is

(ǫc+3

2kBT)n

where nis the density of these electrons and ǫcis the

energy of the bottom of the conduction band.

In short this is just a matter of realizing that the electrons in the

conduction band are essentially classical (are activated with Boltzmann

factors not fermi factors), so classical stat mech applies and one can ap-

ply the equipartition theorem. One obtains ǫcfor each particle excited,

then an extra 3

2kBTfor the translational degrees of freedom as usual in

equipartition theorem.

One can, of course, do the calculation more rigorously writing the

total energy density as (compare problem 17.2)

E/V =Z∞

Ege(E)nF(β(E−µ))dE

where nF(x) = 1/(ex+ 1) is the Fermi occupation factor. As in 17.2 it

145

is acceptable to replace nF(x) by the Boltzmann factor to get

E/V =Z∞

Ege(E)e−β(E−µ)dE

The same manipulation obtains

E/V =eβ(µ−Ec)√2m3/2

~3π2Z∞

[(E−Ec) + Ec]p(E−Ec)e−β(E−Ec)dE

Note the second term in brackets, the Ecterm (compare 17.2) gives the

same integral as the above calcualtion of nin 17.2 so we obtain nEc.

To evaluate the remaining term we redeﬁne variables y=E−Ecand

performing the integral, whixh we recognize as being precisely ∂/∂β of

the prior integral from 17.2. Since the prior integral was proportional to

T(3/2) we obtain (3/2)(kBT) times the prior integral and thus a total

of (3/2)(kBT)nproving the result.

(17.5) Semiconductors

Describe experiments to determine the following prop-

erties of a semiconductor sample: (i) sign of the majority

carrier (ii) carrier concentration (assume that one carrier

type is dominant) (iii) band gap (iv) eﬀective mass (v)

mobility of the majority carrier.

(i,ii) Sign of majority carrier and carrier concentration (assuming there

is only one type of carrier) are both easily measured with Hall eﬀect. (iii)

band gap may be measured optically. Or by carrier concentration (es-

sentially conductance) as a function of temperature. (iv) Eﬀective mass

is measured with cyclotron resonance (v) mobility is easily measured via

resistivity once concentration of carriers is known.

(17.6) More Semiconductors

Outline the absorption properties of a semiconductor

and how these are related to the band gap. Explain the

signiﬁcance of the distinction between a direct and an

indirect semiconductor. What region of the optical spec-

trum would be interesting to study for a typical semicon-

ducting crystal?

Optical absorbtion can occur when a photon is absorbed while exciting

an electron out of the valence band into the conduction band. This

requires a minimum of the gap energy (Small amounts of absorbtion can

occur below the gap for impure semiconductors if there are impurity or

defect states within the gap – also very weak nonlinear processes can

allow multiple photons to be absorbed while exciting a single electron).

In the absorbtion process energy and momentum must both be con-

served. Since photons carry very little momentum given a certain energy

146 Semiconductor Physics

(being that cis very large) one should think of this absorbtion as not

tranferring any momentum to the system. This means that direct gap

absorbtion (where the momentum of the electron does not change) is

highly favored over indirect gap absorbtion. Indirect gap absorbtion

can occur, but it must be assisted by a phonon or some other process

that can account for the necessary momentum.

Semiconductor gaps tend to be in the optical, or infra-red range (some-

where from 400 nm to 3 micron, or roughly 3 eV to .5 eV). Only a very

few wide gap semiconductors reach the optical blue range and UV.

(17.7) Yet More Semiconductors

Outline a model with which you could estimate the en-

ergy of electron states introduced by donor atoms into an

n-type semiconductor. Write down an expression for this

energy, explaining why the energy levels are very close to

the conduction band edge.

One can consider a simple hydrogenic schroedinger equation with an

attractive proton being the ionized donor and the single electron. The

main diﬀerences are that the mass of the electron is replaced by the

band electron mass, and ǫ0is multiplied by the dielectric constant ǫrof

the semiconductor. As a result, the Rydberg becomes replaced by an

eﬀective Rydberg

R∗=R0(m∗/m)(1/ǫ2

This gives us a hydrogenic binding energy that can be extremely small

for typical semiconductors, hence the bound states remain very close the

conduction band edge.

(17.8) Maximum Conductivity*

Suppose holes in a particular semiconductor have mo-

bility µhand electrons in this semiconductor have mobil-

ity µe. The total conductivity of the semiconductor will

σ=e(n µe+p µh)

with nand pthe densities of electrons in the conduction

band and holes in the valence band. Show that, indepen-

dent of doping, the maximum conductivity that can be

achieved is

σ= 2e nintrinic √µeµh

with nintrinsic the intrinsic carrier density. For what

value of n−pis this conductivity achieved?

Actually this is easy. Using law of mass action np =n2

intrinsic. Thus

we write

σ=en µe+nintrinsic

nµh

Now set dσ/dn = 0 to maximize and solve to obtain

n=nintrinsicpµh/µe

Which correspondingly results in

p=nintrinsicpµe/µh

147

plugging into the original expression for σwith a tiny bit of algebra we

obtain

σ= 2e nintrinic √µeµh

as required, and we also obtain

n−p=nintrinsic pµh/µe−pµe/µh

(17.9) Hall Eﬀect with Both n- and p-Dopants*

Suppose a semiconductor has a density pof holes in

the valence band with mobility µhand a density nof

electrons in the conduction band with mobility µn. Use

Drude theory to calculate the Hall resistivity of this sam-

ple.

See also exercise 3.3c. For a single species, we have (See exercise 3.1)

ρ=r BR

−BR r 

where r=nµ and R=q/n with qthe charge on the charge carrier and

nthe carrier density. We deﬁne tensors ρeand ρhfor the two species

with re=nµnand rh=pµhand Re=e/n and Rh=−e/p. The

conductivity tensors are σj=ρ−1

jand then the total conductivity tensor

is σ=σe+σi. Finally this is inverted to give the tensor ρtotal =σ−1.

There is a lot of algebra involved in this. I obtained

ρxx =B2(reR2

h+rhR2

e) + rhre(re+rh)

B2(Re+Rh)2+ (re+rh)2

ρxy =BB2ReRh(Re+Rh) + Rhr2

e+Rer2

h

B2(Re+Rh)2+ (re+rh)2

Semiconductor Devices 18

(18.1) Semiconductor Quantum Well

(a) A quantum well is formed from a layer of GaAs

of thickness Lnm, surrounded by layers of Ga1−xAlxAs

(see Fig. 18.2). You may assume that the band gap of the

Ga1−xAlxAs is substantially larger than that of GaAs.

The electron eﬀective mass in GaAs is 0.068 mewhereas

the hole eﬀective mass is 0.45 mewith methe mass of

the electron.

Sketch the shape of the potential for the electrons

and holes.

What approximate value of Lis required if the band

gap of the quantum well is to be 0.1 eV larger than that

of GaAs bulk material?

(b) *What might this structure be useful for?

(a) This is a particle in a box problem. Both electron and holes are

particles in a box of length L. Thus the lowest lying electron in the well

Ee=Ec+π

L2~2

2m∗

with Ecthe bulk conduction band minimum. Similarly the highest lying

hole state in the well is

Eh=Ev−π

L2~2

2m∗

Thus the diﬀerence in energy is

Ee−Eh=Ebulkgap +~2

2π

L21

m∗

e−1

m∗

h

Setting Ee−Eh−Ebulkgap to 1eV and solving for Lgives 8 nm.

(b) This type of quantum well device is useful to precisely design a

band gap for example for a laser where one wants to ﬁx the emission

wavelength. If one puts donor impurities outside of the well (on both

sides, say) the donated electrons can reduce their energies by falling

into the well, but the ionized dopants remain behind. This is known as

modulation doping. It is useful because one can obtain extremely high

mobility electrons within the quantum well since there are no ionized

dopants in the well to scatter oﬀ of. One uses these structures heavily

for fundamental physics studies of clean (unperturbed) electrons.

150 Semiconductor Devices

(18.2) Density of States for Quantum Wells

(a) Consider a quantum well as described in the previ-

ous exercise. Calculate the density of states for electrons

and holes in the quantum well. Hint: It is a 2D electron

gas, but don’t forget that there are several particle-in-a-

box states.

(b) Consider a so-called “quantum wire” which is a

one-dimensional wire of GaAs embedded in surrounding

AlGaAs. (You can consider the wire cross-section to be a

square with side 30nm.) Describe the density of states for

electrons or holes within the quantum wire. Why might

this quantum wire make a very good laser?

(a) For a 2D electron gas for electrons with mass m, we quickly cal-

cualte the density of states.

N= 2AZkF

(2π)2

with Athe area and the factor of 2 out front for spin and ka two

dimensional vector. This can be converted to

n=N/A =k2

2π

When the energy of an electron is given by

E=~2k2

we have k2= 2mE/~2and we then have a density of states per unit

volume of

g=dn

dE =m

π~2

This is the correct answer for any E≥0 and the density of states is zero

for any E < 0. We then write more precisely that

g(E) = m

π~2Θ(E)

where Θ is the step function which has value 1 for nonnegative argument

and value 0 for negative argument.

In our quantum well we must make a few minor changes. First of

all, we should use the eﬀective mass rather than the actual mass of

the electron. Secondly the energy of the particle in the quantum well

also includes its particle-in-a-box energy for its motion transverse to the

2D quantum well. Thus for an electron in the conduction band of the

quantum well we have

E=Ec+~2π2

2m∗

L2+~2k2

2m∗

where Ecis the bulk conduction band bottom, and a= 1,2,3,... is

the transverse quantum number of the particle in the well. Fixing the

transverse quantum number a, the density of states would be

g(E) = m∗

π~2ΘE−Ec−~2π2

2m∗

L2

151

Now accounting for the fact that there may be many transverse modes

we have

g(E) = m∗

π~2X

a>0

ΘE−Ec−~2π2

2m∗

L2

and this expression remains true up to an energy where the transverse

modes spill out of the box.

Analogously the density of holes in the valuence band in the quantum

well is

g(E) = m∗

π~2X

a>0

ΘEv−~2π2

2m∗

L2−E

(b) First we determine the density of states for a one dimensional

electron gas.

N= 2LZkF

−kF

2π

with Lthe Length of the system and the factor of 2 out front for spin.

This can be converted to

n=N/L =2kF

using k=√2mE/~we then obtain a density of states

g(E) = dn

dE =√2m

π~E−1/2Θ(E)

Now in the seminconductor quantum wire we must consider the trans-

verse modes. In general the energy of an electron in the wire is then given

E=Ec+~2π2

2m∗

1+a2

L2+~2k2

2m∗

where a1and a2are the mode indices (integers greater than zero) in the

two transverse directions. Adding the density of states associated with

all of these modes we obtain

g(E) = √2m∗

π~X

a1,a2>0E−Ec−~2π2

2m∗

1+a2

L2−1/2

ΘE−Ec−~2π2

2m∗

1+a2

L2

and similarly in the valence band

g(E) = p2m∗

π~X

a1,a2>0Ev−~2π2

2m∗

1+a2

L2−E−1/2

ΘEv−~2π2

2m∗

1+a2

L2−E

(18.3) p-nJunction*

Explain the origin of the depletion layer in an abrupt

p-njunction and discuss how the junction causes rectifca-

tion to occur. Stating your assumptions, show that the

total width wof the depletion layer of a p-njunction is:

w=wn+wp

152 Semiconductor Devices

where

wn=2ǫrǫ0NAφ0

eND(NA+ND)1/2

and a similar expression for wpHere ǫris the relative

permittivity and NAand NDare the acceptor and donor

densities per unit volume, while φ0is the diﬀerence in

potential across the p-njunction with no applied voltage.

You will have to use Poisson’s equation to calculate the

form of φgiven the presence of the ion charges in the

depletion region.

Calculate the total depletion charge and infer how

this changes when an additional voltage Vis applied.

What is the diﬀerential capacitance of the diode

and why might it be useful to use a diode as a capacitor

in an electronic circuit?

Let us set the position xto be zero, where we have an ndoped region

to the left (negative x) and a pdoped region to the right (positive x).

Let the depletion widths be wnand wprespectively, and the doping

densitities be NDand NArespectively. Within the depletion widths

there is a net charge built up (see Fig 18.4 of the book for example).

We must solve the Poisson equation ∂2

xφ=ρ/(ǫ0ǫr) where ρis the local

charge density (which is constant in each region). Setting φ(x= 0) = 0

for simplicity, we immediately obtain

φ(x) = −eND

2ǫ0ǫr

x2+CDx x < 0

φ(x) = eNA

2ǫ0ǫr

x2+CAx x > 0

where CDand CDare constants to be ﬁxed here. We have additional

boundary conditions that the electric ﬁeld must go to zero at the edge

of the depletion region, so we have ∂xφ(x=−wn) = ∂x(x=wa) = 0.

This ﬁxes the constants so that we now have

φ(x) = −eND

2ǫ0ǫrx2+ 2wnxx < 0

φ(x) = eNA

2ǫ0ǫrx2−2waxx > 0

The total potential drop across both regions is then

φ0=φ(−wn)−φ(wa) = e

2ǫ0ǫrNDw2

n+NAw2

a(18.1)

We also note that the total charge in the two depletion regions must sum

to zero (since the depletion occurs by annihilation of opposite charges)

so we have

waNA=wnND(18.2)

Plugging Eq. 18.2 into Eq. 18.1 and solving for wnyields the desired

expression.

wn=2ǫrǫ0NAφ0

eND(NA+ND)1/2

and similarly

wa=2ǫrǫ0NDφ0

eNA(NA+ND)1/2

153

When an additional voltage is added, it simply shifts φ0to φ0+eV .

The total depletion charge per unit cross sectional area is then

q=wnND=waNA=2ǫrǫ0NDNA(φ0+eV )

e(NA+ND)1/2

The diﬀerential capacitance per unit cross sectional area is

C=∂q

∂V =ǫrǫ0NDNA

2e(NA+ND)1/2

(φ0+eV )−1/2

This provides a useful circuit element as it allows one to control a ca-

pacitance by applying a DC voltage.

(18.4) Single Heterojunction*

Consider an abrupt junction between an n-doped semi-

conductor with minimum conduction band energy ǫc1and

an undoped semiconductor with minimum conduction

band energy ǫc2where ǫc1< ǫc2. Describe qualitatively

how this structure might result in a two-dimensional elec-

tron gas at the interface between the two semiconductors.

Sketch the electrostatic potential as a function of position.

It should say, ǫc1> ǫc2.

In this problem, nothing interesting happens in the valence band — so

we only draw the conduction band. The situation is quite similar to the

p-n junction. Before the two semiconductors are brought together, there

are electrons in the conduction band on the left (or in dopant orbitals

slightly below the conduction band as shown in the ﬁgure), but empty

states on the right at lower energy because the conduction band energy

is lower. This situation is shown in ﬁgure 18.1



Fig. 18.1 Before electrons are allowed to ﬂow between

the two seminconductors, there are electrons on the left

at higher energy than empty states on the right. The

blue lines indicate the conduction band minima. The

dotted line is the fermi energy (slighly below the conduc-

tion band assuming that electrons are bound to dopants).

As in the p-n junction, electrons want to ﬂow between the two semi-

conductors in order to lower their energies, but in so doing charge builds

up (a postive charge is left on the left and negative charged electrons

accumulate on the right. When equilibrium is established the electro-

chemical potential is as shown in ﬁgure 18.2. The electrons are back

154 Semiconductor Devices

attracted to the positive charges they left behind, thus accumulating in

a roughly triangular well that forms near the interface. If the conﬁne-

ment is suﬃciently strong, this becomes a particle in a box problem and

the electrons in the well may become restricted to a single transverse

wavefunction — thus becoming a strictly two dimensional electron gas.



Fig. 18.2 Electrochemical potential of a semiconductor

heterostructure. There is net postive charge on the left

where electrons have left their donor atoms behind and

net negative charge on the right where electrons have

accumulated without positively charged nuclei.

(18.5) Diode Circuit

Design a circuit using diodes (and any other simple

circuit elements you need) to convert an AC (alternating

current) signal into a DC (direct current) signal.

*Can you use this device to design a radio reciever?

For the purpose of this problem we will make the crude assumption

that a diode (arrow in a circuit diagram) is an ideal circuit element that

allows current ﬂow one way but no current ﬂow the other way.



Fig. 18.3 A half-wave rectiﬁer.

The circuit shown in Fig. 18.3 is known as “half-wave rectiﬁer”. If the

source voltage is sinusoidal, V=V0sin(ωt), the voltage across the load

155

resistor is then

V=V0sin(ωt)Θ(sin(ωt))

where Θ is the Heaviside step function. I.e., the current goes to zero

instead of going negative. Now instead of giving a true DC output, this

gives an output that is nonnegative. In order to smooth the nonnegative

(but ﬂuctuating) voltage into a smooth DC voltage one inserts a capac-

itor into the circuit to act as a “battery” as shown in Fig. 18.4. This

eﬀectively damps the high frequency components of the signal leaving

only the low frequency DC component. One should choose a capacitor

such that RloadC≫1/ω in which case the non-dc component of the

resultant signal is of magnitude VAC ≈V0/(ωRC) .



Fig. 18.4 A half-wave rectiﬁer with a smoothing capac-

itor.

One can do a bit better in circuit design by using more diodes to

construct a “Full wave rectiﬁer” as shown in Figure 18.5.





Fig. 18.5 A full-wave rectiﬁer.

In this case, in the absence of the smoothing capacitor, the voltage

156 Semiconductor Devices

across the load is given by

V=|V0sin(ωt)|

a smoothing capacitor can be used as above. This scheme has the advan-

tage that the resulting DC voltage ends up being of higher magnitude

than that of the half-wave, and the ﬂuctuating DC component ends up

being smaller.

In order to design a radio, we must ﬁrst think about how radios

encode information. The simplest encoding system is AM or amplitude

modulation. In this case a signal A(t) (for simplicity let us assume the

signal is everywhere positive) is multiplied by a high frequency carrier

wave sin(ωt) and the product A(t) sin(ωt) is transmitted and then re-

ceived by an antenna. To convert the received signal back into A(t) one

simply puts it through a rectiﬁer. If one wants to do a slightly bet-

ter job, one wants to recieve only signals where the carrier frequency is

very close to some given frequency ω0. To do this, one needs to build

a resonant (LC) circuit which will respond only to frequencies near its

resonance. A sample 1-diode radio circuit (often known as a “crystal”

radio after the fact that the diode was often made of a small crystal)

is shown in Fig. 18.6. In this ﬁgure the right hand side (D1,C3,E1) is

just a half-wave rectiﬁer as discussed above. Here E1 is the “earphone”

or output load of the circuit. The middle of the circuit L2,C2 is the

resonant LC circuit, which is inductively coupled to the incoming signal

from the antenna. Note that the capacitors are made tunable here so

that the resonance frequencies can be modiﬁed. Many other designs are

possible as well.

Fig. 18.6 A radio circuit. Here E1 is the “earphone”

or output load of the circuit. The incoming antenna is

on the far left. A very crude radio could be made by

with only the rectiﬁer (D1,C3) attached directly to an

incoming antenna and an outgoing earphone E1.

157

(18.6) CMOS Circuit*

Design a circuit made of one n-MOSFET and one p-

MOSFET (and some voltage sources etc.) which can act

as a latch—meaning that it is stable in two possible states

and can act a single bit memory (i.e., when it is turned

on it stays on by itself, and when it is turned oﬀ it stays

oﬀ by itself).

A simple (and very rough) CMOS latch circuit (also known as an

SCR or silicon-controlled rectiﬁed) is shown in Fig. 18.7. When the

switch (S1) is closed no current ﬂows. The reason for this is that there

is no voltage on either of the two gates, so both transistors are “oﬀ”

meaning they prevent current ﬂow (and therefore prevent voltage from

being transmitted). However, when a voltage is momentarily applied

to the ON input, then this activates the P-MOSFET, allowing current

to ﬂow to the gate of the N-MOSFET, which then activates it, allowing

current to ﬂow to the gate of the P-MOSFET even when the ON voltage

is removed. Thus the circuit is latched in the “On” state. To put

the circuit back in the “Oﬀ” state, one must disconnect the switch S1

momentarily.











Fig. 18.7 A CMOS latch circuit.

Magnetic Properties of

Atoms: Para- and

Dia-Magnetism 19

(19.1) ‡Atomic Physics and Magnetism

(a) Explain qualitatively why some atoms are param-

agnetic and others are diamagnetic with reference to the

electronic structure of these materials.

(b) Use Hund’s rules and the Aufbau principle to de-

termine L,S, and Jfor the following isolated atoms:

(i) Sulfur (S) atomic number = 16

(ii) Vanadium (V), atomic number = 23

(iii) Zirconium (Zr), atomic number = 40

(iv) Dysprosium (Dy), atomic number = 66

(a) To a ﬁrst approximation, paramagnetic atoms have net moment

J6= 0 which can be re-aligned, whereas the typical diamagnetic atoms

have no moment.

If an atom has completely ﬁlled shells (say, the noble gases), then J=

L=S= 0, and the atom is diamagnetic due to Larmor Diamagnetism.

If an atom has a net moment J6= 0, from unﬁlled shells, then it is

paramagnetic. This is almost all other atoms.

However (more advanced answer): it is also possible to have J= 0

while having Land Snonzero. In this case, the atom can either be para-

magnetic or diamagnetic. Both para and dia terms are weak and either

one can win in this case. This can occur for atoms that are one electron

short of half-ﬁlled shells. (Known as Van Vleck Paramagnetism).

In metals one can have Pauli paramagnetism associated with re-orientation

of the spins of the conduction electron. One can also have Landau dia-

magnetism (beyond the scope of the course) which is the diamagnetic

response of the orbital motion of the conduction electrons. Pauli para-

magnetism in a metal is much weaker than Curie paramagnetism and

can be roughly the same size as diamagnetic eﬀects.

(b) Shells are ﬁlled in the order

1s,2s,2p,3s,3p,4s,3d,4p,5s,4d ...

with s,p,d shells containing 2,6,10 electrons respectively.

(i) 16S: [Ne]3s23p4

(ii) 23V: [Ar]4s23d3

(iii) 40Zr: [Kr]5s24d2

160 Magnetic Properties of Atoms: Para- and Dia-Magnetism

(iv) 54Xe: [Xe] (All ﬁlled shells)

(v) 66Dy: [Xe]6s24f10

Where ﬁlled shell conﬁgurations [Ne] contains 10 electrons, [Ar] con-

tains 18, [Kr] contains 36 and [Xe] contains 54.

Thus

(i) Sulfer has 4 electrons in a p-shell, these ﬁll lz=−1,0,1 with spin

up and lz= 1 with spin down. Thus L= 1 and S= 1, and since the

shell is more than half full J=L+S= 2.

(ii) Vanadium has 3 electrons in a d-shell, these ﬁll lz= 2,1,0 with

spin up giving L= 3 and S= 3/2. Since the shell is less than half ﬁlled

J=L−S= 3/2.

(iii) Zirconium has 2 electrons in a d-shell, these ﬁll lz= 2,1 with

spin up giving L= 3 and S= 1. Since the shell is less than half ﬁlled

J=L−S= 2.

(iv) Xenon is a noble gas, meaning all shells are ﬁlled, so J=L=

S= 0.

(v) Dysprosium has 10 electrons in an f-shell, these ﬁll all the spin up

state (7 of them) and lz= 3,2,1 for spin down giving L= 6 and S= 2.

Since the shell is more than half ﬁlled J=L+S= 8.

Note that none of these atoms violate the Madelung rule which dic-

tates the ﬁlling order or violates Hund’s rules when the atoms are iso-

lated. (Violations do sometimes occur but these atoms work as they are

supposed to).

(19.2) More Atomic Physics

(a) In solid erbium (atomic number=68), one electron

from each atom forms a delocalized band so each Er atom

has eleven felectrons on it. Calculate the Land´e g-factor

for the eleven electrons (the localized moment) on the Er

atom.

(b) In solid europium (atomic number =63), one elec-

tron from each atom forms a delocalized band so each Eu

atom has seven felectrons. Calculate the Land´e g-factor

for the seven electrons (the localized moment) on the Eu

atom.

Er typically is in a +3 state. 2 of those are from the core s-orbitals.

1 is from the f orbital.

(a) For 11 electrons in an f-shell, using Hund’s ﬁrst rule we obtain

S= 3/2 and Hund’s second rule we have L= 6. Since the shell is more

than half ﬁlled, Jis given by the sum J= 6 + 3/2 = 15/2. Using the

formula for the Lande gfactor (with g= 2) we have

˜g=1

2(g+ 1) + 1

2(g−1) S(S+ 1) −L(L+ 1)

J(J+ 1) = 6/5

This is almost exactly right.

(b) The counting here is messed up. There are 7 electrons before

losing one. For 7 electrons in an f-shell, L= 0 and S= 7/2 and J=S.

is purely spin, so ˜g=g= 2 (it also comes out of the above formula as

161

well). In fact, we should have 6 electrons when one is lost, and we get

L= 3 = Sand J= 0. However, This is a van-vleck ion, so in fact the

result is not what is predicted here.

(19.3) Hund’s Rules*

Suppose an atomic shell of an atom has angular mo-

mentum l(l= 0 means an s-shell, l= 1 means a p-shell

etc, with an lshell having 2l+ 1 orbital states and two

spin states per orbital.). Suppose this shell is ﬁlled with

nelectrons. Derive a general formula for S,L, and Jas

a function of land nbased on Hund’s rules.

The shell has 2l+1 orbital states and 2 spin states per orbital. Hund’s

ﬁrst rule tells us that

S(n, l) = 





20≤n≤2l+ 1

4l+2−n

22l+ 1 ≤n≤4l+ 2

The second rule tells us that for n≤2l+ 1 we have

L(n, l) =

x=l

x=l−n+1

We can do this sum to obtain

L(n, l) = 1

2n(2l+ 1 −n)

For 2l+ 1 ≤n≤4l+ 2 we can consider only the electrons in addition

to the L= 0 half-ﬁlled shell, so L(n, l) = L(n−(2l+ 1), l) So we have

in all,

L(n, l) = 





2n(2l+ 1 −n) 0 ≤n≤2l+ 1

2(n−2l−1)(4l+ 2 −n) 2l+ 1 ≤n≤4l+ 2

And thus we have (using Hund’s third rule) J=L−Sfor less than half

ﬁlled and J=L+Sfor more than half ﬁlled so

J(n, l) = 





2n(2l−n) 0 ≤n < 2l+ 1

2(n−2l)(4l+ 2 −n) 2l+ 1 ≤n≤4l+ 2

‡(19.4) Para and Diamagnetism

Manganese (Mn, atomic number=25) forms an atomic

vapor at 2000K with vapor pressure 105Pa. You can

consider this vapor to be an ideal gas.

(a) Determine L, S, and Jfor an isolated manganese

atom. Determine the paramagnetic contribution to the

(Curie) susceptibility of this gas at 2000K.

(b) In addition to the Curie susceptibility, the man-

162 Magnetic Properties of Atoms: Para- and Dia-Magnetism

ganese atom will also have some diamagnetic susceptibil-

ity due to its ﬁlled core orbitals. Determine the Larmor

diamagnetism of the gas at 2000K. You may assume the

atomic radius of an Mn atom is one ˚

Angstrom.

Make sure you know the derivations of all the formulas

you use!

Using the ideal gas law, the density of Mn atoms is n=P/(kBT). We

will use this below.

(a) Atomic Mn has orbital conﬁguration 25Mn: [Ar]4s23d5, meaning

a half ﬁlled d-shell. This then has L= 0 and S= 5/2. Since this is

purely spin moment the g-factor is 2.

Now we need to determine the paramagnetic susceptibility of a spin

S= 5/2. Let us write the partition function

5/2

m=−5/2

e−βgµBBm

Since we will be concerned with small B, it is probably simplest to

expand the partition function directly

Z≈

5/2

m=−5/21−(βgµBBm) + 1

2(βgµBBm)2+...

= 6 + 35

4(βgµBB)2

Calculating the moment

moment = −∂(kBTlog Z)/∂B =35

12β(gµB)2B

Yielding a susceptibility of the gas

χ=∂M/∂H =nµ0

12β(gµB)2= [P/(kBT)2]35

3µ0µ2

B= 1.6×10−7

(b) Diamagnetism is understood from the schroedinger equation and

expanding for small magnetic ﬁeld

H=(p+eA)2

2m=p2

2m+1

2m(p·A+Ap) + e2A2

The middle term is a L·B, and has zero expectation if there is a ﬁlled

shell such that hLi= 0. The ﬁnal term is the source of diamagnetism.

One can write in circular gauge

A=1

2r×B

and the ﬁnal term ends up being

8m|B|2(r⊥)2=e2

6m|B|2r2

163

where we have used the spherical symmetry of the atom so that x2+y2=

(2/3)r2. We thus have the susceptibility

χ=−µ0dE2/dB2=−µ0

ρe2

6mhr2i

where here ρis the density of electrons. Since there are 20 core electrons

per atom, we have ρ= 20nwith nthe gas density (using 25 here would

also be sensible). Using 1 angstrom for rwe thus obtain

χ=−4.3×10−9

which is much much smaller than the paramagnetic contribution, even

at 2000K!

‡(19.5) Diamagnetism

(a) Argon is a noble gas with atomic number 18 and

atomic radius of about .188 nm. At low temperature it

forms an fcc crystal. Estimate the magnetic susceptibility

of solid argon.

(b) The wavefunction of an electron bound to an im-

purity in n-type silicon is hydrogenic in form. Estimate

the impurity contribution to the diamagnetic susceptibil-

ity of a Si crystal containing 1020 m−3donors given that

the electron eﬀective mass m∗= 0.4meand the relative

permittivity is ǫr= 12. How does this value compare to

the diamagnetism of the underlying silicon atoms? Si has

atomic number 14, atomic weight 28.09, and density 2.33

g/cm3.

(a) First we establish the density of Argon. The radius of the avrgon

atom is r=.188 nm. For an fcc crystal, the nearest neighbor distance is

r=a/(2√2) We then have four atoms per conventional unit cell for an

overall atomic density of 4/a3= 1/(4√2r3). The number of electrons is

18 times the number of atoms, so we have (see 19.4 for derivation)

χ=−µ0

ρe2

6mhr2i=−µ018 e2

(4√2r3)6mhr2i=−µ0

4√2

mr =−1×10−4

This number is actually a bit too big because the average hr2iis smaller

than this number since most of the electrons are further inside the atom

than the full atomic radius. Measurement gives us a suscpetability about

10−5, which would correspond to phr2i ≈ r/3.

(b) For a hydrogenic orbital (recall ψ∼e−r/a0) it is easy to calcualte

that hr2i= 3a2

0. We need only calculate the Bohr radius of an impurity

in a silicon atom. Recalling that the expression for the Bohr radius is

a0=4πǫ0~2

me2

we see that the Bohr radius in silicon should be rescaled by

aSi

0=a0ǫrm/m∗=a0∗12/.4 = 1.6nm.

Again using the result of the prior problem (and using m∗rather than

χ=−µ0

ρe2

6m∗hr2i ≈ −10−11

164 Magnetic Properties of Atoms: Para- and Dia-Magnetism

Finally we would like to compare this to the diamagnetism of the pure

silicon. First let us calculate the density of atoms in the system

n=2.33g

cm3

mol

28.09g

6.02 ×1023atoms

mol

(100cm)3

m3= 5 ×1028 atoms

Now we need to determine the atomic radius of silicon from the data

given. If we didn’t know the crystal structure, we might guess roughly

that

r=1

2n−1/3

which would be exact for a simple cubic lattice. However, we know that

Si is diamond structure with 8 atoms per conventional unit cell, so the

density is n= 8/a3with athe lattice constant so a= 2n−1/3. The

nearest neighbor distance is from [0,0,0] to [a/4, a/4, a/4] or a distance

of a√3/4, so the atomic radius is a√3/8 = 1

4n−1/3= 1.17 Angstrom.

We then have

χ=−µ0

ρe2

6mhr2i ≈ −4×10−6

This is much much greater than the diamagnetism of the few impurities.

(Note: I think the table value is almost exactly this, however, this is

fortuitous as the table value is usually given in cgs which diﬀers from

the SI version by a factor of 4 pi. Again the error is in the estimate of

r2. )

‡(19.6) Paramagnetism

Consider a gas of monatomic atoms with spin S= 1/2

(and L= 0) in a magnetic ﬁeld B. The gas has density

(a) Calculate the magnetization as a function of Band

T. Determine the susceptibility.

(b) Calculate the contribution to the speciﬁc heat of

this gas due to the spins. Sketch this contribution as a

function of µBB/kBT.

“Monatomic atoms” ? What was I smoking when I wrote this? It

should say monovalent. Doh! I should have also speciﬁed the g-factor

(as usual we can take it to be 2).

(a) Same calculation as we have done a million times.

Z=e+βgµB1

2B+e−βgµB

F=−kBTlog Z

m=−∂F/∂B =gµB

2tanh(βgµB

2B)

M=mn

χ= lim

B→0µ0∂M/∂B =µ0n(1

2gµB)2β

(b) Similarly, we have

U=∂log Z/∂β =gµB

2Btanh(βgµB

2B)

C=∂U/∂T =1

kBT2(gµB

2B)2sech2(βgµB

2B)

165

(19.7) Spin JParamagnet*

Given the Hamiltonian for a system of non-interacting

spin-Jatoms

H= ˜gµBB·J

(a)* Determine the magnetization as a function of Band

(b) Show that the susceptibility is given by

χ=nµ0(˜gµB)2

J(J+ 1)

kBT

where nis the density of spins. (You can do this part

of the exercise without having a complete closed-form ex-

pression for part a!)

(a) The eigenenergies of a single spin are

Em= ˜gµBBm m =−J, −J+ 1 ,−J+ 2 ,...J −1, J

So the canonical partition function of the single spin is

m=−J

e−βgµBBm =eβ˜gµbBJ

p=0

e−β˜gµBp

=eβ˜gµbBJ 1−e−β˜gµBB(2J+1)

1−e−β˜gµBB=eβ˜gµBB(2J+1)/2−e−β˜gµBB(2J+1)/2

eβ˜gµBB/2−e−β˜gµBB/2

=sinh(β˜gµBB(2J+ 1)/2)

sinh(β˜gµBB/2)

we construct the free energy per spin F/N =−kBTlog Zand then get

the magnetic moment per spin

m=−∂(F/N)

∂B = (˜gµB/2) [(2J+ 1) coth(β˜gµBB(2J+ 1)/2) −coth(β˜gµBB/2)]

and the total magnetizaton is then the moment per spin times the density

of spins n, so

M=n(˜gµB/2) [(2J+ 1) coth(β˜gµBB(2J+ 1)/2) −coth(β˜gµBB/2)]

(b) To obtain the susceptibility, we want

χ= lim

B→0µ0

∂M

∂B

To take this limit we take the argument of the coth to be small and we

use the expansion

lim

x→0coth x=1

x+x

Note that the two 1/x terms cancel when subtracted leaving

M∼n(˜gµB/2) (2J+ 1)2β˜gµBB/6−β˜gµBB/6

So we have

χ=nµ0(˜gµB)2

J(J+ 1)

kBT

as required.

166 Magnetic Properties of Atoms: Para- and Dia-Magnetism

Note, the question claims that part (b) may be achieved without com-

pleting part (a). To do this, note that we only concerned with small B

so we can expand the partition function directly for small B.

m=−J

e−β˜gµBBm =

m=−J

1−β˜gµBBm +1

2(β˜gµBB)2m2+...

The ﬁrst term in the sum gives (2J+ 1) and the second term gives 0 by

symmetry. The third term is the only hard one. I claim that

G[J] =

m=−J

3(2J+ 1)(J+ 1)J(19.1)

I will derive this below, but for now let us assume it is correct. So we

have

Z= (2J+ 1)[1 + 1

6(β˜gµBB)2J(J+ 1) + ...]

Using the free energy per spin is F/N =−kBTlog Z, since Bis small

we have

F/N =−kBTlog(2J+ 1) −1

6kBT(˜gµBB)2J(J+ 1)

And the magnetic moment per site is

m=−∂(F/N)

∂B =(˜gµB)2BJ(J+ 1)

3kBT

which multiplying by the density of spins nto give the magnetization,

gives the same result as above.

Finally we turn to derive Eqn. 19.1. The result of this sum must be

some polynomial in J. Further, approximating it as an integral, it must

have a maximum power of J3, and the coeﬃcient of the J0term must

be zero since G[0] = 0. Thus we propose

G[J] = aJ3+bJ2+cJ

Then we can also write the diﬀerence of two successive sums as just the

new ad

G[J+ 1] −G[J] = 2(J+ 1)2

which we multiply out to give

a(3J2+ 3J+ 1) + b(2J+ 1) + c= 2(J+ 1)2

matching coeﬃcients then gives the values of a, b, c which proves Eq. 19.1.

Spontaneous Magnetic

Order: Ferro-, Antiferro-,

and Ferri-Magnetism 20

(20.1) Ferromagnetic vs Antiferromagnetic

States

Consider the Heisenberg Hamiltonian

H=−1

hi,ji

JSi·Sj+X

gµBB·Si(20.1)

and for this exercise set B= 0.

(a) For J > 0, i.e., for the case of a ferromagnet, in-

tuition tells us that the ground state of this Hamiltonian

should simply have all spins aligned. Consider such a

state. Show that this is an eigenstate of the Hamiltonian

Eq. 20.1 and ﬁnd its energy.

(b) For J < 0, the case of an antiferromagnet on a cu-

bic lattice, one might expect that (at least for B= 0) the

state where spins on alternating sites point in opposite

directions might be an eigenstate. Unfortunately, this is

not precisely true. Consider such a state of the system.

Show that the state in question is not an eigenstate of the

Hamiltonian.

Although the intuition of alternating spins on alter-

nating sites is not perfect, it becomes reasonable for sys-

tems with large spins S. For smaller spins (like spin 1/2)

one needs to consider so-called “quantum ﬂuctuations”

(which is much more advanced, so we will not do that

here).

It is useful here to recall that

Si·Sj=1

2(S+

iS−

j+S−

iS+

j) + Sz

iSz

(Indeed, students often need to be reminded of this! Maybe it is worth

giving this as a hint!)

(a) If each spin is aligned in the ˆzdirection (it has Sz= 1), then the

energy is −gµbBper spin and for each bond we have energy −JSi·Sj=

−JSz

iSz

j=−JS2since S+on the spins all give zero. The system is in

an energy eigenstate with energy

E=−NgµB|B| − NzJS2/2

with zthe number of neighbors of each site (=6 for a cubic lattice).

(b) Assuming an antiferromagnetic conﬁguration, the key here is to

note that there are terms S+

iS−

jwhich do not vanish (where S+is ap-

plied to a down spin, and S−is applied to an up spin). This means

that when the hamiltonian is applied to the proposed antiferromagnetic

ground states, it generates other spin conﬁgurations. Hence this is not

an eigenstate.

168 Spontaneous Magnetic Order: Ferro-, Antiferro-, and Ferri-Magnetism

(20.2) Frustration

Consider the Heisenberg Hamiltonian as in Exercise 20

with J < 0, and treat the spins as classical vectors.

(a) If the system consists of only three spins arranged

in a triangle (as in Fig. 20.2), show that the ground state

has each spin oriented 120◦from its neighbor.

(b) For an inﬁnite triangular lattice, what does the

ground state look like?

Probably here I should have stated explicitly that all three spins have

the same |S|(might be interesting to consider a case where they don’t

all have the same spin!).

(a) The Hamiltonian is

H=J(S1·S2+S1·S3+S2·S3)

Since Si·Si=S2is a constant we can write

H= (J/2)(S1+S2+S3)2+ constant

To minimize the energy, we must have

S1+S2+S3= 0

which implies that the spins are at 120 degree angles from each other –

but can lie in any plane.

(b) For a triangular lattice, each triangle must have three spins each

at 120 degree angles from its neighbors. So choose three directions all

at 120 degree angles from each other in any given plane. Call these

three directions A, B, C. Now we must assign each site on the triangular

lattice one of the three values A, B, or C in such a way that all triangles

contain one site of type A, one of type B, and one of type C. One can

think of this as being now a crystal whose unit cell has three times the

area of the original unit cell, and now contains one spin of each type A,

B, C.

(20.3) Spin Waves*

For the spin-Sferromagnet particularly for large S, our

“classical” intuition is fairly good and we can use simple

approximations to examine the excitation spectrum above

the ground state.

First recall the Heisenberg equations of motion for any

operator

i~dˆ

dt = [ ˆ

O, H]

with Hthe Hamiltonian (Eq. 20.1 with Sibeing a spin S

operator).

(a) Derive equations of motion for the spins in the

Hamiltonian Eq. 20.1. Show that one obtains

~dSi

dt =Si× JX

Sj−gµbB!(20.2)

where the sum is over sites jthat neighbor i.

In the ferromagnetic case, particularly if Sis large, we

can treat the spins as not being operators, but rather as

being classical variables. In the ground state, we can set

all Si= ˆzS (Assuming Bis in the −ˆzdirection so the

ground state has spins aligned in the ˆzdirection). Then

to consider excited states, we can perturb around this

169

solution by writing

i=S− O((δS)2/S)

i=δSx

i=δSy

where we can assume δSxand δSyare small compared to

S. Expand the equations of motion (Eq. 20.2) for small

perturbation to obtain equations of motion that are linear

in δSxand δSy

(b) Further assume wavelike solutions

δSx

i=Axeiωt−ik·r

δSy

i=Ayeiωt−ik·r

This ansatz should look very familiar from our prior con-

sideration of phonons.

Plug this form into your derived equations of motion.

Show that Sx

iand Sy

iare out of phase by π/2.

What does this mean?

Show that the dispersion curve for “spin-waves” of

a ferromagnet is given by ~ω=|F(k)|where

F(k) = gµb|B|

+JS(6 −2[cos(kxa) + cos(kya) + cos(kza)])

where we assume a cubic lattice.

How might these spin waves be detected in an ex-

periment?

the spectrum you just derived, show that the speciﬁc heat

due to spin wave excitations is proportional to T3/2.

(a) We need the angular momentum algebra

[Sx, Sy] = iSz

and cyclic permutations of this. We then have the Heisenberg equations

i~dSx

dt = [Sx

i,H]

The only terms that this does not commute with are those containing

iand Sz

i. Thus we have

i~dSx

dt = [Sx

i, gµb(BySy

i+BzSz

i)−JX

(Sy

jSy

i+Sz

jSz

i)]

with the sum over jbeing over neighbors of i. Thus we obtain

i~dSx

dt =i

(gµb)(BySz

i−BzSy

i)−JX

(Sy

jSz

i−Sz

jSy

i)



and similar for the other two components of the spin. Thus we conclude

~dSi

dt =Si×

JX

Sj−gµbB



Writing

Si= ˆzS +δSi

we obtain

~dδSi

dt =JS ˆz×X

δSj+δSi×(JZS ˆz+gµb|B|)

where Zis the number of neighbors of a site.

170 Spontaneous Magnetic Order: Ferro-, Antiferro-, and Ferri-Magnetism

(b) Plugging in the wave ansatz (for a cubic lattice) we obtain

i~ωAx=F(k)Ay

i~ωAy=−F(k)Ax

where

F(k) = JS(6 −2[cos(kxa) + cos(kya) + cos(kza)]) + gµb|B|

This system of equations can be solved immediately to give dispersion

~ω=|F(k)|

Spin waves are typically detected by inelastic neutron scattering. This

is like scattering from phonons except that one uses (spin-polarized)

neutrons in order to couple to the spin.

quadratic

E=JSa2k2

We will also need

k= (E/JSa2)1/2

and

dk =dE/(2(EJSa2)1/2)

Calculating the total energy stored in spin waves, we have

U=VZBZ

(2π)3E(k)nB(E(k)) ≈V

2π2Z∞

dkk2E(k)nB(E(k))

with nBthe usual Bose factor and the approximation accurate for low

temperatures. We rewrite this as

U=V

4π2(JSa2)3/2Z∞

dEE3/2nB(E)

=V(kbT)5/2

4π2(JSa2)3/2Z∞

dx x5/2

ex−1

which we diﬀerentiate to get the heat capacity

C=dU/dT =kb

V(kbT)3/2

4π2(JSa2)3/2Z∞

dx x5/2

ex−1

(one does not really need to carry all of the constants to see how it

scales!). For those who are interested though, the integral can be eval-

uated in the usual way

Z∞

dx x5/2

ex−1=

∞

n=1 Z∞

dxx5/2e−nx =

∞

n=1

n7/2Z∞

dyy5/2e−y= Γ(7/2)ζ(7/2)

where Γ(7/2) = 15√π/8 and ζ(7/2) ≈1.1267 ....

171

(20.4) Small Heisenberg Models

(a) Consider a Heisenberg model containing a chain of

only two spins, so that

H=−JS1·S2.

Supposing these spins have S= 1/2, calculate the en-

ergy spectrum of this system. Hint: Write 2S1·S2=

(S1+S2)2−S12−S12.

(b) Now consider three spins forming a triangle (as

shown in Fig. 20.2). Again assuming these spins are

S= 1/2, calculate the spectrum of the system. Hint:

Use the same trick as in part (a)!

Again assuming these spins are S= 1/2, calculate the

spectrum of the system.

Note the obvious typo, it should read 2S1·S2= (S1+S2)2−S12−S22.

(a)

H=−(J/2)[(S1+S2)2−S2

1−S2

Since (S)2=S(S+ 1) for spin-1/2 we have (S)2= 3/4. Further, when

two spin 1/2’s are added they can form either a spin-0 singlet or a spin-

1 triplet (three Szstates). So (S1+S2)2takes the values 0 for the

singlet or S(S+ 1) = 2 for the S= 1 triplet. Thus the Hamiltonian has

eigenstates 3J/4 for the singlet (one eigensate) and −J/4 for the triplet

(three eigenstates).

(b) Similarly

H=−(J/2)[(S1+S2+S3)2−S2

1−S2

2−S2

Here, again for spin 1/2 we have (S)2= 3/4. When adding three spin-

1/2s, we can obtain spin-1/2 in two ways and spin-3/2 in one way. (To

see this, think about adding the ﬁrst two spin-1/2 to get spin-0 or spin-

1. Now adding a spin-1/2 the spin-0 gives spin-1/2 and adding spin-1/2

to the spin-1 gives either spin-1/2 or spin-3/2. Note that counting the

total nbumber of eigenstates we should get 8 since each spin-1/2 has

two possible Szstates. Each of the two possible spin-1/2’s can take two

possible Szstates and the spin-3/2 can take 4 possible Szstates, which

gives a total of 8 possible Szstates.

In the case that the three spins add to spin-1/2, we obtain energy

3J/4 (four eigenstates) whereas if the three spins add to spin-3/2, S2=

S(S+ 1) = 15/4, so the energy is −3J/4 (four eigenstates).

H=−(J/2)[(S1+S2+S3+S4)2−S2

1−S2

2−S2

3−S2

The sum of the four spins can give spin 0 in two ways, spin 1 in 3 ways,

and spin 2 in one way. Again we should add up the total number of

eigenstates to check that it is 24= 16. We have 2 spin 0’s + 3 states for

spin 1 in 3 ways + 5 states in spin 2. So we have 2+9+5 = 16. The spin-

0 singlets (two eigenstates) have energy (−J/2)[0 −4(3/4)] = 3J/2. The

three spin-1 triplets (9 eigenstates) have energy (−J/2)[2 −4(3/4)] =

J/2 and the spin-2 ﬁveplets (5 eigenstates) have energy (−J/2)[2 ×3−

4(3/4)] = −3J/2

172 Spontaneous Magnetic Order: Ferro-, Antiferro-, and Ferri-Magnetism

(20.5) One-Dimensional Ising Model with B= 0

(a) Consider the one-dimensional Ising model with spin

S= 1. We write the Hamiltonian for a chain of Nspins

in zero magnetic ﬁeld as

H=−J

N−1

i=1

σiσi+1

where each σitakes the value ±1. The partition function

can be written as

Z=X

σ1,σ2,...σN

e−βH .

Using the transformation Ri=σiσi+1 rewrite the parti-

tion function as a sum over the Rvariables, and hence

evaluate the partition function.

Show that the free energy has no cusp or disconti-

nuity at any temperature, and hence conclude that there

is no phase transition in the one-dimensional Ising model.

(b) *At a given temperature T, calculate an expres-

sion for the probability that Mconsecutive spins will be

pointing in the same direction. How does this probability

decay with Mfor large M? What happens as Tbecomes

small? You may assume N≫M.

(a) The ﬁrst spin can be either in the spin-up or spin down state, so

we leave σ1as a variable to be summed over. The remaining spins are

deﬁned by Rifor i= 1,...N −1 where each Rcan take the values ±1.

The Hamitonian in terms of the Rvariables is

H=−J

N−1

i=1

So the partition function is

Z= 2(e−βJ +eβJ )N−1= 2(2 cosh(βJ))N−1

with the factor of 2 out front being the sum over the ﬁrst spin. The free

energy is thus

F=−kBTlog Z=−kBT N log 2 −kbT N log cosh(βJ)

which is a completely continuous function with no cusps or discontinu-

ities at ﬁnite β.

(b) The probability that a given Ris in the +1 state is

P(R= +1) = eβJ

eβJ +e−βJ =1

e−2βJ + 1

Having Rin the +1 state tells us that two consecutive spins are pointing

in the same direction. Looking at M−1 consecutive values of R, the

probability that all of these are +1 is then

P(M) = 1

e−2βJ + 1 M−1

which would put Mconsecutive spins in the same direction. This prob-

ability decays exponentially with M. To see this, rewrite

P(M) = exp[−(M−1) log(e−2βJ + 1)]

173

so the decay length is 1/log(e−2βJ + 1). In the low tempearture limit

P(R+ 1) becomes close to unity, and the decay length becomes long.

Actually, we can see that for very large βwe have log(e−2βJ +1) ≈e−2βJ ,

so the decay length is e2βJ lattice sites and is exponentially long.

(20.6) One-Dimensional Ising Model with B6= 0*

Consider the one-dimensional Ising model with spin

S= 1. We write the Hamiltonian (Eq. ??) for a chain of

Nspins in magnetic ﬁeld Bas

i=1 Hi(20.3)

where

H1=hσ1

Hi=−Jσiσi−1+hσifor i > 1

where each σitakes the value ±1 and we have deﬁned

h=gµBBfor simplicity of notation.

Let us deﬁne a partial partition function for the ﬁrst M

spins (the ﬁrst Mterms in the Hamiltonian sum Eq. 20.3)

given that the Mth spin is in a particular state. I.e.,

Z(M, σM) = X

σ1,...,σM−1

e−βPM

i=1 Hi

so that the full partition function is Z=Z(N, +1) +

Z(N, −1).

(a) Show that these partial partition functions satisfy

a recursion relation

Z(M, σM) = X

σM−1

TσM,σM−1Z(M−1, σM−1)

where Tis a 2 by 2 matrix, and ﬁnd the matrix T. (Tis

known as a “transfer matrix”).

(b) Write the full partition function in terms of the

matrix Traised to the (N−1)th power.

limit, can be written as

F/N ≈ −kBTlog λ+

where λ+is the larger of the two eigenvalues of the matrix

(d) From this free energy, derive the magnetization,

and show that the susceptibility per spin is given by

χ∝βe2βJ

which matches the Curie form at high T.

(a) The transfer matrix takes the form

T=e−βH(+,+) e−βH(+,−)

e−βH(−,+) e−βH(−,−)=e−β(−J+h)e−β(J−h)

e−β(J+h)e−β(−J−h)

(b) The full partition function is then

Z=vT N−1vT

where v= (1,1). Although Tis not hermitian, it can still be brought to

diagonal form via a Jordon decomposition

T=JΛJ−1

where Λ is a diagonal matrix of the eigenvalues of Tand Jis a matrix

made of the eigenvectors of T(these are no longer orthonormal!). (c)

Assuming these two eigenvalues are not degenerate, we then have for

large Nthat

Z∼λN−1

174 Spontaneous Magnetic Order: Ferro-, Antiferro-, and Ferri-Magnetism

where λ+is the larger of the two eigenvalues. Thus the free energy per

site is

F/N ∼ −kBTlog λ+

For the record, the value of the larger eigenvalue is

λ+=e−β(h+J)

21 + e2βh +p1−2e2βh +e4βh + 4e4β(h+J)

Just to check that this agrees with the previous problem, note that

setting h= 0 we correctly obtain

λ+=e−βJ +eβJ

(d) Probably we want to derive the magnetic moment per spin which is

m=−∂(F/N)

∂B =gµBkbT∂log λ+

∂h

The result is a bit of a mess. Then once we have this, we want the

susceptibility per spin

χ=µ0∂m/∂B

which one evaluates at B= 0. It is much easier to make expansions for

small hsince this is all we will need. In the end we will obtain

χ=g2µ2

Bµ0kbT∂2log λ+

∂h2=g2µ2

Bµ0βe−2βJ

as claimed.

Domains and Hysteresis 21

(21.1) Domain Walls and Geometry

Suppose a ferromagnet is made up of a density ρof

spins each with moment µB.

(a) Suppose a piece of this material forms a long cir-

cular rod of radius rand length L≫r. In zero external

magnetic ﬁeld, if all of the moments are aligned along the

L-direction of the rod, calculate the magnetic energy of

this ferromagnet. (Hint: a volume of aligned magnetic

dipoles is equivalent to a density of magnetic monopoles

on its surface.)

(b) Suppose now the material is shaped such that

r≫L. What is the magnetic energy now?

where might it go to minimize the magnetic energy in the

two diﬀerent geometries. Estimate how much magnetic

energy is saved by the introduction of the domain wall.

(d) Suppose the spins in this material are arranged in

a cubic lattice, and the exchange energy between nearest

neighbors is denoted Jand the anisotropy energy is very

large. How much energy does the domain wall cost? Com-

paring this energy to the magnetic energy, what should

we conclude about which samples should have domain

walls?

(e) Note that factors of the lattice constant aare often

introduced in quoting exchange and anisotropy energies

to make them into energies per unit length or unit vol-

ume. For magnetite, a common magnetic material, the

exchange energy is JS2/a = 1.33 ×10−11 J/m and the

anisotropy energy is κS2/a3= 1.35×104J/m3. Estimate

the width of the domain wall and its energy per unit area.

Make sure you know the derivation of any formulas you

use!

(a) Following the hint, the magnetic energy between two monopoles

of charge qmand q′

mseparated by distance ris given by

E=µ0

4π

qmq′

Here we have charge of magnitude qm=µBAρ where A=πr2is the

area of the end. Check that this has the right dimensions, recall that

a dipole is a charge times a length, so qmcorrectly has dimension of a

charge.

Thus the energy is

µ0

4π

(µBρ(πr2))2

(b) This problem, on the other hand is analogous to the energy of a

capacitor. The energy stored in a capacitor is q2/(2C) where electrically

the capacitance is ǫ0A/d with Athe area and dthe spacing. The analogy

here is thus A/(dµ0). Thus the total energy stored is

E=q2

mµ0d/(2A) = µ0(µBρ)2(πr2)L

parallel to the Asurface). This puts two magnetic charges of the same

176 Domains and Hysteresis

sign right next to each other and is therefore energetically unfavorable.

Now consider a domain wall perpendicular to this plane — i.e., that runs

along the Laxis. In the (a) case we can view this as, instead of having a

single charge at each end of the rod, now we have two opposite charges

along each end of the rod (each with charge ±qm/2). This completely

kills the leading energy cost and now we have instead some sort of dipolar

interaction between the two ends — with energy typically on the much

smaller order

E=µ0

4π

(µBρ(πr2/2))2r2

Similarly with the capacitor conﬁguration of part (b), we can convert

“charge” energy to “dipole” energy by placing a domain wall in the

diraction along the axis (in the direction of L). However, here we will

not be able to consider the system “dipolar” until the size of the domains

dis smaller than L. Once we do this, the energy will again drop as in

the (a) case proportional to d6/L3

(d) Since the lattice spacing will be a=ρ−1/3, the energy of intro-

ducing a domain wall of area Ais AJρ2/3= 2LrJρ2/3. In the (a) case,

for large enough L, the magnetic energy is not very large to begin with,

so introducing an energy cost proportional to Lis not a very good idea.

Now let us consider the (b) case. Let us consider introducing a domain

wall network spaced by d. The number of domains is πr2/d2and the

total aread of the walls will be πr2L/d and their total domain wall en-

ergy will be πr2LJρ2/3/d. As a function of d, as discussed in part (c)

then magnetic energy drops strongly when d < L. Thus, there will be

some happy medium which optimizes the total energy.

(e) As mentioned in the text, the width of a domain wall should be

pJ/κ which here is 3100 lattice spacings. The energy per unit area is

then √Jκ = 4 ×107Joules per a2area.

(21.2) Critical Field for Crystallite

(a) Given that the energy of a crystallite in a magnetic

ﬁeld is given by

E/V =E0− |M||B|cos θ−κ′|M|2(cos θ)2

show that for |B|< Bcrit there is a local energy minimum

where the magnetization points opposite the applied ﬁeld,

and ﬁnd Bcrit.

(b)* In part (a) we have assumed Bis aligned with the

anisotropy direction of the magnetization. Describe what

can occur if these directions are not aligned.

volume) is the activation barrier for the system to get

from the local minimum to the global minimum.

(d) Can you make an estimate (in terms of actual num-

bers) of how big this activation barrier would be for a fer-

romagnetic crystal of magnetite that is a sphere of radius

1 nm ? You may use the parameters given in Exercise

31.1.e (you may need to estimate some other parameters

as well).

(a) Let cos θ=z. The function E(z) is an upside-down parabola

with maxium at z=−B/(2κ′|M|)<0. If this parameter is z > −1

then there are two minima, one at θ= 0 and the other at θ=π. The

absolute minium is always at θ= 0.

(b) Let us consider Bagain pointing in the ˆzdirection and the anisotropy

direction oriented at an angle φfrom ˆz. In this case we have an energy

177

functional

E/V =E0− |M||B|cos θ−κ′|M|2[cos(θ−φ)]2

If y=κ|M|/(2B) is small, then the minimum will always be close to 0. If

this parameter is greater than 1/2, then there may be a metastable min-

imum as well, although for yclose to 1/2 then there is only a metastable

minimum if φis close to zero. If yis large, then the metastable minima

always exist near to φand φ+π.

E0−κ|M|2and Emax =E0. So the activation barrier is κ|M|2.

(d) The activation barrier is just the total anisotropy energy. This is

1.35 ×104J/m3and the size is (4/3)π×1nm3≈10−26m3, so we have a

total energy of about 10−22J≈10Kelvin activation energy.

(21.3) Exact Domain Wall Solution*

The approximation used in Section 21.1.1 of the en-

ergy of the anisotropy (κ) term is annoyingly crude. To

be more precise, we should instead write κS2(cos θi)2and

then sum over all spins i. Although this makes things

more complicated, it is still possible to solve the problem

so long as the spin twists slowly so that we can replace

the ﬁnite diﬀerence δθ with a derivative, and replace the

sum over sites with an integral. In this case, we can write

the energy of the domain wall as

E=Zdx

a(JS2a2

2dθ(x)

dx 2

−κS2[cos θ(x)]2)

with athe lattice constant.

(a) Using calculus of variations show that this energy

is minimized when

(Ja2/κ)d2θ/dx2−sin(2θ) = 0

(b) Verify that this diﬀerential equation has the solu-

tion

θ(x) = 2 tan−1exp √2(x/a)rκ

J

thus demonstrating the same L∼pJ/κ scaling.

comes Etot/(A/a2) = 2√2S2√Jκ.

It is better to write the ﬁrst equation as + sin2instead of −cos2.

These diﬀer only by a constant, but in the former case, the background

energy is zero whereas as written the background energy is ﬁnite.

(a) This is a trivial exercise in calculus of variations. To clarify it,

write θ=θ0+δθ and expand to linear order in δθ. Isolating the terms

inside the integral linear in δθ, and then integrating by parts to remove

the derivatives gives

−JS2a2d2θ(x)

dx2+ 2κS2cos θ(x) sin θ(x)δθ(x)

Setting this variation to zero immediately gives the desired result.

(b) This is an exercise in nasty algebra. First write b=√2(1/a)pκ

so that we have

θ(x) = 2 tan−1(ebx)

178 Domains and Hysteresis

From this we obtain on the left of the equation

dθ/dx = 2bebx

1 + e2bx =b

cosh(bx)(21.1)

d2θ/dx2=−b2sinh(bx)

cosh2(bx)(21.2)

On the right of the equation we need to evaluate

sin 2θ= 2 sin θcos θ= 4 sin θ

2cos θ

2(cos2θ

2−sin2θ

2)(21.3)

We have θ/2 = tan−1(ebx) so ebx = tan θ

2so we have 1 + e2bx = sec2θ

cos θ

2=1

√1 + e2bx

sin θ

2=ebx

√1 + e2bx

Plugging these results into the prior expression for sin 2θthe gives

sin 2θ= 4 ebx

1 + e2bx

1−e2bx

1 + e2bx =−2 sinh(bx)

cosh2(bx)

Comparing this result to Eq. 21.2 along with the expression for bimme-

diately conﬁrms the desired solution.

E=Zdx

a(JS2a2

2dθ(x)

dx 2

+κS2[sin θ(x)]2)

we have dθ/dx =b

cosh(bx)and

sin θ= 2 cos θ/2 sin θ/2 = 2ebx

1 + e2bx = 1/cosh(bx)

This then naturally gives

E=Zdx

a(JS2a2

2b

cosh(bx)2

+κS21

cosh2(bx))

Plugging in the value of bthis simpliﬁes to

E= 2κS2Zdx

cosh2bx =2κS2

ab Zdx 1

cosh x

plug in the value of bto get a prefactor of S2√2κJ and the integral of

sech2is tanh thus giving a factor of 2, yielding the desired result.

Mean Field Theory 22

(22.1) ‡Weiss Mean Field Theory of a Ferromag-

net

Consider the spin-1/2 ferromagnetic Heisenberg Hamil-

tonian on the cubic lattice:

H=−J

hi,ji

Si·Sj+gµBBX

Si(22.1)

Here, J > 0, with the sum indicated with hi, jimeans

summing over iand jbeing neighboring sites of the cu-

bic lattice, and Bis the externally applied magnetic ﬁeld,

which we will assume is in the ˆzdirection for simplicity.

The factor of 1/2 out front is included so that each pair of

spins is counted only once. Each site iis assumed to have

a spin Siof spin S= 1/2. Here µBis the conventional

Bohr magneton deﬁned to be positive. The fact that the

ﬁnal term has a + sign out front is from the fact that the

electron charge is negative, therefore the magnetic mo-

ment opposes the spin direction. If one were to assume

that these were nuclear spins the sign would be reversed

(and the magnitude would be much smaller due to the

larger nuclear mass).

(a) Focus your attention on one particular spin Si, and

write down an eﬀective Hamiltonian for this spin, treat-

ing all other variables Sjwith j6=ias expectations hSji

rather than operators.

(b) Calculate hSiiin terms of the temperature and the

ﬁxed variables hSjito obtain a mean-ﬁeld self-consistency

equation. Write the magnetization M=|M|in terms of

hSiand the density of spins.

dM/dH =µ0dM/dB in this approximation.

(d) Find the critical temperature in this approxima-

tion.

Write the susceptibility in terms of this critical tem-

perature.

(e) Show graphically that in zero external ﬁeld (B= 0),

below the critical temperature, there are solutions of the

self-consistency equation with M6= 0.

(f) Repeat parts (a)–(d) but now assuming there is an

S= 1 spin on each site (meaning that Sztakes the values

−1,0,+1).

(a) The eﬀective Hamiltonian for one spin-1/2 is

Hi=Si·

−JX

jhSji+gµbB

= +Si·gµbBi,ef f

where

gµbBi,ef f =gµbB−JX

jhSji

with the sums being over sites jneighboring site i. (Note the factor of

1/2 out front is now missing).

(b)Assuming Baligned with ˆztells us that hSkishould be aligned

with −ˆzif nonzero. Thus we can treat these quantities as scalars rather

than vectors. We then have the usual calculation

Zi= exp(βgµbBi,eff (1/2)) + exp(−βgµbBi,eff (1/2)) (22.2)

hSz

ii=−1

2tanh[βg ˜µbBeff (1/2)]

180 Mean Field Theory

which gives us the self consistency equation

hSzi=−1

2tanh[β(gµbB−JzhSzi)(1/2)] (22.3)

where zis the number of neighboring spins for each site (which is z= 6

for a cubic lattice).

for small ﬁeld we can expand the tanh for small argument to obtain.

hSzi=−1

2[β(gµbB−JzhSzi)(1/2)]

hSzi=(−1/4)β(gµbB)

1−βJz/4

The moment per site is thus

m=−gµbhSzi=(1/4)(gµb)2B

kbT−Jz/4

giving the susceptibility

χ=(µ0/4)(gµb)2N

kbT−Jz/4=(1/4)(gµb)2N

kb(T−Tc)(22.4)

refWeissMeanFerro where Nis the density of spins.

(d) the critical temperature is the point where the susceptibility di-

verges, or kbTc=Jz/4.

(e) Graphically we plot the right and left sides of Eq. 22.3 as shown

roughly in Fig. 22.1. The horizontal axis is −hSi. The left side of the

equation is a straight line, the right side is a tanh. At high enough T,

these intersect only at hSi= 0, however at T < Tcthey intersect as well

at a ﬁnite value of hSi. Note that the point at which the curves are

tangent is the critical temperature. We will show in the next problem

that below Tcthe solution with hSi= 0 is unstable.

(f) The procedure is the same for S= 1. In this case, the partition

function Eq. 22.2 is replaced by

Zi= exp(βgµbBi,eff ) + 1 + exp(−βgµbBi,ef f )(22.5)

leading to

hSz

ii=−2 sinh(x)

2 cosh(x) + 1

with x=βgµbBeff =β(gµbB−JzhSz

ii). At high Twe expand for small

xwe obtain the equation

hSzi=−2x

3=−2

3β(gµbB−JzhSzi)

which then gives us

hSzi=(−2/3)βgµbB

1−2

3βJz

181

Fig. 22.1 graphical solution of Eq. 22.3

or a susceptibility of

χ=(2/3)µ0(gµb)2N

kbT−2

3Jz

and a critical temperature of kbTc= (2/3)Jz (where Nis the density of

spins).

(22.2) Bragg-Williams Approximation

This exercise provides a diﬀerent approach to obtain-

ing the Weiss mean-ﬁeld equations. For simplicity we will

again assume spin 1/2 variables on each site.

Assume there are Nlattice sites in the system. Let

the average spin value be hSz

ii=s. Thus the probability

of a spin being an up spin is P↑= 1/2 + swhereas the

probability of a spin being a down spin is P↓= 1/2−s.

The total number of up spins or down spins is then NP↑

and NP↓respectively where there are Ntotal lattice sites

in the system.

(a) Consider ﬁrst a case where sites do not interact

with each other. In the micro-canonical ensemble, we can

count the number of conﬁgurations (microstates) which

have the given number of spin-ups and spin-downs (de-

termined by s). Using S=kBln Ω, calculate the entropy

of the system in the large Nlimit.

(b) Assuming all sites have independent probabilities

P↑and P↓of pointing up and down respectively, calculate

the probability that two neighboring sites will point in the

same direction and the probability that two neighboring

sites will point in opposite directions.

Use this result to calculate an approximation to the

expectation of the Hamiltonian. Note: This is not an ex-

act result, as in reality, sites that are next to each other

will have a tendency to have the same spin because that

will lower their energies, but we have ignored this eﬀect

here.

derive the approximation to the free energy

F=E−T S

=NkBT(1

2+s) log( 1

2+s)

+ ( 1

2−s) log( 1

2−s)

+gµBBzNs −JNzs2/2

where zis the number of neighbors each spin has, and we

have assumed the external ﬁeld Bto be in the ˆzdirec-

tion. (Again we assume the spin is electron spin so that

182 Mean Field Theory

the energy of a spin interacting with the external ﬁeld is

+gµbB·S.)

(d) Extremize this expression with respect to the vari-

able sto obtain the Weiss mean ﬁeld equations.

Below the critical temperature note that there are

three solutions of the mean ﬁeld equations.

By examining the second derivative of Fwith re-

spect to s, show that the s= 0 solution is actually a

maximum of the free energy rather than a minimum.

Sketch F(s) both above and below the critical tem-

perature for B= 0. At non-zero B?

(a) The number of conﬁgurations is

Ω = N

N↑=N!

(N(1/2 + s))!(N(1/2−s))!

giving the entropy (using Stirling’s approximation)

S=kbln Ω = −kbN(1

2+s) log( 1

2+s) + (1

2−s) log( 1

2−s)

(b) The expected energy per bond is −J(1/2)2times the probability of

having like spins on neighboring sites plus J(1/2)2times the probability

of having unlike spins. We thus have

E/bond =−(J/4)[P↑P↑+P↓P↓]+2(J/4)P↑P↓=−(J/4)(P↑−P↓)2=−Js2

If there are N z/2 bonds in the whole system (with zbeing the number

of neighbors of each site) we thus obtain a total energy

E=−JN zs2/2

We must add to this the coupling to the external ﬁeld which is simply

NgµBs. Adding these terms together F=E−T S gives us the desired

result.

(c)

ds =gµbB−sJz +kbTlog( 1

2+s)−log( 1

2−s)

Setting this to zero we obtain

β(−gµbB+sJz) = log( 1

2+s)−log( 1

2−s)

deﬁning the left to be x, we exponentiate both sides to get

ex=1/2 + s

1/2−s

which we rearrange to

s=1

ex−1

ex+ 1 =1

2tanh(x/2) = 1

2tanh(β(−gµbB+sJz)/2)

which is the same self-consistency condition.

183

(d) Look at the second derivative

d2F

ds2=−Jz +kbT1

1/2 + s+1

1/2−s

Now examine this at s= 0, we obtain

d2F

ds2m=0

=−Jz + 4kbT

Note that this changes sign exactly at the critical temperature! Thus

below Tcthe s= 0 solution becomes a maximum of the free energy rather

than a minimum as it is as above Tc. The free energy as a function of s

is plotted in Fig. 22.2.

Fig. 22.2 Free energy (vertical axis) as a function of s(horizontal axis), at, and

below Tc. Left: in zero applied B. Right: In gµbB= 0.02Tc

(22.3) Spin SMean Field Theory

Using the result of Exercise 19.7 use mean ﬁeld the-

ory to calculate the critical temperature for a spin Sfer-

romagnet with a given g-factor g, having coordination

number zand nearest-neighbor exchange coupling Jex.

(It may be useful to re-solve Exercise 19.7 if you don’t

remember how this is done.)

184 Mean Field Theory

From Exercise 19.7, in a ﬁeld Bthe resulting hSiper site (for small

perturbation B) is given by

hSi=S(S+ 1)

3βgµBB

And in mean ﬁeld theory, the eﬀective ﬁeld seen from neighboring spins

gµBBef f =JexzhSi

Thus we have at the crtical point

hSi=S(S+ 1)

3βJexzhSi

which has the solution for the critical temperature

kbTc=S(S+ 1)

3Jexz

(22.4) Low-Temperature Mean Field Theory

Consider the S= 1/2 ferromagnet mean ﬁeld calcula-

tion from Exercise 22.1. At zero temperature, the magnet

is fully polarized.

(a) Calculate the magnetization in the very low temper-

ature limit. Show that the deviation from fully polarized

becomes exponentially small as Tgoes to zero.

(b)* Now consider a spin Sferromagnet. Determine

the magnetization in the low Tlimit. You can express

your result conveniently in terms of the result of Exercise

22.3.

(c)* In fact this exponential behavior is not observed

experimentally! The reason for this has to do with spin-

waves, which are explored in Exercise 20.3, but are not

included in mean ﬁeld theory. Using some results from

that exercise, determine (roughly) the low-temperature

behavior of the magnetization of a ferromagnet.

(a) In absence of external ﬁeld, using the results of 22.1, the self

consistency equation is

Sz=1

2tanh(βJzSz/2)

For large argument xwe have

tanh x=1−e−2x

1 + e−2x≈1−2e−2x

So at low temperature we have

Sz≈1

2−e−βJzSz

Since at low temperature, Szis very close to 1/2, this becomes

Sz≈1

2−e−βJz/2

185

Another way to see this is to note that, if all the neighbors are aligned,

then gµBef f =Jz 1

2. Thus the energy for ﬂipping a given spin in this

eﬀective ﬁeld is Jz/2. So the activated deviation from fully aligned is

e−βJz/2.

(b) Probably the easiest way to do this one is to use the ﬁnal comment

of part (a). If all the spins are aligned to extremize Sz, the eﬀective ﬁeld

is gµBef f =JzS. The low energy excitations excite Szby a single step

at activation energy JzS, thus we obtain

Sz=S−e−βJzS

I have no idea what the comment about expressing the answer in terms

of 22.3!

(in absence of external ﬁeld) is given by

~ω=JSa2|k|2

Each excitation reduces the magnetization by one step. The total num-

ber of spins is V /(a3) and the number of excitations (the number of lost

steps) at ﬁnite Tis

VZdk

(2π)3nB(~ωk)

with nBthe bose factor. Thus, for each spin the reduction in magneti-

zation is

a34π

(2π)3Z∞

dkk21

eβJS2a2k2−1=1

2π2kbT

JS23/2Z∞

dx x2

ex−1

and the ﬁnal integral gives 2ζ(3) ≈2.4.

(22.5) Mean Field Theory for the Antiferromag-

net

For this Exercise we use the molecular ﬁeld (Weiss

mean ﬁeld) approximation for the spin-1/2 antiferromag-

netic model on a three-dimensional cubic lattice. The

full Hamiltonian is exactly that of Eq. 22.1, except that

now we have J < 0, so neighboring spins want to point

in opposite directions (compared to a ferromagnet where

J > 0 and neighboring spins want to point in the same

direction). For simplicity let us assume that the external

ﬁeld points in the ˆzdirection.

At mean ﬁeld level, the ordered ground state of this

Hamiltonian will have alternating spins pointing up and

down respectively. Let us call the sublattices of alternat-

ing sites, sublattice Aand sublattice Brespectively (i.e,

Asites have lattice coordinates (i, j, k) with i+j+kodd

whereas Bsites have lattice coordinates with i+j+k

even).

In mean ﬁeld theory the interaction between neighbor-

ing spins is replaced by an interaction with an average

spin. Let sA=hSziAbe the average value of the spins

on sublattice A, and sB=hSziBbe the average value of

the spins on sublattice B(we assume that these are also

oriented in the ±ˆzdirection).

(a) Write the mean ﬁeld Hamiltonian for a single site

on sublattice Aand the mean ﬁeld Hamiltonian for a sin-

gle site on sublattice B.

(b) Derive the mean-ﬁeld self-consistency equations

sA=1

2tanh(β[JZsB−gµBB]/2)

sB=1

2tanh(β[JZsA−gµBB]/2)

with β= 1/(kBT). Recall that J < 0.

tions to a single self-consistency equation. (Hint: Use

symmetry to simplify! Try plotting sAversus sB.)

186 Mean Field Theory

(d) Assume sA,B are small near the critical point and

expand the self-consistency equations. Derive the critical

temperature Tcbelow which the system is antiferromag-

netic (i.e., sA,B become non-zero).

(e) How does one detect antiferromagnetism experi-

mentally?

(f) In this mean-ﬁeld approximation, the magnetic sus-

ceptibility can be written as

χ=−(N/2)gµ0µBlim

B→0

∂(sA+sB)

∂B

(why the factor of 1/2 out front?).

Derive this susceptibility for T > Tcand write it in

terms of Tc.

Compare your result with the analogous result for

a ferromagnet (Exercise 22). In fact, it was this type of

measurement that ﬁrst suggested the existence of antifer-

romagnets!

(g)* For T < Tcshow that

χ=(N/4)µ0(gµb)2(1 −(2s)2)

kBT+kBTc(1 −(2s)2)

with sthe staggered moment (ie, s(T) = |sA(T)|=

|sB(T)|).

Compare this low Tresult with that of part f.

Give a sketch of the susceptibility at all T.

This follows very much problem 22 (a)

Hi,A =Si·

−JX

jhSjBi+gµbB



Hi,B =Si·

−JX

jhSjAi+gµbB



(b) solving as above gives the desired self-consistency equations.

sA=1

2tanh(βJZsB/2)

sB=1

2tanh(βJZsA/2)

These are solved by

sA=−sB=1

2tanh(βJZsB/2) = −1

2tanh(βJZsA/2)

Recall that J < 0. (Since tanh is an odd function these are the only

possible solutions for J < 0, which one can check graphically by plotting

sAversus sB).

Thus the problem is reduced to

sA=−1

2tanh(βJZsA/2) = 1

2tanh(β|J|ZsA/2)

which is identical to the relation we obtained in the ferromagnetic case.

(d) As in the ferromagnet kbTc=z|J|/4.

(e) This staggered moment is most easily observed with neutron scat-

tering where the scattering will be spin-dependent. Going into the anti-

ferromagnetic phase there is a new unit cell of lattice constant 2a, thus

one sees new diﬀraction peaks at k=n2π/(2a) (with nodd) which are

not present above Tc.

187

(f)The factor of 1/2 is because we have only N/2 A sites or B sites.

Expanding the self consistency equations for high Twe obtain

sA=β[JZsB−gµbB]/4)

sB=β[JZsA−gµbB]/4)

One can solve this system of two equations in two unknowns to obtain

sA=sB=−βgµbB

4−βJz

Thus giving

χ=(1/4)µ0(gµb)2N

kb(T+Tc)

compare to the T−Tcfactor in Eq. 22.4.

(g) This required implicit diﬀerentiation. Diﬀerentiating our two self-

consistency equations gives

∂sA

∂B B=0

=βJZ

∂sB

∂B B=0 −βgµb

4sech2(βJZsB/4)

∂sB

∂B B=0

=βJZ

∂sA

∂B B=0 −βgµb

4sech2(βJZsA/4)

Noting that sA=−sB=sso both sech terms are the same, we add

both equations together to get

∂(sA+sB)

∂B h=0

=βJZ

∂(sA+sB)

∂B B=0 −βgµb

2sech2(βJZs/4)

Making note that sech2+ tanh2= 1, we then have

sech2(βJZs/4) = 1 −tanh2(βJZm/4) = 1 −(2s)2

where we have used the self-consistency equation to replace the tanh by

s. We thus have

∂(sA+sB)

∂B B=0

=βJZ

∂(sA+sB)

∂B B=0 −βgµb

2(1 −(2s)2)

This then can be rearranged into

∂(sA+sB)

∂B B=01−βJZ(1 −(2s)2)/4=−(βgµb/2)(1 −(2s)2)

χ=(N/4)µ0(gµb)2(1 −(2s)2)

kbT+kbTc(1 −(2s)2)

with sthe staggered moment (with Nthe density of spins). Above Tc,

we have s= 0 and we have the same behavior as part f above. Below

Tc, the factor 1 −(2s)2goes quickly down to zero, and the susceptibility

drops rapidly. At Tcthere is a cusp. This behavior is illustrated in the

ﬁgure 22.3

188 Mean Field Theory

Fig. 22.3 Susceptibility of an Antiferromagnet.

(22.6) Correction to Mean Field*

Consider the spin-1/2 Ising ferromagnet on a cubic lat-

tice in ddimensions. When we consider mean ﬁeld theory,

we treat exactly a single spin σiand the z= 2dneigh-

bors on each side will be considered to have an average

spin hσi. The critical temperature you calculate should

be kBTc=Jz/4.

To improve on mean ﬁeld theory, we can instead treat

a block of two connected spins σiand σi′where the neigh-

bors outside of this block are assumed to have the average

spin hσi. Each of the spins in the block has 2d−1 such

averaged neighbors. Use this improved mean ﬁeld the-

ory to write a new equation for the critical temperature

(it will be a transcendental equation). Is this improved

estimate of the critical temperature higher or lower than

that calculated in the more simple mean-ﬁeld model?

We have to consider four conﬁgurations in our partition function:

(↑,↑),(↑,↓),(↓,↑),(↓,↓)

Correspondingly, the partition function is

Z=e−β(Jhσi(z−1)+gµBB)−βJ/4+ 2eβJ/4+eβ(Jhσi(z−1)+gµBB)−βJ/4

From this we calculate the magnetization per site (note that this parti-

tion function represents two sites), and then setting Bto zero we obtain

m=hσi=e−βJ/4sinh(β(Jhσi(z−1))

2e−βJ/4cosh(β(Jhσi(z−1)) + 2eβJ/4

189

Expanding for small hσiwe ﬁnd equality (which ﬁnds Tc) when

kBT=J(z−1)

4e−βJ/4

cosh(βJ/4)

The ﬁnal factor in brackets is always less than unity, so this expression

always gives a lower prediction for Tcthan our prior mean ﬁeld theory.

Magnetism from

Interactions: The Hubbard

Model 23

(23.1) Itinerant Ferromagnetism

(a.i) Review 1: For a three-dimensional tight binding

model on a cubic lattice, calculate the eﬀective mass in

terms of the hopping matrix element tbetween nearest

neighbors and the lattice constant a.

(a.ii) Review 2: Assuming the density nof electrons

in this tight binding band is very low, one can view the

electrons as being free electrons with this eﬀective mass

m∗. For a system of spinless electrons show that the total

energy per unit volume (at zero temperature) is given by

E/V =nEmin +Cn5/3

where Emin is the energy of the bottom of the band.

Calculate the constant C.

(b) Let the density of spin-up electrons be n↑and the

density of spin-down electrons be n↓. We can write these

n↑= (n/2)(1 + α)(23.1)

n↓= (n/2)(1 −α) (23.2)

where the total net magnetization of the system is given

M=−µbnαB.

Using the result of part (a), ﬁxing the total density of

electrons in the system n,

calculate the total energy of the system per unit

volume as a function of α.

Expand your result to fourth order in α.

Show that α= 0 gives the lowest possible energy.

Argue that this remains true to all orders in α

HHubbard =UX

↑Ni

↓

with U≥0 where Ni

σis the number of electrons of spin

σon site i.

Calculate the expectation value of this interaction term

given that the up and down electrons form Fermi seas

with densities n↑and n↓as given by Eqns. 23.1 and 23.2.

Write this energy in terms of α.

(d) Adding together the kinetic energy calculated in

part b with the interaction energy calculated in part c,

determine the value of Ufor which it is favorable for αto

become non-zero.

For values of Unot too much bigger than this value,

calculate the magnetization as a function of U.

Explain why this calculation is only an approxima-

tion.

(e) Consider now a two-dimensional tight binding

model on a square lattice with a Hubbard interaction.

How does this alter the result of part (d)?

(a.i) The tight binding spectrum is

E=E0−2t[cos(kxa) + cos(kya) + cos(kza)]

which we expand to get

E=Emin +ta2|k|2

192 Magnetism from Interactions: The Hubbard Model

where Emin =E0−6tis the bottom of the band. We thus have

ta2|k|2=~2|k|2/(2m∗)

m∗=~2

2ta2

(a.ii) As we have calculated many times (note we are considering spin

polarized electrons here, so there is no factor of 2 out front)

N=V

(2π)3Zkf

dk =V

(2π)3Zkf

4πk2dk =V

(2π)3

4πk3

kf= (6π2n)1/3

The total energy is given by

E−EminN=V

(2π)3Zkf

~2k2

2m∗dk =V

(2π)3Zkf

~2k2

2m∗4πk2dk

(2π)3

2m∗

4πk5

5=V

20π2

m∗k5

Thus we have

E/V =Eminn+Cn5/3

with

C=1

20π2

m∗(6π2)5/3=ta2

10π2(6π2)5/3

(b) Note that nremains ﬁxed. So we have

E/V −Eminn= +Chn5/3

↑+n5/3

↓i

=C(n/2)5/3h(1 + α)5/3+ (1 −α)5/3i

Taylor expanding here, note that the odd terms of the expansion cancel

leaving only even terms

E/V −Eminn= 2C(n/2)5/31 + (1/2!)(5/3)(2/3)α2+

(1/4!)(5/3)(2/3)(−1/3)(−4/3)α4+...

= 2C(n/2)5/31 + (5/9)α2+ (5/243)α4+...

Note that successive terms of the expansion always have positive coeﬃ-

cient.

larly n↑a3and n↓a3are the expected number of spin up and spin down

electrons per site. Thus, the expectation of the hubbard interaction per

unit volume is

Ehubbard/V = (U/a3)(n↑a3)(n↓a3) = Ua3n↑n↓

=Ua3(n/2)2(1 + α)(1 −α) = U a3(n/2)2(1 −α2)

193

(d) We thus obtain a total energy given by

Etotal/V = constant+α2h2C(5/9)(n/2)5/3−Ua3(n/2)2i+α4h2C(n/2)5/3(5/243)i+...

(23.3)

And taking the limit of small αwe see that α= 0 is the solution only

for

U≤2C(5/9)(n/2)−1/3

For Unot too much bigger than this, we can use the quartic form of the

energy given in Eq. 23.3. Minimizing with respect to αgives

α=s−2C(5/9)(n/2)5/3+U a3(n/2)2

4C(n/2)5/3(5/243)

and M=−µbnα.

(e) In the 2d case, the key here is that N∼k2

fand ǫ∼E2

F, so

E/V ∼n2. As a result, we have the kinetic term given by

E=˜

C[(1 + α)2+ (1 −α2)] = 2 ˜

C[1 + α2]

and there is no quartic term. Thus, we have a total energy of the form

Etotal = const + α2(˜

C−˜

KU )

Once Ubecomes large enough that α= 0 is not the lowest energy

solution, then αimmediately goes to its maximum possible value 1.

Thus, the transition is discontinuous going suddenly from unpolarized

spins to fully polarized spins.

(23.2) Antiferromagnetism in the Hubbard

Model

Consider a tight binding model with hopping tand a

strong Hubbard interaction

HHubbard =UX

↑Ni

↓.

(a) If there is one electron per site, if the interaction term

Uis very strong, explain qualitatively why the system

must be an insulator.

(b) On a square lattice, with one electron per site,

and large U, use second-order perturbation theory to de-

termine the energy diﬀerence between the ferromagnetic

state and the antiferromagnetic state. Which one is lower

energy?

(a) If Uis strong enough, there must always be one electron per site.

This makes a traﬃc jam of electrons where no one can move (so long as

there is no doping).

(b) For the antiferromagnet, each spin can make a virtual excursion

to each of the neighboring sites, at an energy cost of U. If the hopping

is tat 2nd order in perturbation theory, this gives a reduction in the

ground state energy, per site, of

−2zt2/U

where zis the number of neighbors. For a ferromagnet, no excursion is

allowed by the Pauli principle, so it is higher energy.

BookSol.dvi Steven H. Simon The Oxford Solid State Basics, Solution Manual University Press (2015)

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