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Condensed Matter Theory

A.Komnik, Institut fur Theoretische Physik, Universitat Heidelberg

Contents

1 Homogeneous Fermi and Bose systems 21.1 Drude theory of metals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21.2 Sommerfeld’s improvements of the Drude theory . . . . . . . . . . . . . . . 61.3 Hartree and Hartree-Fock approximations . . . . . . . . . . . . . . . . . . . 101.4 Screening . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 141.5 Basic principles of Fermi liquid theory . . . . . . . . . . . . . . . . . . . . . 201.6 Tomonaga-Luttinger liquid (TLL) . . . . . . . . . . . . . . . . . . . . . . . 241.7 Conventional superconductivity . . . . . . . . . . . . . . . . . . . . . . . . . 291.8 Degenerate Bose gases and Bose-Einstein condensation (BEC) . . . . . . . . 41

2 Fermions and Bosons on the lattice and in external potentials 492.1 Periodic structures . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 492.2 Bloch theorem and its consequences . . . . . . . . . . . . . . . . . . . . . . 502.3 Electron energy bands in crystals . . . . . . . . . . . . . . . . . . . . . . . . 532.4 Band structure and transport properties . . . . . . . . . . . . . . . . . . . . 582.5 Dynamical properties of the lattice . . . . . . . . . . . . . . . . . . . . . . . 602.6 Phonons . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 662.7 Electron-phonon interaction . . . . . . . . . . . . . . . . . . . . . . . . . . . 692.8 Solids in external fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 752.9 Quantum Hall effect . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 802.10 Magnetic properties of solids . . . . . . . . . . . . . . . . . . . . . . . . . . 88

Suggested reading

Introduction:

• N.W.Ashcroft and N.D.Mermin, Solid State Physics, Saunders College Publishing,1976

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1 Homogeneous Fermi and Bose systems

1.1 Drude theory of metals

1900 Paul Drude presented a theory of metals which was largely inspired by the kinetictheory of gases. The most important postulates of this theory are the following ones:

• electrons are free and scatter on ion impurities and on each other

• they do not interact between the collisions

• collisions are instantaneous

• the probability to be scattered in unit time is proportional to 1/τ , where τ is aphenomenological constant - aka mean free time, relaxation time

• electrons are in thermal equilibrium due to collisions only: the velocity after a collisionis only determined by the local temperature, its direction is random

DC conductivity

Let’s now attempt to compute some measurable quantities. The simplest one is the gener-alisation of the Ohm’s law:

E = ρ j , (1)

where E is the electric field, j is the current density and ρ is the resistivity of the sample1.We can derive this result using the conventional formula for the current density of particlesmoving with uniform velocity v at concentration n:

j = −nev , (2)

the (positive) factor e tells one that we’re dealing with the electric current. A good ap-proximation for v is the average velocity for the electrons found in the following way. Lett be the time since the last collision then:

v(t) = v0 − eEt

me. (3)

We use that 〈v0〉 = 0 (see the last postulate) and that 〈t〉 = τ . Then

v ≈ −eEτ

me, (4)

and thus

j = σ0 E , σ0 =ne2τ

me, (5)

where σ0 is the electrical conductivity. While n is known with a high degree of precision∼ 1022 cm−3 for alkali metals, τ is purely phenomenological. From the knowledge of σ0

an estimation gives τ ∼ 10−14 − 10−15 sec. The mean free path is found from mev20/2 ≈

(3/2)kBT ⇒ lm ≈ v0τ ≈ 0.1 − 1 nm. This is a pretty realistic lengthscale of the order ofthe lattice constant. But our estimation of σ0 is as good as our τ . And in fact both of themappear to be temperature-dependent. In order to better access the quality of the theorywe need quantities, which do not contain τ .

1In this form the formula is independent of the sample geometry. The more conventional form would beV = IR, where V = EL is the voltage drop over the length L, I is the total current and R = ρL/A is theresistance of the sample with cross-sectional area A. Of course one recovers (1).

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Hall effect and magnetoresistance

1879 Edwin Hall was testing whether the Ampere’s force was acting on the current (or itscarriers) or directly on the conductor. As a result he found a transversal voltage drop in ametallic sampled placed in crossed electric and magnetic fields. In order to find this voltagedrop we first look how the momentum of an electron changes in presence of an externalforce f(t):

p(t + dt) = (1− dt/τ)︸ ︷︷ ︸probability of not being scattered

× [p(t) + f(t)dt + O(dt2)]︸ ︷︷ ︸change in momentum

(6)

Thus in the leading order we obtain

dpdt

= −pτ

+ f(t) , (7)

which in the case of the Lorentz force f = −(e/c)v ×H− eE is modified to

dpdt

= −e

(E +

1mec

p×H)− p

τ. (8)

In the steady state we have dp/dt = 0, therefore we obtain two equations

0 = −eEx − ωcpy − px/τ , 0 = −eEy + ωcpx − py/τ , with ωc = eH/(mec) (9)

being the cycloctron frequency . Now we replace the momentum by the relation (2) andobtain

σ0Ex = ωcτjy + jx , σ0Ey = −ωcτjx + jy . (10)

In the absence of transversal current jy = 0

Ey = −ωcτ

σ0jx = − H

necjx . (11)

One usually introduces the Hall coefficient

RH =Ey

jxH= − 1

nec, (12)

which is exactly the quantity we’re looking for – it’s τ -independent. However, a comparisonwith the experimental data still reveals two major problems:

• RH seems to be T - and H-dependent

• RH can even become negative (especially for polyvalent metals such as Al), whichhints that the current carriers here can become positively charged

Nonetheless we proceed and take a look on the

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AC conductivity

Let’s suppose our system is subject to a time-dependent E-field of the form

E(t) = Re[E(ω)e−iωt

], we assume then p(t) = Re

[p(ω)e−iωt

](13)

to be the solution of our fundamental equation of motion (7), having the form

−iωp(ω) = −p(ω)τ

− eE(ω) , then j(ω) = σ(ω)E(ω) , with σ(ω) =σ0

1− iωτ. (14)

Here we assumed that

• there’s no Lorentz force: it’s usually by the factor c/v weaker

• λ of the oscillation À than the mean free path lm so that the force is nearly uniform

Is there any propagation of EM waves possible? Let’s use the Maxwell equation

∇×E = −1c

∂H∂t

, ∇×H =4π

cj +

1c

∂E∂t

. (15)

∇× (∇×E) = ∇ (divE)︸ ︷︷ ︸=4πρ=0

−∇2E =iω

c∇×H =

iω

c

(4πσ

c− iω

c

)E . (16)

That’s why we obtain a wave equation

−∇2E =ω2

c2ε(ω)E with ε(ω) = 1 + i

4πσ

ω. (17)

For high frequencies ωτ À 1 in (14) one obtains

ε(ω) = 1− ω2p

ω2with the plasma frequency ω2

p =4πne2

me. (18)

• ε < 0 for ω < ωp only exp. decaying solutions ⇒ no wave propagation possible (⇒the sample is opaque)

• ε > 0 for ω > ωp wave propagation possible ⇒ the sample gets transparent (alkaliindeed become transparent in the UV range)

Are there any charge oscillations without external fields possible? Use the continuityequation

∇ · j = −∂ρ

∂t⇒ ∇ · j(ω) = iωρ(ω) . (19)

On the other hand from the Gauss law

∇ ·E(ω) = 4πρ(ω) ⇒ iωρ(ω) = 4πσ(ω) ρ(ω) . (20)

It is only solvable for

1 + i4πσ(ω)

ω= 0 ⇒ ω = ωp . (21)

So ωp is the frequency of the charge oscillation eigenmode. This excitation is called plasmon.

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Thermal conductivity

Recall the Fourier’s law: the thermal current density is proportional to the temperaturegradient

jQ = −κ∇T , (22)

where κ is the thermal conductivity. Most metals show good agreement with the phe-nomenological Wiedemann-Franz law (1853)

κ

σ

1T

= Lorenz number = universal constant ≈ 2.2 · 10−8 WΩ/K2 (23)

We can calculate κ employing a simplified 1D model. Let E(T (x)) be the energy perelectron which suffered its last collision at point x, depending on the local temperature only.Thermal current induced by the particles coming from the left/right of x is proportional to±(nv/2)E(T (x∓ vτ)), thus2

jQ =nv

2[E(T (x− vτ))− E(T (x + vτ))] . (24)

Since vτ = lm is small we can expand for small lm and obtain

jQ = nv2τdEdT

(−dT

dx

). (25)

Generalising to 3D we use 〈v2x,y,z〉 = v2/3 and

ndEdT

=N

V

dEdT

=1V

dE

dT= cV , (26)

where E = NE is the total energy of the system. Therefore

jQ =13v2τcV (−∇T ) with κ =

13v2τcV =

13lmvcV . (27)

Now we use the kinetic theory: cV = (3/2)nkB and mev2/2 = (3/2)kBT , hence for the

Lorenz number we obtain

κ

σT=

32

(kB

e

)2

≈ 1.11 · 10−8 WΩ/K2 , (28)

which is after a correction by a factor 2 almost the experimental value. Although long timebeing considered to be a major success of the Drude theory this result comes about as amore or less lucky compensation of two wrong values: (i) in reality v is by a factor of 10larger; (ii) cV is in fact by a factor of 100 smaller. This is much better revealed in the

Seebeck effect

By application of temperature gradient one induces a voltage difference (electric field) acrossthe sample3:

E = Q∇T , (29)2Factor 1/2 is due to the equal numbers of right/left moving particles.3The opposite situation of temperature gradient induced by a finite voltage is the Peltier effect.

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Figure 2: Effective confinement potential for the electrons in a metal. The potential usedin the jellium model is depicted in red.

where Q is the thermopower . Let’s find it using the velocity balance. The average velocityat x due to the T gradient (we again start with the 1D model) is:

vQ =12

[v(x− vτ)− v(x + vτ)] = −τvdv

dx= −τ

d

dx

(v2

2

). (30)

In 3D it modifies to

vQ = −τ

6dv2

dT∇T . (31)

Average velocity vE due to finite field is given by (4). Using the balance condition vQ+vE =0 we immediately obtain

Q = − 13e

d

dT

mev2

2= − cV

3ne. (32)

This is again τ -independent. However, no cancellations take place and Q is overestimatedby a factor of 100 in comparison to the experimental values due to wrong cV . Moreover,there’re materials with even a different sign of Q. So one definitely needs a better theory...

1.2 Sommerfeld’s improvements of the Drude theory

Sommerfeld decided to use the (quantum) Fermi-Dirac statistics instead of the (classical)Maxwell-Boltzmann one. But before we treat electrons as quantum particles let’s take a lookonto their QM states within a sample of metal. Bringing many atoms very close togetherdelocalises the electrons occupying the upper energy levels of individual atoms (shown inblack, see Fig. 2). The delocalised electrons occupy then the energy levels of the effectiveconfinement potential (green). It is fairly complex but can in the first approximation bemodelled by a potential box with uniform potential (red), which describes the confiningproperties of the ion potential. Of course, this works in 3D as well and is known as thejellium model . Let’s now assume that our sample has a cubic form with volume V = L3.Then the Schrodinger equation for single-electron energy levels is

− ~2

2me∇2ψ(r) = Eψ(r) . (33)

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For L → ∞ the bulk properties of the metal should not depend on what happens on thesurface so we concentrate ourselves on the simplest possible periodic boundary conditions(PBC)

ψ(x + L, y, z) = ψ(x, y + L, z) = ψ(x, y, z + L) = ψ(x, y, z) . (34)

(33) is then solved by a plane wave

ψk(r) =1√V

eik·r with E(k) =~2k2

2me. (35)

ψk is an eigenfunction of the momentum operator, that’s why we identify

p = ~k then E = mev2/2 with v = p/me . (36)

PBC require ks to be quantised

kx,y,z =2πnx,y,z

L. (37)

Later we shall need a number of states in unit volume of k-space:

dN = dnxdnydnz =(

L

2π

)3

dkxdkydkz =V

8π3dkxdkydkz , (38)

which is the density of states (DOS) in the k-space.According to Pauli principle such single-particle states are filled one-by-one starting

from the lowest possible energy E = 0. For large N the area which is thereby filled in thek-space is nearly a sphere with a radius kF . Thus the total number of electrons in thesample is given by

N = 2s4π

3k3

F

V

8π3⇒ n =

N

V=

k3F

3π2, (39)

where 2s takes care of the spin degree of freedom. kF is the Fermi wave vector , its associatedmomentum pF and energy EF are Fermi momentum and Fermi energy , respectively. Fermivelocity vF = pF /me plays now the role of v from the previous Section.

Now we’re in a position to calculate the net energy of a filled Fermi sphere (we assumethe temperature to be zero):

E = 2s

∑

|k|<kF

E(k) =~2

me

∑

k<kF

k2 =V

8π3

~2

me

∫ kF

0d3k k2

∣∣∣V→∞

=1π2

~2k5F

10meV , (40)

where we took advantage of the DOS derived above and d3k = 4πk2dk. The energy perparticle is found using the result (39):

E =E

N=

310~2k2

F

me=

35EF =

35kBTF , (41)

where TF = EF /kB is the effective (Fermi) temperature. Surprisingly TF ∼ 104K (ÀTroom). That is the reason why the actual v in the Fermi sea is by a factor of 10 largerthen the one predicted by the kinetic theory. The above relations can also be used to findthe equation of state for the Fermi gas,

p = −(

∂E

∂V

)

N

=23

E

V. (42)

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At finite temperatures the filling of states in k-space occurs according to the Fermi-Diracdistribution

nF (E) =(e(E−µ)/kBT + 1

)−1. (43)

It can be interpreted as the probability to fill the k-state corresponding to the energy E .Then for the total number of particles within the Fermi sea

N = 2s

∑

k

nF (E(k)) =V

4π3

∫d3knF (E(k)) . (44)

At T = 0, of course, one immediately reproduces (39) when µ = EF . For the calculationat arbitrary T it is more convenient to rewrite it as an integration over energy

N = V

∫ ∞

0dk

k2

π2nF (E(k)) = V

∫ ∞

0dE DF (E) nF (E) , (45)

where we introduce the fermionic DOS in the energy space4

DF (E) =me

(π~)2

√2meE~2

=3n

2EF

( EEF

)1/2

. (46)

General structure of the above integral is

I =∫ ∞

0dε g(ε) nF (ε) =

∫ ∞

0dε

g(ε)e(ε−µ)/T + 1

. (47)

First we substitute ε− µ = Tz, where z is a new dimensionless integration variable,

I =∫ ∞

−µ/Tdz T

g(µ + Tz)ez + 1

= T

∫ µ/T

0dz

g(µ− Tz)e−z + 1

+ T

∫ ∞

0dz

g(µ + Tz)ez + 1

. (48)

In the first integral we use1

e−z + 1= 1− 1

ez + 1then (49)

I =∫ µ

0dεg(z)− T

∫ µ/T

0dz

g(µ− Tz)ez + 1

+ T

∫ ∞

0dz

g(µ + Tz)ez + 1

. (50)

In the second integral we can safely assume the upper boundary to be infinity, since thecorrections are exponentially small and expand for small Tz (in comparison to µ)

I =∫ µ

0dεg(z) + T

∫ ∞

0dz

g(µ + Tz)− g(µ− Tz)ez + 1

(51)

≈∫ µ

0dεg(z) + 2T 2g′(µ)

∫ ∞

0dz

z

ez + 1+

T 4

3g′′′(µ)

∫ ∞

0dz

z3

ez + 1. (52)

The remaining integrals are given by5

∫ ∞

0dz

z

ez + 1=

π2

12,

∫ ∞

0dz

z3

ez + 1=

7π4

120. (54)

4To simplify the calculation we’ve set kB = 1.5The calculation for the first one is performed in the following way:∫ ∞

0

dzz

ez + 1=

∫ ∞

0

dz ze−z∞∑

n=0

(−1)ne−zn =

∞∑n=0

(−1)n

[ze−z(n+1)

−(n + 1)

∣∣∣∞

0+

1

n + 1

∫ ∞

0

dz e−z(n+1)

]

=

∞∑n=0

(−1)n 1

(1 + n)2=

1

2ζ(2) = π2/12 . (53)

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Thus for the Sommerfeld expansion we obtain

I =∫ µ

0dε g(ε) +

π2

6T 2g′(µ) +

7π4

360T 4g′′′(µ) + . . . . (55)

Going back to (45) we then obtain

n = N/V = n(µ/EF )3/2 +π2T 2

8n

E3/2F

õ

+ . . . (56)

At T = 0 we immediately recognise the result µ = EF . The solution up to the terms of theorder T 2 (see Problem 2)

µ = EF − π2

6T 2

[∂

∂E ln DF (E)]

E=EF

+ . . . (57)

Let’s now calculate the specific heat. By analogy to (45) the total energy of the system

E = 2s

∑

k

E(k) nF (E(k)) = V

∫ ∞

0dεDF (ε)εnF (ε)

= V

∫ µ

0dε ε DF (ε) +

π2

6T 2

[∂

∂ε(εDF (ε))

]

ε=µ

+ . . .

. (58)

Now we have to derive with respect to temperature. In the first term only µ is temperature-dependent. In the second term we don’t have to derive the square bracket because it wouldproduce terms of higher order in T , thus:

cV =1V

∂E

∂T= µDF (µ)

dµ

dT+

π2

3T

[DF (ε) + ε

∂DF (ε)∂ε

]

ε=µ

+ . . . (59)

Now we group the first and the third terms together:

cV =π2

3TDF (µ) + µDF (µ)

[dµ

dT+

π2

3T

1DF (µ)

∂DF (ε)∂ε

]

ε=µ

+ . . . (60)

Now we observe that due to (57) the term in square brackets is identically zero, thereforethe dominant term in the low-temperature expansion of cV is given by

cV =π2

3TDF (µ) = n

π2

2T

EF. (61)

We observe here that cV ∼ DF (EF ), so that it really is dominated by the DOS at theFermi edge only! Its linear behaviour in T can also be understood qualitatively. At finitetemperature the Fermi distribution is washed out and particles in the shaded area areexcited above the Fermi edge. The number of such electrons is proportional to T whereasthe energies they acquire are also ∼ T , see Fig. 3. That’s why their total energy is ∼ T 2

leading to the ∼ T behaviour of the specific heat.The two most important consequences of (61) are

• cV is linear in T (this is in accordance with the specific heat of metals at low temper-atures)

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Figure 3: Fermi distribution function at zero and finite temperature.

• cV is suppressed by a (quite small) factor T/EF , hence the ratio between the classicalDrude and Sommerfeld results is

cV,SommerfeldcV,Drude

=π2

3T

EF(62)

This has a number of consequences for the quantities we’ve calculated within the Drudemodel.

Thermal conductivity

We still can use (27). While cV is smaller than the classical one by a factor of T/EF , v2 isnot ∼ 2T/me but to v2

F = 2EF /me ⇒ it is larger by a factor of EF /T . Since τ does notchange, κ changes only slightly (but actually to a better value). So Wiedemann-Franz lawremains almost unchanged.

Thermopower

We still can use (32) combining it with (61), then

Q = −π2

6T

eEF, (63)

which is by a factor of 100 smaller than the Drude prediction and is almost ok in comparisonto the experimental data.

Hall coefficient, AC/DC conductivities

do not change since they do not depend on the distribution functions.So the difficulties remain and maybe we’ve got to refine our approach to include electron-

electron interactions.

1.3 Hartree and Hartree-Fock approximations

Thus far we have considered the single-particle Schrodinger equation:[− ~2

2me∇2 + U(r)

]ψ(r) = Eψ(r) . (64)

It turns out that within this picture we are not able to take into account electronic corre-lations whatever the potential U . We have to construct and solve an equation for a morecomplex wave function depending on coordinates and spin variables of all electrons

ψ(r) → Ψ(r1, σ1; r2, σ2; . . . rN , σN ) . (65)

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Then the corresponding equation would look like

HΨ =N∑

i=1

− ~2

2me∇2

i Ψ− Ze2∑

R

1|ri −R|Ψ

︸ ︷︷ ︸ion scattering

+12

∑

i6=j

e2

|ri − rj |Ψ︸ ︷︷ ︸

electron-electron scattering

= EΨ , (66)

where R denotes the positions of the ions. This is a very complex equation and one mightask the question whether it is possible by any choice of U to reduce it to the one of the kind(64). It should of course include the ion-electron interaction as well as electron-electroninteraction via

Uion(r) = −Ze2∑

R

1|r−R| , Uel(r) = −e

∫dr′ ρ(r′)

1|r− r′| , (67)

where ρ(r′) is the charge density distribution due to all the other electrons in the system.It is reasonable to assume it to be the sum of all densities due to individual electrons,

ρ(r) =∑

i

ρi(r) = −e∑

i

|ψi(r)|2 . (68)

Let’s now assume U = Uion +Uel to be the new potential which we can use in (64) in orderto solve it for the wave function of the ith electron:

− ~2

2me∇2ψi(r) + Uion(r)ψi(r) +

e2

∑

j

∫dr′

|ψj(r)|2|r− r′|

︸ ︷︷ ︸Uel

ψi(r) = Ei ψi(r) . (69)

This set of equations is called Hartree equations and it is usually solved in the followingprocedure: (i) start with some simple ψi(r), e.g. with plane waves; (ii) calculate Uion inthe brackets; (iii) solve again (69) for a new set of ψi(r); (iv) if the change to them is notsmall enough repeat from (ii), if it is then the problem is solved.

Although being a quite plausible and very physical scheme the Hartree procedure failsto account for such an elementary thing as the Pauli principle. In order to see that we recallthat the above Schrodinger equation is equivalent to some variational problem, which assertsthat a solution to HΨ = EΨ is given by any state Ψ that makes stationary the quantity

〈H〉Ψ =(Ψ,HΨ)(Ψ,Ψ)

, (70)

where the scalar product is defined in the following way

(Ψ,Φ) =∑

σ1,...σN

∫dr1 · · · drN Ψ∗(r1, σ1; . . . rN , σN )Φ(r1, σ1; . . . rN , σN ) . (71)

It can be shown that the Hartree equations above are the result of the variation with respectto the wave function of the form

Ψ(r1, σ1; . . . rN , σN ) = ψ1(r1, σ1) ψ2(r2, σ2) . . . ψN (rN , σN ) . (72)

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This function does not comply with the Pauli principle requiring it to change sign uponexchange of any two of its arguments:

Ψ(r1, σ1; . . . ri, σi; . . . rj , σj . . . rN , σN ) = −Ψ(r1, σ1; . . . rj , σj ; . . . ri, σi . . . rN , σN ) . (73)

This property has the Slater determinant

Ψ(r1, σ1; . . . rN , σN ) =

∣∣∣∣∣∣∣∣

ψ1(r1, σ1) ψ1(r2, σ2) . . . ψ1(rN , σN )ψ2(r1, σ1) ψ2(r2, σ2) . . . ψ2(rN , σN )

. . . . . . . . . . . .ψN (r1, σ1) ψN (r2, σ2) . . . ψN (rN , σN )

∣∣∣∣∣∣∣∣. (74)

So the next logical step would be the substitution of this equation in (70). Then we obtain

〈H〉Ψ =∑

i

∫d3rψ∗i (r)

[− ~2

2me∇2 + Uion(r)

]ψi(r) +

12

∑

i,j

∫d3rd3r′

e2

|r− r′| |ψi(r)|2|ψj(r)|2

− 12

∑

i,j

∫d3rd3r′

e2

|r− r′|δσiσjψ∗i (r)ψi(r′)ψ∗j (r

′)ψj(r) . (75)

Observe the different signs in front of the interaction terms. After the variation with respectto ψ∗i (r) as above we obtain the following equation,

[− ~2

2me∇2 +

[Uion(r) + Uel(r)

]]ψi(r)−

∑

j

∫d3r′

e2

|r− r′|δσiσjψ∗j (r

′)ψi(r′)ψj(r)

= Ei ψi(r) . (76)

Here Uel is the Hartree term or direct contribution, known from (69). The second termwhich is absent in our first attempt is the exchange term. This Ansatz is known under thename Hartree-Fock approximation. In general these equations are as difficult to solve asthose of the Hartree approximation. However, there’s an exactly solvable case: the jelliummodel. Here we start with the plane waves (35) with an additional prefactor χi for the spinpart, which we immediately discard (a reason will be given later). For convenience we setV = 1. Next we observe that in the completely uniform system Uion + Uel = 0 identicallybecause of the charge neutrality of the system. Then for the exchange correction to theground state energy we obtain

−∑

j

∫d3r′

e2

|r− r′| e−ikjr

′+ikir′+ikjr = −

∑

j

∫d3(r′ − r)

e2

|r− r′| ei(kj−ki)(r−r′) eikir︸︷︷︸

ψi(r)

= δE(ki)ψi(r) . (77)

Let’s now calculate this additional contribution to the energy setting q = kj − ki (for thechoice of polar coordinates see Fig. 4)

δE(ki) = −∑

j

∫d3r

e2

|r|eiqr = −e2

∑

j

∫2πr2drd cos θ

reiqr cos θ

= −2πe2∑

j

∫dr

eiqr − e−iqr

iq= −

∑

j

4πe2

q2. (78)

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Figure 4:

Figure 5:

In the last integration we implemented an exponential cut-off eiqr → eiqr−δr and have takenδ → 0 after the calculation. The last summation over j is performed in the following way(how the coordinates are chosen is shown in Fig. 4)

δE(k) = −∑

k′<kF

4πe2

|k− k′| = −4πe2

8π3

∫

k′<kF

dk′∫ 1

−1

2πk′2d(cos θ)k2 + k′2 − 2kk′ cos θ

=︸︷︷︸k′/kF =x′,k/kF =x,cos θ=y

−4πe2

8π3kF

∫ 1

0dx′2πx′2

∫ 1

−1

dy

x2 + x′2 − 2xx′y

= −e2kF

π

∫ 1

0dx′

x′2

−2xx′ln

(x2 + x′2 − 2xx′

x2 + x′2 + 2xx′

)

= −e2kF

πx

∫ 1

0dx′

[x′ ln(x− x′)− x′ ln(x + x′)

]=

2e2

πkFF(k/kF ) ,

where F(x) =12

+1− x2

4xln

∣∣∣∣1 + x

1− x

∣∣∣∣ . (79)

So that the energy per particle is now given by (see plots Fig. 5)

E(k) =~2k2

2me− 2e2

πkFF(k/kF ) . (80)

The full energy of the Fermi sphere is therefore given by

E = 2s

∑

k<kF

~2k2

2me− e2kF

π

∑

k<kF

[1 +

k2F − k2

2kkFln

∣∣∣∣kF + k

kF − k

∣∣∣∣]

= N

(35EF − 3

4e2kF

π

), (81)

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where the first term is the one from (41). Let’s now measure the energies in RydbergsRd= e2/(2a0) ≈ 13.6eV, where a0 = ~2/(mee

2) is the Bohr’s radius, then

E/N =e2

2a0

[35(kF a0)2 − 3

2π(kF a0)

]≈

(2.21

(rs/a0)2− 0.916

(rs/a0)

)Rd , (82)

where rs is the radius of a sphere, volume of which is equal to the volume per electron.Since in typical metals rs ≈ 2 − 6a0, the second term is of almost the same order as thefirst one! Hartree-Fock is essentially a high density approximation because it gets betterfor smaller rs

6.The above approximation possesses one serious problem. The electron velocity in the

vicinity of the Fermi edge ∼ ∂E/∂k|k→kFis log-divergent. This results in a quite divergent

behaviour of the specific heatcV (low T ) ∼ T/ ln T

towards the low temperatures. This can be traced back to the singular behaviour of the‘naked’ Coulomb potential. In order to understand how we can overcome this difficulty weneed to understand

1.4 Screening

Let’s assume we introduce a test charge ρext(r) into our homogeneous electronic system.Then according to the Poisson equation it produces some potential φext(r), which satisfies

−∇2φext(r) = 4πρext(r) . (83)

On the other hand, there’re also physical charge density as well as potential connected byan identical relation

−∇2φ(r) = 4πρ(r) , (84)

where ρ = ρext + ρind with an induced charge density ρind. The potentials are connectedto each other by a linear relation

φext(r) =∫

d3r′ ε(r, r′)φ(r′) =∫

d3r′ ε(r− r′) φ(r′) , (85)

since our system is translation invariant. Now we go over to the Fourier components definedby

ε(q) =∫

d3r e−iqr ε(r) , ε(r) =∫

d3q(2π)3

eiqr ε(q) , (86)

when we obtain

φ(q) =1

ε(q)φext(q) . (87)

As long as the external charge is very small the induced charge would be in lowest orderapproximation proportional to the full potential,

ρind(q) = χ(q) φ(q) . (88)

6In fact, it is also possible to compute the terms of higher orders aka correlation energy . They are givenby

0.0622 ln(rs/a0)− 0.096 + . . .

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Performing the FT of (84) and (83) and subtracting them from each other we obtain

q2

4π[φ(q)− φext(q)] = ρind(q) = χ(q)φ(q) ⇒ φ(q) =

φext(q)1− (4π/q2)χ(q)

. (89)

Comparing that with (87) we obtain the relation

ε(q) = 1− 4π

q2χ(q) = 1− 4π

q2

ρind(q)φ(q)

. (90)

In order to make progress we now apply the Hartree approximation in some special cases.

Thomas-Fermi approximation (static screening)

Here we solve the single-particle Schrodinger equation with the full φ(r) being the potential

− ~2

2me∇2 ψi(r)− eφ(r) ψi(r) = Ei ψi(r) . (91)

According to Hartree approach we then need the single-particle solutions of this equationto construct the particle density (68) finally using the Poisson equation (83) to computethe new potential. Thomas-Fermi method simplifies this assuming that φ(r) depends onr only weakly so that at any point in the sample one can define a ‘local’ single-particleenergy7

E(k) =~2k2

2me− eφ(r) . (92)

Then the coordinate-dependent particle density is simply given by

n(r) =∫

d3k4π3

exp

[1T

(~2k2

2me− eφ(r)− µ

)]+ 1

−1

. (93)

Obviously, the induced charge is simply −en(r) + en0, where n0 is the above equationcalculated in the absence of any potential. That’s why we obtain

ρind(r) = −e [n0(µ + eφ(r))− n0(µ)] = −e2 ∂n0

∂µφ(r) . (94)

Thus

χ(q) = −e2 ∂n0

∂µand q-independent, which means:

ε(q) = 1 +4πe2

q2

∂n0

∂µ= 1 +

k20

q2with k2

0 = 4πe2 ∂n0

∂µ(95)

defining the Thomas-Fermi wave vector. Let’s apply this formula for an external potentialgiven by pointlike charge

φext(r) =Q

r, φext(q) =︸︷︷︸

see (78)

4πQ

q2. (96)

7The exact requirement is that the potential changes only on the scales much larger than 1/kF . So,everything works well only for q ¿ kF .

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The full potential in the metal is then

φ(q) =1

ε(q)φext(q) =

4πQ

q2 + k20

. (97)

In order to obtain the actual form of the full potential we apply the inverse FT:

φ(r) =∫

d3q(2π)3

eiqr 4πQ

q2 + k20

=Q

2π2

∫ ∞

0dq 2πq2

∫ 1

−1d(cos θ)

eiqr cos θ

q2 + k20

=Q

π

∫ ∞

0dq

q2

q2 + k20

∫ 1

−1dq eiqry =

Q

iπr

∫ ∞

0dq

q

q2 + k20

(eiqr − e−iqr

)

=Q

iπr

∫ ∞

−∞dqeiqr q

q2 + k20

=Q

iπr

∫ ∞

−∞dq

12

(1

q + ik0+

1q − ik0

)eiqr

=Q

i2πr2πi eir(ik0) =

Qe−k0r

r. (98)

(In the last integration we closed the contour in the upper half-plane.) This is, of course,nothing else but the Yukawa potential , or the screened Coulomb potential with the screeninglength 1/k0.

Lindhard or RPA approximation (dynamical screening)

Let’s now consider an oscillating potential

δU(r, t) = Ueiqr+iωt+αt , (99)

where q and ω are the wave vector and the frequency of the oscillation and α > 0 and smalldescribes an adiabatic switching of the interaction from t = −∞ to t = 0. Eigenstates forthe electrons without the potential are the plane waves of the form (set the volume V = 1again)

|k〉0 = eikr+iE(k)t/~ . (100)

The perturbation mixes them with all the others,

ψk(r, t) = |k〉 = |k〉0 + bk+q(t)|k + q〉0 , (101)

where the coefficients of this expansion are found from [see e.g. Landau-Lifshits Vol. III,§41, Eq.(41.1)]

bk+q(t) = − i

~

∫ t

−∞dt 0〈k + q|δU(r, t)|k〉0 = − Ueiωt+αt+i[E(k)−E(k+q)]t/~

E(k)− E(k + q) + ~ω − i~α

≈ − Ueiωt+αt

E(k)− E(k + q) + ~ω − i~α. (102)

Change in the wave function leads to a change in the particle densities,

δρ(r, t) = e∑

k

(|ψk(r, t)|2 − 1) ≈ e

∑

k

[bk+q(t)eiqr + b∗k+q(t)e−iqr

]. (103)

Now we use (102) and take into account that (99) is a real quantity and also contains aterm with −ω as well as −q. Then the correction to the particle density is

δρ(r, t) = e∑

k

[U

E(k)− E(k + q) + ~ω − i~α+

U

E(k)− E(k− q)− ~ω + i~α

]

×eiqr+iωt+αt + compl. conj. (104)

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The sum over k is, of course, only restricted to the states k < kF . At finite temperaturethe Fermi distribution is taken into account just by weighting every single term by nF (k).Performing a variable shift k− q → k in the second term we finally obtain

δρ(r, t) = eU∑

k

nF (k)− nF (k + q)E(k)− E(k + q) + ~ω − i~α

eiqr+iωt+αt + compl. conj. (105)

Now we observe that eδρ(r, t) = ρind and δU/e is the full potential φ(r) in (90). Thereforewe immediately find the dynamic dielectric constant

ε(q, ω) = 1 +4πe2

q2

∑

k

nF (k)− nF (k + q)E(k + q)− E(k)− ~ω , (106)

which is sometimes called Lindhard formula.The static screening case can be recovered in the following way. First we set ω = 0.

Assuming long wavelength limit q ¿ kF we expand

E(k + q)− E(k) ≈ q · ∇kE(k) and nF (k)− nF (k + q) ≈ q · ∇kE(k)(−∂nF

∂E)

. (107)

Then inserted back into the Lindhard formula it yields (we have taken V = 1 again)

ε(q, 0) = 1 +4πe2

q2

18π3

∫d3k

(−∂nF

∂E)

= 1 +4πe2

q2

∫dE DF (E)

(−∂nF

∂E)

. (108)

In the limit T → 0 there’s an important property(−∂nF

∂E) ∣∣∣

T→0→ δ(E − µ) , (109)

which after insertion into the previous result immediately produces the result (95). Forq ∼ kF we have to calculate the integrals in (106) explicitly. At T = 0 we obtain with thenotation k = k + q

∑

k

nF (k)− nF (k + q)E(k + q)− E(k)

=2me

~2

∑

k

nF (k)q2 + 2q · k +

∑

k

nF (k)

q2 − 2q · k

=4me

~2

∑

k

nF (k)q2 + 2q · k . (110)

In the last step we used the fact that ε(−q, 0) = ε(q, 0). Now we can go over to theintegration:

4me

~2

∑

k

nF (k)q2 + 2q · k =

e2me

q2π2~2

∫ kF

0dk 2πk2

∫ 1

−1

d(cos θ)q2 + 2qk cos θ

. (111)

After performing the integrals we obtain for the dielectric constant

ε(q, 0) = 1 +4mekF e2

π~2q2F(q/2kF ) , (112)

where the F function is defined by (79). The singularity around q = 2kF is essential.Among other things by repeating the calculation leading to the Yukawa potential (98) wecan verify that the effective potential behaves like

φ(r) ∼ cos(2kF r)r3

, (113)

i.e. is oscillatory instead of exponentially decaying one. This effect is observable as Friedelor Ruderman-Kittel oscillations.

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Friedel oscillations and Friedel sum rule

Example: Consider free one-dimensional (1D) particles in vicinity of an impenetrable wallat, say, x = 0. Because of the open boundary the boundary condition for the wave functionis ψk(x = 0) = 0, therefore it must be of the kind

ψk(x) ≈ 1√L

(eikx − e−ikx

). (114)

Now we, as always, fill all states up to kF and calculate the density profile

ρ(x) =∑

k<kF

ψ∗k(x) ψk(x) =1L

∑

k<kF

[2− 2 cos(2kx)] =12π

2∫ kF

0dk [1− cos(2kx)]

=kF

π

[1− sin(2kF x)

2kF x

]= ρ0

[1− sin(2kF x)

2kF x

]. (115)

So the ‘screening’ of the boundary occurs on the lengthscale of ∼ 1/kF . We observe,however, that the averaged density

1L

∫ L

0dx ρ(x)

∣∣∣L→∞

→ ρ0 − ρ0

L

π

2kF, (116)

because∫∞0 dx sin(x)/x = π/2. That means that the boundary has a finite charge eπρ0/(2kF ) =

e/2 (we have to restore e by hand since we were talking about the densities). In order tounderstand that we’ve got to develop some general theory of Friedel oscillations.

Let’s consider the system in a spherical potential with infinitely large walls with radiusR. Then the radial wave functions are given by Bessel functions and the quantisationcondition on the surface is ψ(R) = 0. For small l and large r one can obtain by expansion

jl(knr) ∼ sin (knr − πl/2)knr

so from jl(knR) = 0 follows knR = (n + l/2)π . (117)

If we have a scatterer at the origin r = 0, which generates scattering phase shifts δl, thenthe radial parts are

Rl(kr) ∼ sin (kr + δl − πl/2)kr

. (118)

Hence the phase shifts change the quantisation condition to

knR + δl(kn) = (n + l/2)π . (119)

That’s why the density of available particle states is

dn

dk=

R

π︸︷︷︸free case

+1π

dδl

dk︸ ︷︷ ︸scatterer induced

, (120)

with the net change due to the scatterer

d(δn)dk

=1π

dδl

dk

< 0 for repulsive interaction> 0 for attractive interaction

. (121)

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After summation over all available scattering channels

dN

dk=

d

dk

∑

l,ml,mspin

δl,ml,mspinπ

. (122)

The extra states are necessary to screen off the impurity therefore the full integral must beequal to the (excess) impurity charge Z:

Z =∫ kF

0dk

(dN

dk

)=

∑

l,ml,mspin

δl,ml,mspinπ

. (123)

This identity is the Friedel sum rule. It can also be recast in the form

Z = 4π

∫ ∞

0dr r2 [ρ(r)− ρ0] . (124)

The integrand is more or less referring to the Friedel oscillations, which can be calculatedin the following way,

ρ(r)− ρ0 = 2∫

|k|<kF

d3k(2π)3

[|ψk(r)|2 − |ψk(r, δl = 0)|2]

=︸︷︷︸large r

8π

(2π)3r2

∑

l

(2l + 1)∫ kF

0dk

[sin2(kr + δl − πl/2)− sin2(kr − πl/2)

],(125)

where we’ve taken advantage of (118)8. The remaining integration can be recast in theform

∼∫ kF

0dk [cos (2kr − πl)− cos (2kr + 2δl − πl)]

≈∫ kF

0dk (−1)l

︸ ︷︷ ︸cos(πl)

cos(2kr)− cos

[2k(r +

dδ

dk) + 2δl − 2kF

dδ

dk

], (126)

where we assumed the phase shifts to be small so that we can expand

δl(k) = δl(kF ) + (k − kF )dδ

dk. (127)

Keeping only the terms of the leading order (actually only ∼ δ(kF )) we obtain

≈ 12r

[− sin(2kF r) + sin[2kF r + 2δl(kF )]] . (128)

Combining everything together we obtain9

limr→∞[ρ(r)− ρ0] =

14π2r3

∑

l

(2l + 1)(−1)l sin[δl(kF )] cos[2kF + δl(kF )] + o(1/r4) . (129)

Similar calculations can be performed in general dimension d and lead to

δρ(r) ∼ sin(2kF r)rd

.

8k2 gets cancelled through the Jacobian 4πk2.9here we use the identity

sin α− sin β = 2 sin[(α− β)/2] cos[(α + β)/2] .

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1.5 Basic principles of Fermi liquid theory

Despite the quite strong Coulomb interactions between the electrons in a metal the freeelectron model in the form we used above works pretty well. Why is that so? The answerto this question was given by Landau 1957 in his Fermi liquid (FL) theory developed firstfor 3He and later adapted by Landau and Silin to the electron subsystems in metals.

Let’s assume we start with free electrons and adiabatically switch on the interactions.There are then two basic changes:

• the energies of the single-particle states ψi entering the Slater determinant change

• electrons start to scatter between different ψi

While the former process is well described within the HF approximation the latter cannotbe taken care of in any of the approximative schemes we’ve discussed thus far. However, ifthe scattering is weak then we still can rely on the free electron model (Fermi gas). Let’sestimate this scattering rate. Assume an excited electron with energy E1 > EF scatters onone within the Fermi sea E2 < EF thereby exciting it into E3 > EF and going into the stateE4 > EF itself. From the energy balance we have

E1 + E2 = E3 + E4 . (130)

In the special case E1 = EF all E2,3,4 = EF therefore no scattering is possible and the single-particle state lifetime = ∞. For finite E1 − EF ¿ EF the phase space for every resultingparticle is ∼ (E1 −EF ). One of the energies is fixed by (130), therefore the scattering rate(or the inverse relaxation time) 1/τ ∼ (E1 − EF )2. At finite temperature there would beadditional phase space due to the Fermi edge widening. Therefore

1τ≈ a(E1 − EF )2 + bT 2 , (131)

with temperature- and energy-independent a and b. Let’s now access the importance ofelectron-electron interaction at the Fermi edge for E1 = EF . In the Born approximation a isproportional to the FT of the scattering potential: a ∼ |U(q)|2. The latter is the screenedpotential (97), maximal value of which is 4πe2/k2

0, therefore

1/τ ∼ T 2(4πe2/k20)

2 = T 2(π2~2/mekF )2 . (132)

The rhs still has a wrong dimension. In order to cure that we can use the parameters of ourFermi gas kF , me as well as ~. The only correct combination making the rhs to a quantitymeasured in 1/sec is m3

e/~7. That leads to

1/τ = A1~

T 2

EF, (133)

where A is just a number ∼ 1 − 100. Since Troom ∼ 10−2eV and EF ∼ 1 − 10eV therelaxation time is of the order τ ∼ 10−10sec À 10−14sec measured in experiments. Thatmeans that the electronic correlations are very much weaker than the actual processesresponsible for the finite electrical resistivity of the metal. That’s why the free electronmodel works so well.

But let’s now quantify a little bit the Fermi liquid theory by going over to the

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Second quantisation formulation

First we take a look onto the original Hamiltonian:

H = −∑

i

~2

2me∇2

i +12

∑

i6=j

U(ri − rj) . (134)

Its eigenfunctions are complicated objects of the form (73) carrying very inconvenient num-ber of different indices/variables. Let’s introduce by analogy to the second quantisation ofharmonic oscillator fermionic annihilation/creation operators with the properties

a†σ(r), aσ′(r′)

= δ(r− r′)δσσ′ ,

a†σ(r), a†σ′(r

′)

=aσ(r), aσ′(r′)

= 0 . (135)

Then we can introduce coordinate-independent wave functions constructed as

ΨN (E) =∑

σi

∫d3r1 . . . d3rN Ψ(r1, σ1; . . . rN , σN ) a†σ1

(r1) . . . a†σN(rN ) |0〉 , (136)

where the fermionic vacuum |0〉 is defined as aσ(r)|0〉 = 0. The corresponding Hamiltonianacting in the space of ΨN (E)-functions is then

H = −∑

σ

∫d3r a†σ(r)

~2

2me∇2 aσ(r)

+12

∑

σ,σ′

∫d3r d3r′ a†σ(r)aσ(r)U(r− r′)a†σ′(r

′)aσ′(r′) . (137)

It is often more convenient to work in the momentum (or wave vector – keep in mind thatthe only difference is the factor ~, which is often set to be unity) representation (‘planewave expansion’)

aσ(p) =∫

d3r eip·r/~ aσ(r) . (138)

The corresponding anticommutation relations are preserved,

a†σ(p), aσ′(q)

= δp,qδσσ′ ,

a†σ(p), a†σ′(q)

= aσ(p), aσ′(q) = 0 . (139)

Then the Hamiltonian is given by H = H0 + HI10

HI =12

∑

p,q,k,σ,σ′a†σ(p− q/2) aσ(p + q/2) U(q) a†σ′(k + q/2) aσ′(k− q/2)

=1

2V

∑

p,q,k,σ,σ′a†σ(p− q)a†σ′(k) U(q) aσ′(k− q)aσ(p) , (140)

where the free (non-interacting) part can be written as

H0 =∑p,σ

E(p) a†σ(p)aσ(p) =∑p,σ

p2

2menσ(p) with np = a†σ(p)aσ(p) . (141)

10V in the last expression is restored just for convenience.

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The Fermi liquid picture is then the approximation a†σ(k)aσ(k− q) ≈ nσ(k) and a†σ(p− q)aσ(p) ≈nσ(p). So that the effective Hamiltonian is written down as

HFL =∑p,σ

ε(p) nσ(p)− 12V

∑

p,k,σ,σ′nσ(p) fσ,σ′(p,k) nσ′(k) , (142)

where ε(p) as well as the function f is determined by the parameters of the bare Hamiltonian(140). The most important feature of this effective picture is the fact that [nσ(p),HFL] = 0,which means that it can be replaced by numbers. Therefore the Hamiltonian becomes afunctional of momentum distribution function nσ(p). To make further progress one assumesthat the excitations around the groundstate of the system (which is nothing but the filledFermi sphere, of course) carry the same quantum numbers as the original non-interactingelectrons and have thus the same wave functions, but different eigenvalues. These excita-tions are called (Landau) quasi-particles. We postulate the existence of the correspondingcreation/annihilation operators. Then the excitations around the groundstate are

single-particle excitation =

a†p,σ|0〉 for p > pF quasi-electronap,σ|0〉 for p < pF quasi-hole

. (143)

As we have seen above and as can immediately be realised in face of the interaction Hamilto-nian (140) a state a†p,σ|0〉 can decay into two particles and one hole a†q1,σ1 a†q2,σ2a−p+q1+q2,σ′ |0〉,whereby q1,2 > pF and | − p + q1 + q2| < pF . In order to estimate the decay rate of theoriginal excited state we use Fermi’s golden rule,

W =2π

~|〈in|U |out〉|2δ(Ein −Eout) , (144)

with Ein = ε(p) and Eout = ε(q1) + ε(q2)− ε(−p + q1 + q2). Therefore we obtain

W = 2s2π

~

∫d3q1

(2π)3

∫d3q2

(2π)3U2(p− q1)Θ[ε(q1)]Θ[ε(q2)] Θ[−ε(−p + q1 + q2)]

× δ[ε(p) + ε(−p + q1 + q2)− ε(q1)− ε(q2)]︸ ︷︷ ︸energy conservation

. (145)

Let’s now assume that U(p) changes only slightly on the relevant energy scales so that itcan be considered to be a constant ≈ U . Effectively we integrate everything in the vicinityof the Fermi surface. Furthermore we rewrite

∫d3q

(2π)3=

∫dεDF (ε)

∫dn

4π≈ DF (EF )

∫dε

∫dn

4π, (146)

where dn = cos θ dθdφ denotes the angle integration. Therefore we obtain

W =4π

~U2D2

F (EF )∫ ε

0dε1

∫ ε

0dε2

∫dn1dn2

(4π)2Θ[−ε(−p + q1 + q2)]

× δ[ε + ε(−p + q1 + q2)− ε1 − ε2] . (147)

In vicinity of the Fermi edge we can neglect ε, ε1,2 ¿ EF (remember: we count the energiesfrom the Fermi edge) since

ε(−p + q1 + q2) =1

2me

[p2 − 2p · (q1 + q2) + q2

1 + q22 + 2q1 · q2 − p2

F

]

=p2

F

me(1 + n1 · n2 − n · n2 − n · n1) . (148)

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can become as large as EF itself (we use p = npF etc.). For the δ-function we then obtain

δ[ε(−p + q1 + q2)] =me

p2F

δ (1 + n1 · n2 − n · n2 − n · n1) . (149)

Plugging this all back we obtain

W (ε) = 1/τ =2π

~ε2

EFU2DF (EF )2 I , (150)

where the integral

I =∫

dn1dn2

(4π)2δ (1 + n1 · n2 − n · n2 − n · n1) (151)

can be shown (Exercise 6) to be I = 1.Although there’s a one-to-one correspondence between the Fermi gas and Fermi liquid

states the dispersion relation of quasi-particles ε(p), entering (142) is different from thekinetic energy of free electrons. In vicinity of the Fermi surface we can expand:

ε(p) = EF + v∗F (p− pF ) + . . . where v∗F =∂ε(p)∂p

∣∣∣p=pF

≈ pF

m∗ (152)

is different from the Fermi gas value vF = pF /me. The quantity m∗ is called effective mass.One important consequence of that is the different DOS at the Fermi edge.

dN =V

(2π)34πk2 dk =

V

(2π~)34πp2 dp ≈ V

(2π~)34πp2

F

dε

vF, which means that

DFL(EF ) =dN

V dε≈ m∗pF

2π2~3≈ m∗

meDF (EF ) , (153)

contrary to the one calculated in (46). Since the quasi-particles share all the other propertiesof the electrons (Fermi distribution function etc.) the only change to the formulas is thedifferent DOS. For example in order to calculate the corrected cV we can again use (61),then

cV,FL

cV,free=

m∗

me. (154)

Observe that no change to the temperature dependence occurs.An additional change in the effective mass is also incurred by the interaction term in

(142). Unfortunately there’s no consistent procedure to obtain the function f from theoriginal microscopic Hamiltonian, so one has to resort to symmetry considerations. Sincein most cases one is interested in the processes around the Fermi surface, f only dependson the angle θ between the vectors k,p (the length of both is then, of course, pF ). Thenwe can conveniently introduce the following partial wave expansion

fσ,σ′(k,p) =∞∑

l=0

(f

(s)l + σσ′ f (a)

l

)Pl(cos θ) , (155)

where Pl are the Legendre polynomials. The constants f(a),(s)l are usually reworked into

the dimensionless Fermi liquid parameters:

F(s),(a)l = 2DFL(EF ) f

(s),(a)l . (156)

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Figure 6: Left panel: momentum distribution function at T = 0 for: (i) Fermi gas (blackdashed line); (ii) Fermi liquid with the quasi-particle weight Z (black solid line); (iii)Tomonaga-Luttinger liquid (red line). Right panel: energy-dependent DOS for a Fermi gas(black) and a TLL (red).

Among other things it can be shown, that yet another correction to the effective mass isgiven by

m∗∗

m∗ = 1 +F

(s)1

3. (157)

Another important change due to the quasi-particle interaction is in the momentum dis-tribution function for the original electrons, which now acquires a finite step 0 < Z < 1,called quasi-particle weight , at the Fermi edge see Fig. 6

1.6 Tomonaga-Luttinger liquid (TLL)

Spinless case

TLL is a typical example of a non-Fermi liquid, physical properties of which are completelydifferent. The FL can rigorously be shown to be produced by a thorough resummationof the perturbation theory with respect to interaction term (Abrikosov & Khalatnikov).However, the perturbation expansion diverges in 1D. Let’s take the non-interacting part(141) and linearise the dispersion relation around pF , then (see Fig. 7)

H0 =∑

p

p2

2mea†(p)a(p) ≈ vF

∑

k

|k| a†kak , (158)

where we have neglected an unimportant constant term and dropped the spin indices forconvenience. The interaction Hamiltonian (140) can be rewritten in the form

HI =1

2L

∑q

ρ(q) U(q) ρ(−q) , (159)

where the Fourier modes of the new density operators are defined as

ρ(q) =∑

p

a†p−q/2ap+q/2 =∑

p>0

a†p−q/2ap+q/2 +∑

p<0

a†p−q/2ap+q/2 = ρ1(q) + ρ2(q) . (160)

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Figure 7: Left panel: linearised dispersion relation in the Tomonaga model. Right panel:dispersion relation of the Luttinger model. Red and green denotes the right- and left-mover,respectively.

Here in the last step we’ve introduced the left/right moving components of the density.Let’s analyse their commutation relations

[ρ1(k), ρ1(k′)] =∑

p,p′>0

[a†p−k/2ap+k/2, a†p′−k′/2ap′+k′/2

](161)

=∑

p>0

[a†p−k/2ap+k′+k/2 Θ(p + k/2 + k′/2)− a†p−k′−k/2ap+k/2Θ(p− k/2− k′/2)

].

A good approximation for this commutator can be obtained by taking the expectation valuewith respect to H0. Then we use that 〈a†kak′〉 = nF (k)δkk′ and thus k′ = −k and

≈∑

p>0

nF (p− k/2)− nF (p + k/2) =∑

kF−k/2<p<kF +k/2

1 =

k L2π k < 2kF

2kFL2π k ≥ 2kF

, (162)

where we apply the main feature of the Luttinger model and extend the range of allowedk values to −∞ for the right moving branch and to ∞ for the left-moving branch, seeFig. 711(see the sum structure for k > 0 in Fig. 8). So as a good approximation we canassume the following commutators to be exact identities

[ρ1(k), ρ1(−k′)] = δkk′kL

2π, [ρ2(k), ρ2(−k′)] = −δkk′

kL

2π, [ρ1(k), ρ2(−k′)] = 0 . (163)

Therefore we can introduce new bosons according to the scheme

ρ1(k) = bk

(kL

2π

)1/2

, ρ1(−k) = b†k

(kL

2π

)1/2

,

ρ2(k) = b†−k

(kL

2π

)1/2

, ρ2(−k) = b−k

(kL

2π

)1/2

. (164)

11One can perform the same summation within the Tomonaga model as well. Then we would obtain

=

k2

L2π

k < 2kF

kFL2π

k ≥ 2kF

The rest of the calculation can be done in the same way as for the Luttinger model up to some minorchanges in numerical factors.

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Figure 8: Diagram for the summations in Eq. (162). Left panel: the Tomonaga modelcase for k > 0. Only the black shaded area contributes. The green shaded area drops outbecause of the requirement p > 0. Right panel: the Luttinger model case. Again, only theblack shaded area contributes.

Then [bk, b†k′ ] = δkk′ as required. That’s why for the interacting part of the Hamiltonian we

obtain

HI =∑

k

Uk(bk + b†−k)(b†k + b−k) , (165)

where Uk = |k|Uk/(4π). It is also possible to rewrite H0 in terms of new bosons. Take alook onto the commutator

[ρ1(k),H0] = vF

∑

p>0,k′|k′|

[a†p−k/2ap+k/2, a†k′ak′

]

= vF

∑

p>0

a†p−k/2ap+k/2 (|p + k/2| − |p− k/2|) (166)

obviously (|p + k/2| − |p− k/2|) =

k for p > k/22p for p < k/2

. (167)

For small k for most p the first alternative is the leading one so that

[ρ1(k),H0] ≈ vF k∑

p

a†p−k/2ap+k/2 = vF kρ1(k) . (168)

Hence we can write

[bk,H0] = vF kbk = ωk bk , with ωk = vF |k| . (169)

We observe that the same result can be obtained with H0 =∑

k>0 ωkb†kbk. On the other

hand we also can show that

[ρ2(k), H0] = −vF kρ2(k) , (170)

which means that we can safely write

H0 =2πvF

L

∑

k>0

[ρ1(−k)ρ1(k) + ρ2(k) ρ2(−k)] =∑

k

ωk b†kbk . (171)

So that now the full Hamiltonian is given by

H =∑

k

ωk b†kbk +∑

k

Uk (bk + b†−k)(b†k + b−k) . (172)

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This can be diagonalised by a linear transformation to some new operators

Qk =1√2ωk

(bk + b†−k) , Pk = i

√ωk

2(b†k − b−k) , (173)

with canonical commutation relations [Qk, Pk′ ] = iδkk′ . Then the free Hamiltonian lookslike12

H0 =12

∑

k

(P−kPk + ω2

k QkQ−k

), while the full one

H =12

∑

k

[PkP−k + QkQ−k

(ω2

k + 4ωkUk

)]. (174)

So we see that the new eigenvalues are

Ek =√

ω2k + 4ωkUk = vF |k|

√1 + U(k)/(πvF ) = vF |k|/g(k) . (175)

A genuinely diagonal Hamiltonian is obtained by a back transformation of the kind

Qk =1√2Ek

(αk + α†−k

), Pk = i

√Ek

2

(α†k − α−k

),

⇒ H =∑

k

Ek(α†kαk + 1/2) . (176)

This Hamiltonian describes the excitation spectrum of the system around its groundstate,which is still a filled Fermi sea. These excitations, contrary to the Fermi liquid quasi-particles are bosons and describe density waves. These we’ve already encountered in theDrude theory – these are plasmons. Their dispersion relation in different cases is:

• simplest case U(k) ≈ U0 and constant, then

Ek = vF |k| with vF = vF /g , g = 1/√

1 + U0/(πvF )

g is the dimensionless interaction parameter. 0 < g < 1 for repulsive interactions,g > 1 for attractive ones, and g = 1 in the free non-interacting case.

• if we assume

Uk =43

e2k2F

k2then Ek =

√v2F k2 + ω2

p ,

where ω2p = 4πe2n0/me is the plasma frequency, see e. g. (18), then for small k (long

wavelengths) we obtain Ek ≈ ωp and for large k (small wavelengths) we obtain theplasmon with the linear dispersion as above but with g = 1.

System with spin

In this case we can introduce spin-dependent densities according to the prescription:

ρσ=↑,↓(q) =∑

p

a†p−q/2,σap+q/2,σ (177)

12where we’ve added an unimportant term∑

k ωk/2

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and subsequently rework them into the charge and spin densities

ρ = ρ↑ + ρ↓ , σ = ρ↑ − ρ↓ . (178)

These can again be decomposed into the left/right moving components with the following(still approximative) commutation relations13

[σ1(k), σ1(−k′)

]= δk,k′

kL

π,

[σ2(k), σ2(−k′)

]= −δk,k′

kL

π,

[σ1(k), σ2(−k′)

]= 0 ,

[σi(k), σj(−k′)

]= 0 . (179)

In the spirit of the previous Subsection we introduce new creation/annihilation operators

σ1(k) = ck

(kL

π

)1/2

, σ1(−k) = c†k

(kL

π

)1/2

,

σ2(k) = c†−k

(kL

π

)1/2

, σ2(−k) = c−k

(kL

π

)1/2

, (180)

which again satisfy bosonic commutation relations [ck, c†k′ ] = δk,k′ . In addition, the chargeand spin density wave operators commute: [ck, b†k′ ] = 0. We can again calculate thecommutator

[σ1(k),H0] ≈ vF kσ1(k) . (181)

The same properties has also

πvF

L

∑

k>0

σ1(−k)σ1(k) . (182)

On the other hand (168) is still valid14. Since the interaction does not break the SU(2)symmetry of the system – it is not spin-dependent – we immediately can write down theHamiltonian in terms of new particles bk and ck:

H =∑

k

ωk(b†kbk + c†kck) +

∑

k

Uk (bk + b†−k)(bk + b−k) =∑

k

(Ek α†kαk + ωk c†kck

). (183)

So in addition to the plasmons we also have spin density wave excitations – magnons orspinons with ck, c

†k being the corresponding annihilation/creation operators. We also can

identify

Hsw =∑

k

ωk c†kck (184)

as the spin wave Hamiltonian. A very important characteristic of the magnons is theirvelocity, which is exactly equal to the original Fermi velocity, vsw = vF . As a result theplasmon and magnon velocities are different : vF 6= vF . This phenomenon is known underthe name spin–charge separation. It is expected to be observable in all interacting 1Dliquids and is possible in higher dimensional systems under very special conditions only.

13Extra factor 2 is due to spin degree of freedom?14The definition of the new bk bosons has to be adjusted because of the additional 2s prefactor.

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Single-particle properties in the original fermionic basis can be accessed using thebosonization identity

a1/2,σ(x) =1

2πa0e±ikF x±J1/2,σ(x) ,

J1/2,σ(x) = −2π

L

∑

k>0

e−αk

k

[e−ikx ρ1/2,σ(k)− eikx ρ1/2,σ(−k)

]. (185)

Now we summarise the fundamental differences between the FL and TLL:

• spin–charge separation vmagnon 6= vplasmon; it is observable in the Raman spectra of1D systems

• power law singularity of the momentum distribution function in vicinity of the Fermiedge, see Fig. 6; for Uk = U0 =const

nk ∼ |k − kF |(1/g+g)/2−1

• power law singularity of the energy dependent DOS, see Fig. 6

DTLL(E) ∼ |E − EF |α , with αend = 1/g − 1 , αbulk = (g + 1/g)/2− 1 (186)

leads to highly non-Ohmic current-voltage characteristics of contacts between FLand TLL or two TLLs, known as zero bias anomaly (ZBA); was already observed incontacts between normal metals – single-wall carbon nanotubes

1.7 Conventional superconductivity

Cooper instability

Yet another non-Fermi liquid is the superconducting electron system. Let’s consider afilled Fermi sea and two electrons interacting via U(r1 − r2). To find the correspondinggroundstate we need to solve

[H0(r1) + H0(r2) + U(r1 − r2)]Ψ(r1, r2) = EΨ(r1, r2) . (187)

For the free part we obtain simple plane waves

H0(r)ψk(r) = εk ψk(r) ⇒ ψk(r) = V −1/2 eik·r , (188)

which have to be combined into a Slater determinant - like object

Ψ(r1, r2) =∑

k

gk(α1β2 − β1α2) cos[k · (r1 − r2)]

+g′k (α1β2 + β1α2), α1α2, β1β2 sin[k · (r1 − r2)] , (189)

where α1,2 is the spin wave function for electrons 1, 2 with spin up and β1,2 with spin down.The first part then corresponds to the singlet configuration while the second one describesthe triplet state. Since we’re looking for the groundstate it’s instructive to have the centerof mass at rest k1 = −k2. Let’s now consider attractive interaction, then the system’senergy is obviously minimised for the symmetric wave function since then cos 0 = 1, whichmeans that the groundstate corresponds to σ1 = −σ2. Then we can write

Ψ(r1, r2) =∑

k

gk [ψk,↑(r1) ψ−k,↓(r2)− ψ−k,↓(r1) ψk,↑(r2)] . (190)

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In order to find the energy E of this two-particle state we plug this back into the originalHamiltonian (187), then using that

H0(r1)∑

k

gkψk,↑(r1) ψ−k,↓(r2) =∑

k

gk [H0(r1)ψk,↑(r1)] ψ−k,↓(r2)

=∑

k

gkεk ψk,↑(r1) ψ−k,↓(r2)

we find that

(E − 2εk)Ψ(r1, r2) = U(r1 − r2)Ψ(r1, r2) ,∑

k′[Eδ(k′)− 2εk′ ]gk′e

ik′·(r1−r2) = U(r1 − r2)∑

k′′gk′′e

ik′′·(r1−r2) . (191)

Now we multiply both sides with exp[−iq · (r1 − r2)] and integrate over r1 − r2 (this isactually a Fourier transform of both sides) and obtain

(E − 2εq)gq =∑

k′′U(k′′ − q)gk′′ . (192)

Now we model the interaction in the following way:

U(k′′ − q) = −U for − w < εq, εk′′ < w

0 otherwise(193)

Then the solution of the above equation is

gq =U

2εq − EI with I =

∑

|εk′′ |<w

gk′′ . (194)

We are looking for a bound state so E = −2∆ with ∆ > 0. Using the last equation weobtain

I = UI

2

∑

|εk′′ |<w

1εk′′ + ∆

≈ UI

2DF (EF )

∫ w

0dε

1ε + ∆

= UI

2DF (EF ) ln(w/∆) ,

meaning that there’s indeed a bound state with the energy (measured with respect to theFermi surface)

∆ = we− 2

DF (EF )U . (195)

This bound state is called Cooper pair .

Multi-particle groundstate

is now, of course, completely different from that of the Fermi liquid. In order to find it weuse the effective Hamiltonian, derived from (137) where we keep only the interaction of thekind we identified in the previous Subsection:

H =∑

k,σ

εk a†k,σak,σ − U

V

∑

k,k′a†k′,↑a

†−k′,↓a−k,↓ak,↑ . (196)

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We use now the variational method of Bogolyubov and introduce new quasi-particles:

bk,↓ = uk ak,↓ + vk a†−k,↑ , bk,↑ = uk ak,↑ − vk a†−k,↓ . (197)

bs are fermions and thus satisfy anticommutation relations, which enforces the followingrelations

u2k + v2

k = 1 ⇒ uk = cos θk , vk = sin θk , (198)

so that the inverse transformation is given by

ak,↑ = uk bk,↑ + vk b†−k,↓ , ak,↓ = uk bk,↓ − vk b†−k,↑ . (199)

Transforming the effective Hamiltonian into the new basis yields

H = 2∑

k

εkv2k +

∑

k

εk(u2k − v2

k)(b†k,↑bk,↑ + b†k,↓bk,↓

)+ 2

∑

k

εkukvk

(b†k,↑b

†−k,↓ + b−k,↓bk,↑

)

− U

V

∑

k

B†kBk′ , (200)

where

Bk = u2kb−k,↓bk,↑ − v2

kb†k,↑b

†−k,↓ + ukvk

(b−k,↓b

†−k,↓ − b†k,↑bk,↑

). (201)

Let’s fix the occupation numbers of the individual states15

b†k,σbk,σ = nk,σ , bk,σb†k,σ = 1− nk,σ .

Then the full energy is just the expectation value of the above H, where we take into accountonly terms diagonal in the quasi-particle occupation operators (since no transitions betweendifferent states are allowed because of fixed nk,σ condition),

E = 2∑

k

εkv2k +

∑

k

εk(u2k − v2

k)(nk,↑ + nk,↓)− U

V

[∑

k

ukvk(1− nk,↑ − nk,↓)

]2

. (202)

Now we vary it with respect to uk using the relation (198) (which means that δvk =−(uk/vk)δuk) and obtain

δE

δuk= − 2

vk(1− nk,↑ − nk,↓)

[2εkukvk − U

V(u2

k − v2k)

∑

k′uk′vk′(1− nk′,↑ − nk′,↓)

](203)

At the stationary point we obtain

2εkvkuk = ∆(u2k − v2

k) with ∆ =U

V

∑

k′uk′vk′(1− nk′,↑ − nk′,↓) . (204)

15Which means that the entropy of the system

S = −∑

k,σ

[nk,σ ln nk,σ + (1− nk,σ) ln(1− nk,σ)]

is kept konstant.

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From the first of these relations we obtain

u2k, v

2k =

12

1± εk√

∆2 + ε2k

. (205)

The substitution into the second one yields

U

2V

∑

k

1− nk,↑ − nk,↓√∆2 + ε2k

= 1 . (206)

At this point we should remember that our quasi-particles are: (i) fermions; (ii) excitationsaround the filled Fermi sphere, which is the vacuum of the theory. Therefore the distributionfunction is the Fermi distribution with zero chemical potential:

nk,↓ = nk,↑ = nF (εk) =(eεk/T + 1

)−1. (207)

Now we go over from the summation to integration, then our equation for ∆ transforms to

U

2

∫d3k

(2π)31− 2nF (k)√

∆2 + ε2k

= 1 . (208)

At T = 0 there are no quasi-particles that’s why nF = 0. Then the calculation of theintegral can be performed in the following way16

∫dk

k2

√∆2

0 + v2F (kF − k)2

≈ k2F

vF

∫dε

1√∆2

0 + ε2≈ 2k2

F

vFln(W/∆0) . (209)

Here we used the assumption that only the immediate vicinity of the Fermi surface isimportant (therefore k2 → k2

F in the infinitesimal of the integration). Moreover, ∆0 isincorporated as the lowest boundary (or IR cut-off) since a posteriori it is the smallestenergy scale in the problem. W is the UV cut-off and is of the order of the bandwidth orEF in the case of free Fermi system or just W ∼ w (the range of energies at which theinteraction is attractive, see above). The factor 2 comes about as a result of integration overthe states just above the Fermi surface as well as those just below it. Hence the solution

UkF me

2π2ln(W/∆0) = 1 (210)

is given by

∆0 = We− 2π2

UmekF = We−2/[UDF (EF )] , where we use DF (EF ) = mekF /(π2~3) . (211)

Calculation of the single-particle excitation spectrum (the dispersion relation of the quasi-particles) can also be performed with the full energy of the system at hand. Then

E(k) =δE

δnk,σ= εk(u2

k − v2k) + 2

U

Vupvp

∑

k′uk′vk′(1− nk′,↑ − nk′,↓)

= εk(u2k − v2

k) + 2∆ukvk =√

∆20 + ε2k . (212)

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Figure 9: Left: dispersion relation of the elementary excitations in a superconductor. Cen-ter: temperature dependence of the gap. Right: temperature dependence of the specificheat.

So single-particle excitations are only possible at energies > ∆0, which means there’s a gapin the excitation spectrum and thus the energy dependent DOS in the system is zero forenergies between EF −∆0 and EF + ∆0.

Let’s now turn to the temperature-dependent case. Then for ∆(T ) we first rewrite (208)

−1 +U

2

∫d3k

(2π)3E(k)= U

∫d3knF (k)(2π)3E(k)

. (213)

Using the calculation above and its result (210) we get for the lhs (UkF me) ln(∆0/∆)/(2π2~3).Then

UkF me

2π2~3ln

(∆0

∆

)=

U

(2π)3

∫d3k

1E(k)(eE(k)/T + 1)

≈ U

8π3

∫4πk2

F dk1

E(k)[eE(k)/T + 1]. (214)

In the rhs we can go over to the integral over dεk = πvF dk (because of the original dispersionεk = vF |k − kF |) and get

ln(

∆0

∆

)=

∫dε

√ε2 + ∆2

[e√

ε2+∆2/T + 1] = 2I(∆/T ) ,

I(u) =∫ ∞

0dx

1√

x2 + u2(e√

x2+u2 + 1) . (215)

We consider different temperature regimes:

• low temperatures T ¿ ∆0 ⇒ large u, therefore after the expansion we’re confrontedwith a Gaussian integration:

I(u) ≈∫ ∞

0dx

1u

e−u(1+ x2

2u2 ) = e−u√

π/(2u) , (216)

16Here we observe that since the integral involves positive quantities only, no solutions exist for negativeU (repulsive interactions).

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the equation is thenln(∆0/∆) = e−∆/T

√2πT/∆ .

Its rhs is very small so ∆ is close to ∆0 and hence we expand

ln(∆0/∆) = − ln(

1 +∆−∆0

∆0

)≈ −∆−∆0

∆0and this is ≈ e−∆0/T

√2πT/∆0

producing the following asymptotics:

∆(T ) = ∆0

(1−

√2πT

∆0e−∆0/T

). (217)

• almost critical temperature T ≈ Tc, where ∆ is small. Here we use the followingtechnology: add and subtract an auxiliary integral, so that I = I1 + I2 and

I1 =12

∫ ∞

0dx

(1√

x2 + u2− 1

xth(x/2)

),

I2 =12

∫ ∞

0dx

(1x

th(x/2)− 1√x2 + u2

th√

x2 + u2

2

).

The first contribution of I1 is elementary and needs a UV regularisation. The sec-ond contribution is done by parts, with the first resulting part needing the sameregularisation. Then we obtain17

I1 =12

(− ln(u/2) +

12

∫ ∞

0dx

ln x

ch2(x/2)

). (218)

the last integral is equal to 2 ln(π/(2γ)) where ln γ = C = 0.577 is the Euler constant.That means that I1 = ln(π/(γu))/2. On the other hand I2(0) = 0. The first term ofthe expansion in u2 is

I2 = −u2

4

∫ ∞

0

dx

x

(1x

thx

2

)′.

th is a meromorphic function that’s why there’s an expansion of the kind

th(x/2) = 4x

∞∑

n=0

1π2(2n + 1)2 + x2

then (219)

I2 = 2u2∞∑

n=0

∫ ∞

0

dx

[π2(2n + 1)2 + x2]2=

u2

2π2

∞∑

n=0

1(2n + 1)3

= u2 7ζ(3)16π2

.

Hence, going back to (215):

ln(∆0/∆) = ln(πT/(γ∆)) +7ζ(3)8π2

∆2

T 2. (220)

17Here we proceed in the following way: we introduce a UV cut-off D À u and an additional substitutionx = ushα. Then the integral is

∫ D

0

dx1√

x2 + u2=

∫ Arsh(D/u)

0

dα = Arsh(D/u) = ln[D/u +

√(D/u)2 + 1

]→ ln(2D/u) .

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Here one immediately observes that ∆ is zero for all temperatures above the criticaltemperature

Tc = γ∆0/π ≈ 0.57∆0 . (221)

Below Tc in the lowest order in Tc − T we can verify that

∆ = Tc

[8π2

7ζ(3)(1− T/Tc)

]1/2

≈ 3.06Tc

√1− T/Tc . (222)

To summarise the temperature-dependence of the gap (see Fig. 9): (i) ∆ = 0 above Tc;(ii) finite and growing for 0 < T < Tc; (iii) finite at T = 0 and equal ∆0. This is just likein a system undergoing a phase transition. Then ∆ corresponds to the order parameter ,which is zero in the disordered phase and finite in the ordered phase. The critical exponentβ = 1/2 indicates that we’re working here within a mean field theory or MFT . Before weidentify the operator which corresponds to the order parameter we calculate some

Thermodynamic properties

In the low temperature regime T ¿ Tc the specific heat can be calculated from the variationof the full energy:

δE =∑

k

E(k)(δnk,↑ + δnk,↓) = 2∑

k

E(k)δnk . (223)

The specific heat is proportional to δE/δT . Going over to the integration over energies:

C = V DF (EF )∫

dεk

√∆2

0 + ε2k∂nF (E(k))

∂T, (224)

and using that for low density of quasi-particles nF (εk) ≈ exp(−εk/T ) and E(k) =√

∆20 + ε2k ≈

∆0 + ε2k/(2∆0) we immediately find

C = V

√2mekF ∆5/2

0

~2(πT )3/2e−∆0/T . (225)

At higher temperatures it is more convenient to work with the grand canonical poten-tial18. It is a function of T and may be found using the Feynman-Helmann relation (seee. g. LL V Statistical Physics)

(∂Ω∂U

)

T,V,µ

= 〈∂H

∂U〉U ≈ ∂E

∂U. (226)

In our case the correction to the energy is given by the last term of (202), then we have(derive first and insert ∆ afterwards)

∂Ω∂U

= −V∆2

U2. (227)

18We also keep the chemical potential constant, instead of the total number of particles.

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We now integrate between U = 0, which corresponds to the normal phase, and some finiteU corresponding to the superconducting phase,

Ωsc − Ωn = −V

∫ U

0du(∆2/u2) . (228)

At T = 0 we can use the relation between U and ∆0 from (211):

d∆0

dU=

2π2~2

mekF

∆0

U2

and plug it into the above integral, which then yields

Ωsc − Ωn = −VmekF

4π2~2∆2

0 . (229)

According to the theorem of infinitesimal corrections [see e. g. LL Volume 5, §24], whichstates (δE)S,V,N = (δΩ)T,V,µ we now have for the difference in energies between the super-conducting and normal state

Esc − En = −VmekF

4π2~2∆2

0 , (230)

which is definitely negative. In the opposite case T → Tc we differentiate (220) with respectto U and arrive at

7ζ(3)4π2T 2

∆d∆ =d∆0

∆0=

2π2~2

mekF

dU

U2. (231)

Then plugging dU/U2 into (228) and declaring it to be the free energy

Fsc − Fn = −V7ζ(3)mekF

8π4~2T 2

∫ ∆

0d∆∆3 . (232)

The following result is then obtained by integration and using the gap temperature depen-dence (222),

Fsc − Fn = −V2mekF T 2

c

7ζ(3)~2(1− T/Tc)

2 . (233)

Since dF = −SdT − pdV then S = −(∂F/∂T )V :

Ssc − Sn = −V4mekF Tc

7ζ(3)~2(1− T/Tc) , furthermore cV = T

(∂S

∂T

)

V

,

therefore at T → Tc we obtain a jump in the specific heat:

Csc − Cn = V4mekF Tc

7ζ(3)~2= V

4π2

7ζ(3)TcDF (EF ) . (234)

If we recall that in the Fermi gas case CV = V (π2/3)TDF (EF ), see (61), then we obtainfor the jump

Csc(Tc)Cn(Tc)

=12

7ζ(3)+ 1 ≈ 2.43 . . . (235)

So the specific heat as function of temperature has a discontinuity at Tc thus we’re dealingwith a phase transition of a second kind. The resulting plot is depicted in Fig. 9.

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Mean field theory

We have not yet a physical picture for the order parameter ∆. Let’s get back to our originalHamiltonian (196). For the further calculations we can even assume the interaction to bemore general:

H =∑

k,σ

εk a†k,σak,σ − 1V

∑

k,k′Uk,k′ a

†k′,↑a

†−k′,↓a−k,↓ak,↑ . (236)

Let’s set

δk = a−k,↓ak,↑ − 〈a−k,↓ak,↑〉 , (237)

and assume it to be very small. If we now solve for the operator product, insert itback into (236) and neglect terms of the second order in δk as well as a constant ∼∑

k,k′ Uk,k′ 〈a†k′,↑a†−k′,↓〉〈a−k,↓ak,↑〉 we obtain

H ≈∑

k,σ

εk a†k,σak,σ − 1V

∑

k,k′Uk,k′

[a†k′,↑a

†−k′,↓〈a−k,↓ak,↑〉+ 〈a†k′,↑a†−k′,↓〉a−k,↓ak,↑

]. (238)

Now we define

∆k =∑

k′Uk,k′ 〈a−k′,↓ak′,↑〉 . (239)

In this way we obtain the effective Hamiltonian in the mean field approximation or MFA

HMFA =∑

k,σ

εk a†k,σak,σ −∑

k

∆k a†k,↑a†−k,↓ −

∑

k

∆∗k a−k,↓ak,↑ , (240)

which is also known under different other names: Bogolyubov-de Gennes, Gorkov , orBardeen-Cooper-Schrieffer (BCS) Hamiltonian (1957-1960). This can again be diagonalisedby the Bogolyubov transformation (199)

(bk,↑b†−k,↓

)=

(uk −vk

v∗k u∗k

)(ak,↑a†−k,↓

). (241)

The self-consistency condition for ∆k is, of course, directly related to the equation (206).This order parameter can be interpreted as being related to the density of some superfluidphase – of a Cooper pair condensate. In the coordinate space ∆(r) ∼ Ψ(r), where Ψ(r) isthe Cooper pair wave function, which is the same for all pairs. For a perfectly homogeneoussystem ∆ in equilibrium is just a complex constant. In finite fields it acquires a coordinatedependence.

Let’s assume we subject our system to a vector potential A(r), then the change in thekinetic energy can be written as

Ψ∗[

12m

(−i~∇− (e/c)A)2 −(− ~

2

2m∇2

)]Ψ = −1

cJ(r) ·A(r) . (242)

Deriving with respect to A(r) we obtain an expression for the current density:

J(r) = −ie~2m

(Ψ∗∇Ψ−Ψ∇Ψ∗)− e2

mcΨ∗ΨA(r) . (243)

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If the field is small then since there’s a gap in the excitation spectrum, the chances that Ψchanges significantly are small (Ψ is ‘robust’). Without field the current is zero:

J0(r) = −ie~2m

(Ψ∗0∇Ψ0 −Ψ0∇Ψ∗

0) = 0 . (244)

Therefore in the finite field as Ψ ≈ Ψ0 we get

J(r) ≈ − e2

mcΨ∗ΨA(r) = − e2

mcneA(r) , (245)

when all electrons are in the condensate Ψ∗Ψ ≈ ne at e. g. T ¿ Tc. (245) does not containthe wave function any more and is known under the name London equation. This is the keyfor understanding various superconductor properties. The simplest example is the perfectconductivity. Since A is static the electric field in the sample is

E = −1c

∂A∂t

= 0 ⇒ from J = σE ⇒ σ →∞ (246)

so the resistivity of the sample is zero! Let’s combine (245) with the Maxwell equation

4π

cJ(r) = ∇×H = ∇× (∇×A(r)) = ∇ (∇ ·A)−∇2A = −∇2A ,

for the gauge ∇ ·A = 0. Plugging this back into the London equation (245) yields

∇2A =1λ2

A or ∇2H =1λ2

H , ∇2J =1λ2

J . (247)

The solutions of these equations are of the kind H = H0 e−z/λ, where λ =√

mc2/(4πnee2)is the penetration depth. This is the physical reason for the Meissner-Ochsenfeld effect– the magnetic field and currents cannot penetrate into the superconductor. It is thus aperfect diamagnetic material.

We’re dealing with a phase transition. They usually can be associated with a brokensymmetry . That means that the symmetry of the Hamiltonian is not shared by the ground-state. Let’s consider a non-interacting fermion system in a field. Then the Hamiltonian isgiven by

∑σ

∫d3r a†σ(r)

12me

[−i~∇+ (e/c)A]2 aσ(r) . (248)

This Hamiltonian is invariant under the U(1) gauge transformation

A → A +~c2e∇χ , aσ(r) → aσ(r) e−i(e/~c)χ . (249)

While the interacting Hamiltonian (236) is U(1)-symmetric as well, the effective (240) isnot because the order parameter transforms according to the prescription

ak,σ → ak,σe−i(e/~c)χ ⇒ ∆k =∑

k,k′Uk,k′ 〈a−k′,↓ak′,↑〉 ⇒ ∆k → ∆ke−i2(e/~c)χ , (250)

which in general corresponds to a different groundstate. So we’re confronted with a brokengauge symmetry because the groundstate is characterised by a fixed phase (in a homoge-neous system).

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Figure 10: Left: integration path for flux quantization calculation. Right: Josephsoncontact.

Flux quantization and Josephson effect

The order parameter can be written down as Ψ = |ψ0|eiϕ. In a junction of two supercon-ductors with an insulating layer |Ψ0|2 is proportional to the condensate density on bothsides. The two components do not overlap, so |Ψ0| has not coordinate dependence. Weobtain for the current (243) in the system

J =e~m|ψ0|2

(∇ϕ− 2e

~cA

). (251)

Let’s now consider a superconductor with a hole and integrate the current density alongthe closed contour which lies much further away from the boundary than the penetrationdepth λ, see Fig. 10. Then the current is zero. For the rest we obtain

0 =∮

dl · J =e~m|ψ0|2

∮dl ·

(∇ϕ− 2e

~cA

). (252)

The first integral on the rhs is obviously equal to 2πn. For the second we obtain 2πΦ/Φ0,where Φ0 = hc/2e is the flux quantum. Thus we obtain a flux quantisation condition

Φ = nΦ0 . (253)

The phase difference between two superconductors can in fact be observed in the Joseph-son effect . Consider a weak contact between two superconductors. The energy gain/lossdue to the presence of a contact is proportional to the wave function overlap:

E = C︸︷︷︸constant

∫

interfacedydz (Ψ1Ψ∗

2 + Ψ∗1Ψ2) = 2C

∫dydz |Ψ1Ψ2| cos(ϕ1 − ϕ2) . (254)

Let’s assume some B in plane with the interface. From the gauge transformation we knowthat ϕ and A enter in a combination

∇ϕ− 2e

~cA ,

which means that a phase difference along the simple line integration between the points 1and 2, see Fig. 10 is proportional to (we can choose A such that there’s only an x-component

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Figure 11: a) Current-voltage characteristics of a Josephson junction. b) Integration pathfor the Onsager-Feynman quantization condition.

Ax)

ϕ2 − ϕ1 − 2e

~c

∫ 2

1dxAx

Therefore E in the field is given by

E = 2C

∫dydz |Ψ1Ψ2| cos

(ϕ2 − ϕ1 − 2e

~c

∫ 2

1dxAx

). (255)

We calculate the variation with respect to the field Ax, then

δE =4Ce

~c

∫dydz

∫ 2

1dx |Ψ1Ψ2| sin

(ϕ2 − ϕ1 − 2e

~c

∫ 2

1dxAx

)δAx . (256)

From a general formula of electrodynamics

δE = −1c

∫dV J · δA . (257)

we immediately obtain

J =4e

~C|Ψ1Ψ2| sin

(ϕ2 − ϕ1 − 2e

~c

∫ 2

1dxAx

). (258)

Or, in a general set-up

J = Jc sin(ϕ1 − ϕ2) with Jc =π∆(T )2eR

th[∆(T )2T

], (259)

where R is the bare contact resistance in the normal phase and ∆(T ) is the temperature-dependent gap. This perpetual current is called Josephson current or supercurrent (1962).This can be seen experimentally in the current-voltage relation of a tunneling contact, seeFig. 11, and is the basic operating principle of such sensitive devices as SQUIDs.

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1.8 Degenerate Bose gases and Bose-Einstein condensation (BEC)

Bose-Einstein condensation

Let’s now turn to the systems of non-interacting bosons. Just as electrons in the metal inthe jellium model we assume them to be confined in some 3D potential (preferrably of acubic form) with volume V 19.

At low temperatures all bosons attempt to populate the lowest lying available state.Then they all are described by the same wave function. A typical spacial spread of awavepacket for an individual particle is given by the de Broglie wave length

λ = 2π/k = h/p , p2/2m = kBT ⇒ λ = h/√

2mkBT . (260)

When it becomes comparable with average interparticle distance ∼ n1/3 there’s a chancethat they all end up in a single state. It has a chance to happen at the temperature

T ∼ h2n2/3

2mkB, (261)

which is a very low. In order to understand the physical properties of such a collective stateone needs more precise calculations.

We start again with a particle number as a function of temperature and chemical po-tential and use (44) changing only the distribution function,

N =∑

k

nB(E(k)) =V

(2π)3

∫dk

4πk2

e(E−µ)/T − 1

=V

4π2

(2m

~2

)3/2 ∫ ∞

0dε

√ε

e(ε−µ)/T − 1=

V

4π2

(2mT

~2

)3/2 ∫ ∞

0d(ε/T )

√ε/T

eε/T−µ/T − 1, (262)

where we neglect the spin degree of freedom. First of all, since the population probabilitymust be positive the chemical potential is always negative. If we keep the particle densityN/V constant and lower the temperature µ would grow and becomes equal to zero at thetemperature

Tc =~2

2m

(4π2n

g3/2

)2/3

≈ 3.31~2

mn2/3 , (263)

where

g3/2 =∫ ∞

0dx

√x

ex − 1=

∫ ∞

0dx√

xe−x∞∑

n=0

e−xn =∞∑

n=0

∫ ∞

0dx√

xe−x(n+1)

=∞∑

n=0

√π

2(n + 1)3/2=√

π

2ζ(3/2) ≈ 2.32 (264)

The integral is mapped onto a Gaussian one by the substitution t =√

x.For T < Tc the equation (262) does not have any negative solutions. This apparent

problem does not come up when one performs the transition from the summation to inte-gration more carefully. In fact, for T < Tc the particles with energies ε > 0 are distributedaccording to the distribution

dNε =V

4π2

(2m

~2

)3/2

dε

√ε

eε/T − 1. (265)

19The relevant systems might be 4He (Kapitsa, Allen, Misner; 1938) (below λ-transition at 2.17K), 87Rb(Cornell, Wieman; 1995), 23Na (Ketterle; 1995) (below 170nK).

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Thus their number is given by

Nε =∫

dNε =V

4π2

(2mT

~2

)3/2 ∫ ∞

0dx

√x

ex − 1= N

(T

Tc

)3/2

. (266)

The number of particles in the lowest lying state is now found easily

N0 = N

[1−

(T

Tc

)3/2]

. (267)

This behaviour suggests that one’s dealing with some kind of phase transition with N0

being the order parameter. At T = 0 all particles are in the energetically lowest lying stateand the Bose-Einstein condensation (BEC) occurs. It does not really mean that the bosons‘condense’ and localize in space in the conventional sense. It rather means a condensation inthe momentum space: all k become equal and all bosons share the same dynamics. There’sone aspect in which BEC is profoundly different from classical phase transitions: it doesnot require interactions.

For the calculation of the total energy we only need to take into account particles withenergies ε > 0, therefore

E =∫

dNε ε =V

4π2

(2m

~2

)3/2

T 5/2 g5/2 , (268)

where

g5/2 =∫ ∞

0dx

x3/2

ex − 1=

3√

π

4ζ(5/2) ≈ 1.78 . (269)

That means that the specific heat is simply

cV =5E

2T∼ T 3/2 . (270)

This we can integrate in order to obtain the entropy:

S =5E

3Tthen from F = E − TS we get F = −2

3E . (271)

Using this we can calculate the pressure

p = −(

∂F

∂V

)

T

=1

6π2

(2m

~2

)3/2

g5/2 T 5/2 ≈ 0.085m3/2

~3T 5/2 , (272)

that is volume-independent. This is not surprising since at T < Tc most particles are inthe 0-state with zero momentum and thus are unable to oppose the decreasing volume.

At exactly T = Tc all above quantities are continuous. However, it can be shown, thatthe derivative of the specific heat has a jump (Exercise 13). That means that BEC is aphase transition of the third order. However, in presence of even weak interactions thesituation changes dramatically.

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BEC order parameter

Before we proceed to the interacting case we need a useful definition of the BEC state (or ofthe order parameter). Below Tc the momentum distribution function can be written downas

n(k) = N0 δk,0 + δk, 6=0V

(2π~)3(eεk/T − 1

)−1. (273)

Here one implicitly assumes that the lowest lying state with E = 0 corresponds to k = 0,which is only true for translationally invariant systems. How can we generalise that toother systems, e. g. traps or confining potential? Let’s assume N bosons can be describedby the (particle exchange symmetric) wavefunction

Ψ(r1, r2, ...rN ; t) . (274)

The single-particle density matrix is defined by (factor N compensates the normalisationof Ψ itself?)

ρ1(r, r′; t) = N

∫dr2...drN Ψ∗(r, r2, ..., rN ; t)Ψ(r′, r2, ..., rN ; t) . (275)

Obviously, it is Hermitean, ρ1(r, r′; t) = ρ∗1(r′, r; t) and is thus diagonalisable:

ρ1(r, r′; t) =∑

i

ni(t) χ∗i (r, t) χi(r′, t) , (276)

where the χi(r, t) are a complete orthogonal set at any given t and are not necessarilyeigenfunctions of the original Hamiltonian. Then:

• if all ni are of the order unity, then the system is in the normal state

• if there’s only one eigenvalue of order N , then the system exhibits a simple BEC

• if there’re two or more eigenvalues of order N and the rest of order unity, then wedeal with the fragmented BEC

In the two last cases one speaks of a non-vanishing off-diagonal long range order or ODLRO .The order parameter can then in the simple BEC case be defined as20

Ψ0(r, t) =√

N0(t) χ0(r, t) . (277)

Just as in the BCS it is a solution of no Hamiltonian. Similarly, we can define a condensatephase ϕ

χ0(r, t) = |χ0(r, t)|eiϕ(t) or Ψ0(r, t) = |Ψ0(r, t)|eiϕ(t) . (278)

The condensate density and current are

ρc(r, t) = N0(t) |χ0(r, t)|2 = |Ψ0(r, t)|2

jc(r, t) = N0(t)(−i

~2m

χ∗0(r, t)∇χ0(r, t) + c.c.)

= N0(t) |χ0(r, t)|2 ~m∇ϕ(r, t)

= |Ψ0(r, t)|2 ~m∇ϕ(r, t) . (279)

20Sometimes it is convenient to keep only the χ0 part without the normalisation.

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We can define the ratio of those two quantities as a superfluid velocity ,

jc(r, t)ρc(r, t)

= vs(r, t) =~m∇ϕ(r, t) . (280)

From this equation follows that everywhere where χ0 6= 0 we have

• vs is curl-free

∇× vs(r, t) = 0 (281)

• if we integrate the above equation along a contour C on which χ0 6= 0 and vanisheswithin C then we obtain the Onsager-Feynman quantization condition

∮

Cdl · vs =

nh

m. (282)

BEC in interacting systems

The full treatment of this problem is very complex. We’ll try to give some qualitativearguments. Let’s assume for definiteness that we’ve got a contact interaction of the form

U(r) = U0δ(r) , (283)

where U0 is some constant. Consider two interacting particles described by the wavefunctionΨ(r1, r2). The correction to the energy due to interactions can be written down as

Eint ≈ 〈U〉 = U0

∫d3r1 d3r2 δ(r1 − r2) |Ψ(r1, r2)|2 = U0

∫d3r |Ψ(r, r)|2 . (284)

Next we consider two options:

• In the case of both particles being in the same state χ(r) 21 we have Ψ(r1, r2) =χ(r1)χ(r2) and thus

Esameint ≈ U0

∫d3r |χ(r)|4

• for particles in different states χ1,2 we have

Ψ(r1, r2) =1√2

[χ1(r1)χ2(r1) + χ2(r1)χ1(r1)]

and therefore for the energy correction we get

Ediffint ≈ 2U0

∫d3r |χ1(r)|2 |χ2(r)|2 .

If χ1,2 differ only slightly and ≈ χ we obtain that definitively Esameint < Ediff

int for repulsiveinteractions. So that even the interacting system favors the constellation in which all par-ticles in the lowest lying state. However, the situation is completely different for attractiveinteractions. This kind of reasoning can easily be generalised to the arbitrary number ofparticles.

21χ may be but is not necessarily the eigenstate of the full Hamiltonian. Important point is the existenceof a full orthonormal set of states.

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Bogolyubov presented the first microscopic quantitative theory for interacting Bosegases. One starts with a Hamiltonian similar to the one we used in the BCS theory (236):

K = H − µN =∑

k

(εk − µ)b†kbk +1

2V

∑

k′,k,q

Uq b†k+qb†k′−qbk′bk , (285)

where Uq is the Fourier transform of the interaction potential. As one can easily see thefundamental interaction process is scattering of two bosons on each other with an exchangein momentum q. In contrast to the fermionic case before we keep µ explicitly.22 At T < Tc

N0 is of the order N that’s why the most important processes take place around ε ≈ 0. Thenthe most important processes are those involving b0 and b†0. They can be approximatelytreated as scalars of magnitude ∼ √

N0:

b†0|N0〉 =√

N0 + 1|N0 + 1〉 ≈√

N0|N0〉 ,b0|N0〉 =

√N0|N0 − 1〉 ≈

√N0|N0〉 . (286)

Let’s now concentrate on the interaction term of the form (283), when Uq = U0 for all mo-menta. In the next step we identify all processes which involve states with zero momentum,

k = k′ = q = 0 : U02V N2

0 k = k′ = 0 , q 6= 0 : U02V N0b

†qb†−q

q = −k , k′ = 0 : U02V N0b

†kbk q = k′ , k = 0 : U0

2V N0b†k′bk′

k = −k′ = −q : U02V N0bkb−k q = k = 0 , k′ 6= 0 : U0

2V N0b†k′bk′

q = k′ = 0 , k 6= 0 : U02V N0b

†kbk

(287)

All other processes are of the order lower than N0 and can be neglected. Then the aboveHamiltonian in this approximation is

K = −µN0 +U0N

20

2V+

′∑

k

(εk − µ +

U0N0

V

)b†kbk

+U0N0

2V

′∑

k

(b†kb†−k + b−kbk + b†kbk + b†−kb−k

), (288)

where the sums go over all k 6= 0. The chemical potential is found by minimisation of Kwith respect to N0. Then since all sums produce terms of the order N − N0 ¿ N0 weobtain

δK =(−µ +

U0N0

V

)δN0 . (289)

Therefore addition of a particle to the condensate is most energy efficient when the bracketis zero and thus for µ = U0N0/V .23 After plugging that back into K we obtain

K = −U0N20

2V+

12

′∑

k

εk

(b†kbk + b†−kb−k

)+

U0N0

2V

(b†kb†−k + b−kbk + b†kbk + b†−kb−k

).(290)

This can be diagonalised with help of the yet another Bogolyubov transformation for bosons,defined as

(bkb†−k

)=

(chθk −shθk

−shθk chθk

)(ηk

η†−k

), (291)

22There’s no EF any more from which were our zero energy – our groundstate is completely different.23Of course, for U0 = 0 we have our non-interacting BEC µ = 0.

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where ηk are the new bosonic operators for the excitations around the ground state. Forthe special choice of the angle θk:

thθk =U0N0/V

εk + U0N0/V. (292)

the Hamiltonian is diagonal in ηk:

K = −U0N20

2V+

′∑

k

E(k) η†kηk + C , (293)

Notice that the equation (292) has a solution for all k only if U0 > 0 i. e. only for repulsiveinteractions. C is a numerical constant.24

The dispersion relation of the new particles is very interesting:

E(k) =

√ε2k + 2

U0N0

Vεk . (294)

With εk = ~2k2/2m we have two different limits.

• for U0 → 0 or large enough k we again obtain E(k) = εk, as expected

• for small k, on the contrary, we obtain a linear dispersion

E(k) ≈ ~(

U0N0

mV

)1/2

|k| = ~vsound |k| , (295)

which describes nothing else but sound with velocity vsound! It has very interestingconsequences, which we’re immediately going to discuss.

The transition between these two regimes occurs at p ≈ mvsound. The associated typicallength is the healing length

ξ =~p

=~

mvsound. (296)

Now we take a look onto the Hamiltonian (285) in the coordinate space. Similar to thefermionic case (138) we can introduce the plane wave expansion for the bosonic fields

bk =∫

d3r1√V

e−ik·r ψ(r) . (297)

Upon insertion of them into (285) and calculation of the sums over wave vectors we obtain

H =∫

d3rψ†(r)(~2∇2

2m− µ

)ψ(r) +

12

∫d3rd3r′ψ†(r)ψ†(r′)U(r− r′)ψ(r)ψ(r′) , (298)

if we treat ψs as operators. In the next step we can derive the exact equation for the fieldoperators in the Heisenberg representation:

i~∂tψ(r, t) = [ψ(r, t),H] =[−~

2∇2

2m+ Uext(r, t) (299)

+∫

d3r′ ψ†(r′, t)U(r′ − r)ψ(r′, t)]

ψ(r, t) ,

24In fact, it is divergent. This divergency can be, however, traced back to our special choice of q-independent interaction potential Uq = U0 and is thus unphysical.

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which is valid for arbitrary external potentials Uext25. Finally, in the last step we again can

employ the Bogolyubov approximation and substitute the quantum field ψ by the classicalfield Ψ0 [which is, of course, our order parameter (277)]. Then in the case of contactinteraction (283)

i~∂tΨ0(r, t) =[−~

2∇2

2m+ Uext(r, t) + g|Ψ0(r, t)|2

]Ψ0(r, t) , (300)

where g =∫

d3rU(r). The above substitution has exactly the same physical sense as thetransition from quantum electrodynamics to classical electrodynamics. The result is theGross-Pitaevskii equation (1961), widely used for description of the inhomogeneous Bosesystems.

Superfluidity

Let us consider a flow of a Bose condensate in a capillary. Let v be the velocity flow in thelaboratory frame of reference. The slow down of the flow occurs by creation of an excitationwith momentum p in the Fluid frame of reference. Than the full energy in the L system:

EL = (N0 − 1)mv2

2+

(p + mv)2

2m. (301)

The excitation lowers the energy in the laboratory frame when EL < Nmv2/2, which canhappen when

p2

2m+ p · v < 0 . (302)

For antiparallel p and v we obtain p < 2mv, or, for the excitation with dispersion relationE(p)

v > E(p)/p . (303)

That means that if v is smaller than the rhs, then a dissipationless flow is possible and onemay observe the phenomenon of superfluidity . Geometrically it means that if one can drawa tangent to the dispersion curve starting from the origin which has a finite slope then thesuperfluidity is possible. This is the content of the Landau criterion.

First of all we observe that the superfluidity is not possible in the free system withthe dispersion p2/2m, since here the smallest tangent slope is zero. On the other hand,in the interacting system with the dispersion (295) we should observe this phenomenon.However, the critical velocity in real Helium-II below the λ-point is much smaller than thesound velocity because of the additional complication of the excitation spectrum – there’sa local minimum, known as the roton minimum, see Fig. 12. How can one experimentallyverify the superfluidity? From (281) one might follow, that it is not possible to rotate asuperfluid since for ‘normal’ rotating bodies v = ω×r and thus ∇×v = 2ω 6= 0. Thereforeif we rotate a bucket with liquid Helium-II the shape would be given by

z(r) =ω2r2

2g

ρn

ρn + ρs, (304)

where ρn,s are the densities of the normal/superfluid components, respectively. This is notobserved though. The reason is that although (281) holds, the circulation (282) is not

25We’ve included µ into it as well.

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Figure 12: Left: Dispersion relation of excitations in Helium-II and Landau criterion. Right:Profile of a liquid in a rotating bucket.

necessarily zero when the region of the singularity (r = 0) does not contain the superfluiddensity, see the geometry on Fig. 11. So there exist new excitations – vortices, consistingof a core of normal fluid surrounded by circulating superfluid. It is widely believed thatthe destruction of a superflow occurs through excitation of such vortices.

BEC-BCS crossover

In the BCS theory we’ve learned that the Cooper pairs are relatively large, ξ À n−1/3

and weakly bound. This can be immediately traced back to weak attractive interaction.One might ask the question, whether the fermion pairs can be tightly bound by a strongattractive interaction, so that their typical size ¿ n−1/3. Can such pairs then undergo aBEC? If so, would it be possible to observe the BEC-BCS crossover just by changing theinteraction strength?

The answer to both questions is yes: both theoretically and experimentally. It has beenobserved in the gas of 6Li atoms which are prepared in two hyperfine states mJ = 1/2,mI = ±1/2 as well as in similar constellations of 40K. The interaction strength can betuned around the Feshbach resonance just by changing the magnetic field. It allows one totune the scattering length as for the interaction process.

The ‘naive’ approach to the crossover problem would then be the direct extension ofthe BCS-type analysis of Section 240 to a system with strong interactions (Eagles, 1969;Leggett, 1980). It leads to two relations for the gap (or binding energy of the BEC‘molecule’) ∆ and the chemical potential µ:

∆ = EF f(ζ) , µ = EF g(ζ) with ζ = −1/(kF as)

being the dimensionless interaction strength parameter. The numerical behaviour of thefunctions is depicted in Fig. 13. One recognizes two limits

• ζ →∞ (BCS limit) µ → EF and ∆ → EF exp[−π/(2kF |as|)]• ζ → −∞ (BEC limit) µ → −~2/(2mas), which is half of the binding energy of the

molecule, ∆ ∼ (n/as)1/2

From the picture we see, that in the BEC limit the scattering length is negative. Normallyit corresponds to repulsive interactions, which is not exactly what we need. The point is

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Figure 13:

that the the scattering length diverges and thereby changes sign (starting with the weakattractive interaction) when a bound state emerges. So, on the BEC side we indeed have arepulsive interaction, but also an additional bound state describing our diatomic moleculemade of fermions.

The above theoretical picture is a very approximative one. The more detailed theoriesare not that numerous, but their development is a very active area of research. The biggestchallenge is the analytical treatment of the actual crossover – the unitarity regime, whichis the area of very large scattering lengths (the area around zero in the Fig. 13). Here thereare no small parameters and one needs more sophisticated methods.

2 Fermions and Bosons on the lattice and in external poten-tials

2.1 Periodic structures

From now on we return to our model of the metal of Fig. 2 and look onto the arrangementof the background ions. In the ideal case they are arranged in a perfectly periodic structure.Each point/ion can then be addressed in the 3D case by a vector

R = n1a1 + n2a2 + n3a3 , (305)

where n1,2,3 are integers and a1,2,3 are non-collinear vectors, which do not belong to a singleplane and which are called primitive vectors. They generate the Bravais lattice. Typicalexamples are the simple cubic, body centered cubic (bcc) and face centered cubic (fcc)lattices. However, there’re also non-Bravais lattices, e.g. the honeycomb lattice (graphene),which is a lattice with basis because it contains non-equivalent sites.

Every site of the lattice has a number of nearest neighbours. It is called coordinationnumber . For example: simple cubic case = 6, bcc = 8, fcc = 12.

With every site, which can be addressed by (305), one can associate a cell with a finitevolume – primitive unit cell . One obvious choice are the points r, which satisfy

r = x1a1 + x2a2 + x3a3 , (306)

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where 0 ≤ x1,2,3 < 1. Of course, this is not a unique way. Another possibility would bea primitive cell which has the same symmetry as the lattice itself. Then it is called theWigner-Seitz primitive cell .

Now we would like to generalize the Fourier expansion (FE) to 3D periodic structures.From the physical point of view an FE is nothing else but expansion in plane waves. Thelatter are not always lattice-periodic. But they are for a special choice of the wave vectors:

eiK·(r+R) = eiK·r only when eiK·R = 1 . (307)

The set of vectors K, which satisfies the above condition spans the reciprocal lattice. It canin general be written down as

K = k1 b1 + k2 b2 + k3 b3 with integers k1,2,3 and the basis vectors

b1 = 2πa2 × a3

a1 · (a2 × a3), b2 = 2π

a3 × a1

a1 · (a2 × a3), b3 = 2π

a1 × a2

a1 · (a2 × a3). (308)

The reciprocal lattice has the following important properties:

• reciprocal lattice of a Bravais lattice is a Bravais lattice itself

• the reciprocal lattice of a reciprocal lattice is the direct one

•bi · aj = 2πδij

• if v is the volume of the primitive cell of the direct lattice, then the volume of theprimitive cell of the reciprocal lattice is (2π)3/v

The primitive cell of the reciprocal lattice is called first Brillouin zone.

2.2 Bloch theorem and its consequences

The eigenstates for the electrons on a lattice are solutions of the equation

(H − E)ψ =(−~

2∇2

2me+ U(r)− E

)ψ = 0 , (309)

where U(r + R) = U(r) is a periodic potential of the background ions. This fact mightsuggest that the solutions ψ would be lattice-periodic as well. This is wrong though becauseof the Bloch theorem, which states that the eigenstates ψ of the above Hamiltonian can bechosen to have the form of a plane wave times a function unk(r) which is lattice-periodic,

ψnk(r) = eik·r unk(r) , (310)

where n denotes all other quantum numbers (spin etc.). This means that

ψnk(r + R) = eik·R ψnk(r) . (311)

This allows one to reformulate the Bloch theorem in a slightly different form: the eigenstatesof H in (309) can be chosen in such a way that with each of them one can associate a wavevector k such that

ψ(r + R) = eik·R ψ(r) (312)

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for every R in the Bravais lattice. Let’s now consider a crystal which has a length/height/widthof N1,2,3 cells. Then applying the periodic boundary conditions as in (34) we have

ψ(r + Niai) = ψ(r) , on the other hand ψnk(r + Niai) = eiNik·ai ψnk(r) , (313)

which immediately leads to

eiNik·ai = 1 ⇒ ei2πNixi = 1 , (314)

for k =∑

i xibi and finally to the quantization condition

xi = mi/Ni with an integer mi . (315)

That’s why the allowed values for the Bloch wave vector are

k =3∑

i=1

mi

Nibi . (316)

The volume per k of the reciprocal space is then

∆k =b1

N1·(

b2

N2× b3

N3

)=

1N

(2π)3

v=

(2π)3

V. (317)

So the density of states in the k-space is exactly the same as in the case of free electrons![compare with (38)].

We’ve just learned that any electron wave function can be ascribed a quantum numberk. However, one should keep in mind, that contrary to the free case ~k is not the momentumof the electron. The reason for that is that H does not possess the translation invarianceof the full space and thus does not commute with the p = −i~∇ operator. This is easilyseen from

−i~∇ψnk = −i~∇[eik·r unk(r)] = ~kψnk − i~eik·r∇unk(r) 6= pψnk (318)

for any p. p = ~k is called crystal momentum of the electron and sometimes is very useful– more later.

The wave vectors k can be restricted to the first Brillouin zone because if k fulfills theBloch theorem so does k′ = k + K, where K is any vector of the reciprocal lattice. In thecase when eigenfunctions ψnk and eigenvalues Enk with wave vectors in the first Brillouinzone one is talking about the restricted zone scheme. However, one can define periodicfunctions

ψnk+K(r) = ψnk , Enk+K = Enk .

Then one deals with the extended zone scheme. Then Enk has an upper and lower boundand describes an nth energy band .

That this index n really exists can be realized from the following argument. Let’slook for a solution of (309) of the form ψ(r) = eik·ru(r), where k is fixed and u(r) has aperiodicity of the lattice. Substituting this into the Hamiltonian gives

[~2

2me(−i∇+ k)2 + U(r)

]uk(r) = Hkuk(r) = Ekuk(r) . (319)

This has to be solved with a boundary condition uk(r + R) = uk(r). This corresponds to asolution of a Schrodinger equation with an operator Hk in a finite volume of the primitive

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cell. This gives us an infinite set of solutions with discretely spaced eigenvalues, which wenumber by n.

Finally, the most interesting fact about the Bloch electrons is their velocity, which isgiven by (derivation in Exercise 17)

vn(k) =1~∇kEn(k) . (320)

It asserts that the electron energy levels are stationary and are characterized by a constantvelocity. It does not degrade in any way despite the presence of the periodic potentialcontrary to the key assumption of the Drude and Sommerfeld theories of metals.

We fill the energy levels just as in the free electron case one by one. Since we wantto count every state only once we fill them up in the reduced zone scheme and restrictourselves to the 1st Brillouin zone. As before there’s a maximal energy EF , which theelectrons can have. Since the dispersion relation is not quadratic any more, the equationfor the Fermi surface has the following form:

Enk = EF . (321)

Since the Fermi surface is now not spherical it is not trivial to go over from the k-dependentto E-dependent DOS. Let’s take a look onto the sum

F = 2s

∑

nk

fn(k) ≈ 2sV∑

n

∫d3k

(2π)3fn(k) . (322)

If fn(k) depends only on energy E then

F =∫

dE g(E)f(E) with g(E) =∑

n

gn(E) and gn(E) =∫

d3k4π3

δ[E − En(k)] . (323)

On the other hand

gn(E)dE =2s

V(#k vectors between E and E + dE) =

2s

V

1(2π)3/V

∫d3k . (324)

Define Sn(E) as the surface of constant energy. Then d3k = dSndk(k), where dk(k) is thenormal to the surface. It can be found from

E + dE = E + |∇En(k)|dk(k) ⇒ dk(k) =dE

|∇En(k)| . (325)

Going back to (324) we obtain the relation between the DOS in the energy representationand zone structure:

gn(E) =∫

Sn(E)

dSn

4π3

1|∇En(k)| . (326)

Since En(k) is a periodic function and is bounded from both below and above then there’sat least one point where ∇En(k) = 0. These are the van Hove singularities.

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Right: periodic potential of the Kronig-Penny model.

2.3 Electron energy bands in crystals

Let’s now find the electron eigenstates on the lattice. We try to expand them in planewaves

ψ(r) =∑

k

ck eik·r =∑

k

ck |k〉 , (327)

where we use PBC and thus allowed k are given by (316). The correction to the energiesof free particles is found from the elementary perturbative expansion of the kind

Ek = E(0)k + 〈k|U |k〉+

∑

k′

|〈k|U |k′〉|2E(0)k − E(0)

k′+ . . . with E(0)

k =~2k2

2m. (328)

The periodic potential can be written down as

U(r) =∑

K

UK eiK·r , UK =1v

∫

celld3rU(r) e−iK·r , (329)

where K are from the reciprocal lattice because U(r) is very naturally lattice-periodic.Then

〈k|U |k〉 =∫

d3r e−ik·r U(r) eik·r = const and

〈k|U |k′〉 =∫

d3r e−ik·r ∑

K

UK eiK·r eik′·r = (2π)3∑

K

UKδK,k−k′ . (330)

Therefore

Ek = E(0)k + const +

∑

K6=0

|UK|2E(0)k − E(0)

k−K

+ . . . (331)

The last sum is uncontrolled when: (i) |UK| grows fast; (ii) E(0)k → E(0)

k−K. While the firstalternative almost never takes place, the second situation needs to be analyzed closer. Ithappens when k = k−K, exactly for the states in vicinity of the zone boundary, see Fig. 14.In order to make further progress we derive a Schrodinger equation for the electrons on thelattice.

Now we act with the kinetic energy operator on ψ:

−~2∇2

2mψ(r) =

∑

k

~2k2

2mck eik·r (332)

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as well as with the potential energy operator:

U(r)ψ(r) =∑

K,k

UK ck ei(K+k)·r =∑

K,k

UK ck−K eik·r . (333)

Then for the single-particle energy E we obtain the following equations:

∑

k

eik·r[(~2k2

2m− E

)ck +

∑

K

UK ck−K

]= 0

⇒(~2k2

2m− E

)ck +

∑

K

UK ck−K = 0 . (334)

We now want to choose k = k1 −K′, so that k1 is in the first Brillouin zone,(E(0)k1−K′ − E

)ck1−K′ +

∑

K

UK ck1−K′−K = 0 (335)

and subsequently shift K → K−K′:(E(0)k1−K′ − E

)ck1−K′ +

∑

K

UK−K′ ck1−K = 0 , (336)

which has an infinite number of solutions labelled by K′. Here we immediately see that fora given k1 this equation relates only ck1−K with different K to each other. Therefore for(327) we obtain

ψk(r) =∑

K

ck−K ei(k−K)·r . (337)

Let’s now look onto (336) in vicinity of the zone boundary. We keep only two coefficientsck1−K and ck1 , then after an energy shift by U0 we obtain the following equations

(E(0)k1−K − E

)ck1−K = U−K ck1(

E(0)k1− E

)ck1 = UK ck1−K (338)

The solutions of this equation are found after zeroing the determinant of the respectivematrix26,

E± =E(0)k1

+ E(0)k1−K

2± 1

2

√(E(0)

k1− E(0)

k1−K)2 + 4|UK|2 . (339)

Thus we have a mixing of two plane waves eik1·r and ei(k1−K)·r ∼ ψ±, which correspond tothe eigenvalues E±.

• zone center k1 ≈ 0, then |E(0)k1

− E(0)k1−K| À |UK| and we still have approximately

parabolic dispersion of the free particles

• at the zone boundary k1 ≈ K/2

E± ≈ E(0)K/2 ± |UK| (340)

and there’s an avoided crossing and thus a gap 2|UK| between the upper and thelower zone!

26Here we use that for real potential U(r) holds the relation U−K = U∗K.

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Let’s assume the potential to be negative, which corresponds to the attraction of the elec-trons, then27

ψ− =eiK·r/2 + e−iK·r/2

√2

=√

2 cos(K · r/2) ψ+ =eiK·r/2 − e−iK·r/2

√2

= i√

2 sin(K · r/2) .

While in the first case the particles are concentrated around the ions, in the second casethey are indeed mostly between the sites.

Band structure calculation methods

One of the simplest models which allows exact band structure calculations is the 1D Kronig-Penny model (or Dirac comb potential) (1931). Here the ions are considered to be δ-shapedpotentials for the otherwise freely moving electrons. Since the electrons do not interact wecan solve a single-particle problem with the given potential:

U(x) = λ~2

me

∞∑n=−∞

δ(x + na) , (341)

where a is the lattice constant, see Fig. 14. The solution for the wave function betweenx = 0 and a can be written as

u(x) = Aeikx + Be−ikx from the BT we know that there (342)

exists a K such that for the solution between x = a and 2a we have

u(x) = eiKa(Aeik(x−a) + Be−ik(x−a)

). (343)

Please keep in mind that k is the wave vector of the free electrons without U(x)! Thecrystal momentum is now K. The smoothness of the wave function and of its derivativerequires that

u(a + 0) = u(a− 0) , u′(a + 0) = u′(a− 0) + 2λu(a) ,

which leads to the following equations

eiKa(A + B) = Aeika + Be−ika ,

ikeiKa(A−B) = ik(Aeika −Be−ika) + 2λ(Aeika + Be−ika) . (344)

This can be represented in form of a matrix(

eiKa − eika eiKa − e−ika

(ik + 2λ)eika − ikeiKa (−ik + 2λ)e−ika + ikeiKa

)(AB

)= 0 (345)

The determinant of this matrix must vanish, therefore we’ve got the condition

det(...) = −4ieiKa [k cos(ak)− k cos(aK) + λ sin(ak)] = 0

cos(aK) = cos(ak) +λ

ksin(ka) =

=√

1 + (λ/k)2[

cos(ak)√1 + (λ/k)2

+(λ/k) sin(ak)√

1 + (λ/k)2

]

=√

1 + (λ/k)2 cos(ak − ϕ) with ϕ = arctg(λ/k) . (346)27Because in vicinity of the boundary the eigenvectors are (1,±1)/

√2

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Figure 15: Left panel: graphical solution of the equation (347) for λ = 3a, the variable ofthe x-axis is ka. Right panel: zone scheme of the Kronig-Penny potential.

Since cos must always be smaller or equal to unity the allowed states have to satisfy thefollowing relation

|cos(ak − ϕ)| ≤ 1√1 + (λ/k)2

, (347)

graphical solution of which is presented on Fig. 15. The allowed states are indeed organizedin bands. Please note that although not all k are allowed, K run through all possible values⇒ this is indeed the crystal momentum.

Up to now we’ve assumed the electrons ‘living’ freely between the ions and being scat-tered by the latter. There’s also the opposite limit when the electrons only can occupyorbitals localized on the ions and are only allowed to tunnel or hop between them. Thisapproach is called tight-binding model .

We assume the lattice to be an arrangement of ‘sites’ with indices m = 0, . . . , N ,whereby the sites m = N and m = 0 are the same by virtue of PBC. Then we canintroduce the electron creation/annihilation operators c†m and cm, which satisfy the usualanticommutation relation. We introduce a phenomenological parameter t representing theoverlap integral between the electron orbitals localised on the adjacent ions. The Hamiltonoperator of the system can then be written down as

H = −tN∑

m=0

(c†mcm+1 + c†mcm−1

)with PBC . (348)

It is diagonalised in the k-space with the help of the following FT (e. g. for an even N):

cm =1N

N/2∑

n=−N/2

eikn(am) c(kn) with kn = 2πn/(Na) . (349)

Plugging this back into the Hamiltonian H we get

H = −t∑

n,n′

[c†(kn)c(kn′) e−ikn′a + c†(kn)c(kn′) eikn′a

] 1N2

∑m

ei(kn′−kn)am

︸ ︷︷ ︸=Nδnn′

(350)

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Figure 16: Zone structure of the elementary tight-binding model.

For large N À 1 we can go over to the continuum limit∑

n → (Na/(2π))∫

dk and obtain

H =∫

dk

2πEk c†(k)c(k) with Ek = −2at cos(ka) . (351)

Without explicitly using the BT we found the dispersion relation Ek to be periodic in theextended zone scheme. Of course, we can restrict ourselves to the first BZ only, see Fig. 16.There’re two interesting limiting cases:

• ka ¿ 1, which is always the case when the band is barely filled: kF ¿ 1/a, then thedispersion is almost quadratic

Ek = −2at cos(ka) ≈ const + ta3k2 (352)

so that the electrons are almost free

• when we’re dealing with the half-filled band kF ≈ π/(2a) then in vicinity of the Fermiedge we have a linear dispersion relation:

Ek ≈ 2a2t k . (353)

In this case the Fermi velocity is approximately vF = 2a2t.

Thus far we did not included into our analysis any details of the ion/atom potentialU(r). It can be done in the following (and quite general) way. Let φa(r−R) be thepotential of the atom at the site R. Then the trial wave function

ψk(r) =∑

R

φa(r−R) eik·R (354)

fulfills the BT on the one hand. On the other hand it is more or less plane wave in betweenthe lattice sites and more like a localised wave function in vicinity of a lattice site. Let’snow calculate the energy for a single particle:

Ek =

∫d3rψ∗k(r)

[−~2∇2

2me+ U(r)

]ψk(r)

∫d3rψ∗k(r)ψk(r)

(355)

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If we assume that the atomic orbitals of adjacent sites overlap only slightly then denomi-nator is ≈ 1. For the energy we then obtain

Ek =∑

R,R′eik·(R−R′)

∫d3rφ∗a(r−R)

[−~

2∇2

2me+ U(r)

]φa(r−R′)

=∑

h

eik·h Eh , (356)

where we’ve used the fact that the overlap integral can only depend on the distance betweenthe sites h = R−R′. For the overlap matrix element we obtain

Eh =1v

∫d3rφ∗a(r + h)

[−~

2∇2

2me+ U(r)

]φa(r) . (357)

Let va(r) be the atomic potential of the individual site, then for the energy Ea of thelocalised state holds [

−~2∇2

2me+ va(r)

]φa(r) = Ea φa(r) .

Then for the expansion of (356) we obtain

E(0) = Ea , Eh =1v

∫d3rφ∗a(r + h) [U(r)− va(r)]φa(r) , . . . (358)

It is very natural to assume that only the overlap between the next neighbours is the leadingterm, that’s why in e. g. the case of the simple cubic lattice we can restrict ourselves to thevectors h = (a, 0, 0), (0, a, 0), (0, 0, a) only. Hence we arrive at the following approximationfor the band structure28

Ek = Ea + 2E100 [cos(akx) + cos(aky) + cos(akz)] + . . . (359)

Of course, there is a lot of space for improvements. The band structure calculationmethods include much better approximations such as e. g. orthogonalised plane waves(OPW), augmented plane waves (APW), method of pseudopotentials, Green’s functionsmethod and many others. For details see e. g. Ashcroft-Mermin or Ziman. Apart of someinteresting details the main structure remains the same: there’re energy bands with gapsbetween them.

2.4 Band structure and transport properties

As we’ve learned in (316) there’re exactly 2sN states within the BZ of each zone. Dependingon how many electrons per atom get delocalised one has different number of zones filled.Correspondingly varies also the Fermi energy. In most cases one is confronted with followingdifferent cases:

• the uppermost zone is not fully filled ⇒ it’s a metal. Example: alcali metals: onlythe uppermost s−shell electros are delocalised (only 1 electron per elementary cell)⇒ the uppermost band is half-filled

• the uppermost zone is full and the gap to the next one is ≥ 4 eV ⇒ it’s an insulator.Here the chemical potential (Fermi energy) lies in the gap. Example: carbon in thediamond allotropic form, 4 electrons are delocalised and fill two bands. The filledzone is called valence band and the next one conductance band .

28E100 is negative, it follows from the definition (358).

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Figure 17: Left panel: Fermi surface of a 1D system carrying finite electric current. Rightpanel: schematic representation of the Fermi surface shift in a 2D system carrying finitecurrent.

• the same as above but the gap is much smaller ∼ 1meV - 1 eV it’s a semiconductor.Example: germanium Ge, silicon Si, carbon in graphite form.

How can we understand the conductance properties? Let’s take a look on a 1D case ofa metal with half-filled band. Finite current in the system would mean that we’ve got moreright- than left-movers. That means that their chemical potentials are different EL,R. Theparticle density is 1/L, electron velocity v = ~k/me, therefore the current is given by

J = nev =1L

kR∑

k=kL

e~kme

=1L

L

2π

∫ kR

kL

dke~kme

=12π

e~me

∫ kR

kL

12dk2

=e

h

∫ ER

EL

d

(~2k2

2me

)=

e

h(ER − EL)︸ ︷︷ ︸

=eV

⇒ G = I/V = e2/h , (360)

where eV = EL − ER is the voltage bias applied to the system.What is important for the finite conductivity of the system is a room for a shift of the

Fermi surface within the zone. In 1D we’ve shifted the Fermi surface – the points ±kF

to the new positions ±kF + eV/2. If the band were full, which is the case for kF = π/a,the left-/right-mover balance could only be destroyed by excitation of the electrons in theupper band which would cost finite energy! That’s why systems with full bands – insulatorsand semiconductors do not conduct ‘easily’. In the similar way, in 2D we should be able tomove the circle with radius kF and in 3D the sphere etc29.

Summary of the consequences of band structure

• it explains the existence of metals, semiconductors and insulators

• it explains the anomalous Hall effect , which Hall constant is positive – here the holesin the conductance band are the current carrier

29Of course, in reality the Fermi surface is never a circle or a sphere, but rather has a more complicatedstructure. The analogy of the shift within the BZ works in the same way though.

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• it does not considerably change the specific heat calculation in metals since still onlythe states in vicinity of the Fermi edge make the dominant contribution and the bandstructure only affects the corresponding DOS

All other transport properties are going to be addressed later . . .

2.5 Dynamical properties of the lattice

Up to now we considered the lattice to be rigid. It can oscillate though and therebybe responsible for e. g. a nonvanishing contribution to the specific heat. The simplestapproach is the Einstein model according to which all atoms oscillate independently in anuncorrelated fashion with some frequency ωE . For the canonical partition function of oneoscillator we then obtain

Z =∞∑

n=0

e−1T~ωE(n+1/2) =

12sh[~ωE/(2T )]

. (361)

The average energy per atom is obviously

E = −∂β lnZ =~ωE

2cth (~ωEβ/2) . (362)

Therefore for the full specific heat we obtain

cV = 3N∂E

∂T= 3N

(~ωE

2T

)2 1sh2 (~ωE/2T )

= 3N

(θE

2T

)2 1sh2 (θE/2T )

, (363)

where 3 counts the possible polarisation directions of the oscillatory mode and N is the fullnumber of elementary cells in the crystal. θE = ~ωE/kB is some effective temperature

• high temperature expansion T À θE

cV = 3N(θE/T )2(1 + θE/T + . . . )

(1 + θE/T + · · · − 1)2≈ 3NkB , (364)

which is nothing but the Dulong-Petit law and is confirmed by the experimental data

• low temperature expansion T ¿ θE

cV ≈ 3NkB (θE/T )2e−θE/T , (365)

an exponential decrease, contradicts the experiments

An improvement is achieved by taking into account the correlations between the atommovements at different sites. A general procedure is the following one. Let’s assume wepossess the potential energy of the lattice U(uR), where uR is the displacement of the atomfrom the equilibrium position of the site at R.30 For small us we then can expand:

U = U0 +∑

R,i=x,y,z

uiR

∂U

∂uiR

+12

∑

R,R′,i,j=x,y,z

uiRuj

R′∂2U

∂uiR∂uj

R′+ . . . (366)

30In the case of a lattice with basis we just need another index labelling the atoms within the basis.

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On the other hand the atoms also possess kinetic energy

Ekin =∑

R

M

2|uR|2 .

Now from the knowledge of the Lagrange function L = T − Π = Ekin − U using theEuler-Lagrange equations31 we can derive the equations of motion (EOMs):

MuiR = −

∑

R′,j=x,y,z

∂2U

∂uiR∂uj

R′ujR′ = −

∑

R′,j=x,y,z

GijR,R′u

jR′ . (367)

The matrix G can only depend on the distance h = R−R′ between the atoms, thus

M uR = −∑

h

Gh uR+h . (368)

Now we use the BT according to which there exist some q so that

uR(t) = eiq·R u0(t)

We plug this into the previous equation and arrive at

M u0eiq·R = −

∑

h

Gh u0 eiq·h eiq·R .

We see that the exponent eiq·R cancels. By abuse of terminology we relabel u0 → uq andintroduce a ‘Fourier transform’ of the matrix G as

Gq =∑

h

Gh eiq·h . (369)

After these manipulations we now succeeded in reduction of a system of 3N equations to asystem of just 3 (!) harmonic equations

M uq = −Gq · uq . (370)

They are solved by the substitution

uq(t) = uq(0) eiνt , (371)

which generates the following eigenvalue problem for the lattice oscillations:∑

j=x,y,z

[Gij

q − ν2 M δij

]ujq(0) = 0 . (372)

Example: linear harmonic chain in 1D, Bravais lattice. Then the potential energy can bewritten down as32

U =α

2

∑

l

(ul − ul+a)2 . (373)

31We remind that they’re given byd

dt

∂L

∂qi− ∂L

∂qi= 0

32We only consider the interactions between the nearest neighbours.

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Figure 18: Schematic representation of harmonic chains and corresponding dispersion re-lations. Left panel: a homogeneous chain (Bravais lattice), right panel: an inhomogeneouschain with (lattice with basis M1 −M2).

Now we construct the G matrix. The derivatives are

∂U

∂ul= −α(ul−a − ul) + α(ul − ul+a) ,

∂2U

∂ul∂ul′= 2αδl,l′ − α δl,l′±a = Gll′ ,

⇒ Gh = 2αδh,0 − αδh,a − αδh,−a . (374)

For the Fourier transform

Gq =∑

h

Gh eiqh = 2α− αeiqa − αe−iqa = 2α[1− cos(qa)] .

From (372) we then get [Gq − ν2M ]uq(0) = with

ν2(q) = Gq/M = (2α/M)[1− cos(qa)] = (α/M)4 sin2(qa/2)

and thus ν(q) = 2√

α

M|sin(qa/2)| . (375)

As expected the dispersion relation has a periodicity of the reciprocal lattice, see Fig. 18.Not far away from the zone centre when aq ¿ 1 the dispersion is almost linear

ν(q) ≈ qa√

α/M = vsoundq ,

and describes sound waves.Example: linear harmonic chain in 1D, lattice with basis.

The EOMs in the previous case can also be derived directly by analysis of the forcesacting on every site. It immediately gives

Mul = −α(2ul − ul+a − ul−a) , (376)

which is solved by the substitution ul = uq eiql leading to

Muq = −(2α− αeiqa − αe−iqa)uq = −2α[1− cos(qa)]uq .

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Here we immediately recognise the dispersion relation. This philosophy can be applied toan inhomogeneous chain with two different kinds of atoms with masses M1,2. We denotethe corresponding displacements by u1,2. The potential energy has then the form

U = · · ·+ α

2[(u1l − u2l)2 + (u2l − u1l+a)2 + (u2l−a − u1l)2

]+ . . . . (377)

The EOMs are then

M1u1l = −α(u1l − u2l + u1l − u2l−a) , M2u2l = −α(u2l − u1l + u2l − u1l+a) . (378)

Here we can again reduce the full large equation system to only two equations by thesubstitution u1,2l = u1,2e

iql, then

M1u1 = −α(2u1 − u2 − u2e−iqa) = −2αu1 + u2α(1 + e−iqa) ,

M2u2 = −2αu2 + u1α(1 + eiqa) . (379)

The eigenvalues ν are then found from∣∣∣∣

2α−M1ν2 −α(1 + e−iqa)

−α(1 + eiqa) 2α−M2ν2

∣∣∣∣ = 0 , (380)

which leads to the quadratic equation

M1M2ν4 − 2α(M1 + M2)ν2 + 4α2 sin2(qa/2) = 0 . (381)

It’s solutions are

ν2± = α

(1

M1+

1M2

)± α

√(1

M1+

1M2

)2

− 4 sin2(qa/2)M1M2

. (382)

This time we obtain two different branches, see plot. To see the difference we expand forsmall qa ¿ 1. Then

ν− ≈√

α

2(M1 + M2)|qa| , (383)

which effectively corresponds to a lattice with a large elementary cell with the mass M1+M2

and halved spring constant α. The linear dispersion again indicates sound wave that’s whythis branch of the spectrum is called acoustical branch. For the other possibility we obtain

ν+ ≈√

2α

(1

M1+

1M2

)+ o((qa)2) . (384)

This corresponds to the oscillation mode when both members of the single elementarycell M1,2 oscillate in the opposite direction. Since often they are differently charged33

it produces strong dipole momenta and thus can emit/absorb electromagnetic radiation.That’s why it’s called optical branch.

How can we obtain the original homogeneous lattice? For M = M1 = M2 we obtain

ν2± =

2α

M[1± cos(qa)] ⇒ ν+ = 2

√α

M|cos(qa/2)| , ν− = 2

√α

M|sin(qa/2)| , (385)

which are plotted as red lines in Fig. 18. The lattice constant is now halved and the opticalbranch maps onto the acoustical branch between ±2π/a and ±π/a.

In 3D one usually obtains the following additional details:33A typical example is the NaCl crystal with Na+ positively and Cl− negatively charged.

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Figure 19: Dispersion relations of harmonic waves in realistic crystals.

• in general a cubic equation for ν2 (for Bravais lattices)

• 3 acoustical modes with different polarisations: one longitudinal (usually the fastest)and 2 transversal

• in most cases the dispersion relation is anisotropic (see Fig. 19).

Specific heat of the lattice in the Debye model

Let now return to the calculation of the specific heat. The first step of the specific heatcalculation within the Einstein model can be understood as the calculation of the averageenergy,

E =∑

R

~ωE(nB + 1/2) where nB =1

e~ωE/T − 1(386)

is the Bose distribution function, compare to (362). Back then we had only one singleoscillation mode, now we have a number of them labelled by ν(q). Therefore the calculationis now modified to

E =∑q

~ν(q)e~ν(q)/T − 1

, (387)

where we neglect the contribution of the 1/2 because it is temperature-independent andvanishes in the specific heat34

cV =1V

∂E

∂T=

1T 2

18π3

∑

diff polarisations

∫d3q

(~ν(q))2e~ν(q)/T

(e~ν(q)/T − 1

)2 . (388)

The integrand is obviously energy-dependent rather than momentum-dependent, that iswhy it is instructive to go over to the energy integration instead. We introduce a (boson)DOS DB(ν) so that DB(ν)dν is the number of modes between ν and ν + dν normalised tothe full number of modes 3N in the crystal35. Then in the isotropic (?) case we have

cV = 3N

∫dν DB(ν)

(~ν)2e~ν/T

(e~ν/T − 1

)2

1T 2

. (389)

34In fact, it is a divergent contribution.35If we are dealing with a lattice with basis, we have to multiply it with the number of basis atoms s.

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For the calculation of the DOS we make a simplifying assumption: we model the BZ by asphere with the same volume ⇒ there exists a maximal possible q, called Debye frequencyqD, which can be found from

N =V

(2π)343πq3

D ⇒ qD =(

6π2

vc

)1/3

. (390)

Therefore for the DOS we get (see also Fig. 20)

DB(ν)dν =4πq2dq43πq3

D

=3ν2

ν3D

dν , (391)

where νD = vsoundq if we assume to dealing with simple acoustical modes only not far awayfrom the zone centre. Then for the specific heat we get

cV = 3N

∫ νD

0dν

3ν2

ν3D

(~ν/T )2e~ν/T

(e~ν/T − 1

)2 = 3N3

ν3D

(T

~

)3 ∫ ~νD/T

0dx

x4ex

(ex − 1)3, (392)

after we perform a rescaling ν = xT/~. Similar to the Einstein temperature we can intro-duce the Debye temperature ΘD = ~νD/kB and obtain the Debye formula (1912)

cV = 9N(T/ΘD)3∫ ΘD/T

0dx

x4ex

(ex − 1)2. (393)

Let’s analyse some special cases.

• high temperature limit T À ΘD, then one expands the integrand

x4ex

(ex − 1)2→ x4(1 + . . . )

(1 + x− 1 + . . . )2→ x2 ⇒

∫ ΘD/T

0dxx2 =

13(ΘD/T )3

and thus obtain the Dulong-Petit law

cV = 3NkB . (394)

• low temperature limit T ¿ ΘD we use∫ ∞

0dx

x4ex

(ex − 1)2=

4π4

15

and get the T 3-law

cV =12π4

5NkB(T/ΘD)3 . (395)

This is completely different from the behaviour in the Einstein model. One might ask aquestion whether one really needs the Bose distribution function in order to get the T 3-law.Let’s use the simple Boltzmann weight instead36:

cV = 9N(T/ΘD)3∫ ΘD/T

0dxx4e−x ≈ 9N(T/ΘD)3

∫ ∞

0dxx4e−x

︸ ︷︷ ︸=24

= 9 · 24N(T/ΘD)3 , (396)

36The only difference is the absence of the −1 in the denominator.

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Figure 20: Left panel: dashed line is the dispersion relation in the Debye model. Thesolid line represents ν(q) in a realistic material. Right panel: Specific heat as a function oftemperature.

which is indeed almost the ‘quantum’ result because 12π4/5 ≈ 233, 78 and 9 · 24 = 216.How can we get the classical Dulong-Petit result? One can see that in the expansion of

the average energy (387) for small ~:

E =∑q

~ν(q)e~ν(q)/T − 1

→~→0

∑q

kBT . (397)

Interestingly, we need the Bose-Einstein distribution in order to correctly reproduce thisresult because using the simple Boltzmann weight leads to

∑q

~ν(q)e−~ν(q)/T →~→0

∑q

~ν(q) → 0 . (398)

2.6 Phonons

Let’s again consider scattering of electrons (or equally neutrons or photons) on the latticelike we did in the beginning of Section 2.3, but not looking on the energy structure but onthe scattering probabilities between the states |k〉 and |k′〉. The scattering potential (thelattice) can be written down as

U(r) =∑

R

vatom(r−R) . (399)

In the Born approximation the scattering matrix element (scattering amplitude) is simply

Mkk′ = 〈k|U(r)|k′〉 =1V

∫d3r e−ik′·r ∑

R

vatom(r−R)eik·r (400)

=1V

∑

R

ei(k−k′)·R∫

d3r ei(k−k′)·(r−R) vatom(r−R) = vatom(g)1N

∑

R

e−ig·R ,

where g = k− k′. The first part is called atomic factor defined as

vatom(g) =1vc

∫d3r eig·r vatom(r) , (401)

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Figure 21: Umklapp scattering.

and the second one is the structure factor . For the lattice in equilibrium (uR = 0) since

1N

∑

R

eig·R = δK,g (402)

we obtain

Mkk′ = vatom(k− k′) δK,k−k′ . (403)

Assuming an elastic scattering |k| = |k′| and denoting the scattering angle by θ one obtains

|K| = 2|k| sin θ . (404)

The distance d between planes orthogonal to K is proportional to d = 2πn/|K|, n = 1, 2, . . .(Exercise?) Then we obtain the Bragg’s constructive interference condition,

nλ = 2d sin θ . (405)

Let’s now consider oscillating lattice, then

R → R + uR = R +∑

|q|>0

(uq eiq·R + u∗q e−iq·R)

. (406)

Plugging this into the structure factor (402) yields

1N

∑

R

eig·R ≈ δK,g +1N

∑

R

e−ig·R ∑q

(−ig · uq) eiq·R + c. c. , (407)

where we have expanded for small uq. After performing the lattice sum we obtain

= · · ·+∑q

(−ig · uq) δg−q,K + c. c. + . . .

So that the following momentum conservation condition holds

q = k− k′ −K . (408)

Here q is the (quasi)momentum of the phonon. In case a finite K is necessary one dealswith the umklapp scattering , see Fig. 21. It is indeed a particle when there’s some kind

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of energy conservation. This can be shown when we include the time dependence into thecalculation of the scattering amplitude (400). For the plane wave one has

ψk(r, t) =1√V

eik·r−iEkt/~ . (409)

On the other hand, the dynamics of the lattice is described by the time-dependent oscilla-tions

R(t) = R +∑

q>0

(uq eiq·R−iν(q)t + u∗q e−iq·R+iν(q)t

). (410)

Expanding like before for small displacements uq and calculating the scattering amplitudeaccording to

Mk,k′ ∼∫

dt

∫d3rψ∗k(r, t) U(r, t) ψk′(r, t) ∼

∫dteiEkt/~−iEk′ t/~±iν(q)t

∼ δ [Ek − Ek′ ± ~ν(q)] , (411)

we obtain the analogue of the energy conservation relation:

Ek − Ek′ ± ~ν(q) = 0 . (412)

So the phonon is indeed a particle. How can we write down the corresponding Hamilton-operator? Let’s get back to our Example 1 of a homogeneous harmonic chain. Here thecan write down the Hamilton function as

H =12

∑q

(1M

p∗q pq + G(q) u∗quq

), (413)

where pq = Muq. Important fact is that every single q mode is independent. Thereforefor every single mode we can introduce creation/annihilation operators just like we did inquantum mechanics:

bq =1√

2~ν(q)M

[pq − iMν(q)u∗q

], b†q =

1√2~ν(q)M

[p∗q + iMν(q)uq

]. (414)

It can be easily checked, that they are bosons since we find

[bq, b†q′ ] = δq,q′ , [bq, bq′ ] = 0 , [b†q, b

†q′ ] = 0 . (415)

Inverting (414) and substituting the result back into (413) we obtain the Hamiltonian forthe phonons:

H =∑

q

~ν(q)(b†qbq + 1/2) . (416)

Generalisation to the 3D case is rather straightforward. If we have a Hamilton function

H =1

2M

∑

R

p∗RpR +12

∑

R,R′u∗R ·GR−R′ · uR′ . (417)

In the momentum space it therefore looks like

H =12

∑q

1M

p∗qpq + u∗q ·Gq · uq . (418)

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Unfortunately, Gq is not always diagonal and first must be diagonalised. This is of courseequivalent to the procedure of finding the eigenmodes. In 3D for the Bravais lattice we e. g.have i = 3 of them ν(i)(q). Then Gq = diag

M(ν(1)(q))2,M(ν(2)(q))2,M(ν(3)(q))2

and

we can introduce 3 sets of creation/annihilation operators

b(i)q =

1√2~ν(i)(q)M

[p(i)q − iMν(i)(q)u(i)∗

q

], (419)

with the help of which we write down the Hamiltonian:

H =∑

q,i

~ν(i)(q)[b(i)†q b(i)

q + 1/2]

. (420)

Later we shall also need uR in the second quantised form. Using the fact that since thedisplacement is a real quantity we find that pq = p∗−q as well as uq = u∗−q hold. Using itwe can invert (419) and obtain

u(i)q = −i

√~

2Mν(i)(q)(b(i)†

q − b(i)−q) . (421)

On the other hand, from (420) we know that for the time dependence of the annihilationoperator (in Heisenberg picture)

b(i)q (t) = b

(i)q (t = 0) e−iν

(i)q t . (422)

Therefore for the time-dependent displacement of the ion/atom at point R0 we obtain

uR0(t) = −i∑

q,i

√~

2MNν(i)(q)u(i)

q (0)(b(i)†q eiν

(i)q t − b

(i)−qe−iν

(i)q t

)eiq·R0 , (423)

where the exact form of the oscillation mode u(i)q (0) is found from solution of the eigenvalue

problem (372).

2.7 Electron-phonon interaction

Here we would like to take a closer look onto the full Hamiltonian of a solid:

H = Helectrons + Hphonons +∑

ij

U( ri︸︷︷︸electrons

− Rj︸︷︷︸ions

) .

In the last term we take into account the possibility of the lattice to oscillate Rj → Rj +uj .Performing the expansion in small displacement we obtain

U(ri −Rj − uj) = U(ri −Rj) + uj · ∇U(ri −Rj) + o(u2) . (424)

The first term of the rhs is exactly the term we’ve already discussed – it is responsible forthe band structure of the solid and its eigenvalues are found from Eq. (309). The gradientof this potential can be written down in the following form,

∇U(r) =i

N

∑

k

kU(k) eik·r . (425)

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Figure 22: Diagram for the electron-phonon scattering (a). Effective electron-electroninteraction as an exchange by a phonon (b).

Then the electron-phonon potential (not energy) is

Ue−ph(r) =∑

j

uj · ∇U(r−Rj) =i

N

∑

k

U(k)eik·r k ·∑

j

uj e−ik·Rj . (426)

We can now introduce in the last sum (423) in the Schrodinger representation (without thetime dependence) and perform the sum using (402). The result is

Ue−ph(r) =∑q

U(q−K)ei(q−K)·r√

~2NMν(i)(q)

(q−K) · u(i)q (0)

(b(i)†q − b

(i)−q

). (427)

The potential energy (and thus the contribution to the Hamiltonian) is the given by theproduct of the above potential with the probability to find an electron at (r), which isproportional to ρ(r) = Ψ†(r)Ψ(r), and integrated over whole crystal. This procedure yieldsthe electron-phonon interaction or Frohlich Hamiltonian:

He−ph =∑

q,i

Mi(q−K)ρ(q−K)(b(i)†q − b

(i)−q

),

Mi(q−K) =

√~

2ρ0ν(i)(q)U(q−K) (q−K) · u(i)

q (0) , (428)

where ρ0 = MN/V is the material density. In general there also must be a sum over allpossible vectors of the reciprocal lattice K. In reality only very few of them around K = 0make a dominant contribution. The electron density can, of course, be rewritten in termsof its Fourier components, which, in turn, can be rewritten in the second quantised form.Then the el-ph interaction Hamiltonian appears in the most convenient shape:

λ∑

k,q

a†k+qak (bq + b†−q) , (429)

and is usually represented by a diagram in Fig. 22.

BCS attraction mechanism

El-ph interaction plays decisive role in the BCS theory discussed before. Let’s estimate theeffective el-el interaction mediated by the el-ph scattering. There’re two processes: scat-tering of two electrons with momenta k,k′ into k− g,k′ + g as well as into k + g,k′ − g.

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The corresponding amplitudes in the second order perturbation theory37 are (q = g + K)

〈k− g,k′ + g|Ue−ph|k,k′〉 =Mk,k−gM∗

k′,k′+g

Ek − Ek−g − ~ν(q)

〈k + g,k′ − g|Ue−ph|k,k′〉 =Mk,k+gM∗

k′,k′−g

Ek − Ek+g − ~ν(q)=

Mk′′−g,k′′M∗k′,k′−g

Ek′′−g − Ek′′ − ~ν(q), (430)

where in the last term we made a substitution k + g = k′′. Then after summing them weobtain

Veff(g) ∼ |Mk,k−g|2 2~ν(q)(Ek − Ek−g)2 − [~ν(q)]2

. (431)

This can be regarded as the effective scattering amplitude of electrons on each other i. e.the effective amplitude of electron-electron interaction. One immediately realises that thescattering processes with

|Ek − Ek−g| < ~ν(q) , (432)

when the denominator is negative, corresponds to electron-electron attraction. Maximalvalue of ν(q) is the Debye frequency νD, that’s why the energy interval in which we haveeffective attraction is |Ek − Ek−g| < ~νD. That means that in our analysis of Chapter 1w = ~νD.

The amplitude can even be estimated for the case when M depends only on g:

|M(g)|2 ∼ |vatom(g)|2 |g · uq(0)|2︸ ︷︷ ︸Debye−Waller factor

∼ |vatom(g)|2 ~g2

2NMν(q). (433)

Therefore the effective el-el attraction is

Veff(g) ∼ −|vatom(g)|2q2

NMν2(q). (434)

The superconductivity even survives inclusion of screened Coulomb interaction φ(r) e.g.of the form (97). Since it is cut off by its maximal value 4πQ/k2

0, where k−10 is the screening

length and Ve−ph is divergent there’s always room in the energy space for the attractiveinteraction to emerge.

Peierls instability

During the 1970s a number of mostly organic materials has been discovered (such as e. g.TTF-TCNQ38), which were (half)metallic at room temperatures and against all expecta-tion became insulators at lower temperatures (for TTF-TCNQ below ∼ 73K). They wereessentially 1D and their electronic degrees of freedom can be described by a lineariseddispersion around ±kF ,

H =∫

dxψ∗(x)(−~

2∇2

2me− p2

F

2me

)ψ(x) ≈ −ivF

∫dx

(ψ†1(x)∂xψ1(x)− ψ†2(x)∂xψ2(x)

),(435)

37Lowest order PT gives zero contribution since all expectation values of the kind 〈b†q − b−q〉 vanish.38This is one of the first organic conductors discovered in 1973.

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where ψ1,2(x) are the wave functions or field operators for the right/left moving electrons.Then the full operators of arbitrary chirality can be written down as

ψ(x) = ψ1(x) eikF x + ψ2(x) e−ikF x . (436)

Let’s assume there’s some static modulation in the lattice u(x) = u0 cos(2kF x + ϕ). Thenthere’s some distortion potential proportional to its gradient U(x) = λ∂xu(x). Correctionto the Hamiltonian due to this static phonon field is given by

Hint =∫

dxU(x) ψ†(x)ψ(x) ≈∫

dx(ψ†1, ψ2†

) (0 ∆

∆∗ 0

)(ψ1

ψ2

), (437)

where ∆ = ikF λu0eiϕ and we’ve kept only the non-oscillating terms since the oscillatory

terms average away after the integration over the whole x-axis. Therefore in the momentumrepresentation the Hamiltonian can be written down as a matrix

H =(

vF k ∆∆∗ −vF k

). (438)

Its eigenvalues and eigenvectors are

ε± = ±√

∆2 + v2F k2 , |±, k〉 =

(u±−v±

)

u2± =

12

(1 +

vF k

ε±

), v2

± =12

(1− vF k

ε±

). (439)

So there’s a gap for the states around the Fermi edge. At half-filling all − states are filledand + states are empty, that’s why the correction to the energy due to the deformationalpotential is

δEel = Eel(∆)− Eel(∆ = 0) = −2s

∑

k

(√∆2 + v2

F k2 − vF |k|)

. (440)

Going over to the momentum integration and taking into account the fact that the resultingintegral is log-divergent we have to cut off at some finite energy vF kF = EF (i. e. bandwidthin TB model). Then we obtain

δEel = −L∆2

πvF

[ln

(2EF

|∆|)− 1

2

]. (441)

This estimation is, of course, correct as long as |∆| ¿ EF . So the energy of the electronicsubsystem is lowered. What about the positive energy correction due to the deformationpotential? We can estimate it in the following way39:

δElatt ≈ρ0v

2sound

2

∫dx (∂u(x)/∂x)2 = ρ0v

2soundk

2F u2

0L =L|∆|2

λ2. (442)

39The length of the curve is given by the integral∫ b

adx

√1 + (∂y/∂x)2. The profile of our curve is up to

the prefactor u(x). Taking into account that the elastic energy is

α

2x2 ≈ α

2

∫dx(∂u/∂x)2

and using that vsound ≈ a√

α/M we obtain this formula. Remember that the material density is ρ0 = M/a.

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Of course, at sufficiently small |∆| the gain in energy by opening a gap is larger than theelastic energy. The optimal gap is found by minimising the full system energy varying ∆.Then one obtains

∆0 ≈ 2EF e−πvF /λ2. (443)

The most important difference to the SC is the fact that while in the present case the gapopens around ±π/(2a) and its position does not change with the filling factor in the caseof the SC the gap is glued to the Fermi edge of the metal in the normal state. Please alsoobserve the quite large prefactor 2EF À ΘD of the formula. In different materials thisPeierls instability can take place at different temperatures up to several hundreds of K.

Thermal expansion of solids and phonon-phonon interaction

With a growing temperature most of the solids expand. That means that the new equilib-rium ion positions R change. It turns out that in order to cover this phenomenon one needsat least cubic terms in the expansion (366). The full free energy of the crystal contains thentwo contributions: (i) one from the changing volume (higher order terms); (ii) the harmoniccontribution where the frequency depend on volume. For the latter one can assume that

δν

ν= −γ

δV

V, (444)

where γ is some constant. Then the free energy is

F =12κ

(δV

V

)2

︸ ︷︷ ︸(i)

+ T∑q

ln[2sh

(~ν(q)2T

)]

︸ ︷︷ ︸(ii)

, (445)

where κ is the compressibility of the crystal. Correspondingly after the variation withrespect to δV we obtain the following equation

1κ

δV

V=

∑q

γ ~ν(q)12cth

(~ν(q)2T

)= γE(T ) , (446)

where E(T ) is the average energy density in the system. Thus we obtain the Gruneisenformula: the relative thermal expansion at temperature T is proportional to the averageenergy density:

δV

V= κγE(T ) . (447)

The thermal expansion coefficient , which is the derivative of the above formula with respectto temperature is then proportional to the specific heat cV . The Gruneisen parameter γcan then be found in the language of the Debye temperature as

γ = −∂ lnΘD

∂ ln V. (448)

A very important fact is that within the harmonic approximation, as long as we consideronly quadratic in uR terms in (366) the eigenfrequencies of the crystal does not change andthus no thermal expansion is possible. This is easy to see for e. g. a small change in theequilibrium positions ε of the ions: R → (1 + ε)R+uR. Then we effectively have a change

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of displacements uR → εR + uR. It can be inserted back into the quadratic term of theoriginal expansion of the potential energy:

U = U0 +12

∑

R,R′uRuR′ GR−R′ + . . .

→ U0 +12ε2

∑

R,R′RR′GR−R′ +

12

∑

R,R′uRuR′ GR−R′ + . . . (449)

The first two terms do not depend on the displacement and the quadratic in displacementterm has still the same coefficients ⇒ the frequencies do not change [BUT the volume ofthe crystal is now, of course, larger by a factor of (1 + ε)3!].

In order to describe the frequency change we need terms of higher orders – the anhar-monic terms, primarily the cubic ones,

U (3) =13!

∑

R,R′,R′′,j,j′,j′′=x,y,z

ujRuj′

R′uj′R′′

∂3U

∂ujR∂uj′

R′∂uj′′R′′

=∑

R,R′,R′′uRuR′uR′′ GR,R′,R′′ . (450)

We can rewrite it in Fourier components uR =∑

q uq eiq·R,

=∑

R,h′,h′′

∑

q,q′,q′′G(h′,h′′)uquq′uq′′e

i(q+q′+q′′)·R eiq′·h′ eiq′′·h′′ . (451)

After performing the sum over R and introducing the Fourier transform of the G matrixwe obtain

∑

q,q′,q′′G(q′,q′′)uquq′uq′′ δq+q′+q′′,K , (452)

where the last δ-function generates a (quasi)momentum conservation relation similar to thatof (408), which suggests that now phonons are scattered on phonons. In order to verifythat picture we can use the quantisation prescription (421). Then we would obtains termsof the kind b†q′+q′′ bq′ bq′′ , which is nothing but a phonon-phonon scattering, see Fig. 23.We have thus established that the anharmonicity of the ionic potential is responsible forphonon-phonon interaction, which, in turn is related to thermal expansion properties ofcrystals.

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2.8 Solids in external fields

Boltzmann equation

Now we would like to develop an approach which would allow us to calculate the trans-port properties of metals taking the band structure, phonon effects as well as scatteringon impurities into account. One very efficient method to do so is the Boltzmann equationapproach. The principal quantity we want to calculate is the change in the particle dis-tribution function fk(r). In the uniform equilibrium case it is not r-dependent any moreand we have fk(r) = nF (k). How can the distribution function change? There’re manypossibilities:

• diffusion of charge carriers with velocity vk, then

fk(r, t) = fk(r− vkt, 0) ,

which means that

∂fk

∂t

∣∣∣diff

= −vk · ∇fk (453)

• electromagnetic fieldsp = ~k = eE +

e

cvk ×H

then since fk(r, t) = fk−kt(r, 0) we then obtain

∂fk

∂t

∣∣∣field

= −k∂fk

∂k= − e

~

(E +

1cvk ×H

)∇kfk . (454)

• scattering processes

∂fk

∂t

∣∣∣scatt

=∫

d3k′ [fk′(1− fk)− fk(1− fk′)]Q(k,k′) , (455)

where Q(k,k′) is the transition probability.

In the stationary state (which is not necessarily an equilibrium one) we must have

∂fk

∂t

∣∣∣field

+∂fk

∂t

∣∣∣scatt

+∂fk

∂t

∣∣∣diff

= 0 (456)

so that

−vk · ∇fk − e

~

(E +

1cvk ×H

)∇kfk = −∂fk

∂t

∣∣∣scatt

(457)

We now introduce the difference between the actual distribution function and that of theequilibrium:

gk = fk − f0k = fk − f0

k[T (r)] , (458)

which allows us to rewrite the equation in the following form

− vk ·(

∂f0k

∂T∇T

)− e

~

(E +

1cvk ×H

)∇kf0

k

= −∂fk

∂t

∣∣∣scatt

+ vk · ∇gk +e

~

(E +

1cvk ×H

)∇kgk . (459)

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In most cases we can safely assume that the equilibrium configuration of the system isdescribed by the Fermi distribution

f0k = nF (k) =

[e(Ek−µ)/T − 1

]−1and use that vk =

1~

∂Ek∂k

.

Then we can write the Boltzmann equation in the following form(

∂f0k

∂E)

vk ·[−(Ek − µ)

T∇T + (eE−∇µ)

]

= −∂fk

∂t

∣∣∣scatt

+ vk · ∇gk +e

~c(vk ×H) · ∇kgk , (460)

where we have omitted the term E · ∇kgk which is of the order E2 and thus is responsiblefor non-linear corrections beyond the Ohm’s law40.

DC conductivity

is the first quantity we would like to evaluate with the help of the Boltzmann equation. Weassume ∇T = 0, H = 0 and homogeneity of the system ∇gk = 0. Then the above equationreduces to

(∂f0

k

∂E)

vk · eE = −∂fk

∂t

∣∣∣scatt

=∫

d3k′ (fk − fk′)Q(k,k′)

=∫

d3k′ (gk − gk′)Q(k,k′) . (461)

Now we need some information about the scattering in the system. We make the simplestpossible assumption which is called relaxation time approximation:

−∂fk

∂t

∣∣∣scatt

=gk

τ⇒ −∂gk

∂t

∣∣∣scatt

=gk

τ, (462)

where we again recognise the relaxation time τ we have introduced in the Drude model.The solution is then simple

gk = −(

∂f0k

∂E)

τvk · eE . (463)

The electric current can then be calculated by the analogy with the prescription (2):

J = 2s

∫d3k evkfk = 2s

∫d3k evkgk , (464)

because there’s no current without E and thus for fk = f0k the integral is identically zero.

Now we rewrite the k-summation in terms of the integration over the Fermi surface asdiscussed in (326) and obtain

J =1

4π3

∫dE

∫dSF e2τvk (vk ·E)

(∂f0

k

∂E)

1|∇kEk| =

e2τ

4π3~

∫dSF

vk

|vk| vk ·E .

It yields then for the conductivity tensor (observe that it is not a scalar product in theintegrand, but a tensor one)

σ =e2τ

4π3~

∫dSF

vk

|vk| vk . (465)

40Please observe that because vk · (vk×H) = 0 the magnetic field does not enter the lhs of the equation.

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In the most symmetric situation E ‖ J, that’s why vk vk ·E = v2xE = v2E/3, therefore

σ =e2τ

12π3~

∫dSF v =

e2

12π3~

∫dSF lm , (466)

where lm = vτ is the mean free path. Let’s now take a look onto the new fk. gk is big invicinity of the Fermi edge E = EF and it is: (i) positive for vk ·E > 0 (e−s are accelerated);(ii) negative for vk ·E < 0, see Figure ... Actually from (463) we find

fk = f0k −

∂f0k

∂E∂E∂k︸︷︷︸vk

eτ

~E = f0

k −∇kf0k ·

eτE~

= f0(k− eτ

~E

). (467)

This is nothing else but the shifted distribution function just as we have discussed before.Here we can immediately read off the average drift velocity as

vdrift ≈ δpme

=eτEme

. (468)

Residual resistivity

The Boltzmann equation is not yet solved. The most general solution in the lowest orderin the applied electric field is given in the form

gk = −∂f0k

∂E eE ·Λ(k) , (469)

where Λ(k) is some vectorial mean free path. It can alternatively be written via direction-dependent relaxation time Λ(k) = vkτ(k). But how can we calculate these quantities? Wefirst assume the scattering to be elastic and its amplitude to depend on the scattering angleonly. Then

Q(k,k′)d3k = δ(E − E ′)L(k,k′) dΩ′︸︷︷︸solid angle in direction k′

dE ′ , (470)

where L(k,k′) describes scattering processes. Now we substitute (463) into (461) using theabove relation and obtain the self-consistency condition41

vk ·E = τ

∫dΩ′ L(k,k′) (vk − vk′) ·E (471)

for the function L(k,k′). For the most symmetric case it only depends on the angle θbetween the vectors k and k′. Moreover we can assume that vk ‖ E. Then we obtain

Λ−1 =∫

dΩ′ L(θ)(1− cos θ) . (472)

The last factor (1 − cos θ) reflects the fact that the backscattering, i. e. processes withθ = π are most effective mechanisms for generation of finite resistivity. L(θ) is very oftennothing but the scattering cross section σ(θ). If we denote the impurity concentration byni then we obtain for the inverse relaxation time42

Λ−1 = ni 2π

∫ π

0dθ σ(θ) sin θ (1− cos θ) . (473)

41The δ-functions as well as derivatives of f0k vanish after integration over energy.

42Here we use dΩ = dϕ dθ sin θ and calculate the integral over ϕ.

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To make further progress let’s make some simplifying assumptions. First of all we con-sider scattering among the electrons in vicinity of the Fermi edge, so that |k| = |k′| = kF .In the Born approximation up to the prefactor σ(θ) = |Vg|2, where g = k− k′ and g =2kF sin(θ/2). Thus we get for the scattering cross section in the Thomas-Fermi approxima-tion,

σ(θ) =(2meZe2/~2

)2 (g2 + k2

0

)−1, (474)

where k−10 is the screening length. Gathering everything we can calculate the resistivity

due to static scatterers,

ρ = σ−1 = me/(ne2τ) =mevF

ne2Λ(475)

=mevF

ne2ni2π

∫ π

0dθ sin θ(1− cos θ)

(2meZe2/~2

)2 (g2 + k2

0

)−1

Now we make a substitution z = sin(θ/2) and calculate the integral

=mevF

ne2ni2π

(2meZe2

~2k20

)2 ∫ 1

0dz

8z3

[1 + (2kF /k0)2z2]2

=mevF

ne2ni2π

(2meZe2

~2k20

)2 (1 + a) ln(1 + a)− a

2a2(1 + a), (476)

where a = (2kF /k0)2 is just a numerical constant. For the most metals it’s of the orderunity. This is a fundamental result and has a number of universal properties:

• it is temperature-independent and persists towards T = 0, that’s why it’s calledresidual resistivity

• ρ ∼ Z2, where Z is valence difference between the charge of host ions and impurities.This law is aka Linde rule

• the structure of the result is such that it looks like the conductance electrons arescattered by hard spheres with an effective radius

R∗ ∼ 2meZe2

~2k20

.

In general, however, this formula overestimates the resistivity because it does not take intoaccount the Friedel oscillations.

Phonon contribution to resistivity

Let’s now consider how the electron-phonon scattering affects the conductance propertiesof the metal. We need to modify the result (473) and use the corrected scattering crosssection

σ(g) = σion(g) N |g · uq|2 , (477)

where σion is the contribution of the atomic factor and the rest is coming from the structurefactor after expansion in the lowest order in electron-phonon coupling (see the discussion

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in Section 2.6). The latter can be estimated with the help of Debye-Waller factor43, whichat high temperatures (T À ΘD) yields

N |g · uq|2 ≈ Tg2

Mν(q)2≈ T

Mv2sound

≈ ~2q2DT

MΘ2D

. (478)

This result does not depend explicitly on g any more therefore the angle integration is triv-ial. The ion contribution can be calculated as before and gives a temperature-independentfactor

σion = 2π

∫ π

0dθ sin θ σion(θ)(1− cos θ) . (479)

So instead of the result (475) we now obtain

ρph =mevF

ne2

1Λi≈ mevF

ne2Nσion

~2q2DT

Mθ2D

. (480)

So that at high temperatures we have: (i) ρ ∼ T and (ii) ρ ∼ Θ−2D .

At low temperatures the calculation changes a little bit. The linear T dependence in(478) is the high-T expansion of the Bose distribution

kBT

~ν(q)=

T

ν(q)=

1eν(q)/T − 1

. (481)

Thus we change the phonon contribution to the scattering cross section to

N |g · uq|2 ≈ kBT

Mv2sound

~ν(q)/kBT

e~ν(q)/kBT − 1. (482)

The effect of the factor containing the exponential is to cut off all processes with momentumtransfer larger than that for ~ν(q) > kBT . Plugging the above formula into (477) and thenusing (473) we obtain44

Λ−1 = N2π

∫ π

0dθ sin θ(1− cos θ)σion(θ)

~ν(q)/kBT

e~ν(q)/kBT − 1kBT

Mv2sound

. (483)

We now assume that for the relevant scattering angles σion(θ) is only weakly θ-dependentand thus we can set σion(θ) ≈ σion. Furthermore, due to the fact that qD is the largest mo-mentum possible to be absorbed by phonons there’s a maximal scattering angle sin(θmax/2) =qD/2kF . We now go over to integration over q using that cos θ = 1− 2(q/2kF )2, then

Λ−1 = −N2πσion

∫ qD

0d

[1− 2(q/2kF )2

]2(q/2kF )2

kBT

Mv2sound

~vsoundq/kBT

e~vsoundq/kBT − 1

=πNσion

k4F

∫ qD

0dq q3 ~vsound

Mv2sounde

~vsoundq/kBT − 1. (484)

Now we make a substitution z = qΘD/qDT (here we again use the fact that kBθD =~vsoundqD) and obtain

Λ−1 =πN

k4F

(T

ΘD

)5 vsound

Mσion

∫ ΘD/T

0dz

z4

ez − 1. (485)

This integral is very similar to the one we’ve already encountered in (393). We can discussthe limiting cases in the analogous way.

43We use that ν(q) = ~vsoundq and that θD ≈ qDvsound.44Here we’ve got a different prefactor N = N/V with V = 1, which is proportional to the material density,

instead of the impurity concentration ni in the formula for the residual resistivity.

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• high-temperature limit T À ΘD, expand the integrand in z and integrate:

Λ−1 ∼ (T/ΘD)5∫ ΘD/T

0dz z3 ∼ T/ΘD

this is the already known result

• low-temperature limit T ¿ ΘD. We can use that∫ ∞

0dzz4/(ez − 1) = 24ζ(5) ≈ 24.89 . . . ,

then we obtain the Bloch-Gruneisen law

ρph ≈ 24ζ(5)σionmevF

ne2

πN

k4F

(T

ΘD

)5

. (486)

Both temperature regimes adequately describe the temperature dependence of resistivityof many metals and alloys45. There are a number of possibilities of how to improve theabove analysis:

• include inelastic processes

• use the correct phonon spectrum beyond the Debye model

• use realistic θ-dependent σion

• include the effect of umklapp scattering

2.9 Quantum Hall effect

2D electron gas in magnetic field

Let’s now turn to the situation of a finite magnetic field. While from the classical pointof view the finite constant electric field induces a movement of charge carriers along itselfin an effective 1D geometry (in a clean system) the magnetic field forces the particles tomove in a 2D plane perpendicular to B. So let’s analyse the dynamics of electrons in such ageometry. Assume there’s some kind of confinement U(y) in y-direction, see Fig. 24. Thenthere’s a following Schr”odinger equation for a single electron in this 2D plane:

[(i~∇+ eA)2

2me+ U(y)

]ψ(x, y) = Eψ(x, y) . (487)

Let’s use the Landau gauge for the magnetic field in the z-direction:

Ax = −By , Ay = 0 or A = ex By .

Then [(px + eBy)2

2me+

p2y

2me+ U(y)

]ψ(x, y) = Eψ(x, y) , (488)

which can be reduced to a 1D equation by a separation ansatz

ψ(x, y) =1√L

eikxχ(y) ,

45One of the best matchings is achieved for MgB2.

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Figure 24: Geometry of the quantum Hall effect sample.

where L is the length of our sample in the x-direction:[

(~k + eBy)2

2me+

p2y

2me+ U(y)

]χ(y) = Eχ(y) . (489)

We first consider a parabolic confinement U(y) = meω20y

2/2 and zero field B = 0. Thenwe are dealing with a harmonic oscillator Hamiltonian because

[~2k2

2me+

p2y

2me+

12meω

20y

2

]χ(y) = Eχ(y) . (490)

It has following energy levels

Ek,n =~2k2

2me+ ~ω0(n + 1/2) , n = 0, 1, 2 . . . (491)

The wavefunctions are correspondingly

χn(y) = e−q2/2 Hn(q) , where q =√

meω0/~ y , (492)

and Hn are the Hermite polynomials. The particle velocity can easily be found as before

vk =1~

∂Ek,n

∂k=~kme

. (493)

Now we move to the opposite case: set confinement potential to zero U(y) = 0 andconsider finite field B 6= 0. Then (490) can be rewritten in the following form

[12meω

2c (yk + y)2 +

p2y

2me

]χ(y) = Eχ(y) , yk = ~k/(eB) , ωc = eB/me (494)

is nothing but the cyclotron frequency . Effectively we now have the same situation as beforebut with the center of the oscillator potential being shifted to −yk. Correspondingly weobtain now

χn,k(y) = χn(q + qk) with qk =√

meωc/~ yk (495)

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Figure 25: Stratification of the fermion continuum into Landau levels.

with the energy levels

Ek,n = ~ωc(n + 1/2) . (496)

These energy levels are the Landau levels (LLs) (see Fig. 25) and they have the followingproperties:

• their group velocity vk = ∂Ek,n/∂k = 0, thus there’s no net movement of the particles!

• the spacial extent of these levels is about√~/meωc which is the scale for y above

and, on the other hand, it coincides roughly with the radius of the classical particletrajectory

• different k correspond to different positions of levels along the y-axis

• LLs are highly degenerate. The number of electrons which fit into the level with aspecific n in the sample of width W and length L is given by

ζ = 2sW

δyk, since δyk = ~δk/eB and δk = 2π/L (497)

we obtain

ζ = 2sWeBL

h=

2eBS

h. (498)

The higher the magnetic field the higher the level degeneracy.

We conclude that the original Fermi continuum stratifies in a number of energeticallylocalised highly degenerate LLs separated by ~ωc, see Fig. 25. In reality these new en-ergy bands have finite width due to residual scattering on disorder impurities. When themagnetic field is high enough they can be made well separated from each other.

Integer Quantum Hall Effect (IQHE)

The experimental evidence presented by von Klitzing and colleagues 1980 is depicted onFig. 26. There are plateaus in the transversal resistivity, on which there’s virtually nolongitudinal resistivity. The number of filled LLs on the plateaus is

ν =nS

ζ=

nh

2eB(499)

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Figure 26: Experimental evidence for the IQHE.

an integer. If we remember the definition (253) of the flux quantum Φ0 = hc/2e we canmeasure the magnetic flux through the sample in these units Nflux = BS/Φ0. Then thefilling fraction is just

ν =Nelectrons

Nflux. (500)

So, on the plateaus we have an integer number of LLs full and the chemical potential liesbetween them. Since there’s a gap to the next empty LL the Hall state is an insulator.Sometimes the QH liquid is called incompressible, because there’s a gap. Why there’s afinite conductivity then? The answer to that are the edge states.

Let’s consider a system of a finite size in a confinement potential of a general form U(y),see e. g. Fig. 24. The lowest order correction to the dispersion relation of the electrons isthen very simple:

Ek,n = ~ωc(n + 1/2) + 〈k, n|U(y)|k, n〉 . (501)

If U(y) changes only slightly on the typical geometrical scale√~/meωc of an LL, then we

obtain

Ek,n ≈ ~ωc(n + 1/2) + U(yk) . (502)

Calculation of the wave packet velocity then gives

vk,n =1~

∂Ek,n

∂k=

1~

∂U(yk)∂k

=1~

∂U(y)∂y

∂yk

∂k=

1eB

∂U(y)∂y

(503)

The last derivative has, however, different signs on different edges of the sample ⇒ theelectrons on the opposite edges move in the opposite directions!

Since the electrons of each edge are chiral the only backscattering mechanism is viatunneling of the electron across the whole sample to the opposite edge. This is exponentiallysuppressed, therefore the longitudinal conductance is almost perfect.

We also observe that the precise geometrical shape of the edge is not important. If wechip away a part of the sample than the electron movement would not be much affectedby it. Important is only the fact of different signs of the confinement potential on the

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Figure 27: Non-equilibrium transport in a QH sample.

opposite edges. The edge states (channels) are thus topologically protected. In the bulk thesample is an insulator on the edge a conductor. Such materials/setups belong to the classof topological insulators.

Let’s discuss what happens with the transversal conductivity. We apply the voltage inthe following way: on both sides we couple the edges to two reservoirs kept at chemicalpotentials µL,R, see Fig. 27. These reservoirs fix the chemical potential of all outgoing(injected) particles and absorb everything which comes in. Then the upper edge carries thechemical potential µR and the lower one µL. The electric current is then calculated like

J = 2se1L

∑

k,n

vk,n = 2se∑

n

∫ µL

µR

dk

2πvk,n = 2se

∑n

∫ µL

µR

dk

2π~∂Ek,n

∂k

=2e

h

∑n

∫ µL

µR

dE =2e2

h︸︷︷︸G0

MV , (504)

where M is the number of edge states with energies between µL and µR. Here we observethe conductance quantisation.

There’s also an alternative explanation of the conductance quantisation, constructedby Laughlin and Halperin. Let’s assume we’ve deformed our sample into an annulus,see Fig. 28. The applied voltage is now modelled by an electromotive force induced bya magnetic flux Φ′ produced by a field B′ localised in the annulus. Because of the gaugeinvariance changing Φ′ by Φ0 should not affect the properties of the system. As the magneticfield grows the energies of the LLs at the inner edge of the annulus grow while those on theouter edge decrease. In fact, after changing the flux by Φ0 n LLs move above the Fermiedge at the inner edge while n of them dive below EF at the outer edge. To compensatefor that the system brings n electrons from the inner edge to the outer one. When t0 isthe time Φ changes by Φ0, then the transverse current is Iy = ne/t0. On the other hand,the electromotive driving force is Vx = Φ = Φ0/t0. Thus for the transverse conductance weobtain σxy = Iy/Vx = (2e2/h)n, as before.

Now we would like to take a look onto how the multi-particle wavefunction looks like.In order to obtain it we go over to the very convenient disc geometry and use the symmetricgauge, defined as

A = B× r/2 = (−y, x, 0)Bz/2 . (505)

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Figure 28: Laughlin-Halperin construction (Corbino geometry).

Then the Hamiltonian can be rewritten in the form:

H =12

[(−i∂x − y/2)2 + (−i∂y + x/2)2

], (506)

where everything has been made dimensionless by using the magnetic length

` =√~c/(eB) , (507)

as the length unit and the cyclotron frequency ~ωc as the energy unit. Now we introducethe following transformation to the complex plane

z = x− iy = re−iθ , z = x + iy = reiθ . (508)

The derivatives change according to

∂x = ∂z + ∂z and ∂y = −i(∂z − ∂z) . (509)

The full Hamiltonian in these new operators looks like

H =12

(−4

∂2

∂z∂z+

14zz − z∂z + z∂z

). (510)

In analogy to the harmonic oscillator we now introduce bosonic operators:

b = (z/2 + 2∂z)/√

2 , b† = (z/2− 2∂z)/√

2 ,

a† = (z/2− 2∂z)/√

2 , a = (z/2 + 2∂z)/√

2 , (511)

because they satisfy the usual bosonic commutation relations [a, a†] = 1, [b, b†] = 1. In thislanguage the Hamilton operator is then

H = a†a + 1/2 . (512)

The other operators define the angular momentum of the states:

L = −i~∂θ = −~ (z∂z − z∂z) = −~(b†b− a†a) . (513)

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H and L commute, that’s why the eigenstates carry the angular momentum index m. Soas before E = n + 1/2 and

|n,m〉 =(b†)m+n

√(m + n)!

(a†)n

√n!|0, 0〉 . (514)

The vacuum state is trivially

〈r|0, 0〉 =1√2π

e−zz/4 . (515)

In the LLL we then get

|0,m〉 =(b†)m

√m!

|0, 0〉 =zme−zz/4

√2π 2m m!

. (516)

We can now write down the multi-particle state of a fully filled LLL, which is, of course,given by the Slater determinant

ΦLLL =

∣∣∣∣∣∣∣∣

z01 z0

2 z03 . .

z11 z1

2 z13 . .

z21 z2

2 z23 . .

. . . . .

∣∣∣∣∣∣∣∣exp

(−1

4

∑

i

|zi|2)

=∏

j<k

(zj − zk) exp

(−1

4

∑

i

|zi|2)

,(517)

where we have used the fact that this is the Vandermonde determinant . This is not yettranslationally invariant. But it can be done so (as did Laughlin):

ΦLLL =∏

j<k

(zj − zk) exp

(− 1

4N

∑

i

|zj − zk|2)

exp

−N

4

∣∣∣∣∣1N

∑

i

zi

∣∣∣∣∣2 (518)

The crucial feature of this result is its dependence on∏

j<k(zj − zk), which automaticallysatisfies the Pauli exclusion principle.

IQHE relies heavily onto the presence of disorder. This can be shown in the followingway. Assume we have a totally clean system without disorder. It is fully translationallyinvariant. In the reference frame of the system when E = 0 then J = 0 even in presenceof some finite field B = Bez 6= 0. We now look onto the system from the frame movingwith velocity v = vex 6= 0. Then there’s a current density J = −env. Thus there’s a finiteelectric field E = vex × B/c = −vBey/c. From Jx = σxyEy we find σ12 = n/B, whichis independent on v. So there shouldn’t be any oscillations or QHE at all. That meansthat our assumption of the translation invariance of the system is wrong. Disorder is alsoresponsible for the finite width of the plateaus: the dirtier the sample the wider are theplateaus.

Fractional quantum Hall effect (FQHE)

In cleaner samples at slightly higher magnetic fields one can observe resistivity plateaus,which correspond to fractional filling ν = 1/3, 1/5, 1/7, . . . as well as much weaker ones atν = 2/5, 3/7, 2/3, 3/5, 4/7, . . . . In this case all electrons in the system occupy the lowestlying LL (LLL). According to (496) they all have the same kinetic energy, which is thusirrelevant and can be disregarded. For this reason the Hamilton operator of the system isvery simple:

H = −N∑

i<j

1|zi − zj | . (519)

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Although this Hamiltonian looks very simple, there’re major problems solving it. The mainreasons for that are:

(i) the above Hamiltonian is parameter-free

(ii) the degeneracy problem: the number of ground states without the magnetic field isastronomically large (because of the huge degeneracy of the LLs)

(iii) there’s no ‘normal state’ (this is the state in which the interaction is switched off46)

Some very interesting features of the FQHE can already be seen in the wavefunctionfor two interacting particles. If m is their relative angular momentum and M is center ofmass angular momentum, then the unique analytic wave function is47

ΨmM (z1, z2) = (z1 − z2)m (z1 + z2)M e−(|z1|2+|z2|2)/4 . (520)

We observe that the precise form of the interaction potential does not matter – this solutionis the same for any central potential V (|z1 − z2|)! It actually describes a bound state oftwo particles. In the absence of magnetic field the electrons could convert their potentialinto the kinetic energy and fly apart. Here it is not possible because the kinetic energy iscompletely quenched. The precise form of the potential only enters the matrix elements

vm =〈mM |V |mM〉〈mM |mM〉 , (521)

which are called Haldane pseudopotentials. Despite the simplicity of the above wavefunc-tion it is not possible to construct an arbitrary multi-particle state. However, due to theingenious heuristic idea of Laughlin one can write it down for filling fraction ν = 1/n, wheren = 3, 5, 7, . . . odd integer: up to the normalisation

Ψ(z1, z2, . . . , zN ) =N∏

i<j

(zi − zj)n . (522)

Of course, it is totally antisymmetric as required. Surprisingly, according to numericalsimulations it represents the correct wavefunction whatever the interaction potential! [justas (520)]. It describes an incompressible QH liquid – there’s a gap within the LLL. X.-G.Wen has also shown (in 1991) that there’re edge states. They are, however, not non-interacting electrons but are rather described by Tomonaga-Luttinger liquid theory withinteraction parameter equal to the filling g = ν = 1/n. These are the elementary excitations– Laughlin quasiparticles in the systems and they carry fractional charge equal to νe. Thishas been shown in a number of different experiments.

The most successful paradigm explaining many of these effects in the composite fermiontheory (CF), put forward by Jain, Halperin, Lee and Read (1993). The idea is the following.In the state with filling fraction ν there are 1/ν magnetic flux quanta per electron. It issuggested, that every 1/ν−1 of them participate in a bound state with an electron forminga composite fermion, see Fig. 29. This novel kind of quantum liquid is then subject tothe reduced magnetic field with precisely one flux quantum per CF, which means a ‘fillingfraction’ of unity. So effectively one maps the problem to the IQHE for CFs. Microscopicderivation of the CF theory does not yet exist though.

46We have learned e. g. in the case of BCS theory, that SC state can be understood as some kind of aninstability of the normal state, which was in that case the Fermi liquid.

47We neglect the LL mixing, which might be important for strong interaction potentials.

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Figure 29: Composite fermion scenario of the FQHE.

2.10 Magnetic properties of solids

Here we would like to discuss how the solids response to the application of external magneticfields. In general there’re two properties:

• (specific) magnetisation M(H), which at T = 0 is found from the field dependence ofthe ground state

M(H) = − 1V

∂E0(H)∂H

(523)

and at finite temperature from the thermal average

M(H, T ) =∑

n Mn(H) e−En/T

∑n e−En/T

, Mn(H) = − 1V

∂En(H)∂H

(524)

the last two formulas can be rewritten via the free energy F as

M = − 1V

∂F

∂H, F = −T lnZ = −T ln

∑n

e−En(H)/T . (525)

• magnetic susceptibility

χ =∂M

∂H= − 1

V

∂2F

∂H2, (526)

describes how easy/difficult it is to induce a finite magnetisation of the system48

There’re three different contributions: the one coming from the atoms/ions of the lattice– atomic contribution, the other one from the delocalised electrons in the bands – bandcontribution, and, finally, the nuclear contribution. However, due to much larger nuclearmasses their contribution is by a factor 106 − 108 that that of the electrons. Therefore inthe rest we neglect it.

Atomic contribution

In order to analyse it we take a closer look on the corresponding Hamiltonian for a onesingle atom with i = 1, . . . , N electrons:

H =1

2me

∑

i

[pi +

e

cA(ri)

]2+ µBg0S ·H , (527)

48Very often (526) is taken in the limit H → 0.

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where S is the full electronic spin of the atom. g0 ≈ 2 is the gyromagnetic ratio (Landefactor) of the free electron and µB = e~/(2mc) is the Bohr’s magneton. Now we use thesymmetric gauge (505) and rewrite it in the following form

=1

2me

∑

i

p2i

︸ ︷︷ ︸H0

+µB( L︸︷︷︸~L=

∑i ri×pi

+g0S) ·H +e2

2mec2H2

∑

i

(x2i + y2

i ) = H0 + HH . (528)

As a rule the correction to the energies ∆En to the energy level En due to HH is small, sothat we can use perturbation theory and find after keeping only terms of first and secondorder in H

∆En = µBH · 〈n|L + g0S|n〉+∑

n 6=n′

|〈n|µBH · (L + g0S)|n′〉|2En −En′

+e2H2

8mcc2〈n|

∑

i

(x2i + y2

i )|n〉 . (529)

These three terms have the following orders of magnitude:

(i) µBH · 〈n|L + g0S|n〉 ∼ µBH = (~e/(mec))H ∼ ~ωc, which for a field of 1 T is about10−4 eV

(ii)

e2H2

8mcc2〈n|

∑

i

(x2i + y2

i )|n〉 ∼(

eH

mec

)2

mea20 ≈ ~ωc

(~ωc

e2/a0

), (530)

where a0 is the Bohr’s radius and thus e2/a0 is about 1Rd ⇒ (ii) ¿ (i)

(iii)

∑

n6=n′

|〈n|µBH · (L + g0S)|n′〉|2En −En′

∼ (~ωc)2

∆, (531)

where ∆ = En − En′ is of the order of eigenenergies of electrons and thus ∼ 1Rd aswell, so (iii) ¿ (i)

Let’s now analyse different situations:

• simplest case: dielectric with completely full shells (He, H, Ar, Ne solids; alcali-halidese. g. LiF, NaCl). Here J|0〉 = S|0〉 = L|0〉 = 0 and only term (ii) survives. Then forthe correction to the ground state energy

∆E0 =e2

8mec2H2〈0|

∑

i

(x2i + y2

i )|0〉 =e2

12mec2H2〈0|

∑

i

r2i |0〉 therefore

χ = −N

M

∂2∆E0

∂H2= − e2

6mec2ρ 〈0|

∑

i

r2i |0〉

︸ ︷︷ ︸>0

, (532)

meaning that we’re dealing with χ < 0 called Larmor diamagnetism

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• J = 0 but L 6= 0 and S 6= 0, then the term (i) vanishes and the terms (ii) and (iii)are non-zero. This leads to

∆E0 =e2

12mec2H2〈0|

∑

i

r2i |0〉 −

∑n

|〈n|µBH · (L + g0S)|0〉|2En −E0

(533)

and thus to two different contributions to magnetic susceptibility

χ = −N

V

[e2

6mec2〈0|

∑

i

r2i |0〉 − 2µ2

B

∑n

|〈n|(L + g0S)|0〉|2En − E0

]. (534)

Since En > E0 then the second term has the opposite sign and is responsible for thevan Vleck paramatnetism.

• J 6= 0, then the term (i) is the dominant one. Now the ground state is (2J + 1)-folddegenerate though. According to Wigner-Eckart theorem49 we then obtain

〈JLSJz|Lz + g0Sz|JLSJ ′z〉 = g(JLS)Jz δJz ,Jz′ . (536)

Here the calculation of the magnetic susceptibility cannot be done in the way we didit before just taking the free energy to be the ground state energy because in finitesmall field the splitting of the levels can become smaller than T . That’s why we haveto calculate the full free energy for finite T . We’ve got 2J + 1 levels with energiesEJz = γHJz in the field H oriented along the z-axis and γ = µBg(JLS). Thereforefor the free energy we obtain

F = −T lnJ∑

Jz=−J

e−γHJz/T = −T ln[sh (γH(J + 1/2)/T )

sh (γH/2T )

]. (537)

The magnetisation is then given by

M = −N

V

∂F

∂H= −Nγ

2V

[coth

(γH

2T

)− (2J + 1) coth

(γH(J + 1/2)

T

)]

=NγJ

VBJ(γH/T ) with

BJ(x) =2J + 1

2Jcoth (x(J + 1/2))− 1

2Jcoth (x/2) . (538)

This function saturates for x → ∞, which corresponds to full magnetisation of thesystem in very strong fields/at low temperatures. In the opposite limit of vanishingfields we can expand using cothx → 1/x + x/3 and obtain

BJ(x) =J + 1

3x + . . .

which immediately produces the Curie’s law :

χ =N

Vγ2 J(J + 1)

3T, (539)

which describes a paramagnet.49It states that matrix elements of any vector operator in the 2J + 1-dimensional space of eigenfunctions

to Jz and J2 is proportional to the matrix elements of the operator J. Then

〈JLSJz|L + g0S|JLSJ ′z〉 = g(JLS) 〈JLSJz|J|JLSJ ′z〉 . (535)

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Band contribution

The simplest approach we can employ is the one which disregards the interaction effectsas well as the influence of the magnetic field on the orbital movement of electrons. Sowe look on the spin contribution first. Every spin contributes ∓µB/V if it is orientedparallel/antiparallel to the magnetic field. That’s why

M = µB(n+ − n−) ,

where n± is the corresponding particle density. This can be described by a field-dependentDOS

D±(E) =12DF (E ± µBH) =

12DF (E)± µB

2H D′F (E) .

Now we remember that n =∫

dE DF (E) nF (E), which, again using the fact that µ =EF (1 + O(T/EF )2), leads to

M = µ2BH

∫dE D′F (E) nF (E) = µ2

BH

∫dE DF (E)

(−nF (E)

∂E)

, (540)

where in the last step we performed a partial integration. At T = 0 the derivative ofthe Fermi distribution is a δ-function and after the integration we obtain for the magneticsusceptibility

χP = µ2B DF (EF ) . (541)

As DF (EF ) = mekF /(~2π2) we obtain for the Pauli paramagnetic susceptibility followingvery simple formula

χP =( α

2π

)2(a0kF ) , (542)

where α = e2/(~c) is the fine structure constant. χP is very small in comparison to theCurie law. It can be understood to be given by the Curie law evaluated at the effectivetemperature T ∼ TF .

The orbital contribution due to the free electrons is very difficult to evaluate. It dependsvery strongly on the disorder of the sample. In general for ωcτ À 1 one obtains an oscillatorybehaviour (de Haas – van Alfven effect) while for ωcτ ¿ 1 one does not observe anyoscillations any more and one obtains a universal value

χL = −13

χP , (543)

which is called Landau diamagnetism.

Interactions and magnetic properties of solids

The effects of interactions is very interesting. Let’s consider a system of two electrons whichmay be localised on two different ions of a lattice. The wavefunction for this small systemdepends on the mutual orientation of the spins:

• singlet S = 0, then the spin part of the wave function is

|S2, Sz〉 = |0, 0〉 =1√2

(| ↑↓〉 − | ↓↑〉) ,

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which is odd with respect to the electron exchange. The orbital part of the wavefunc-tion is even

ψe =1√2

[ϕ1(r1)ϕ2(r2) + ϕ1(r2)ϕ2(r1)] .

• triplet S = 1, then the spin part of the wave function has the following possibilities

|1, 1〉 = | ↑↑〉 , |1,−1〉 = | ↓↓〉 , |1, 0〉 =1√2

(| ↑↓〉+ | ↓↑〉) (544)

The orbital part of the wavefunction is odd

ψo =1√2

[ϕ1(r1)ϕ2(r2)− ϕ1(r2)ϕ2(r1)] .

There’s a difference in energies due to the Coulomb interaction because (compare to thetreatment in Section 1.3)

∆E =∫

d3r1 d3r2 ψ∗e,oV (|r1 − r2|)ψe,o = A± J where

A =∫

d3r1 d3r2 |ϕ1(r1)|2 |ϕ2(r2)|2V (|r1 − r2|)

J =∫

d3r1 d3r2 ϕ∗1(r1)ϕ∗2(r2)V (|r1 − r2|)ϕ1(r2)ϕ2(r1) . (545)

This effect can be modelled by the operator50

HJ = −J

2(1 + 4S1 · S2) . (546)

This kind of interaction is traced back to the statistical properties of the electrons and iscalled exchange interaction. It is usually much larger than the direct magnetic dipole inter-action but decays exponentially with distance. Therefore effectively only nearest neighboursinteract. If J > 0 then we’re dealing with the ferromagnetic coupling because the systemprefers to align all spins and to maximise the overall magnetisation. For J < 0 the oppositealignment of the spins is chosen and in this case the coupling is antiferromagnetic.

A large system of spins interacting in such a way is described by the Heisenberg model ,

H = −∑

R,R′JRR′ SR · SR′ − γH ·

∑

R

SR , (547)

where we’ve neglected the constant terms from (546) and used the notation γ = µBg. Thesolution of this Hamiltonian is very difficult and in general case impossible. However, tomake progress we can use the mean field approximation. We rewrite the above Hamiltonianas

H = −γ∑

R

SR ·Heff where

Heff = H + γ−1∑

R′JRR′SR′ (548)

50See e.g. Landau-Lifshits Vol III, §62, problem 2.

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Figure 30:

is the effective magnetic field. In the next step we make the MF-approximation: we replacethe effective field by the mean value of the magnetisation:

〈SR〉 =V

Nγ−1M , then Heff = H + λM and

λ =V

Nγ−2J0 with J0 =

∑

h=R−R′Jh , (549)

since 〈SR〉 is R-independent. The next step of the MFA is to use the already calculatedformula for the magnetisation (538), which we now denote M0 and insert Heff for the‘regular’ magnetic field M = M0(Heff/T ). In zero field we then obtain a self-consistencycondition for the temperature-dependent magnetisation

M(T ) = M0 (λM/T ) . (550)

This is a transcendental equation, the solution of which cannot be given in a closed form.However, it’s possible to analyse the conditions, at which there is a solution. We can plotM(T ) as a function of the dimensionless variable λM/T for rhs and lhs of (550), see Fig. 30:

M(T ) = M0(x) , M(T ) = Tx/λ . (551)

Obviously there is a solution only when the slope of the magnetisation curve M0 is largerthan the slope of the line T/λ. The critical value is achieved when they are equal, e. g. for

(∂M0

∂H

)

H=0

= χ0 =Tc

λ. (552)

Here we can use the Curie’s law for magnetic susceptibility (539) and find

Tc = λN

Vγ2 J(J + 1)

3T=

J(J + 1)3

J0 . (553)

Below this temperature the system undergoes a spontaneous magnetisation transition andis thus a ferromagnet . This is a phase transition with the magnetisation being the orderparameter. Its behaviour in vicinity of Tc and T = 0 is given for J = 1/2 by

M(T ) =γ

2

√3

√Tc − T

Tcand M(T ) =

γ

2− γ e−2Tc/T , (554)

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respectively. It’s a second order phase transition with the critical exponent β = 1/2.51

Since the transition takes place at Tc it’s exactly here where the magnetic susceptibilitynow diverges. The corrected Curie law is thus

χ =∂M

∂H=

∂M

∂Heff

∂Heff

∂H= χ0(1 + λχ) ⇒ χ =

χ0

1− λχ0. (555)

Although the above MFA treatment of the problem describes the ferromagnetic phasetransition fairly well qualitatively, it has a number of limitations:

• it predicts the phase transition in any spacial dimension and any coordination number(number of neighbours of a given spin)

• it yields the same critical exponents whatever the details of geometry and strengthof the couplings

The failure of the above picture can already be seen in the simple 1D Ising model .

Ising model

Here the simplifying assumptions in comparison to the Heisenberg model are: (i) oneassumes only the next neighbours to interact; (ii) the coupling is highly anisotropic so thatonly the z-components of the spins are coupled; (iii) all coupling constants are equal. In1D the Hamiltonian is given by

H = −J∑

〈ij〉σz

i σzj − γH

∑

i

σzi , (556)

where σi is the z-Pauli matrix at the site i, J is the coupling constant and H the magneticfield as above. Since all Pauli matrices commute we’re dealing with a classical system, inwhich simply σz

j can be substituted by classical numbers σj = ±1.52 If we assume PBC wecan write down the partition function of the system (with N spins) in the following form

Z =∑

σ1=±1

∑

σ2=±1

· · ·∑

σN=±1

exp[

J

T(σ1σ2 + σ2σ3 + · · ·+ σNσ1) +

γH

T(σ1 + σ2 + · · ·+ σN )

]

=∑

σ1=±1

∑

σ2=±1

· · ·∑

σN=±1

Vσ1σ2 Vσ2σ3 . . . VσNσ1 where

Vσlσl+1= exp

1T

[Jσlσl+1 +

γH

2(σl + σl+1)

]. (557)

This is a matrix with l-independent components

V =(

e(J+γH)/T e−J/T

e−J/T e(J−γH)/T

). (558)

This immediately suggests that for the partition function we have Z = Tr(V N ). For V inthe diagonal form Z = λN

+ + λN− , where λ± are the eigenvalues. In the limit N → ∞ weonly need the larger eigenvalue |λ+| > |λ−| though.

51It’s, of course, the same exponent which we found in the mean field treatment of the superconductivity.52This model also describes binary alloys with two different atom species A and B are described by ±1.

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0.2 0.4 0.6 0.8 1.0 1.2

J

T

1

2

3

4

Figure 31: The partition function (562) (dashed line) and its first (solid line) and second(dot-dashed line) derivatives as functions of J/T .

The eigenvalues of the matrix (558) are

λ± = eJ/T ch (γH/T )±√

e2J/T sh2 (γH/T ) + e−2J/T . (559)

Obviously, |λ+| > |λ−|, that’s why the free energy is

F = −N

VT ln

[eJ/T ch(γH/T ) +

√e2J/T sh2 (γH/T ) + e−2J/T

]. (560)

For the magnetic susceptibility and susceptibility we then obtain

M =γN

V

eJ/T sh(γH/T )√e2J/T sh2 (γH/T ) + e−2J/T

,

χ =γ2N

V

e−J/T ch(γH/T )(e2J/T sh2 (γH/T ) + e−2J/T

)3/2→

H→0

γ2N

V

e2J/T

T. (561)

So χ satisfies a Curie’s law at high temperatures T À J and diverges towards T → 0. Veryimportant point is that there’s no phase transition at finite temperature!

There’s a highly non-trivial extension of the above procedure to 2D case due to Onsager(1944). The most fundamental result of his solution is the partition function53

1N

ln Z(H = 0, T ) = ln[√

2 ch(2J/T )]

+∫ π/2

0

dϕ

πln

[1 +

√1−K2 sin2 ϕ

],

with the parameter K = 2 sh(2J/T )/ch2(2J/T ) . (562)

The second derivative of this function with respect to T (which is, of course, up to anumerical prefactor nothing but the specific heat) diverges at sinh(2J/T ) = 1 (see Fig. 31),which gives a critical temperature

Tc = 2J/arcsinh(1) ≈ J/0.4407 , (563)

which can be compared to the MFA result Tc = J/0.25. So this is indeed a phase transition53See e. g. Pathria, “Statistical mechanics”, p. 383.

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of second order. The behaviour of spontaneous magnetisation is (Young, 1952)

M =

(1 + z2)1/4 (1− 6z2 + z4)1/8 for T ≤ Tc

0 for T > Tc(564)

with z = exp(−2J/T ). In vicinity of the transition point

M ∼ 1.22(

Tc − T

Tc

)1/8

, (565)

wo that the critical exponent β = 1/8 in contrast to 1/2 predicted by MFA.

Quantum Ising chain and quantum phase transitions

The quantum Ising chain (in 1D) is defined in the following way (Pfeuty, 1970)

H = −J

2

N∑

i=1

(σz

i σzi+1 − λσx

i

), (566)

with a PBC as before. It is a quantum model because now the Hamiltonian contains twonon-commuting parts. Thus no classical treatment of the problem is possible any more.54

λ is the transversal field measured in units of J . This system has two completely differentgroundstates depending on λ:

• for λ → 0 only interacting part plays a role and at strictly T = 0 we have a sponta-neously magnetised system and thus a ferromagnet

• for λ → ∞ the interaction part becomes completely insignificant and the system isequivalent to free spins in strong field λJ and is thus a paramagnet

The transition between these two groundstates is called a quantum phase transition.But at which value of λ does this transition occur? In order to get an idea we perform

a duality transformation. Let’s introduce an auxiliary lattice with sites just in between ofthe sites of the original one and introduce new operators

µzn+1/2 =

n∏

i=1

σxi , µz

n−1/2 µzn+1/2 = σx

n , σzn =

n−1∏

j=1

µxj+1/2 , σz

n σzn+1 = µx

n+1/2 . (567)

Then we can rewrite the Hamiltonian in terms of these new operators

H = −J λ

2

N∑

n=1

(µz

n−1/2µzn+1/2 −

1λ

µxn+1/2

). (568)

Here one immediately realises that H(λ) = λH(λ−1). We see here already that λ = 1 is aself-duality point and is a good candidate for the transition point.

The Hamiltonian (566) can be diagonalised using the Jordan-Wigner transformation.Let’s introduce spinless fermions on every lattice site an and a†n, which satisfy the anticom-mutation relations

an, a†m

= δnm , an, am = 0 . (569)

54Sometimes it’s also called Ising chain in the transversal field.

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It turns out that the spin operators of the Ising chain can be written down in the followingform:

σxn = 2a†nan − 1 , σz

n = (−1)n−1 e±iπ∑n

j=1 a†jaj (a†n + an) . (570)

In terms of new fermions the Hamiltonians becomes quadratic:

H =J

2

N∑

n=1

[−(a†n − an)(a†n+1 + an+1) + λ (a†n − an)(a†n + an)

]. (571)

The notation and further calculations can be done more efficient by introduction of latticeMajorana fermions according to the prescription

ζ1(n) =1√2(a†n + an) , ζ2(n) =

−i√2(a†n − an) . (572)

Their most prominent property is that they are their own antiparticles ζ†j (n) = ζj(n).55

Apart of that they still fulfill the usual anticommutation relationsζj(n), ζj′(n′)

= δjj′ δnn′ , as well as ζ2

j (n) = 1/2 . (573)

Then we obtain (extraction of the (n + 1)− (n) term is done on purpose! see later)

H = −iJ∑

n

ζ2(n) [ζ1(n + 1)− ζ1(n)]− (λ− 1)ζ2(n)ζ1(n) . (574)

Let’s now go over from the problem on the lattice with the constant a0 to the continuumlimit. The normalisation of the fermion field is now on

√a0: ζj(n) → √

a0 ζj(x),

H =∫

dx [−ic ζ2(x) ∂x ζ1(x) + im ζ2(x) ζ1(x)] , (575)

where c = Ja0 and m = c(λ− 1)/a0. These parameters have to be fixed in the continuumlimit which is defined by a0 → 0, J → ∞, and λ → 1. If we introduce a two-componentspinor ζ = (ζ1, ζ2) and introduce the notation γ0 = σy, γ1 = iσx and γ5 = σz, then theequation of motion can be written down in form of a 1D Dirac equation56

(iγµ ∂µ −m)ζ(x) = 0 , (576)

which is a relativistic equation for a (Majorana) fermion with mass m and new speed of lightc. Let’s now use another representation of the γ-matrices, where γ5 is diagonal, γ0 = σy,γ1 = iσx, γ5 = σz. Then the Hamiltonian can be written down as

H =∫

dx ξ(x)(−i

c

2σz ∂x + mσy

)ξ(x)

=∫

dx[−i

c

2(ξR∂xξR − ξL∂xξL)− im ξR ξL

], (577)

where ξ and ζ spinors are related by a chiral rotation,

ξR,L = (±ζ1 + ζ2)/√

2 . (578)

55They can also be understood as being the real and imaginary parts of the conventional fermions.56In D dimensions there’re 2(D+1)/2 γ-matrices. Of course, this number makes sense only for D + 1 even.

The 1 + 1 (one spacial plus one time dimension) electrodynamics is also known as Schwinger model .

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Figure 32: Phase diagram of the typical SC perovskite material.

In the last step we go over to momentum representation and obtain

H =∫

dk

2πξ

12

(ck −imim −ck

)ξ . (579)

Its diagonalisation gives the following dispersion relation:

εk = ±√

c2k2 + m2 . (580)

The gap is equal to |m| and vanishes precisely at λ = λc = 1. Here the correlation lengthdiverges and the quantum phase transition takes place.

Quantum phase transitions and high temperature superconductivity

Up to now following facts about high-Tc are well established:

• Tc > 30− 40K [1986 - LaBaCuO (30 K), 2009 - HgBa2Ca2Cu3Ox (135 K)]

• the isotopic effect is very weak

• the gap is highly anisotropic

• this phenomenon is mostly observed in perovskite structures with CuO4 planes atmoderate doping

• the electron density profile shows star/stripe-like patterns just above Tc

• the phase diagram see Fig. 32

With these facts at hand one can make a number of conclusions:

• no isotopic effect ⇒ el-ph coupling as a mechanism can be ruled out

• anisotropic gap ⇒ s-wave SC improbable, maybe dx2−y2-wave instead

• CuO4 planes are antiferromagnets ⇒ spin-spin interaction relevant

• at strictly T = 0 one can tune through different phases just by changing the dopingx, see Fig. 32. It is believed to occur by a number of phase transitions

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One attempt to explain the high-Tc is the resonating valence bond (RVB) model (An-derson et al., 1973-1987, 2004). The basis of the model is the 2D Hubbard model , describingthe electrons of the CuO4 planes by the Hamiltonian

H = −t∑

〈i,j〉(c†icj + c†jci) + U

∑

i

ni↑ni↓ . (581)

At half-filling (one electron per site, it corresponds to zero doping) and U → ∞ it’s al-most impossible for the electrons to travel around ⇒ there’s effectively only spin exchangeinteraction with an antiferromagnetic constant. So the groundstate is indeed an antiferro-magnet, see Fig. 32. At higher doping the electrons are starting to move around and areeffectively forming singlets with their neighbours. They are formed dynamically and havefinite lifetime, that’s why they are called resonating valence bonds. This kind of pairingproduces the Cooper pairs. The most fundamental difficulty is that the above model doesnot explain the strange metal and pseudogap phases found in the high Tc materials.

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Index

acoustical branch, 63anharmonic terms, 74annihilation

fermionic operator, 21anticommutation

relations, 21antiferromagnetic coupling, 92approximation

Hartree-Fock, 12APW method, 58atomic contribution

magnetic susceptibility, 88atomic factor, 66

band contribution, 88BCS Hamiltonian, 37BEC, 41, 42Bessel functions, 18Bloch theorem, 50, 61Bloch-Gruneisen law, 80Bogolyubov method, 31Bogolyubov quasi-particles, 31Bogolyubov transformation, 45Bogolyubov-de Gennes Hamiltonian, 37Bohr’s radius, 14Boltzmann equation, 76Born approximation, 66, 78Bose distribution, 64Bose gas, 41Bose-Einstein condensation, 42boson, 27, 68bosonization identity, 29bound state, 87boundary condition

open, 18boundary conditions

periodic, 7Bragg’s condition, 67Bravais lattice, 49, 61Brillouin

zone, 50broken symmetry, 38BT, 61BT, Bloch theorem, 57BZ, Brillouin zone, 57

cell

primitive unit, 49chiral fermions, 83chiral rotation, 97composite fermion, 87compressibility, 73conductance band, 58conductance quantisation, 84Cooper

pair, 30Cooper instability, 29Cooper pair, 48, 99

condensate, 37coordination number, 49Corbino geometry, 85correlation energy, 14creation

fermionic operator, 21critical

exponent, 35critical temperature

superconductivity, 35critical temperature, 34crystal momentum, 51Curie’s law, 90current-voltage characteristics, 29cyclotron frequency, 3, 81

DC conductivity, 76de Broglie wavelength, 41de Haas - van Alfven effect, 91Debye formula, 65Debye frequency, 65, 71Debye temperature, 65, 73Debye-Waller factor, 79density matrix, 43density of states

in k-space, 7density operator, 24diamagnetic, 38dielectric constant

dynamic, 17Dirac comb potential, 55Dirac equation, 97direct contribution, 12dispersion relation, 23DOS, 7, 8, 29

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Landau quasi-particle, 23Drude model, 76Drude theory, 27duality transformation, 96Dulong-Petit law, 60, 65

edge states, QHE, 83effective

Hamlitonian, 22effective Hamiltonian, 30effective mass, 23eigenfunction, 21Einstein model, 60, 64electron-phonon interaction, 70energy band, 51EOM, 61equation

Hartree, 11equation of motion, 61exchange correction, 12exchange interaction, 12, 92exchange term, 12excitation spectrum, 27expansion

in plane waves, 21partial wave, 23

Fermidistribution, 8edge, 14, 24energy, 7gas, 7, 36golden rule, 22liquid, 20, 30momentum, 7sea, 29surface, 22, 23wave vector, 7

Fermi edge, 20, 77Fermi gas, 20Fermi liquid

parameters, 23ferromagnet, 93ferromagnetic coupling, 92Feshbach resonance, 48fine structure constant, 91FL, 20flux quantum, 39, 83Fourier components

potential, 14Fourier transform, 45, 61Fourier’s law, 5Frohlich Hamiltonian, 70fractional charge, 87fractional filling, 86fractional quantum Hall effect, 86fragmented BEC, 43Friedel

oscillations, 17, 18sum rule, 19

Friedel oscillations, 78

gapsuperconductor, 33

gauge transformation, 38golden rule, 22Gorkov Hamiltonian, 37Gruneisen formula, 73Gruneisen parameter, 73Green’s functions method, 58Gross-Pitaevskii equation, 47gyromagnetic ratio, 89

Haldane pseudopotentials, 87Hall

coefficient, 3Hall effect

anomalous, 59Hall constant, 59Hartree approximation, 15Hartree equations, 11Hartree term, 12Hartree-Fock approximation, 12healing length, 46Heisenberg model, 92Hermite polynomials, 81HF

approximation, 20Hubbard model, 99

incompressiblequantum liquid, 83

Ising modelin 1D, 94

jellium model, 6, 41exchange contribution, 12

Jordan-Wigner transformation, 96Josephson current, 40

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Josephson effect, 39

Kronig-Penny model, 55

Landau criterion, 47Landau diamagnetism, 91Landau gauge, 80Landau levels, 82Larmor diamagnetism, 89lattice

body centered cubic, 49Bravais, 49face centered cubic, 49honeycomb, 49reciprocal, 50simple cubic, 49with basis, 49

lattice with basis, 62Laughlin quasiparticles, 87Legendre polynomials, 23Linde rule, 78Lindhard approximation, 16Lindhard formula, 17London equation, 38Luttinger model, 25

magnetic length, 85magnetic susceptibility, 88magnetisation, 88magnon, 28Majorana fermions, 97Maxwell equations, 38mean field approximation, 37, 92mean field theory, 35, 37mean free path, 77Meissner-Ochsenfeld effect, 38MFA, 37, 92MFT, 35, 37momentum

distribution function, 29

normal state, 43

ODLRO, 43off-diagonal long range order, 43Onsager-Feynman quantization condition, 44optical branch, 63OPW method, 58order parameter, 35, 43

parameters

Fermi liquid, 23Pauli paramagnetic susceptibility, 91PBC, 53, 94PBC, periodic boundary condition, 56Peierls instability, 71, 73Peltier effect, 5penetration depth, 38phase

condensate, 43phase transition, 42

superconducting, 36plane wave

expansion, 21plasma frequency, 4, 27plasmon, 4, 27, 28primitive vectors, 49pseudopotentials method, 58

quantization conditionOnsager-Feynman, 44

quantum Hall effect, 82quantum Ising chain, 96quantum phase transition, 96quasi-particle

Landau, 22weight, 24

quasi-particles, 27Bogolyubov, 31

random phase approximation, 16reciprocal lattice, 50relaxation time, 20, 76relaxation time approximation, 76resonating valence bond, 99RPA, 16Ruderman-Kittel

oscillations, 17RVB, 99Rydberg, 14

scatteringphase shift, 18

scattering cross section, 78scattering rate, 20Schwinger model, 97screened Coulomb potential, 16screening

dynamical, 16length, 16static, 15

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Seebeck effect, 5simple BEC, 43singularity

power law, 29Slater determinant, 12, 29, 86Sommerfeld expansion, 9specific heat

of electrons, 9superconductor, 35, 36

spin wave, 28spin–charge separation, 28spinon, 28spontaneous magnetisation, 93structure factor, 67supercurrent, 40superfluid, 37superfluid velocity, 44superfluidity, 47symmetric gauge, 84symmetry breaking, 38

temperatureeffective, 7Fermi, 7

thermal expansion coefficient, 73thermopower, 6Thomas-Fermi

approximation, 15wave vector, 15

Thomas-Fermi approximation, 78tight-binding model, 56TLL, 24Tomonaga-Luttinger liquid, 24topological insulator, 84

umklapp scattering, 67unitarity regime, 49

vacuumIQHE state, 86fermionic, 21

valence band, 58van Hove singularity, 52van Vleck paramatnetism, 90Vandermonde determinant, 86vortices, 48

wave vectorBloch, 51Fermi, 7

Wiedemann-Franz law, 5, 10Wigner-Eckart theorem, 90Wigner-Seitz cell, 50

Yukawa potential, 16

ZBA, 29zero bias anomaly, 29

103