LectureNotes_SolidStateTheory_Kehrein

8/11/2019 LectureNotes_SolidStateTheory_Kehrein

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Introduction to

Solid State Theory

Lecture Notes

Professor Dr. Stefan Kehrein

Insitute for Theoretical PhysicsPhysics Department

University Goettingen, Germany


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Contents

1 Introduction and Motivation 1

1.1 High-energy physics vs. solid state physics . . . . . . . . . . . . . 1

1.2 Reminder: Quantum mechanics of more than one particle . . . . 2

1.3 Fundamental Hamiltonian and Born-Oppenheimer approximation 5

2 Homogeneous Electron Gas 10

2.1 Quantization conditions and ground state . . . . . . . . . . . . . 10

2.2 Thermal properties . . . . . . . . . . . . . . . . . . . . . . . . . . 13

2.3 Homogeneous electron gas in a magnetic field . . . . . . . . . . . 17

3 Crystal Structure 23

3.1 Crystal lattices . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23

3.2 Reciprocal lattice . . . . . . . . . . . . . . . . . . . . . . . . . . . 24

3.3 Classification of Bravais lattices and crystal structures . . . . . . 25

3.4 Scattering by crystals . . . . . . . . . . . . . . . . . . . . . . . . 27

4 Band Structure 36

4.1 Bloch theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . 37

4.2 Nearly free electron gas . . . . . . . . . . . . . . . . . . . . . . . 43

4.3 Tight-binding ansatz . . . . . . . . . . . . . . . . . . . . . . . . . 48

4.4 Real band structures and symmetry properties . . . . . . . . . . 54

5 Lattice Vibrations and Phonons 57

5.1 Classical theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . 57

5.2 Quantum theory of the harmonic crystal . . . . . . . . . . . . . . 65

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Chapter 1

Introduction and

Motivation

This is a 3rd year undergraduate course which only requires a good workingknowledge of quantum mechanics. It does not use or introduce methods ofmany-body theory.

1.1 High-energy physics vs. solid state physics

The essential difference between the goals of high-energy physics and solid statephysics can be viewed in the following manner:

In high-energy physics we know the correct model (the so called standardmodel) to describe nature up to an energy scale of at least O(100Gev).What we are looking for is the correct fundamental model up to the Planckscale O(1019Gev) where we know that the standard model breaks downand gravity needs to be quantized as well (in a way currently unknown).One can also state this by saying that we know an effective low-energytheory (well, ofO(100Gev)) of an unknown grand unified theory that wetry to find using experiments and theoretical considerations.

In solid state physics the fundamental Hamiltonian describing nuclei andelectrons with a Coulomb interaction is perfectly well known. What weare interested in here is to find the emergent behavior of many (meaning at

least of order 1000) constituents at low energies of order room temperatureO(20mev).

Notice that reductionism is not very helpful to understand the types of behaviorthat we are interested in in this course. For example, it does not really help to

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1 Introduction and Motivation

know what a block of iron is made of in order to understand ferromagnetism:Ferromagnetism occurs when many constituents work together.

Also, while knowing the fundamental Hamiltonian of solid sate physics givesus a recipe to solve for the behavior that we are interested in (namely put theHamiltonian for sufficiently many constituents on a computer and evaluate theeigenenergies and eigenfunctions numerically), this is often only of limited usefor understanding the behavior that we are interested in. Due to the exponentialgrowth of the dimension of the Hilbert space with the number of particles, onecan nowadays solve problems with about 20 particles (spin-1/2), but 1000 parti-cles are even in principle out of reach since dim H becomes of order the numberof particles in the universe.

The strategy in solid state theory is therefore to perform suitable systematicapproximations in our fundamental Hamiltonian that render the resulting modelsolvable using analytical or numerical methods. A key role in this strategy

is played by symmetries, which allow us to simplify models without makingapproximations. This strategy determines the structure of this course.

1.2 Reminder: Quantum mechanics of more than

one particle

Let us first look at the case of N = 2 particles. The following fundamentalaxiom of quantum mechanics sets the stage:

The total Hilbert space H of two particles (1 and 2) is the tenor product of theirindividual Hilbert spaces,

H = H1 H2 . (1.1)

For indistinguishable fermions (e.g. electrons) we further need to restrictHto only those states that are antisymmetric under exchange of the particles.Likewise, for indistinguishable bosons (e.g. photons) we need to restrictH toonly those states that are symmetric under exchange of the particles.

So for two distinguishable particles with quantum numbers q(particle 1) and q

(particle 2) the total state is

| = |q |q (1.2)

But if these particles are indistinguishable bosons we have for q=q

| = 12

(|q |q + |q |q) (1.3)

and for q= q

| = |q |q (1.4)

Introduction to Solid State Theory (Kehrein) 2 University of Goettingen


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3. We consider electrons (spin-1/2). Then

x1, s1; x2, s2| = (x1, s1; x2, s2) (1.13)= (x1, x2) (s1, s2) (1.14)

where(x1, x2) is the orbital wavefunction and (s1, s2) is the spin wave-function. From the antisymmetrization postulate we know that the totalwavefunction is antisymmetric under exchange of the particles

(x1, s1; x2, s2) = (x2, s2; x1, s1) (1.15)and therefore we find the following identity

(x1, s1; x2, s2) = 1

2 ((x1, x2) (s1, s2) (x2, x1) (s2, s1))

= 1

4((x1, x2) + (x2, x1)) ((s1, s2) (s2, s1))

+((x1, x2) + (x2, x1)) ((s1, s2) (s2, s1))

(1.16)

This means that the total wavefunction of two electrons can always bewritten as a superposition of

a) symmetric orbital wavefunction, antisymmetric spin wavefunctionand

b) antisymmetric orbital wavefunction, symmetric spin wavefunction.

Notice that this cannot be generalized to N >2.

a)

forn =n : (x1, x2) = 12

(n(x1) n(x2) + n(x1) n(x2))

forn =n : (x1, x2) = n(x1) n(x2) (1.17)with the spin wavefunction

= 1

2(| | | | ) (1.18)

that is total spin S= 0 (spin singlet).

b) (x1, x2) = 1

2(n(x1) n(x2) n(x1) n(x2)) (1.19)

The possible spin wavefunctions are

= | | , | | , 12 (| | + | | ) (1.20)

that is total spin S= 1 and therefore threefold degenerate. Noticethat this state does not exist for n = n, which again just exemplifiesthe Pauli principle. The Pauli principle turns out to be the keyingredient for understanding the structure of matter.



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We now generalize to N >2 particles. The total Hilbert space is

H = H1 . . . HN . (1.21)

For identical bosons the total state vector must be symmetric under exchangeof any two particles, for identical fermions the total state vector must be anti-symmetric under exchange of any two particles.

In the sequel we will only discuss the case of identical fermions since this courseis mainly concerned with the electronic structure. The normalized basis of the1-particle Hilbert space for the electrons is denoted by |, wherecan be somemulti-index (like|n,l,m,s for the bound states of the hydrogen atom). Onecan show that a normalized basis of the N-fermion Hilbert space is given by

|1, . . . , N

def=

1

N! SN(sgn) |(1) |(2) . . . |(N) (1.22)where the sum runs over all elements of the permutation groupSN. The signof the permutation is defined as +1 for an even number of transpositions, and-1 for an odd number of transpositions making up . One can show that themany-particle wavefunction corresponding to these states can be written as adeterminant

(r1, s1; . . . ; rN, sN) = r1, s1; . . . ; rN, sN|1, . . . , N (1.23)

= 1

N!

1 (r1, s1) 1 (r2, s2) . . . 1 (rN, sN)2 (r1, s1) 2 (r2, s2) . . . 2 (rN, sN)...

...

N(r1, s1) N(r2, s2) . . . N(rN, sN)

Here(r, s) = r, s| is the 1-particle wavefunction with respect to the basis.The above determinant is called Slater determinant. From its structure onecan see immediately that the state|1, . . . , N only exists if all the quantumnumbers are different, otherwise two rows in the determinant are identical and itvanishes. This is the general expression of the Pauli principle that two identicalfermions cannot be in the same quantum state, which follows from the generalantisymmetrization postulate.

1.3 Fundamental Hamiltonian and Born-Oppenheimer

approximation

For simplicity we consider a monoatomic solid consisting ofNnnuclei of mass M,nuclear charge Ze, and Ne = Z Nn electrons. The position coordinates of the



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electrons will always be denoted with ri, the position coordinates of the nucleiwithRi.

1

Then we can write down the fundamental Hamiltonian that we need to solve:

H= He+ Hn+ Hen (1.24)

Its various parts are

Electronic contribution:

He = Te+ Vee (1.25)

with the total kinetic energy of the electrons

Te = 2

2m

Nei=1

2ri (1.26)

and the Coulomb interaction energy of the electrons

Vee =1

2

i=j

e2

|ri rj | (1.27)

Contribution from the nuclei:

Hn= Tn+ Vnn (1.28)

with the total kinetic energy of the nuclei

Tn= 2

2m

Nei=1

2Ri (1.29)

and the Coulomb interaction energy of the nuclei

Vnn=1

2

i=j

(Ze)2

| Ri Rj |(1.30)

Coulomb attraction between nuclei and electrons:

Hen=i=j

Ze2

|ri Rj | (1.31)

Essentially there are only two approximations that we have made in writingdown our fundamental Hamiltonian:

1This section closely follows lecture notes by Thomas Pruschke.



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1. We have neglected relativistic effects. This is usually a very good approx-imation because the relevant velocities in a solid are typically at most of

ordercvac/100.

2. Related to this we have neglected all spin effects, especially spin-orbitinteractions. This is not always a good approximation, especially for heavynuclei. However, one can easily build in the spin-orbit interaction into theHamiltonian, but it makes the notation more cumbersome. That is whywe ignore it for the time being.

It is an empirical observation that (usually) in a solid the nuclei move only

slightly around their equilibrium positions R(0)i . We will give a theoretical rea-

son underlying this approximation further below (Born-Oppenheimer approxi-mation). For now we want to plug it into our fundamental Hamiltonian (1.24)because it motivates the outline of this course:

H= E(0)n + Te+ Vee+ V(0)

en + Hph+ Heph (1.32)

1. E(0)n =1

2

i=q

(Ze)2

| R(0)i R(0)j |(1.33)

is the potential energy of the equilibrium configuration of the nuclei, thatis a constant without further importance for the dynamics.

2. V(0)en =i=j

Ze2|ri R(0)j |

(1.34)

is the Coulomb interaction between the electrons and the nuclei at their

equilibrium positions. Notice that this is simply a potential term for theelectrons.

3. Hph= Hn E(0)n (1.35)describes the dynamics of the nuclei around their equilibrium positions,that is phonons.

4. Heph= Hen V(0)en (1.36)describes the interplay of electron and lattice dynamics.

The outline of this course follows the various terms above, taking more andmore of them into account to arrive a more and more realistic picture of a solid:

1. Chapter 2, Homogeneous electron gas: We study only the kinetic energyof the electrons, Te. This is a minimal model for the electrons in a solidthat already introduces important concepts like Fermi surface, etc.

2. Chapter 3, Crystal structure: The crystal structure determinesE(0)n .



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3. Chapter 4, Band structure: We put the above pieces together and inves-tigate the following terms in the fundamental Hamiltonian

H0 = E(0)n + Te+ V

(0)en (1.37)

This is the most importance approximation for understanding solids andexplains a multitude of phenomena. A thorough understanding of (1.37)is the key goal of this course.

4. Chapter 5, Lattice dynamics: Hph is very important for describing e.g.the specific heat of solids.

5. Chapter 6, Transport: In addition toH0 we need to study deviations fromthe perfect crystal structure and scattering processes due to Heph.

6. Chapter 7, Magnetism: We will study magnetic phenomena based on avery simplified treatment ofVee.

7. Chapter 8, Superconductivity: In most materials this is due to electron-phonon scattering, Heph.

8. Chapter 9, Outlook: Quasiparticles and Fermi liquid theory: This gives afirst glimpse at correlations effects due to Vee. A thorough investigationofVee is the content of the Masters course on solid state theory since itrequires many-body techniques.

Next we want to theoretically motivate the above treatment of the nuclei dynam-ics as a (small) perturbation. This is a consequence of the Born-Oppenheimerapproximation, which is also one of the key approximations in quantum chem-

istry. The Born-Oppenheimer approximation starts from the observation thatfor typical nuclei one has mM = O(10

4), the nuclei are much heavier than theelectrons. If we would set M= , thenTn 0 and every eigenfunction of thefull Hamiltonian can be written in the following way:

({r}, { R}) = ({r}) ({ R}) (1.38)

Here ({ R}) is a wave function describing the spatial structure of the nucleiand({r}) is an eigenfunction of

He+

d{ R}

Ven(ri Rj ) ({ R})

({r}) = Ee ({r}) (1.39)

For mM > 0 the ansatz (1.38) becomes an approximation, the so called Born-Oppenheimer approximation. Another way of viewing (1.38) is the time scaleseparation between fast electron dynamics and slow nuclei dynamics: thereforethe electrons move in a quasistatic background provided by the instantaneouspositions of the nuclei, which motivates the ansatz (1.38). The full problem



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then amounts to a self-consistent determination of({r}) and({ R}) such thatthe total energy becomes minimal. However, for the purposes of this course

we will take the position of the nuclei as given and strongly localized at thepositions R

(0)i :

({ R}) = ( R1 R(0)1 ). . . ( RNnR(0)Nn) (1.40)

The remaining eigenvalue problem for the electrons is then effectively parametrized

by R(0)i .

A very rough estimate for the validity of the Born-Oppenheimer approximationfollows from the mass dependence of the kinetic energy in the ground state ofq harmonic oscillator,

T =1

4

with M1/2

(1.41)

Therefore

TnTe

m

M (1.42)

which is typically still of order 102.2

Another frequently employed approximation is to only look at the dynamics ofthe valence electrons, and to take into account the core electrons by workingwith nuclei with a reduced charge and an effective potential due to the screeningfrom the core electrons. This is sometimes a very good approximation, but canbe problematic for electrons in d and f-shells where a clear separation of valenceand core electrons is not possible. We will say more on this topic in the chapteron band structure.

2It should be noted that correlation effects can lead to effective electron masses that aremuch larger than the bare electron mass, for example in heavy fermion systems. In such casesone needs to re-think the validity of the Born-Oppenheimer approximation.



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Chapter 2

Homogeneous Electron Gas

The homogeneous electron gas is a minimal model for describing the electronsin a solid. In (1.32) it only takes into account the kinetic energy of the electrons,that is we are studying the Hamiltonian

H= Te= 2

2m

Nei=1

2ri (2.1)

Additionally, we need to make sure that the electrons in a given volume Vcompensate the positive charge from the nuclei in V. Therefore the electrondensity n = Ne/V is taken as a given parameter in the homogeneous electrongas. Notice that we can equally only consider the valence electrons, thenNe is

just the number of valence electrons inV .

2.1 Quantization conditions and ground state

Assume that we already know the eigenfunctions of

2

2m2r

(r) = (r) (2.2)

Then the Slater determinant

1N!

1 (r1) 1 (r2) . . . 1 (rN)2 (r1) 2 (r2) . . . 2 (rN)...

...N(r1) N(r2) . . . N(rN)

(2.3)

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2 Homogeneous Electron Gas

is an eigenfunction ofH=Te with the eigenenergy

E=Nei=1

i (2.4)

Therefore we can restrict ourselves to the solution of the 1-particle problem(2.2). Such reductions to effective 1-particle problems will play a key role inall future chapters and simplify the solution enormously, i.e. we do not needmany-body techniques. Notice that we have ignored the electron spin degreeof freedom in writing down the Slater determinant since this will just lead to adouble degeneracy of all energy levels.

In order to set the electron density n, we look at the homogeneous electron gasin a cube with length L, V =L3. This leads to a number of possible boundaryconditions:

The wave function (r) vanishes outside the cube. We identify opposite sides of the cube:

(x, y, z) = (x + L, y, z) = (x, y+ L, z) = (x, y, z+ L) (2.5)

These periodic boundary conditions are also called Born-von Karmanboundary conditions.

The differences between these boundary conditions become negligible in thebulk for sufficiently large volume V. For our purposes the Born-von Karmanboundary conditions turn out to be most convenient, therefore we will use them

in the sequel. However, if one is specifically interested in boundary effects likein two dimensional electron gases or topological materials the other boundaryconditions are more adequate (since they are more realistic).

The normalized eigenfunctions of (2.2) are just the plane waves

k(r) = 1

Vei

kr (2.6)

with eigenenergy

k =2k2

2m (2.7)

Notice that the plane waves are also eigenfunctions of the momentum operatori r with eigenvaluep= k (corresponding to velocity v= k/m).The periodic boundary conditions lead to the quantization condition

eikxL =eikyL =eikzL = 1 (2.8)



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and therefore

kx= 2L

nx , ky = 2L

ny , kz = 2L

nz (2.9)

with integersnx, ny, nz Z. These allowedk-values form a lattice with spacing2/L. The number ofk-values in a large volume ink-space is then to a verygood approximation given by

(2/L)3 =

V

(2)3 (2.10)

The ground state for N electrons follows from filling up all one-particle statesstarting from k = 0 since the energy is proportional tok2. The Pauli principleneeds to be respected (otherwise the Slater determinant vanishes), therefore

each discretek-value can be filled with two electrons: one electron with spin upand one electron with spin down. The occupied states in k-space then form asphere with radiuskF(Fermi wavevector). Its volume is= (4/3) k

3Fand the

corresponding number of electrons is

Ne= 2 43

k3FV

83 =

k3F32

V (2.11)

The additional factor 2 counts the spin degeneracy. Hence the electron densityis

n=Ne

V =

k3F32

(2.12)

Here n has to be interpreted as a given fixed (and temperature-independent)number ensuring overall charge neutrality of our solid, which therefore deter-mines the Fermi wavevector kF.

The surface of the Fermi sphere plays a very important role in determiningthe low-energy properties of solid. It is called Fermi surface and separatesthe occupied states from the unoccupied states (in the ground state). Otherimportant quantities are the Fermi momentum pF = kF, the Fermi energyF =

2 k2F/2m, the Fermi velocity vF = pF/m and the Fermi temperatureTF =F/kB . For example for Cu one has:

F = 7.0 eV

TF = 8.16

104K

kF = 13.6 nm

vF = 1.57 106ms



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2.2 Thermal properties

The (internal) energy of the homogeneous electron gas at zero temperature is

U= 2

|k|


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Integrals like (2.19) and (2.20), where the integrand depends only on the energy,appear very often in solid state physics and can be simplified by a change of

variables dk

43F((k)) =

1

2

dk k2 F((k)) (2.21)

=

d () F() (2.22)

where the density of states() comes from the change of measure:

() = k2

21ddk

(2.23)=

m

222m2

(2.24)

One can rewrite this as

() =

32

nF

F

for >0

0 for


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Sincen =

F0

d () one concludes

0 = ( F) F+ 26

(kBT)2 (F) + O(T4) (2.38)

(T) = F 2

6 (kBT)

2 (F)F

(2.39)

= F

1 1

3

kBT

2F

2+ O

T

TF

4 (2.40)

For typical temperatures the chemical potential only deviates very little fromthe Fermi energy.

With this result we can write for the internal energy for given electron density n

u(T) = F0 d () (2.41)

+F

( F) F+

2

6 (kBT)

2 (F)

+2

6 (kBT)

2 F+ O(T4)

From (2.38) we know that the second term on the right hand side vanishes,hence

u(T) = u0+2

6 (kBT)

2 F+ O

T

TF

4 (2.42)

The electronic contribution to the specific heat is therefore always linear intemperature and directly proportional to the density of states at the Fermisurface

cV =

u

T

n

=2

3 k2BT F (2.43)

Specifically for the homogeneous electron gas in d = 3

cV =2

2

kBT

Fn kB (2.44)

In chapter 5 we will find another contribution T3 to the specific heat of acrystalline solid from lattice vibrations

cV = T+ A T3 (2.45)

For low temperaturesT 0 the electronic contribution always dominatescVT

=+ A T2 (2.46)



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is called Sommerfeld coefficient and proportional to the electron density ofstates at the Fermi energy.

2.3 Homogeneous electron gas in a magnetic field

We consider a homogeneous magnetic field in z -direction

B= (0, 0, B0) (2.47)

Remember that a classical particle with chargeqmoves uniformly inz-direction,while its trajectory projected onto the x y-plane is a circle that the particletransverses with the cyclotron frequency

c=|q| B0m c (2.48)

The quantum mechanical analysis for electrons starts from the Hamiltonian

H= 1

2m

p +

e

cA2

g B

S B (2.49)

withB =e/2mcand Lande factor g = 2. We choose a vector potential A(r)in Landau gauge

A(r) = (0, x B0, 0) (2.50)

Since [H, pz] = 0 the momentum in z-direction is still a good quantum number,

pz = kz. However, the other components of def= p + ec

A do not commute

[x, y] = iec

B0 (2.51)

therefore we can only pick one direction as another quantum number. We choosethey-direction and have the following separation ansatz

(r) = ei(kyy+kzz) (x) (s) (2.52)

We plug this into the stationary Schrodinger equation and find

2

2m+m2c

2 (x x0)2

=

E2k2z

2m 1

2 gBB0

(2.53)

= 1 correspond to spin wavefunction = | , | . Here

x0 = kymc

(2.54)



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hence the remaining problem in x-direction is a harmonic oscillator shifted by x0.The eigenenergies are then given by

E(n, kz, ) = c

n +

1 +

2

+2k2z

2m (2.55)

with n= 0, 1, 2, . . .. These discrete levels are called Landau levels. The eigen-functions are given by

(r) = ei(kyy+kzz) n(x x0) (s) (2.56)

where n(x) is the eigenfunction of the harmonic oscillator corresponding toquantum number n. Since the eigenenergies do not depend on ky, the Landaulevels are strongly degenerate. In order to calculate observable quantities, weneed the density of states, and therefore the degeneracyN

Lof the Landau levels.

We take periodic boundary conditions iny-direction leading to

ky = 2

Lyny (2.57)

with ny Z. In x-direction we use hard wall boundary conditions: Since thewavefunctions inx-directions are exponentially localized around

x0 =

mc

2

Lyny (2.58)

we request 0< x0 Lx leading to

0< ny mc2

LxLy = NL (2.59)

The degeneracy of the Landau level (for fixed kz, ) can also be expressed as

NL=

0(2.60)

where = B0 LxLy is the magnetic flux through the surface and the flux quan-tum

0def=

2c

e (2.61)

The degeneracy of the Landau level just counts the number of flux quantathrough the surface. This massive degeneracy also explains the asymmetry ofthe eigenfunctions in x and y-directions: Suitable combinations of our eigen-functions are plane waves in x-direction and harmonic oscillator eigenfunctionsiny -direction.



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In the next step we calculate the density of states

() = 1V

NL

n,kz,

( E(n, kz, )) (2.62)

= NL

V

Lz2

dkzn,

( E(n, kz, )) (2.63)

= mc

42

n,

dkz(kz k0)

E(n, kz, )kz

k0

1(2.64)

where = E(n, k0, ). The final result takes the form

(, B) = 1

82 2m2 3/2

c n,

c n + 1+2

c n + 1+2 (2.65)and displays characteristic square root singularities equally spaced at energies0, c, 2c, 3c, . . .Since we have already seen that the density of states at theFermi energy plays an important role in physical observables, these singularitieslead to characteristic periodic effects as the magnetic field is varied for fixedFermi energy (sincec B0).The observable that we will look at in more detail is the magnetization. Wecan find the operator Mcorresponding to magnetization by noticing that bydefinition the interaction energy of magnetization and external magnetic field is

M B (2.66)

therefore

H

B = M (2.67)

Once we know the ground state energy depending on the external magnetic field

H(B) |GS(B) =EGS(B) |GS(B) (2.68)

we can deduce the expectation value of the magnetization via

EGS(B)

B

=

GS(B)

B |H(B)

|GS(B)

+

GS(B)

|H(B)

|

GS(B)

B +GS(B)|H(B)

B |GS(B) (2.69)

= EGS(B)

BGS(B)|GS(B) GS(B)| M|GS(B)

= GS(B)| M|GS(B) (2.70)



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field. Also notice that for typical magnetic fields (e.g. B = 1T correspond-ing to c

0.1meV) there are a few 10000 Landau cylinders inside the

Fermi sphere.



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Chapter 3

Crystal Structure

3.1 Crystal lattices

The key building block of crystal structures is the Bravais lattice. We give twoequivalent definitions:

1. A Bravais lattice is an infinite set of points in space (or in the planefor two-dimensional lattices) that looks the same seen from each of thesepoints.

2. A Bravais lattice is given by the points

R = n1a1+ n2 a2+ n3 a3 (d= 3) (3.1)R = n1a1+ n2 a2 a3 (d= 2) (3.2)

with ni Z. The so called primitive lattice vectorsai must be linearlyindependent (i.e. not lie in a plane for d = 3) and are said to generate orspan the Bravais lattice.

Remarks:

1. Notice that the choice ofai is not unique.

2. A honeycomb lattice is not a Bravais lattice. One can see easily that itviolates definition 1 above, it does not appear identical from all points.

We will describe this situation below as a lattice with a (nontrivial) basis.

We define the primitive unit cell as an ob ject that covers the whole space withoutoverlaps or voids when tranlsated by vectors of the Bravais lattice. Notice thatthe choice of the primitive unit cell is not unique:

23


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3 Crystal Structure

1. Ifai are the shortest possible primitive lattice vectors, then the pointsr=i xi aiwith xi [0, 1) define the so called conventional unit cell. Theconventional unit cell is very convenient for the graphical representationof lattices, but it does usually not display the full symmetry of the lattice.Also it is not intuitive which lattice points belong to which unit cell.

2. The Wigner-Seitz cell is defined as all the points that are closer to a givenlattice point than to any other point of the lattice. It displays the fullsymmetry of the lattice.

Now we can define a general crystal structure, which is a Bravais lattice witha basis. The crystal structure consists of identical copies of the same physicalobject in the unit cell, which is translated by vectors of the Bravais lattice.

For example the honeycomb lattice is a lattice with a basis consisting of twoatoms (see problem sets). Other example are all crystals that are non monoatomic,since then clearly the unit cell must contain different atoms.

3.2 Reciprocal lattice

The reciprocal lattice plays a fundamental role in the study of crystalline solids.We can think of it as a generalization of the Fourier transform for functionswith the periodicity of the lattice. It will turn out to be the domain on whichthe energy bands of a solid are defined.

The reciprocal lattice of a Bravais lattice is defined as the set of all points Kthat fulfill

ei KR = 1 R Bravais lattice (3.3)

Remarks:

1. Plane waves with wavevectors Kfrom the reciprocal lattice have the pe-riodicity of the Bravais lattice

eiK(r+R) =ei

Kr (3.4)

2. The reciprocal lattice is itself a Bravais lattice as one verifies easily:

ei(K1+ K2)R =ei

K1R ei K2R = 1 (3.5)

Conventionally the term Bravais lattice is used to refer to direct space.

3. The reciproal lattice of a lattice with a nontrivial basis always refers tothe underlying Bravais lattice.



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The primitive lattice vectors of the reciprocal lattice can be constructed explic-itly from the primitive lattice vectors ai in direct space:

b1 = 2 a2 a3

a1 (a2 a3)b2 = 2

a3 a1a1 (a2 a3)

b3 = 2 a1 a2

a1 (a2 a3) (3.6)

One verifies this immediately from bi aj = 2 ij .The Wigner-Seitz cell of the reciprocal lattice is called the first Brillouin zone.Conventionally the term Wigner-Seitz cell refers to direct space, and first Bril-louin zone to reciprocal space. The first Brillouin zone will turn out to be thedomain in which the energy bands of a solid are defined.

For example the reciprocal lattice of a simple cubic lattice with primitive latticevectors

a1 = a ex , a2 = a ey , a3 = a ez (3.7)

is again a simple cubic lattice with primitive lattice vectors

b1 =2

a ex , b2 =

2

a ey , b3 =

2

a ez (3.8)

However, usually the type of the reciprocal lattice is different from the direct

lattice: e.g. the reciprocal lattice of a body-centered cubic lattice is a face-centered cubic lattice, and vice versa.

3.3 Classification of Bravais lattices and crystal

structures

So far we have only looked at the translation symmetry of a lattice. However,there are additional symmetry operations like rotations that will turn out to beof importance for understanding the band structure of solids.

We will first investigate the classification of Bravais lattices, that is we have atrivial basis.

Bravais lattices:

All rigid (= distance conserving) mappings of a Bravais lattice onto itself aredenoted the symmetry group or space groupR of the Bravais lattice. One canshow (nontrivial) that every element ofR can be written in one of the followingways:



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1. Translation by a Bravais lattice vector.

2. A symmetry operation which leaves at least one point of the lattice in-variant: These are i) rotations, ii) reflections on a plane and iii) inversionat a point.

3. Succesive application of 1 and 2.

Symmetry operations of the Bravais lattice, which leave at least one point in-variant, form a group, the so called point group P.Pand the translation groupare subgroups of the space groupR(this is generally only true for lattices witha trivial basis, see below).

One can show (nontrivial) that there are exactly seven different point groups ofBravais lattices, which one calls crystal systems: cubic, tetragonal, orthorombic,monoclinic, triclinic, trigonal, hexagonal. The following operations are possible

elements of the point group:

1. Rotation by an integer multiple of 2n around an axis (so called n-foldrotation axis). One can verify (problem set) that only the values n =2, 3, 4, 6 are possible in Bravais lattices.

2. Rotation-reflection: A rotation by 2n followed by a reflection at a planeperpendicular to the axis of rotation.

3. Rotation-inversion: A rotation by 2n followed by inversion at a point onthe axis of rotation.

4. Reflection at a plane.

5. Inversion at a point.

The largest point group is the one belonging to the cubic crystal system: it iscalled the full octahedral group Oh and has 48 elements. The smallest pointgroup is the one of the triclinic crystal system: it has only two elements, theidentity operation and inversion.

Taking into account the full space group including translations, one can classifya total of 14 different Bravais lattices. In order to do this one needs to explainwhat one means by different or same: Two Bravais lattices are of the samekind if they can be continuously deformed into one another without violatingthe point group symmetry anywhere along the deformation path.

For example two tetragonal lattices with different ratios c/a belong the samekind of Bravais lattice by this definition, whereas a simple cubic and a body

centered cubic lattice are different (although both belong to the cubic crystalsymmetry class).

Crystal structures

We are now ready to look at the classification of Bravais lattices with a nontriv-ial basis, that is general crystal structures. Since the basis need not have the



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3 Crystal Structure

full symmetry of the point group of the underlying Bravais lattice, the overallsymmetry of the space group (rigid symmetry transformations of the crystal

structure) can be reduced in many different ways.1 One finds a total of 230 dif-ferent space groups or crystal structures in this manner. Likewise, there arenow 32 different point groups of a Bravais lattice with a basis, the so calledcrystallographic point groups.

The elements of the space group can be expressed as either

1.{g|a} (so called Wigner notation) corresponding to the transformation

r= g r+ a (3.9)

where g is either rotation, reflection or inversion, and a =

ni ai is thetranslation by a Bravais lattice vector

2. a screw axis: The crystal structure remains invariant by translation througha non-Bravais lattice vector followed by a rotation around the axis of thisvector

3. a glide plane: The crystal structure remains invariant by translationthrough a non-Bravais lattice vector followed by reflection at a plane thatcontains this vector

The space groups where all elements have the structure{g|a} are called sym-morphic space groups. There are a total of 73 symmorphic space groups and wewill usually simplify things by only considering such symmorphic space groups(since then the point group is a subgroup of the space group, which makes iteasier to classify the band structure). For the remaining 197 non-symmorphicspace groups I refer to the advanced literature on this topic: non-symmorphicspace groups also contain screw axes or glide planes.

Notice that the existence of screw axes or glide planes symmetries requiresspecific relations between the dimensions of the basis and the Bravais latticevectors. An important example is the monoatomic hexagonal closed packedstructure (hcp) that contains both a screw axis and a glide plane.

3.4 Scattering by crystals

We consider the problem of wave scattering by a sample. The three by farmost important examples in solid state physics are the scattering of electromag-

netic waves (x-ray scattering), of electrons or neutrons (matter waves). For thetime being we focus on matter wave scattering and briefly review the relevantconcepts from quantum mechanics.

1 Imagine for example a cube with its two halves colored in different ways: this object nolonger displays the symmetry of the full octrahedal group.



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3 Crystal Structure

Potential scattering in d = 3

Our starting point is the stationary Schrodinger equation 2

2m + V(r)

(r) = E (r) (3.10)

We rewrite it as

+ k2 U(r) (r) = 0 (3.11)

whereE= 2k2

2m is the kinetic energy of the incoming particle (we assume thatthe potential falls off sufficiently quickly as r and U(r) = 2m

2 V(r).

As the boundary condition for the incoming wave we use (remember the d =1 case from your quantum mechanics course) a plane wave with wave vectorki, which we take in z-direction. We assume that the potential is localizedsomewhere around the origin. Then asymptotically for large |r| the solutions of(3.11) take the following form

(r)r eikz + f(, )e

ikr

r (3.12)

since the probability current needs to be conserved ( 1r -decay of the scatteredwaves). and are the spherical coordinates ofr, and can therefore also be

thought of as the spherical coordinates of the wave vector kfunder which thescattered particle is detected. The number of particles dn scattered into ddefines the differential cross section

dn= Fi (, ) d (3.13)

whereFi is the flux of incoming particles. One can show easily

(, ) = |f(, )|2 (3.14)

so the problem is reduced to calculating the scattering amplitude f(, ).

The most convenient way to do this is by using an equivalent integral equa-tion formulation of (3.11). LetG(r) be a Greens function of the homogeneousequation (we only need to specify it later)

(

+k

2)G

(r

) =

(r

) (3.15)

and0(r) be a solution of the homogeneous equation

( + k2) 0(r) = 0 (3.16)



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3 Crystal Structure

Then every(r) which fulfills the integral equation

(r) = 0(r) +

d3rG(r r) U(r) (r) (3.17)

is a solution of the stationary Schrodinger equation (3.11). One can verify thiseasily by applying (r + k

2) on both sides of (3.17) and making use of (3.15)and (3.16).

One can iterate the integral equation (3.17)

(r) = 0(r) +

d3rG(r r) U(r) (r) (3.18)

= 0(r) +

d3rG(r r) U(r) 0(r) (3.19)

+ +

d3r d3rG(r r) U(r) G(r r) U(r) (r)= . . .

which immediately sorts the various terms in powers of the (small) scatteringpotentialU(x). The first Born approximation retains only the first power in thescattering potential and yields

(r) = 0(r) +

d3rG(r r) U(r) 0(r) + O(U2) (3.20)

Next we specifically choose

0(r) = eikir (3.21)

that is an incoming plane wave in directionki, and the Greens function

G(r) = 14

eik |r|

|r| (3.22)

where k =|ki| (property (3.15) can be verified easily). The advantage of theintegral formulation now becomes apparent since (3.20) generates an eigensolu-tion with the boundary condition set by 0(r). Inserting everything into (3.20)we arrive at

(r) = eikir 1

4 d3r

eik |rr|

|r

r

| U(r) ei

kir (3.23)

In order to identify the scattering amplitudes we can analyze this expression forlarge distances|r|. In this limit one verifies immediately

|r r| r r r+ O

r

r2

(3.24)



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3 Crystal Structure

and hence

eik |rr

||r r|

r eikrr

eikrr

(3.25)

= eik|r|

r ei

kfr (3.26)

plus terms that decay more quickly for large distances. Here we have introducedthe final scattering wavevector via

kfdef= r |ki| (3.27)

Putting everything together

(r) = eikir 14 eikr

r

d3r U(r) ei(kfki)r

(3.28)

and by comparison with (3.12) we can read off the scattering amplitude

f(, ) == 14

d3r ei(kfki)r

U(r) (3.29)

and the differential cross section

(, ) = m2

424

d3r ei Kr

V(r)2

(3.30)

where K

def

= kf

ki is the momentum transfer of the scattering process. Here

, enter via r = r(, ) from (3.27), i.e. they determine the direction ( kf)of the scattered particle. The important observation is that in the first Bornapproximation the scattering amplitude is simply the Fourier transform of thescattering potential at a wavevector given by the momentum transfer K.

This has immediate consequences when we are considering scattering by a pe-riodic structure. While we do not know exactly the form of the scatteringpotential in a crystal (e.g. for neutron scattering), we can still say that it musthave the translation symmetry of the underlying Bravais lattice

V(r) = V(r+ R) (3.31)

for all Bravais lattice vectors R. The scattering amplitude is therefore

f( K) = m22

d3r ei Kr V(r) (3.32)

= m22

R

xu.c.

d3x ei K(R+x) V( R + x) (3.33)



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3 Crystal Structure

where the sum runs over all lattice vectors R and the integration over the unitcell (u.c.). Because of (3.31) this can be written as

f( K) = m22

R

ei KR

xu.c.d3x ei Kx V(x) (3.34)

Apart from the Fourier transform of the scattering potential within the unit cell(we will say more about this later), the characteristic feature for scattering bya periodic potential is the sum

R

ei KR (3.35)

over all Bravais lattice vectors. Clearly, when the momentum transfer K is a

reciprocal lattice vector, all terms in the sum are 1 because of (3.3). HenceR

ei KR = number of unit cells in sample (3.36)

that is effectively infinite. Likewise, when Kdoes not belong to the reciprocallattice one is summing up numbers distributed over the unit circle that effec-tively average to zero.

We will take a small moment to check this statement in d = 1. From themathematical point of view it is the special case

n=

e2ikn =

m=

(k

m) (3.37)

of the Poisson sum rule which has to be understood as an equality for distri-butions: both sides need to be multiplied by sufficiently nice test function ofkand integrated over k . Notice that (3.37) is equivalent to the above statement:Whenk is a reciprocal lattice vector (here: any integer number m) the sum overthe phases becomes a -function, otherwise it vanishes.

A nice proof of the Poisson sum rule uses the Dirac comb

C(x) =

n=(x n) (3.38)

For convenience we use the following definitions of the Fourier transform

f(k) def

=

dx e2ikx f(x) (3.39)

f(x) =

dk e2ikx f(k) (3.40)



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3 Crystal Structure

One verifies easily that the Dirac comb is identical with its Fourier transform

C(k) =

m=(k m) (3.41)

for example by realizing that C(x) is a periodic function with period 1, that cantherefore be expressed as a Fourier series

C(x) =

m=am e2imx (3.42)

with some coefficients am. Therefore

C(k) =

m=

am (k

m) (3.43)

and applying this procedure another time leads to a2m= 1, and hence am= 1.

If one now looks at the convolution of an arbitrary functionf(x) with the Diraccomb one has

(f C)(x) =

dy C(y) f(x y) =

f(x n) (3.44)

(f C)(k) =

dx e2ikx

f(x n) (3.45)

xx+n=

n=e2ikn f(k) (3.46)

From the convolution theorem we also know

(f C)(k) = f(k) C(k) = f(k)

m=(k m) (3.47)

and comparison of both expressions shows (3.37)

n=

e2ikn =

m=(k m) (3.48)

The structure factor f( K) from (3.34) can therefore only be nonzero if the

momentum transfer K is a reciprocal lattice vector

K = kf ki (3.49)= hb1+ kb2+ lb3 (3.50)

This is called the Laue condition. Two remarks:



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3 Crystal Structure

1. Here we are only considering elastic scattering, i.e.|ki| = |kf|. Especiallyin neutron scattering experiments one makes use of inelastic scattering for

studying the phonon dispersion as will be discussed later in Chapter 5.

2. One might wonder what happens to overall momentum conservation ifK= 0 in a scattering process. The answer lies in our simplified modelingof the crystal as a rigid potential without kinetic energy for the nuclei. Inreality the momentum transfer Kcontributes to the center of mass mo-tion of the crystal (and is therefore completely negligible for macroscopicsamples).

For (3.50) one also writes K = [h k l] with the so called Miller indices [h k l].In order to understand the picture behind Miller indices, we first introduce theconcept of a lattice plane: A lattice plane goes through at least three non-collinear points of the Bravais lattice. A family of lattice planes are lattice

planes with a fixed distance from one another. One can now easily show thatthere is a one to one correspondence between families of lattice planes andreciprocal lattice vectors: these lattice planes are perpendicular to Kand havea separationd = 2/| K|.One can verify this immediately by constructing planes from the condition K r = 2 m, wherem is some integer. Because ofeiKr = 1 for all Bravais latticevectors this construction assigns each Bravais lattice point to one of the planes.The Miller indices [h k l] therefore designate a family of lattice planes.

The Bragg scattering condition starts from the picture of specular reflection fromsuch a family of lattice planes with distance d. Ifis the angle between incomingray and plane, then the path distance between two neighboring reflected rays is2d sin and constructive interference only occurs if

2d sin = n (3.51)

where n is an integer and the wavelength of the scattering wave. (3.51) iscalled the Bragg condition and one can show (tutorial) that Bragg condition(3.51) and Laue condition (3.50) are in fact equivalent.

Neutron scattering:

In the case of neutrons the potential V(x) describes the very short range nuclearforces:

V(x) =

j in ucbj(x xj ) (3.52)

Here the sum runs over all the atoms in a unit cell. bj is the nuclear scatteringlength of the atom located at xj . The structure factor is then

S( K) =

j in uc

bjei Kxj (3.53)



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3 Crystal Structure

x-ray scattering:

Photons are not described by the Schrodinger equation that we have used inthe derivation of the Laue condition. On the other hand, the Bragg derivationemploys the very simplified picture of geometrical optics. It is not trivial toshow that a more microscopic derivation still arrives at the same final result forthe scattering of electromagnetic waves as for matter waves. This was first doneby P. Ewald in the so called dynamic theory of x-ray scattering. In essence oneneeds to model the generation and superposition of waves emitted by oscillatingdipoles, which in turn stem from the electrons driven by electromagnetic radi-ation. The final result is such that one can think of the scattering potential inthe above expression as being proportional to the density of electrons

V(x)

j in uc

Zjgj (x xj ) (3.54)

Zj is the nuclear charge of ion j and gj is a short-range function that modelsits electron cloud. However, for the wave length of x-rays this function is notwell described by a -function and the structure factor

S( K) =

j in uc

Zjfj ( K) ei Kxj (3.55)

contains the Fourier transform ofgj .

Notice that the structure factor (3.53) or (3.55) can vanish for certain direc-

tions K for a monoatomic basis that is commensurate with the Bravais lattice.One calls thesesystematic absencessince it implies that certain Bragg peaks aremissing.

The following reformulation of the Laue condition is useful for analyzing scat-tering experiments: For elastic scattering one has k = k, therefore

k = |k K| (3.56) k2 = k2 2k K+ K2 (3.57)

k K = 12

K (3.58)

This expresses the possibility for elastic scattering solely through the incomingwave vector. The geometric meaning of (3.58) is that the tip of the incoming

wave vector kmust be located on a Bragg plane, that is a plane perpendicular toa given reciprocal lattice vector Kand bisecting it. Bragg planes are not dense,therefore in general the scattering condition is not fulfilled and experiments

make use of the following procedures to ensure a nonzero result:

1. Laue method: One uses a single crystal with a fixed orientation, but theincoming x-rays are not monochromatic. Therefore one can fulfill thescattering condition (3.58) for some k in the spectrum and finds Braggpeaks.



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Chapter 4

Band Structure

In the fundamental Hamiltonian (1.32) of solid-state physics we now considerthe following parts

H0 = E(0)n + Te+ V

(0)en (4.1)

We do not seriously address the question where the crystal structure underlying

E(0)n andV

(0)en originates,1 but use it as an input for our subsequent investigation.

ClearlyH0therefore has the symmetry {g|a} of the underlying crystal structure.H0 is the minimum model for making any material specific statements in solidstate physics and therefore of fundamental importance. Also, it serves as astarting point for investigating all other neglected terms using perturbation

theory, and its properties must therefore be very well understood. One propertythat makes H0 very appealing is that it can be thought of as a sum of one-electron Hamiltonians, just like in the homogeneous electron gas. Hence it willbe sufficient to solve a one-particle problem and we can again find the many-particle ground sate by constructing the Slater determinant like in Chapter 2.

In writing down (4.1) we have made the following three approximations:

1. We have neglected all perturbations of the periodic crystal structure (whichalways occur in real materials) and surfaces.

2. We have neglected the lattice dynamics.

3. We have neglected the electron-electron interaction.

We will say more about these approximations in subsequent chapters. Especiallyfor transport properties it will be essential to lift the approximations 1 and 2. 2

1This topic will be dealt with in the Masters course on Solid-State Theory.2Otherwise transport will be either infinite or zero as will be discussed in Chapter 6.

36


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4 Band Structure

Putting everything together we have shown

TR (r) = (r+ R) (4.16)!

= c( R) (r) = eikR (r) (4.17)

Defining

u(r)def= ei

kr (r) (4.18)

we have therefore proven the Bloch theorem since u(r) has the periodicity ofthe Bravais lattice

u(r+ R) = eik(r+R) (r+ R) = ei

k(r+R) eikR (r) (4.19)

= eikr (r) (4.20)

= u(r) (4.21)

Before looking at the constructive proof of the Bloch theorem we will addressthe question which wave vectors k are allowed in a large crystal. Like for thehomogeneous electron gas we will use Born-von Karman boundary conditions.In order to do this we consider a large volume Vwhich is commensurable withthe primitive unit cell, i.e. we take N1 unit cells in directiona1,N2 unit cells indirectiona2, etc. The total number of unit cells in V isN=N1N2N3 and theBorn-von Karman boundary condition implies

(r+ Ni ai) = (r) (4.22)

fori = 1, 2, 3. From the Bloch theorem we now know

n,k(r+ Ni ai) = eiNikai n,k(r) (4.23)

!= n,k(r) (4.24)

Hence fork= x1b1+ x2b2+ x3b3 we find

e2iNixi = 1 (4.25)

and therefore xi = mi

Ni, mi Z. The allowed values ofk (we will call these the

Born-von Karman set) are then

k=

3

i=1miNi

bi , mi Z (4.26)

For later considerations we work out the volume kassociated with each allowedvalue ofk: this is simply

k=b1N1

b2N2

b3N3

=

1

Nb1 (b2 b3) (4.27)



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4 Band Structure

U (r) = K

UKei Kr

q

cqeiqr

=K,q

UKcq Keiqr (4.35)

where one can verify easily that q = q+ K is also an element of the Born-von Karman set. Comparing the coefficients in the Schrodinger equation weconclude

2

2mq2 E

cq+

K

UKcq K = 0 (4.36)

for all qin the Born-von Karman set. Now one can always write

q= k K (4.37)

where klies in the first Brillouin zone and Kis a reciprocal lattice vector. Thenwe arrive at the key result

2

2m(k K)2 E

ck K+

K

UK Kck K = 0 (4.38)

which must hold for all reciprocal lattice vectors K. For any givenk in the firstBrillouin zone one observes that (4.38) only couples coefficients ck, ck K, . . .that differ by a reciprocal lattice vector. Hence the initial Schr odinger equationdecouples into N independent problems, one for each Born-von Karman valuein the first Brillouin zone, and the solutions labelled by k can be written as

k(r) =

K

ck Kei(k K)r (4.39)

= eikr uk(r) (4.40)

where we have defined

u(r)def=

K

ck Kei Kr (4.41)

Clearlyu(r +R) = u(r) due to the basic properties of reciprocal lattice vectors.This completes our second proof of the Bloch theorem.

Remarks:



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4 Band Structure

1. The good quantum number kis calledcrystal momentum. Proof b demon-strated thatk can be restricted to the first Brillouin zone. Therefore the

first Brillouin zone emerges as the definition range of the electronic bandstructure in a crystal, which makes it one of the most important conceptsin solid state theory. Notice that k is not the physical momentum, butfor the time being should be thought of as a good quantum number ina periodic potential. We will say more about the physical meaning ofthe crystal momentum when we study the dynamics of Bloch electrons inchapter 6.

2. The role of the discrete quantum number n (band index) becomes clearwhen we insert the Bloch form (4.3) of the eigenfunctions into the Schrodingerequation. This leads to

Hkuk(r) = Ekuk(r) (4.42)

with periodic boundary conditions uk(r+R) = uk(r) and

Hk = 2

2m(i + k)2 + U(r) (4.43)

This constitutes a hermitean eigenvalue problem in the finite volume C(primitive unit cell). Therefore we expect a discrete spectrum labelled bya discrete quantum number n leading to

En,k = En(k) (4.44)

Upon variation ofk in the first Brillouin zone the energy En(

k) will varycontinuously for fixed band indexn. This yields theelectronic band struc-

ture.

3. If one builds up a localized electron from eigenfunctions in bandn withwavevectors aroundk, its group velocity is given by

vn(k) = 1

kEn(k) (4.45)

according to quantum mechanics. It is quite remarkable that one canconstruct such solutions with constant nonzero group velocity in spite ofthe periodic scattering in the crystal. Notice that (4.45) also shows that fora general dispersion relation (

= free electrons) there is no simple relation

between the crystal momentum and the physical momentum mvn(k).



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4 Band Structure

This yields a consistent expansion in powers ofUas long as the denominator isnot degenerate

2

2m

(k K)2 (k K0)2 |U| (4.55)and one shows easily

ck K0 = 1 + O(U2) (4.56)

K= K0 ck K = O(U) (4.57)

E = 2

2m(k K0)2 + O(U2) (4.58)

Therefore in the non-degenerate case (4.55) eigenenergies and eigenfunctionsare only slightly changed from the free electron case. Notice that according to

(4.54) the periodic perturbation only couples coefficients with the same goodquantum numberk. Hence a breakdown of non-degenerate perturbation theoryonly occurs if there exists K1= K0 with5

(k K1)2 = (k K0)2 (4.59)

With the wave vector q= k K0 of the free electron we can rewrite this as(q K)2 =q2 (4.60)

where K= K1 K0 is a nonzero reciprocal lattice vector. The condition (4.60)is identical to the Laue condition for scattering of a periodic potential (3.58).We conclude that non-degenerate perturbation theory breaks down (howeverweak the potential) when the wave vector qof the free electron lies on (near) a

Bragg plane

q=1

2K+ K (4.61)

This is intuitively obvious from the Laue picture: For incoming wavevectorson a Bragg plane the periodic potential has a particularly strong effect due tointerference.

We now assume that qlies on exactly one Bragg plane.6 This means that thereare exactly two degenerate states with the same quantum numberk and (4.48)yields

E 2

2m

(k

K0)2 ck K0 = UK1 K0 ck K1

E 2

2m(k K1)2

ck K1 = UK0 K1 ck K0 (4.62)

5The same observations hold for near degeneracy.6The situation of multiple Bragg planes can be discussed analogously, one just needs to

take more coefficients into account.



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4 Band Structure

where we have usedU0 = 0, plus terms in higher order inU. A nonzero solutionof this linear system of equations exists only if

det

E 22m (k K0)2 UK1 K0

UK1 K0 E

2

2m (k K1)2

= 0 (4.63)

This yields the two solutions

E= 2

2m(k K0)2 |UK1 K0 | (4.64)

where we have made explicit use of the degeneracy condition.

We observe that the periodic potential splits the dispersion relation in order Uif the wave vector of the free electron lies on a Bragg plane, which therefore

leads to band gaps in the dispersion relation (forbidden energy regions in thefirst Brillouin zone). These band gaps will turn out to be the key ingredient forinsulating behavior.

From (4.62) we can now also find the structure of the Bloch eigenfunctions atto these band gaps:

ck K0 = sgn(UK1 K0 ) ck K1 (4.65)

E.g. for UK1 K0 >0 (the other sign just leads to an exchange of + and -) onefinds:

E = 2

2m(k K0)2 UK1 K0 (4.66)

(r) = eikr

ei K1r ei K0r

(4.67)

|i(r)|2 sin2

1

2( K1 K0) r

(4.68)

E+ = 2

2m(k K0)2 + UK1 K0 (4.69)

+(r) = eikr ei

K1r + eiK0r (4.70)

|i(r)|2 cos2

1

2( K1 K0) r

(4.71)

One observes the characteristic change of the Bloch eigenfunctions at a bandgap: A node at the lattice points turns into a maximum, and vice versa.



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4 Band Structure

We now consider the many-electron problem in the periodic potential. Justas in the case of the homogeneous electron gas, a certain electron density n is

fixed and determined by overall charge neutrality. The zero temperature groundstate is therefore given by a Slater determinant consisting of the 1-particle Blocheigenfunctions |n, k, , where =, is the spin: all such 1-particle states up toan energyFcontribute,En(k) F. The Fermi energy F itself is determinedby the electron density n.

We will later see (Chapter 6) that a Fermi energy F in a band implies metallicbehavior. Likewise, F in a band gap means that one has an insulator.

7 Aninsulator with a band gap of less than approximately 2eV is also called a (intrin-sic) semiconductor since thermal excitations into the unoccupied band becomepossible.

The Fermi surface is defined as the surface of constant energy with En(k) = F.In later chapters it will turn out to play a key role for describing the low-energy

behavior of materials. It can have a very complicated topology as we will see atthe end of this section. Notice that by definition a Fermi surface only exists inmetals.

A first important property that we want to look at is the electron contributionto the specific heat. The calculation runs along identical lines as for the homo-geneous electron gas and one finds that the Sommerfeld coefficient is given by(2.45)

cV =2

3 k2BT F (4.72)

with the total density of states Fat the Fermi energy: In general the Fermisurface can lie in various bands and we need to sum over these contributions

F =

n

n(F) (4.73)

with the contribution of band n to the density of states defined as (compare(2.26)

n() =

1.BZ

dk

43( En(k)) (4.74)

The integration runs over the first Brillouin zone.

Now one can learn a lot about n() by just looking at the dispersion relation.

One deduces this by first defining Sn() as the surface with En(k) = in thefirst Brillouin zone. The the volume between Sn() andSn( + d) is just() d

() d=

Sn()

dS

43k(k) (4.75)

7Strictly speaking, in an insulatorFis defined as the zero temperature limit of the chemicalpotential in the grand canonical ensemble. With this definition Fturns out to lie in the middleof band gaps.



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4 Band Structure

wherek(k) is the length of a vector perpendicular to Sn(En(k)) that connects

Sn() andSn( + d). Since

En(k) is normal to the surface, we can also write

+ d = + |En(k)| k(k) (4.76) k(k) = d

|En(k)|(4.77)

and we have the result

n() =

Sn()

dS

431

|En(k)|(4.78)

Clearly, an insulator has no electronic contribution to the specific heat linearin temperature (in fact it is exponentially suppressed). If En(k) = 0 onespeaks of a van Hove-singularity that yields a particularly large contribution to

the density of states. In three dimensions a van Hove-singularity is integrableand (F) remains finite. However, the higher order terms of the Sommerfeldexpansion need to be reconsidered since they now contain (singular) derivativesof(). In one or two dimensions a van Hove-singularity at the Fermi energyleads to a diverging Fand a different low temperature behavior of the specificheat (see tutorials).

Structure of Fermi surface:

We now investigate the structure of the Fermi surface in a nearly free electrongas. One constructs it by first drawing the simple Fermi surface (sphere) of thefree electron gas with the same density of electronsn. If this Fermi surface liescompletely in the first Brillouin zone of the periodic potential, we are nearly donewith our construction: The only additional effect is the characteristic lowering

of the band energies at points closer to the boundary of the first Brillouin zonethat we have derived in perturbation theory. Hence the Fermi surface of the freeelectron gas at points that are close to the boundary moves even closer to theboundary. Since the total volume enclosed by the Fermi surface has to remainconstant this implies that the Fermi surface at points that are further from theboundary moves even further from the boundary.

If the Fermi surface of the free electron gas lies (at least partly) outside the firstBrillouin zone the topology can become quite involved. Whenever it touches aBragg plane we have the shifts/opening of gaps from degenerate perturbationtheory. This e.g. often leads to characteristic neck-like structures of the Fermisurface. In general the Fermi surface of the free electron gas will touch multipleBragg planes. In order to study the emerging structure we first define theconcept of a general Brillouin zone:

1. The first Brillouin zone is the set of all points in k-space that one canreach from the origin without crossing a Bragg plane.

2. The second Brillouin zone is the set of all points that one can reach fromthe first Brillouin zone by crossing exactly one Bragg plane.



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4 Band Structure

3. Then + 1 Brillouin zone is the set of all points not contained in then 1Brillouin zone that one can reach from the n Brillouin zone by crossing

exactly one Bragg plane.

One can show easily that all these Brillouin zones have the same volume (of thefirst Brillouin zone).

In the second step of our construction we now deform the free Fermi surface inthe characteristic manner prescribed by perturbation theory whenever it touchesa Bragg plane. Therefore the different Brillouin zones naturally correspond tothe different bands n.

In the last step of the construction one moves the contribution of the Fermisurface in then Brillouin zone back into the first Brillouin zone by shifting withsuitable reciprocal lattice vectors.8 This gives bandn in the first Brillouin zone.Notice that in general the Fermi surface will consist of various bands, which can

make the topology very complicated.

4.3 Tight-binding ansatz

The tight-binding ansatz starts from the opposite limit than the nearly freeelectron gas: In zeroth order one considers isolated atoms, whose overlap leadsto corrections in a suitable perturbative expansion around this limit.

We first need to know the spectrum of the atomic Hamiltonian9

Hat = 2

2m + Uat(r) (4.79)

Hat n(r) = En n(r) (4.80)

The full crystal Hamiltonian is

H= 2

2m + U(r) (4.81)

with the potential generated by a sum over all Bravais lattice vectors

U(r) =

R

Uat(r R) (4.82)

If we simply plug the atomic eigenfunctions into the full Hamiltonian we find

H n(r) = Hat n(r) + U(r) n(r) (4.83)

8One also often sees the so called repeated zone scheme where one draws the periodicallyrepated picture in reciprocal space.

9More generally, this would be the Hamiltonian in a primitive unit cell.



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4 Band Structure

where

U(r) def=R=0

Uat(r R) (4.84)

can be thought of as the perturbation of the atomic potential in the unit cellat the origin by the atomic potentials generated in the other unit cells. If theatomic wavefunctions are sufficiently localized, they will mainly contribute inthe unit cell at the origin, where U(r) is small. In zeroth order one thereforesimply neglects the product U(r) n(r) and finds

H n(r) = En n(r) (4.85)

or likewise for all Bravais lattice vectors

H n(r R) = En n(r R) (4.86)

So we have foundNdegenerate eigenfunctions in zeroth order. Both for makingcontact with the nearly free electron gas and for looking at corrections to thisresult, it will be useful to bring these N eigenfunctions into Bloch form. Wedefine for allk from the Born-von Karman set in the first Brillouin zone

nk(r)def=

R

eikR n(r R) (4.87)

This has Bloch form because

n

k

(r+ R) = R

eikR n(r+ R

R) (4.88)

= eikR

R

eik(RR) n(r ( R R)) (4.89)

= eikR nk(r) (4.90)

In zeroth order of the tight-binding approximation one therefore simply finds aflat band

En(k) = En (4.91)

of N degenerate Bloch eigenstates with an eigenenergy given by the atomic

orbital.Next we want to take the influence ofU(r) into account in leading order. Forthe eigenfunctions of the full Hamiltonian we make the following ansatz

k(r) =

R

eikR (r R) (4.92)



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4 Band Structure

The same derivation (4.90) holds to verify that this ansatz is of Bloch form. Still,one can wonder if all Bloch eigenfunctions can be written in the form (4.92).

The answer to this question is affirmative as will be seen in the subsection onWannier functions at the end of this section. So the only approximation comesfrom making the additional ansatz

(r) =

n

bn n(r) (4.93)

where we only want to take a small number of atomic orbitals into account.In particular, one neglects scattering states in the decomposition (4.93) whichwould be necessary to have a complete basis. Because of (4.93) the tight-bindingansatz is also called LCAO method (linear combination of atomic orbitals).

Plugging this into the stationary Schrodinger equation with the full Hamiltonian(4.81) yields

Hk(r) =R,n

eikR bn

2

2m + Uat(r R) + U(r R)

n(r R)

=R,n

eikR bn En n(r R) (4.94)

+R,n

eikR bn U(r R) n(r R) (4.95)

!= E(k) k(r) (4.96)

= E(k) R,n

eikR bn n(r R) (4.97)

One can rewrite this asR,n

(E(k)En) eikR bn n(r R) =R,n

eikR bn U(r R) n(r R)(4.98)

We multiply this equation bym(r) and integrate over rusing the orthogonalityof the atomic orbitals

dr m(r) n(r) = mn (4.99)

which leads to

(E(k) Em) bm =

R=0,n(E(k) En) eikR bn

dr m(r) n(r R)

+

n

bn

dr m(r) U(r) n(r)



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4 Band Structure

+ R=0,n

eikR bn

dr m(r) U(r R) n(r R)

(4.100)

where we have always separated the sums into terms R = 0 and R= 0. Nowall the terms on the right hand side of this equation are small: The first andthe third term because the atomic orbitals are localized in a unit cell, and thesecond term becauseU(r) is small in the unit cell at the origin. Hence we can

conclude that|(E(k) Em) bm| is small for all m. The only possible solutionis that the eigenenergy E(k) lies close to the energy E0 of some atomic orbital,and that bm is only then not small when Em E0. In the first nontrivialorder of the LCAO method one therefore takes only the terms with En E0on the right hand side of (4.100) into account, since otherwise a small term bnis multiplied by another small integral.

Ifm is an s-orbital there are no degeneracies and (4.100) reduces to anexplicit equation for the energy of the s-band. We will study this situationexplicitly below.

For a p-orbital we have threefold degeneracy. Hence (4.100) is a linearsystem of three equations, which yields E(k) for the three p-bands.

For a d-orbital we likewise have five bands, etc.

Another important approximation is to restrict the sums over R to sums overnearest neighbors (n.n.) since the overlap of the atomic orbitals decays rapidlyfor larger separation.

For an s-band (4.100) becomes

E(k) Es= (E(k) Es)R=0

eikR ( R)

R=0

eikR ( R) (4.101)

where we have defined

( R) =

dr s (r) s(r R) (4.102)

=

dr U(r) |s(r)|2 (4.103)

( R) =

dr s (r) U(r

R) s(r

R) (4.104)

This gives

E(k) = Es +

R=0 eikR ( R)

1 +

R=0 eikR ( R)

(4.105)



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4 Band Structure

The wave function for an s-orbital is real, also s(r) = s(r) and because ofinversion symmetry likewiseU(r) = U(

r). This yields( R) = (

R) and

thereforeR=0

eikR ( R)

n.n.

( R) cos(k R) (4.106)

where we have restricted the sum to nearest neighbors. Since ( R) in thedenominator of (4.105) only gives a small correction, we have the followingresult for the energy of an s-band in the first nontrivial order of the tight-bindingapproximation

E(k) = Es n.n.

( R) cos(k R) (4.107)

One makes the characteristic observation that the band width is set by theoverlap integral (4.104).

We now look explicitly at an s-band in an fcc-lattice. a is the size of the unitcell. One sees immediately that the origin has 12 nearest neighbors with

R=a

2(1, 1, 0),a

2(1, 0, 1),a

2(0, 1, 1) (4.108)

which leads to 12 possible values ofk R

k R= a2

(ki kj) (4.109)

with i, j = x, y, i, j = x, z or i, j = y, z. AlsoU(r) has the full symmetry of

the cubic lattice and therefore ( R) = for all R from (4.108). Then (4.107)becomes

Es(k) = Es

2 cos(a

2(kx+ ky)) + 2 cos(

a

2(kx ky)) + . . .

= Es

4

cos

akx

2

cos

aky

2

+ cos

aky

2

cos

akz

2

+cos

akx

2

cos

akz

2

(4.110)

with the overlap integral

=

dr s (x, y, z) U(x a/2, y a/2, z a/2) s(x a/2, y a/2, z a/2)

(4.111)



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4 Band Structure

Wannier functions:

We will prove the following theorem: Every Bloch function for every band canbe written in the form

nk(r) =

R

eikR n(r R) (4.112)

n(r) is called the Wannier function for band n. Notice that we have implicitlyalready used this theorem in writing down the ansatz (4.92).

The proof proceeds by considering nk(r) as a function ofk for fixed r. Since

nk(r) is periodic as a function ofk in the reciprocal lattice, we know that it

can be expanded in a Fourier series in the reciprocal reciprocal lattice, which isof course just the direct lattice:

nk(r) =

R

fn(r,R) ei Rk (4.113)

where the Fourier coefficients are given by

fn(r, R) = 1

C

dk ei Rk nk(r) (4.114)

HereC is the volume of the first Brillouin zone. Also, the Fourier coefficientsdo not really depend both on r and R since for every Bravais lattice vector R0

fn(r+ R0,R + R0) = 1

C d

k ei(R+R0)k nk(r+R0) (4.115)

= 1C

dk ei(R+R0)k eikR0 nk(r) (4.116)

= fn(r, R) (4.117)

where we have explicitly used the Bloch form in the last step. Therefore

n(r R) def= fn(r,R) (4.118)

and we have proven the above statement.

Remarks:

1. For narrow bands the Wannier functions are very similar to the atomic

orbitals.

2. For wider bands the Wannier functions are still somehow localized. Theyare therefore a convenient starting point for investigations where localproperties are important, like in semiclassical transport theory or for dop-ing in semiconductors. We will say more about this in later chapters.



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4 Band Structure

Hence

Pg

(k )f(r)= (gk) Pgf(r) (4.125)and therefore putting everything together:

PgHk = HgkPg (4.126)

We now need to distinguish two cases:

k remains invariant under a certain point group operation, gk =k. Be-cause of

[Pg, Hk] = 0 (4.127)

Pguk is again an eigenfunction (but not necessarily a new one) with thesame eigenenergy and crystal momentum.

For gk=k we conclude that Pguk is an eigenfunction ofHgk, thereforewe have degeneracies at different points of the Brillouin zone, En(k) =

En(gk).

For any given k one defines the star of k as the set of all vectors that aregenerated by acting with the point group on k. For a generic value ofk its starwill contain as many elements as there elements in the point group. However,whenk lies on a symmetry line the number of elements in the star is reduced.The subgroup of

Pwhich leavesk invariant is called the small group ofk. The

extreme case is the -point (k= 0) in the Brillouin zone where the small groupis obviously always identical with the full point group of the crystal structure.10

While the case gk= k leads to a somehow trivial (expected) symmetry, theelements of the small group allow nontrivial insights into degeneracies for givenvalues of the crystal momentum. In order to see this we first distinguish be-tween normal and accidental degeneracies: A normal degeneracy occurs whenall the degenerate eigenvectors can be connected by symmetry operations. Bydefinition this implies that one has an irreducible representation of the respec-tive group. The situation with additional accidental degeneracies that cannotbe explained in this way will not concern us in the sequel.

Therefore all possible (non-accidental) degeneracies of energy bands are related

to irreducible representations of the respective small group ofk. By looking up

the possible dimensions of such irreducible representations11 one can therefore

10A similar situation can occur at edges of the Brillouin zone since momenta gk which onlydiffer by reciprocal lattice vectors can be identified. Therefore the symmetries/degeneraciesin a band structure generically increase as one approaches the border of the Brillouin zone.

11For example one can consult M. Tinkham, Group Theory and Quantum Mechanics, Ap-pendix B.



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4 Band Structure

say something about possible degeneracies along symmetry lines of the Bril-louin zone. If one works harder one can even say something about how the

degeneracies split if one moves away from symmetry points.

Optical properties:

From the electronic band structure one can deduce some basic optical propertiesof crystals. The most important process for the absorption of a photon is theexcitation of an electron from an occupied to an unoccupied state. Thereforeinsulators with a band gap larger than 3.2eV appear transparent, only lightwith UV frequencies or higher could possibly be absorbed by this process (thereare other weaker absorption processes like 2-photon processes, photon-phononscattering, etc.). Examples for such insulators are diamond, quartz and AlO.

Semiconductors will strongly absorb photons with an energy larger than thedirect bandgap, but less below. Hence they often appear colored. An exampleis cadmium sulfid with a bandgap of 2.6ev, which therefore absorbs violet andblue, and appears red.

The optical properties of metals are more complicated due to their large conduc-tivity, which leads to reemission processes by the conduction electrons (thereforethey often appear shiny). This will be a topic for the Masters course in solidstate theory.



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Chapter 5

Lattice Vibrations and

Phonons

In the fundamental Hamiltonian (1.32) of solid state theory

H = E(0)n + Te+ V(0)

en

+Hph+ Heph+Vee (5.1)

we have so far only considered the first line. Specifically, we have ignored thedynamics of the lattice ions Hphin the second line and simply froze them at the

points specified by the Bravais lattice. This leads to some severe shortcomings:

1. We cannot explain the observed specific heat for temperature exceeding afew K.

2. WithoutHephwe cannot explain the temperature dependent conductiv-ity of metals.

3. Conventional superconductivity relies onHeph.

In this chapter we now set out to include the dynamics of the lattice ions.

5.1 Classical theory

We first consider a classical theory, that is we think of the lattice ions as classicalparticles. In the next section we will then quantize the theory, based on theresults of the classical treatment in this section.

We make two basic assumptions about the lattice dynamics:

57


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5 Lattice Vibrations and Phonons

1. The average position of each lattice ions is still the (before) frozen Bravaislattice (plus possibly basis) vector.

2. We only consider small deviations from equilibrium (in a sense to be mademore precise later).

With these assumptions we make the following ansatz for the position of ion R:

r( R) = R + u( R) (5.2)

whereu( R) is the deviation from equilibrium. The interaction potential of theions will only depend on the distance between two ions and can be expanded inthe small parameter u( R)

U = 1

2R, R

(r( R)

r( R)) (5.3)

= N

2

R

( R)

+1

2

R, R

[u( R) u( R)] ( R R)

+1

4

R, R

[u( R) u( R)]

2( R R)

+O(u3) (5.4)

The first line on the rhs is just the constant contributionE(0)n in the fundamental

Hamiltonian which plays no role in the dynamics. The second line vanishes sinceby definition Rare the equilibrium positions: Therefore the force on a given ionR exerted by all the other ions vanishes

R

( R R) (5.5)

For the so called harmonic crystal we only take into account the quadratic termin the third line of (5.4), especially we neglect all terms in O(u3) and higher.Therefore we have the Hamilton function

H= R

P( R)2

2M +

1

2 R, R,

u( R) D( R R) u( R) (5.6)

with

D( R R) = R, RR

2(r)

r rnu

RR

2(r)

r rnu

RR

(5.7)



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5 Lattic

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