Solid State Theory I Alexander Shnirman Institute TKM, Karlsruhe Institute of Technology, Karlsruhe, Germany (Dated: January 10, 2019)

1 Contents

I. General Information 7

II. Born-Oppenheimer approximation 7

III. Bravais and Reciprocal Lattices 9

IV. Bloch Theorem 10 A. 1-st proof 10 B. Born-von Karmann boundary conditions 12 C. 2-nd proof 13 D. Properties of the Bloch states 14

V. Almost free electrons. 16 A. Example in 1D 18 B. Lattice with basis, structure factor 19

VI. Bands, Fermi surface, Isolators, Semiconductors, Metals. 20

VII. Tight binding 21 A. Wannier functions 21 B. Schr¨odinger equation for Wannier functions 21 C. Linear Combination of Atomic Orbitals (LCAO) 22 D. Single orbital (s states), one band 24 E. Alternative formulation of tight-binding method 24

VIII. Dynamics of Bloch electrons 26 A. Semi-classical equation of motion of Bloch electrons 26 B. Wave packet argument 26 C. Proof for potential perturbation (not for vector potential) 27 D. Eﬀective mass 29 1. Example 30 E. Classical equations of motion 30 F. Only electric ﬁeld 31

2 G. Concept of holes 32

IX. Bloch electrons in magnetic field 33 A. Closed and open orbits 35 B. Cyclotron frequency 35 C. Semiclassical quantization (Bohr-Sommerfeld) of orbits 36 D. Berry phase and modification of Bohr-Sommerfeld 38 1. Transformation to the instantaneous basis 38 2. Geometric Phase 39 3. Example: Spin 1/2 39 4. Geometric interpretation 41 5. Modification of the Bohr-Sommerfeld quantization of cyclotron orbits 42 E. Magnetic susceptibility 42 1. Grand canonical ensemble 43 F. Bohr-van-Leeuven Theorem 43

X. Paramagnetism Pauli and Diamagnetism Landau 44 A. Pauli paramagnetism 44 B. Landau levels 45 C. Degeneracy of the Landau Level 45 D. Landau diamagnetism 46 E. van Alphen - de Haas eﬀect 47

XI. Boltzmann equation, elastic scattering on impurities. 47 A. Kinematics 47 B. Collision integral for scattering on impurities 48 C. Relaxation time approximation 50 D. Condutivity 50 E. Determining the transition rates 51 F. Transport relaxation time 53 G. Local equilibrium, Chapman-Enskog Expansion 54 H. Onsager relations 57

3 XII. Magneto-conductance, Hall effect 58 A. Hall effect 58 B. Magnetoresistance 60 1. Closed orbits 60 2. Open orbits 61 C. Quantum Hall Effect (QHE) 61 1. Relation to the Berry phase, Berry curvature, anomalous velocity, top. insulators 63

XIII. Fermi gas 65 A. One-particle correlation function 65 B. Two-particle correlation function 66 C. Jellium model, energy of the ground state 67 1. Kinetic energy 68 2. Potential energy 68 3. Interaction energy 68 D. Hartree-Fock equations as a variational problem 69

XIV. Fermi liquid 70 A. Spectrum of excitations of the ideal Fermi gas 70 B. Landau hypothesis 71 1. Implications 72 C. Gas model 72 D. Landau function f 74 E. Zero sound 75

XV. Phonons 76 A. Quantization of phonon modes. 78 B. Phonon density of states 80 C. Speciﬁc heat 80 1. High temperatures 81 2. Low temperatures 81 D. Debye and Einstein approximations 82

4 1. Debye 82 2. Einstein 83 E. Neutron scattering 83 F. Results 86

XVI. Plasma oscillations, Thomas-Fermi screening 87 A. Plasma oscillations 87 B. Thomas-Fermi screening 88 C. Dielectric constant of a metal 89 D. Eﬀective electron-electron interaction 91

XVII. Electron-Phonon interaction, Fr¨olich-Hamiltonian 91 A. Derivation without taking into account screening 91 B. Including screening 93 C. Direct derivation with screening 93 D. Phonon induced interaction between electrons 94 1. Comparison 95

XVIII. BCS theory of superconductivity 95 A. Phonon induced interaction between electrons 95 B. Cooper problem (L. Cooper 1955) 96 1. Symmetry 98 C. BCS state (J. Bardeen, L. Cooper, and R. Schrieffer (BCS), 1957) 98 1. Averages 99 2. Total energy 101 D. Excitations 102 1. Mean field 103 2. Nambu formalism 104 E. Finite temperature 105 1. More precise derivation 106 F. Heat capacity 107 G. Isotope effect. 109

5 XIX. Electrodynamics of superconductors. 109 A. London equations 109 1. Time-independent situation 109 B. Another form of London equations 110 C. Microscopic derivation of London equation 110 D. Pippard vs. London, coherence length. 112 E. Superconducting density 113 F. Critical field 113 G. Order parameter, phase 114 H. Ginsburg-Landau Theory 115 1. Landau Theory 115 2. Ginsburg-Landau Theory, equations 115 3. Coherence length 116 4. External field 117 5. Reduced Ginsburg-Landau equations 117 I. Surface energy 118 J. Higgs mechanismus 119 K. Flux quantization 120 L. Josephson effect 121 M. Macroscopic quantum phenomena 123 1. Resistively shunted Josephson junction (RSJ) circuit 123 2. Particle in a washboard potential 124 3. Quantization 126 4. Phase and Number of particles (Cooper pairs) 127 5. Josephson energy dominated regime 128 6. Charging energy dominated regime 128

References 129

6 I. GENERAL INFORMATION

Literature: 1) G. Czycholl, Theoretische Festk¨orperphysik. 2) C. Kittel, Quantum Theory of Solids. 3) N.W. Ashcroft and N.D. Mermin, Solid State Physics. 4) A. A. Abrikosov, Fundamentals of the theory of metals. 5) J.M. Ziman, Principles of the Theory of Solids. Superconductivity: 6) J.R. Schrieﬀer, Theory of Superconductivity. (Chapters 1-4) 7) M. Tinkham, Introduction to Superconductivity.

II. BORN-OPPENHEIMER APPROXIMATION

If we are interested in not very high energies it is meaningful to split an atom into, on one hand, an ion, which contains the nucleus and the strongly coupled electrons and, on the other hand, the weakly coupled electrons. For simplicity we consider a situation when there is one weakly coupled electron per atom. Then ions have charge +e and the electrons −e. The Hamiltonian of N ions and N electrons reads:

H = Hel + Hion + Hel−ion , (1) where

2 2 X ~pi X e Hel = + , (2) i 2m i

~ ~ Here ~pi ≡ −ih∂/∂~r¯ i and Pn ≡ −ih∂/∂¯ Rn. ~ ~ e2 ~ Rather simple-mindedly we could assume Vion(Rn −Rm) = and Vel−ion(~ri −Rn) = |R~ n−R~ m| − e2 . These interaction potentials are, of course, a bit naive. There will be corrections |~ri−R~ n| due to the fact that ions have structure (are not point-like particles).

7 Ions are much heavier than electrons: m/M 10−3. Thus electrons move much faster. The Born-Oppenheimer approximation is appropriate. First one solves the problem for the electrons only considering the positions of ions ~ el fixed. The wave function of the electrons (the coordinates Rn are parameters) ψ = el ~ ~ ψ (~r1, . . . , ~rN ; R1,..., RN ) satisfies the Schrödinger equation

el el el [Hel + Hel−ion]ψα = Eα ψα . (5)

α numbers all the eigenstates. We only need the ground state. el ~ ~ The eigenenergies are functions of ions’ coordinates: Eα (R1,..., RN ). The wave function of the total system is assumed to have a form (this is an ansatz): P ~ ~ el ~ ~ ψ = α φα(R1,..., RN )ψα (~r1, . . . , ~rN ; R1,..., RN ). The Schr¨odinger equation reads:

X el X el X el Hψ = [Hel + Hion + Hel−ion] φαψα = [Hel + Hel−ion] φαψα + Hion φαψα (6) α α α X el el X el X X 1 h ~ 2 el ~ ~ el i = Eα φαψα + (Hionφα)ψα + φα(Pn ψα ) + 2(Pnφα)(Pnψα ) (7) α α α n 2M

el We project (E − H)ψ = 0 upon ψβ . This gives

el,∗ el X ψβ Hψ = Eφβ = Eβ φβ + Hionφβ + Aβα , (8) α where

X 1 h el,∗ ~ 2 el ~ el,∗ ~ el i Aβα = φα(ψβ Pn ψα ) + 2(Pnφα)(ψβ Pnψα ) (9) n 2M

Our aim is to argue that the terms Aβ,α can be neglected due to the smallness of the ratio m/M. Since the interaction between ions and electrons depends only on the distance ~ el el el between them: Vel−ion(~ri − Rn), so does the wave function ψ . Thus |P ψ | ∼ |pψ |. So, for example, we can estimate the ﬁrst term of Aβα as

2 2 X Pn el 1 X pi el m el ψ ≈ ψ ≈ Ekin . (10) n 2M M i 2 M The electronic kinetic energy can be, in turn, estimated from the characteristic atomic energy

2 4 e me −10 0 = = 2 ≈ 13, 6eV ≈ 0.2 · 10 erg , 2a0 h¯

¯h2 −8 el where a0 = me2 ≈ 0.5 · 10 cm is the Bohr radius. The estimate reads Ekin/N ∼ 0. Thus the ﬁrst term of (9) is estimated as N(m/M)0.

8 el To estimate the second term of Aβα as well as the energy per ion stored in Eβ φβ +Hionφβ we think of each ion as of a harmonic oscillator in a potential well described by the spring 2 constant, K, of the order K ∼ 0/a0. This is because shifting an ion by a distance of order a0, i.e., by a distance of order of the distance between the ions and the electrons should cost approximately one electronic energy. Then the characteristic energy of an ion is given by q q i ∼ h¯ K/M ∼ 0 m/M. Since the potential and the kinetic energy of an oscillator are 2 q equal we obtain an estimate for the momentum of the ion: P /M ∼ i ∼ 0 m/M. Thus √ q √ 1/4√ 1/4 P ∼ Mi ∼ 0 Mm ∼ (M/m) 0m ∼ (M/m) p, where p is the characteristic √ electron momentum p ∼ 0m ∼ h/a¯ 0. 3/4 Thus, we can estimate the second term of (9) as NpP/M ∼ N(m/M) 0. Finally, the el 2 1/2 ionic energy stored in Eβ φβ + Hionφβ is estimated as NP /M ∼ N(m/M) 0. This energy is larger than both terms of (9). Neglecting (9) we obtain the approximate Schr¨odinger equation for the ions:

el ~ ~ [Hion + E (R1,..., RN )]φ = Eφ (11)

Thus the total interaction potential for the ions reads

total X ~ ~ el ~ ~ Vion = Vion(Rn − Rm) + E (R1,..., RN ) . (12) n

el ~ ~ 0 Hion + E (R1,..., RN ) = Hion + Hphonon . (13) while for the electrons one has

0 Hel−ion = Hel−ion + Hel−phonon (14)

III. BRAVAIS AND RECIPROCAL LATTICES

To be written.

9 IV. BLOCH THEOREM

Potential of ions are periodic with periods being the vectors of the Bravais lattice.

h¯2 H = − ∆ + U(~r) , (15) 2m with U(~r + R~) = U(~r) and R~ ∈ Bravais Lattice. We look for eigenstates: Hψ = Eψ (16)

Bloch Theorem: eigenstates have the following form:

i~k~r ψn.~k(~r) = e un,k(~r) , (17) ~ ~ where un,k(~r) is periodic, i.e., un,~k(~r + R) = un,~k(~r). In addition k ∈ ﬁrst Brillouin zone while n ∈ Z.

A. 1-st proof

~ We deﬁne translation operator TR~ so that TR~ f(~r) = f(~r + R).

−1 † 1) TR~ is unitary. (Unitary operators satisfy U = U ). We have obviously

T −1 = T (18) R~ −R~

To obtain T † we note the following R~ Z 3 ∗ hφ1| TR~ |φ2i = d rφ1(~r)TR~ φ2(~r) Z 3 ∗ ~ = d rφ1(~r)φ2(~r + R) Z 3 ∗ ~ = d rφ1(~r − R)φ2(~r) Z 3 ∗ = d r(T−R~ φ1(~r)) φ2(~r) (19)

Thus T † = T = T −1. R~ −R~ R~

2) TR~ commutes with H,[TR~ ,H] = 0.

~ ~ ~ TR~ Hψ = H(~r + R)ψ(~r + R) = H(~r)ψ(~r + R) = HTR~ ψ (20)

10 3) All operators TR~ form an Abelian group, i.e., commute with each other.

T T ψ = T T ψ = ψ(~r + R~ + R~ ) (21) R~ 1 R~ 2 R~ 2 R~ 1 1 2

T T = T T = T (22) R~ 1 R~ 2 R~ 2 R~ 1 R~ 1+R~ 2

This means that the set of operators H, TR~ (all of them) have common eigenstates (a full set of them).

Hψ = Eψ (23)

TR~ ψ = cR~ ψ (24)

From unitarity follows |c | = 1. From commutativity of T : c c = c . R~ R~ R~ 1 R~ 2 R~ 1+R~ 2

~ ~ R are the vectors of Bravais lattice. Thus R = n1~a1 + n2~a2 + n3~a3. This gives

n1 n2 n3 cR~ = (c~a1 ) (c~a2 ) (c~a3 ) (25)

2πixj We deﬁne c~aj = e . Then

2πi(n1x1+n2x2+n3x3) cR~ = e (26)

~ P ~ ~ Now we start using the reciprocal lattice. We deﬁne k = xjbj where bj are the elementary vectors of the reciprocal lattice. Then we can rewrite as follows

i~kR~ cR~ = e (27)

~ ~ P ~ P ~ Indeed, k · R = jl xjnlbj · ~al = 2π j xjnj (for reciprocal lattice we have bj · ~al = 2πδjl). Thus we obtain i~k·R~ TR~ ψ = e ψ , (28) i.e., each eigenvector is characterized by a vector ~k. Thus we have

ψ = ei~k·~ru(~r) , (29) where u(~r + R~) = u(~r). (We can deﬁne u as e−i~k·~rψ).

11 Thus all the eigenstates are split into families characterized by different vectors ~k. Actu- ally only ~k belonging to the first Brillouin zone (Wigner-Seitz unit of the reciprocal lattice) (or any other primitive unite of the reciprocal lattice) give different families. This follows from ei(~k+K~ )R~ = ei~kR~ . Indeed, if ~k is outside the first Brillouin zone, then we can find K~ in the reciprocal lattice so that ~q = ~k − K~ is in the first Brillouin zone. Then we use

ψ = ei~k·~ru(~r) = ei~q·~re−iK~ ·~ru(~r) = ei~q·~ru˜(~r) , (30) whereu ˜(~r) ≡ e−iK~ ·~ru(~r) andu ˜(~r + R~) =u ˜(~r). In each family introduce index n counting the states of the family. The functions u depend, then, on ~k ∈ ﬁrst B.Z. and on n. Thus, ﬁnally

i~k·~r ψ = e un,~k(~r) (31)

B. Born-von Karmann boundary conditions

The B-v-K conditions read:

ψ(~r + Nj~aj) = ψ(~r) , (32) for j = 1, 2, 3 and N1,N2,N3 1. The total number of primitive cells is then N1N2N3. ~ ~ iNj k·~aj ~ P ~ This limits the possible values of k. Namely we must have e = 1. With k = xjbj ~ where bj are the elementary vectors of the reciprocal lattice we obtain xj = mj/Nj. Although it would be better to chose all allowed values of ~k within the ﬁrst Brillouin zone it is simpler here to use a diﬀerent primitive cell in the reciprocal lattice. Namely we can chose mj = 0, 1, ..., Nj − 1. This gives

~ X mj~ k = bj , (33) j Nj

~ for mj = 0, 1, ..., Nj − 1. There are N = N1N2N3 allowed vectors k. The volume in the reciprocal lattice per one vector ~k:

~  ~ ~  ~ ~ ~ b1 b2 b3 ∆k1∆k2∆k3 = ∆k1 · (∆k2 × ∆k3) = ·  ×  N1 N2 N3 1 (2π)3 = , (34) N v

V where v ≡ N = ~a1 · (~a2 × ~a3).

12 To calculate a sum over the whole primitive cell (1-st B.Z.) we use

Z 3 Z Z X d k Nv 3 V 3 = = 3 d k = 3 d k (35) ∆k1∆k2∆k3 (2π) (2π) ~k

C. 2-nd proof

We expand both the wave function and the potential in the basis of momentum states, i.e., plane waves. Thus: X i~q·~r ψ(~r) = c~q e (36) ~q The boundary conditions, e.g., those of Born-von Karmann make the set of ~q-vectors discrete:

X mj~ ~q = bj , (37) j Nj where mj ∈ Z. The sum is not limited to the ﬁrst Brillouin zone. The potential energy is a periodic function (Bravais-lattice). Thus it can be expanded as

X iQ~ ·r U(~r) = UQ~ e , (38) Q~ where Q~ runs over the reciprocal lattice. We have

Z 1 −iQ~ ·~r UQ~ = dV U(~r)e , (39) v P.U. where the integration is over a primitive unit of the Bravais lattice and v is the volume of the primitive unit. Since U is real (hermitian) we have U = U ∗ . −Q~ Q~ The Schr¨odinger equation now reads

2 ! X i~q·~r h¯ 2 Eψ = E c~q e = − ∇ + U ψ ~q 2m h¯2q2 = X c ei~q·~r + X U c ei(Q~ +~q)·~r 2m ~q Q~ ~q ~q Q,~q~ h¯2q2 = X c ei~q·~r + X U c ei~q·~r , (40) 2m ~q Q~ ~q−Q~ ~q Q,~q~ where in the last line we substituted ~q → ~q − Q~ . The coeﬃcients in front of each harmonics must sutisfy this equation. Thus h¯2q2 ! E − c = X U c . (41) 2m ~q Q~ ~q−Q~ Q~

13 We see that only ~q’s related by a vector of the reciprocal lattice inﬂuence each other. Each such family can be characterized by a vector in the 1-st Brillouin zone. Thus, in each family we introduce ~k and all the ~q’s in the family are given by ~k + K~ , where K~ runs over the reciprocal lattice. This gives

 2 2  h¯ (~k + K~ ) X E −  c~ ~ = U ~ c~ ~ ~ . (42) 2m k+K Q k+K−Q Q~

The number of equations for each ~k ∈ 1-st B.Z. is inﬁnite as K~ runs over the whole reciprocal lattice.

We will use index n to count solutions of Eq. (42). The solution number n is a set cn,~k+K~ for all vectors K~ ∈ reciprocal lattice. Since Eq. (42) is a Schr¨odinger equation and the sets cn,~k+K~ are the wave functions, they are orthonormal, i.e.,

X ∗ c c ~ ~ = δn ,n , (43) n1,~k+K~ n2,k+K 1 2 K~ and complete X ∗ c c ~ ~ = δ ~ ~ . (44) n,~k+K~ 1 n,k+K2 K1,K2 n (Note that K~ serves here as coordinate of the wave function.) The eigenstates in the coordinate representation then read

X i(~k+K~ )·~r i~k·~r X iK~ ·~r i~k·~r ψn,~k(~r) = cn,~k+K~ e = e cn,~k+K~ e = e un,~k(~r) , (45) K~ K~ where X iK~ ·~r un,~k(~r) ≡ cn,~k+K~ e . (46) K~ Now, if we slightly change ~k, only the LHS of the equation (42) slightly changes. One can expect that in each family n the states and the eigen-energies change smoothly. We obtain bands.

D. Properties of the Bloch states

• Bloch states are orthonormal.

14 We obtain Z Z 3 ∗ X ∗ 3 i(~k2+K~ 2−~k1−K~ 1) d r ψ (~r)ψ ~ (~r) = c c ~ ~ d r e n1,~k1 n2,k2 n1,~k1+K~ 1 n2,k2+K2 K~ 1,K~ 2 X ∗ = V c c ~ ~ δ~ ~ ~ ~ , (47) n1,~k1+K~ 1 n2,k2+K2 k2+K2,k1+K1 K~ 1,K~ 2

Since ~k and ~k both are in the 1-st B.Z. we have δ = δ δ . Thus 1 2 ~k2+K~ 2,~k1+K~ 1 ~k1,~k2 K~ 1,K~ 2 Z 3 ∗ X ∗ d r ψ (~r)ψn ,k (~r) = V δ~ ~ c c ~ ~ n1,k1 2 2 k1,k2 n1,~k1+K~ 1 n2,k1+K1 K~ 1 = V δ δ . (48) ~k1,~k2 n1,n2

In the thermodynamic limit V → ∞ we have V δ → (2π)3δ(~k − ~k ). ~k1,~k2 1 2

• Basis of Bloch states is complete.

X X ∗ ψn,~k(~r1)ψn,~k(~r2) n ~k∈1.B.Z

X X X ∗ −iK~ 1·~r1 iK~ 2·~r2 i~k·(~r2−~r1) = c c ~ ~ e e e n,~k+K~ 1 n,k+K2 n ~k∈1.B.Z K~ 1,K~ 2

X X i(~k+K~ )·(~r2−~r1) = e = V δ(~r2 − ~r1) . (49) ~k∈1.B.Z K~

• Crystal momentum

The vectorh ¯~k is not the momentum and the Bloch states are not eigenstates of the momentum operator. Indeed

~ ~ i~k·~r ~ ~pψn,~k = −ih¯∇ψn,~k =h ¯kψn,~k + e ∇un,~k . (50)

The vectorh ¯~k is called ”crystal momentum”.

• Discreetness of states indexed by n.

The Schr¨odinger equation for a given ~k

 2 2  h¯ (~k + K~ ) X E −  c~ ~ = U ~ c~ ~ ~ (51) 2m k+K Q k+K−Q Q~ can be rewritten for the function

X iK~ ·~r u~k(~r) ≡ c~k+K~ e . (52) K~

15 as   h¯2(~k − i∇~ )2 E −  u~ (~r) = U(~r)u~ (~r) , (53) 2m k k ~ accompanied by the periodic boundary conditions u~k(~r + R) = u~k(~r). The problem thus must be solved in one primitive unit of the Bravais lattice and can give only discreet spectrum.

This equation is the starting point of the so called kp-approximation or kp- Hamiltonian. Indeed we can rewrite (53) as follows   ~ 2 2 2 ! (~p − h¯k0) h¯ ~ h¯ δk  + U(~r) + δk~p u~ (~r) = E − u~ (~r) , (54) 2m m k 2m k

~ ~ ~ ~ where ~p = −ih¯∇ and δk ≡ k − k0. Assuming we have the solution for some value of ~ ~ the crystal momentum k0 we can then use the operator (¯h/m)δk · ~p as a perturbation.

~ • The eigenenergies E~n,~k are continuous (and analytic) functions of k. Quite clear from the Schr¨odinger equation (53). No proof provided.

• Extension to the whole reciprocal lattice. ~ One can extend the deﬁnition of Bloch states ψn,~k for k not necessarily being in the 1-st. B.Z. Then this function is a periodic function of ~k with the periods given by the reciprocal lattice.

V. ALMOST FREE ELECTRONS.

We start from the Schr¨odinger equation

 2 2  h¯ (~k + K~ ) X E ~ −  c ~ ~ = U ~ c ~ ~ ~ (55) n,k 2m n,k+K Q n,k+K−Q Q~ for the coeﬃcients of the function

X iK~ ·~r un,~k(~r) ≡ cn,~k+K~ e . (56) K~ ~ ~ ~ ~ ~ Renaming K1 ≡ K and K2 ≡ K − Q we obtain   2 ~ ~ 2 h¯ (k + K1) X E ~ −  c ~ ~ = U ~ ~ c ~ ~ (57) n,k 2m n,k+K1 K1−K2 n,k+K2 K~ 2

16 We start from the limit of free electrons U = 0. The solutions of (57) are trivial: for each ~ ~ ~ k from the 1-st. B.Z. we can ﬁnd a set of vectors Kn(k) such that

h¯2(~k + K~ )2 E = (0) ≡ n (58) n,~k n,~k 2m ~ and c ~ ~ = δ . The vectors K~ (k) should be chosen so that the discrete set of energies n,k+Kl n,l n

En,~k is monotonically growing as a function of n. This leads to the ”reduced” scheme of zones, in which pieces of the parabola are shifted into the 1-st. B.Z. (see Fig. 1).

250

200

150

100

-3 -2 -1 1 2 3

FIG. 1: Reduced zones scheme.

Now consider U 6= 0. First, UQ~ =0 gives a total shift of energy. Thus, we take it into

account and put UQ~ =0 = 0. There are two possibilities: 1) For a given ~k there are no other vectors of the reciprocal lattice K~ such that (0) ≈ (0) l l,~k n,~k (more precisely the diﬀerence of the two energies of order or smaller than U). Then we are in the situation of the non-degenerate perturbation theory. This gives for l 6= n

U ~ ~ Kl−Kn 2 c ~ ~ = + O(U ) (59) n,k+Kl (0) − (0) n,~k l,~k and for the band energy we obtain

U ~ ~ U ~ ~ (0) X Kn−Kl Kl−Kn 3 E ~ = ~ + + O(U ) (60) n,k n,k (0) − (0) l6=n n,~k l,~k The bands repel each other. 2) There are some (at least one in addition to K~ ) vectors K~ 6= K~ such that (0) ≈ (0) . n l n l,~k n,~k ~ ~ We denote all m such vectors (incluing Kn) by Kl with l = 1, . . . , m. The degenerate perturbation theory tells us to solve the following system of m equations (j = 1, . . . , m):   h¯2(~k + K~ )2 m E − j c = X U c (61)  ~k  ~k+K~ j K~ j −K~ i ~k+K~ i 2m i=1

Double degeneracy. Consider a special (but probably the most important) case when

17 K

q q−K

FIG. 2: Bragg plane.

~ ~ the degeneracy is between two energies corresponding to vectors K1 and K2. First we note that the condition on ~k for this to happen coincides with the one for the Bragg scattering ~ ~ ~ ~ ~ ~ of the X-rays. Namely, the condition of degeneracy reads |k + K1| = |k + K2| = |k + K1 − ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ (K1 − K2)|. Introducing ~q1 ≡ k + K1, ~q2 ≡ k + K2, and K ≡ K1 − K2 (K ∈ reciprocal lattice) we see that the relation between the wave vectors in the expanded band picture ~q1 and ~q2 is like between the wave vectors of the incident and the reﬂected waves in the Bragg ~ scattering, ~q1 − ~q2 = K. Both have to end at the so called ”Bragg plane” as depicted in ~ 1 ~ Fig. 2. In particular the condition on ~q1 reads |~q1 · K| = 2 |K|. The eigenvalues are determined as zeros of the determinant of the following matrix

 (0)  E~ − −U ~  k 1,k K  (62)  (0)  −U−K~ E~k − 2,k

The solutions read v u 2 (0) + (0) u (0) − (0) 1,k 2,k u 1,k 2,k 2 E~ = ±   + |U ~ | (63) k 2 t 2 K

(0) (0) In particular, the splitting exactly at the Bragg plain, where 1,k = 2,k is given by

E2,k − E1,k = 2|UK~ |.

A. Example in 1D

Extended, reduced, and periodic zone schemes.

18 FIG. 3: Extended zone scheme in 1D.

B. Lattice with basis, structure factor

~ Assume there are identical ions in positions dj (basis) in each primitive cell of the Bravais ~ ~ ~ lattice characterized by vector R. Each ion creates a potential φ(~r − R − dj), so that

X X ~ ~ U(~r) = φ(~r − R − dj) . (64) R~ j The resulting potential U(~r) is still periodic with periods being the vectors of the Bravais lattice, U(~r + R~) = U(~r). We need

Z Z 1 −iK~ ·~r 1 −iK~ ·~r X X ~ ~ UK~ = dV U(~r)e = dV e φ(~r − R − dj) v P.U. v P.U. R~ j Z 1 −iK~ ·~r X ~ 1 ~ ∗ = dV e φ(~r − dj) = φ(K)SK~ , (65) v all space j v where Z φ(K~ ) ≡ φ(~r)e−iK~ ·~r , (66) all space and

X iK~ ·d~j SK~ = e . (67) j The structure factor provides a nice way to recover the proper Bravais lattice. This is important both in Bragg scattering and in the Bloch theory. Assume we have a ”bcc” lattice a with basis vectors ~a1 = a~x, ~a2 = a~y, and ~a3 = 2 (~x + ~y + ~z). We (wrongly) decide to treat

19 FIG. 4: Only black points are allowed.

it as a ”cs” lattice with ~a1 = a~x, ~a2 = a~y, and ~a3 = a~z and with two identical ions per ~ ~ a unit cell with d1 = 0 and d2 = 2 (~x + ~y + ~z). The (wrong) reciprocal lattice is given by ~ ~ ~ ~ ~ ~ ~ K = m1b1 + m2b2 + m3b3, where b1 = (2π/a)~x, b2 = (2π/a)~y, and b3 = (2π/a)~z. We obtain for the structure factor

X iK~ ·d~j π(m1+m2+m3) SK~ = e = 1 + e . (68) j Possible results are either 0 or 2. The correct reciprocal lattice is only those K~ for which S = 2. It is easy to see that it is an ”fcc” lattice (Fig. 4). Indeed no Bragg scattering appear for points (vectors K~ ) with S = 0. Also no zone gap appears at Bragg planes corresponding to such vectors!

VI. BANDS, FERMI SURFACE, ISOLATORS, SEMICONDUCTORS, METALS.

Index n counts bands. Number of state in a band is 2× the number of vectors ~k in the 1-st. B.Z., i.e., 2N. Bands can overlap in energy.

20 VII. TIGHT BINDING

A. Wannier functions

One can show that the Bloch states can be presented in a diﬀerent form:

X i~k·R~ ~ ψn,~k(~r) = e wn(~r − R) , (69) R~ where

3 1 X V Z d k wn(~r) = ψn,~k(~r) = ψn,~k(~r) N N 1. B.Z. (2π)3 ~k∈1. B.Z. Z d3k = v ψn,~k(~r) . (70) 1. B.Z. (2π)3

By operation of translation we obtain

1 1 w (~r − R~) = X ψ (~r − R~) = X e−i~k·R~ ψ (~r) . (71) n N n,~k N n,~k ~k∈1. B.Z. ~k∈1. B.Z. Indeed, substituting Eq.(71) into Eq.(69) we obtain

1 ψ (~r) = X ei~k·R~ X e−i~q·R~ ψ (~r) = X δ ψ (~r) = ψ (~r) . (72) n,~k N n,~q ~k,~q n,~q n,~k R~ ~q∈1. B.Z. ~q∈1. B.Z. It is easy to check that the Wannier functions of diﬀerent bands n are orthogonal. Also orthogonal are the Wannier functions of the same band but shifted to diﬀerent R~’s.

B. Schr¨odinger equation for Wannier functions

Assume the total potential is a sum of atomic ones (for a simple Bravais lattice with one atom per unit): X ~ U(~r) = Ua(~r − R) . (73) R~ Then from  2  h¯ ∆ X ~ Hψ ~ = − + Ua(~r − R) ψ ~ = E ~ ψ ~ (74) n,k 2m n,k n,k n,k R~ we obtain

 2  X i~k·R~ ~ h¯ ∆ X ~ X i~k·R~ ~ E ~ e wn(~r − R) = − + Ua(~r − R1) e wn(~r − R) . (75) n,k 2m R~ R~ 1 R~

21 ~ ~ ~ ~ In the r.h.s. we separate the terms with R1 = R from those where R1 6= R: h¯2∆ ! E X ei~k·R~ w (~r − R~) = X − + U (~r − R~) ei~k·R~ w (~r − R~) n,~k n 2m a n R~ R~ X X ~ i~k·R~ ~ + Ua(~r − R1)e wn(~r − R)

R~ R~ 16=R~ h¯2∆ ! = X − + U (~r − R~) ei~k·R~ w (~r − R~) 2m a n R~ X ~ i~k·R~ ~ + ∆U(~r, R)e wn(~r − R) , (76) R~ where ∆U(~r, R~) ≡ P U (~r − R~ ) = U(~r) − U (~r − R~). R~ 16=R~ a 1 a

C. Linear Combination of Atomic Orbitals (LCAO)

P Simplest approximation for the Wannier function w = m bmφm where φm are the atomic orbitals, such that Haφm = Ea,mφm. This can be, e.g., a multiplet of the orbital momentum L with 2L + 1 degenerate states (we omit the band index n). This gives

X X i~k·R~ ~ X X ~ i~k·R~ ~ bm(E~k − Ea,m) e φm(~r − R) = bm ∆U(~r, R)e φm(~r − R) , (77) m R~ m R~ ~ We have restricted our Hilbert space to linear combinations of atomic orbitals φm(~r − R) shifted to all vectors of the Bravais lattice. While we cannot guarantee that Eq. (77) holds exactly (in the whole Hilbert space) we can choose the coeﬃcients bm and the energy E~k so that Eq. (77) holds in our restricted space. That is we demand that Eq. (77) projected ~ on all φm(~r − R) holds. Due to the periodicity of the l.h.s. and the r.h.s. of Eq. (77) it is suﬃcient to project only on φm(~r).

Projecting on φl(~r) we obtain Z X X i~k·R~ 3 ∗ ~ (E~k − Ea,l)bl + bm(E~k − Ea,m) e d rφl (~r)φm(~r − R) m R~6=~0 Z X X i~k·R~ 3 ∗ ~ ~ = bm e d rφl (~r)∆U(~r, R)φm(~r − R) , (78) m R~ Introducing Z ~ 3 ∗ ~ Il,m(R) ≡ d rφl (~r)φm(~r − R) (79) and Z ~ 3 ∗ ~ ~ hl,m(R) ≡ d rφl (~r)∆U(~r, R)φm(~r − R) (80)

22 we obtain

X X i~k·R~ ~ (E~k − Ea,l)bl + bm(E~k − Ea,m) e Il,m(R) m R~6=~0 X X i~k·R~ ~ = bm e hl,m(R) , (81) m R~

This is a homogeneous matrix equation on coeﬃcients bm. To have solutions one has to demand that the determinant of the matrix vanishes, This determines the band energies

En,~k. The number of bands is equal to the number of states in the multiplet. It is necessary to note that we actually failed to produce a valid Wannier (n) function in this procedure. Indeed, in general, the eigenvectors of (81) bm are ~k-dependent (n is the band index counting diﬀerent eigenvectors). Thus, for our ”Wannier function” we obtain

X (n) wn(~r) = bm (k)φm(~r) . (82) m

As it is ~k-dependent, it cannot be the Wannier function. Of course, there is no problem if we have only one atomic orbital per unit cell. For the Bloch function we get

X i~k·R~ X (n) ~ X (n) X i~k·R~ ~ ψn,~k(~r) = e bm (k)φm(~r − R) = bm (k) e φm(~r − R) , (83) R~ m m R~ and for the periodic part of the Bloch function we obtain

X i~k·(R~−~r) X (n) ~ X (n) X −i~k·(~r−R~) ~ un,~k(~r) = e bm (k)φm(~r − R) = bm (k) e φm(~r − R) . (84) R~ m m R~ Thus our method has actually reduced to building quasi-Bloch states from the atomic orbitals: X i~k·R~ ~ φm,~k(~r) ≡ e φm(~r − R) , (85) R~ and then looking for optimal linear combinations of these states

X (n) ψn,~k(~r) = bm (k) φm,~k(~r) . (86) m The proper Wannier functions are in general a bit more inviolved.

23 D. Single orbital (s states), one band

We obtain P i~k·R~ ~ R~ e h(R) E~ = Ea + , (87) k P i~k·R~ ~ 1 + R~6=~0 e I(R) Assume that only nearest neighbors matrix elements, do not vanish (and also h(0)). It is important to note that I(R~) 1. Thus

X ~ ~ i~k·R~ E~k ≈ Ea + h(0) + (h(R) − h(0)I(R))e , (88) R~∈n.n. Then, for diﬀerent Bravais lattices with one ion per primitive cell we obtain: 1) 1-D lattice with lattice constant a.

Ek = Ea + h(0) + 2W cos(ak) , (89) where W = h(a) − h(0)I(a).

2) sc-lattice, ~a1 = a~x,~a2 = a~y,~a3 = a~z,

E~k = Ea + h(0) + 2W (cos(akx) + cos(aky) + cos(akz)) , (90) where W = h(a) − h(0)I(a).

3) bcc-lattice. One of the possible choices of the primitive basis is: ~a1 = a~x,~a2 = a~y,~a3 = 1 a(~x + ~y + ~z), however the nearest neighbors are at R~ = a (±x ± y ± z). Altogether 8 2 √ 2 neighbors each at distance 3a/2. We obtain

E~k = Ea + h(0) + 8W cos(akx/2) cos(aky/2) cos(akz/2) , (91) √ √ where W = h( 3a/2)−h(0)I( 3a/2). (Interesting exercise: show that the reciprocal lattice in fcc).

E. Alternative formulation of tight-binding method

E ~ Each primitive cell is characterized by states R, m . Index m can count either states of the same atom or states of diﬀerent atoms in the cell. For example in graphen we would have m = A, B, where A and B denote sub-lattices. The overlaps of diﬀerent states vanish: hR , m | |R , m i = δ δ . One postulates a tunneling Hamiltonian 1 1 2 2 R~ 1,R~ 2 m1,m2 ED X X ~ ~ ~ ~ H = tm1,m2 (R1 − R2) R2, m2 R1, m1 (92) R~ 1,m1 R~ 2,m2

24 ~ ∗ ~ The Hamiltonian is hermitian, i.e., tm1,m2 (R) = tm2,m1 (−R). The Bloch states: E X i~k·R~ X ~ ψ~k = e bm R, m . (93) R~ m The ”Wannier function” (see discussion above why it is not a valid Wannier function): E P ~ w = m bm R, m .

The energies and the coeﬃcients bm are determined by substituting the Bloch wave function into the Schr¨odinger equation: Hψ~k = E~kψ~k. We obtain

ED E X X ~ ~ ~ ~ X i~k·R~ X ~ Hψ~k = tm1,m2 (R1 − R2) R2, m2 R1, m1 e bm R, m m R~ 1,m1 R~ 2,m2 R~ ~ ~ E X X ik·R1 ~ ~ ~ = e bm1 tm1,m2 (R1 − R2) R2, m2 R~ 1,m1 R~ 2,m2 ~ ~ E X ik·R2 X ~ = E~k e bm2 R2, m2 (94) m2 R~ 2 E ~ Comparing coeﬃcients in front of R2, m2 we obtain

~ ~ ~ ~ X ik·R1 ~ ~ ik·R2 e bm1 tm1,m2 (R1 − R2) = E~ke bm2 . (95) R~ 1,m1

~ ~ ~ With R ≡ R1 − R2 X i~k·R~ ~ e bm1 tm1,m2 (R) = E~kbm2 . (96) R,m~ 1 We again have reduced the problem to a matrix equation. Examples: 1) 1-D, m=0 (1 state per primitive cell)

X i~k·R~ E~k = tR~ e (97) R~

For nearest neighbors tunneling E~k = 2t(1) cos(ka). 2) Exercise: graphen.

25 VIII. DYNAMICS OF BLOCH ELECTRONS

A. Semi-classical equation of motion of Bloch electrons

We want to describe the evolution of electron’s wave function when a weak and slowly changing external ﬁeld is added. That is the Hamiltonian now reads

2 −ih¯∇~ − e A~(~r) H = c + U(~r) + eφ(~r) , (98) 2m where for electrons e = −|e| < 0. The potential U(~r) is periodic while A~ and φ change little on the scale of primitive cell of the Bravais lattice (slow fields). Our aim to prove that the electrons in the band n are governed by the following effective Hamiltonian e H = −i∇~ − A~ + eφ . (99) eff,n n hc¯

B. Wave packet argument

We localize the electron of a certain band n into a wave packet:

Z Z 3 ~ 3 ~ i~k·~r Φ(~r) = d k g(k)ψn,~k(~r) = d k g(k)un,~k(~r)e . (100)

~ ~ The function g(k) is centered around a certain quasi-impuls k0 and has a width ∆k such that the width of the wave packet in the real space ∆r is small enough. The two are related as ∆k∆r ∼ 1. The time evolution of the wave packet is given by

Z ~ 3 ~ ik·~r−in,~kt/¯h Φ(~r, t) = d k g(k)un,~k(~r)e . (101)

We expand around ~k and . We assume one can approximate u (~r) ≈ u in the 0 n,~k0 n,~k n,~k0 whole interval of ∆k. (This works in many cases but sometimes it is very important to take into account the so called Berry curvature.) Then

∂ ~ ~ Z iδ~k· ~r− n,k t/¯h ik0·~r−i ~ t/¯h 3 ~ Φ(~r, t) ≈ u (~r)e n,k0 d δk g(~k)e ∂k . (102) n,~k0

Thus we conclude that the wave packet propagates with the velocity

∂~r 1 ∂ ~ ~v = = n,k . (103) ∂t h¯ ∂~k

26 Assume now the electron is influenced by an electric field E~ . The work done by the field pro unit of time is eE~ · ~v. This work is ”used” to change the energy of the electron. Thus we obtain ∂ ∂ ~ d~k d~k = n,k =h~v ¯ = eE~ · ~v. (104) ∂t ∂~k dt dt Thus we obtain d~k h¯ = eE.~ (105) dt The quasi-momentumhk ¯ satisfies the same equation as the usual momentum for free electrons!

C. Proof for potential perturbation (not for vector potential)

We consider the following problem

2 −ih¯∇~ H = + U(~r) + U (~r) = H + U (~r) , (106) 2m ext 0 ext

Here U is the periodic lattice potential and Uext = eφ is the external and weak potential. ~ More precisely what has to be weak is the external electric ﬁeld, i.e., ∼ ∇Uext. We want to solve the time-dependent Schrdinger equation: ∂ψ ih¯ = Hψ (107) ∂t We expand ψ(t) in basis of Wannier functions

X ~ ψ(t) = an,R~ (t)wn(~r − R) (108) n,R~ Recall the representation of a Bloch wave function

X i~k·R~ ~ ψn,~k(~r) = e wn(~r − R) . (109) R~

i~k·R~ In this case an,R~ = e . Wannier functions are given by 1 w (~r) = X ψ (~r) . (110) n N n,~k ~k∈1. B.Z. and 1 w (~r − R~) = X e−i~k·R~ ψ (~r) . (111) n N n,~k ~k∈1. B.Z.

27 First we investigate how H0 acts on the (shifted) Wannier functions using the fact

H0ψn,~k = n,kψn,k.

1 1 H w (~r − R~) = X e−i~k·R~ H ψ (~r) = X e−i~k·R~ ψ (~r) . (112) 0 n N 0 n,~k N n,k n,~k ~k∈1. B.Z. ~k∈1. B.Z. We use now the Wannier expansion (109) and obtain

X 1 X ~ ~ ~ H w (~r − R~) = eik·(R1−R)w (~r − R~ ) 0 n N n,k n 1 R~ 1 ~k∈1. B.Z. X ~ ~ ~ = n(R1 − R)wn(~r − R1) , (113)

R~ 1 where 1 (R~) ≡ X ei~k·R~ . (114) n N n,k ~k∈1. B.Z. The Schr¨odinger equation now reads: ∂ψ ih¯ = ih¯ X a˙ (t)w (~r − R~) ∂t n,R~ n n,R~ = Hψ = (H + U )ψ = X a (t)(H + U )w (~r − R~ ) 0 ext n2,R~ 2 0 ext n2 2 n2,R~ 2 = X a (t) X (R~ − R~ )w (~r − R~ ) + X a (t)U w (~r − R~ ) . n2,R~ 2 n2 1 2 n2 1 n2,R~ 2 ext n2 2 n2,R~ 2 R~ 1 n2,R~ 2 (115)

The Wannier functions form a complete orthonormal basis. Thus we just compare the coeﬃcients:

ih¯a˙ = X a (R~ − R~ ) n,R~ n,R~ 2 n 2 R~ 2 Z + X a d3r w∗ (~r − R~)U (~r)w (~r − R~ ) . (116) n2,R~ 2 n ext n2 2 n2,R~ 2 The ﬁrst term in the r.h.s. of (116) is rewritten as follows

X X X ~ ~ a (R~ − R~ ) = (R~ )a = (R~ )e−iR1·(−i∇)a n,R~ 2 n 2 n 1 n,R~−R~ 1 n 1 n,R~ R~ 2 R~ 1 R~ 1 ~ ~ = n(k → −i∇)an,R~ . (117)

~ ~ Here we have used a = e−R1·∇a . That is already here we consider a as a ”good” n,R~−R~ 1 n,R~ n,R~ function in all the space, i.e., an,~r.

28 The second term of the r.h.s. of (116) is approximated as Z X a d3r w∗ (~r − R~)U (~r)w (~r − R~ ) ≈ U (R~)a . (118) n2,R~ 2 n ext n2 2 ext n,R~ n2,R~ 2

That is only diagonal matrix elements of Uext are left. Since Uext is slowly changing in space, i.e,. it changes very little on the scale of a promitive cell, while the Wannier functions are localized on the scale of the cell this approximation is justiﬁed.

Thus, the Schrödinger equation for the ”envelope” wave function an,R~ reads h ~ ~ i ih¯a˙ n,R~ = n(−i∇) + Uext(R) an,R~ . (119) ~ If we now ”forget” that an,R~ is defined only in the locations R and define it in the

whole space, an,~r we obtain a Schrödinger equation with the effective Hamiltonian Heff,n = ~ n(−i∇) + Uext(~r). In presence of vector potential it becomes (with no proof given here) e H = −i∇~ − A~ + eφ . (120) eff,n n hc¯

D. Eﬀective mass

~ We see that the operator n(−i∇) plays the role of the kinetic energy. The free kinetic ¯h2(~k)2 energy reads free = 2m . In many cases the relevant values of crystal momenta lay around ~ ~ ~ an extremum of n(k) at k = k0. Then one can expand to the second order 2 ! ~ ~ 1 ∂ n n(k) ≈ (k0) + (kα − k0,α)(kβ − k0,β) . (121) 2 ∂kα∂kβ ~k=~k0 In analogy to the free case the tensor 2 ! ∗−1 1 ∂ n m = 2 (122) α,β h¯ ∂kα∂kβ ~k=~k0 is called the eﬀective mass tensor. In the simplest case when the tensor is proportional to ∗−1 ∗ the unity matrix, i.e., m = (1/m )δα,β one cane introduce the eﬀective (band) mass α,β m∗. For example in a simple cubic lattice we have ~ (k) = const. − 2W (cos(akx) + cos(aky) + cos(akz)) , (123)

where in comparison to what we did earlier we change the deﬁnition of W so that W > 0. ~ 2 2 The energy has a minimum at k0 = 0 and the eﬀective mass is obtained from W a k = h¯2k2/(2m∗) and is given by h¯2 m∗ = . (124) 2W a2

29 1. Example

Consider a semiconductor with fully occupied valance band and an empty conductance band. The conductance band is characterized by an eﬀective mass m∗ ≈ 0.1m. That is ~ 2 2 ∗ Econd(k) = Econd +h ¯ k /(2m ), where Econd is the bottom of the conduction band. An impurity (dopant ion) creates a potential

e2 U = − , (125) ext |r| where ≈ 10 is the dielectric constant. The dynamics of an extra electron added to the conduction band will be governed by an eﬀective Hamiltonian which has the form of the Hydrogen atom one, but with diﬀerent parameters:

h¯2∇~ 2 e2 H = E − − . (126) eﬀ cond 2m∗ |r|

The energy levels are known m∗e4 1 E = E − (127) n cond 22h¯2 n2 The binding energy m∗e4 m∗ 1 = Ry , (128) 22h¯2 m 2 where Ry ≈ 13eV. Thus we obtain a binding energy of order 10−3Ry. The size of he bound ¯h2 2 state, i.e., the new Bohr radius is given by e2m∗ ≈ 10 a0. Thus we obtain Hydrogen-like bound states in the energy gap.

E. Classical equations of motion

Considering now the eﬀective Hamiltonian classically, i.e., replacing −ih¯∇~ with ~p we obtain ~p e ! H = − A~ + eφ(~r) (129) n h¯ ch¯ This Hamiltonian is useful for description of dynamics of wave packets. The equations of motion read

d ~r = ∇~ H dt p d ~p = −∇~ H (130) dt r

30 The ﬁrst equation gives  ~  d ∂H 1 ∂n(k) vj = rj = =   . (131) dt ∂p h¯ ∂k j j ~ ~p e ~ k= h¯ − ch¯ A The second equation reads

d e X ∂Ai pj = vi − e∇jφ dt c i ∂rj ! e X ∂Ai ∂Aj ∂Aj = vi − + − e∇jφ c i ∂rj ∂ri ∂ri e ~ e X ∂Aj = ~v × B + vi − e∇jφ (132) j c c i ∂ri Using d ∂Aj X ∂Aj Aj = + vi (133) dt ∂t i ∂ri we obtain d e e ~ e ∂Aj pj − Aj = ~v × B − − e∇jφ . (134) dt c c j c ∂t e ~ ~ 1 ∂A~ ~ Introducing ~pkin = ~p − c A and recalling that E = − c ∂t − ∇φ we obtain d e ~p = ~v × B~ + eE.~ (135) dt kin c

F. Only electric ﬁeld

The equations of motion read  ~  1 ∂n(k) vj =   . (136) h¯ ∂k j ~ ~p k= h¯ d ~p = eE.~ (137) dt This gives     ~ 2 ~ dvj 1 d ∂n(k) 1 X ∂ n(k) dki =   =   dt h¯ dt ∂kj h¯ i ∂kj∂ki dt   2 ~ 1 X ∂ n(k) X ∗−1 = 2   (eEi) = (m )ji(eEi) . (138) h¯ i ∂kj∂ki i Or, one can invert the eﬀective mass tensor and obtain

X ∗ (m )ijv˙j = eEi = Fi . (139) j

31 These relations once again show the role of the eﬀective mass. The mass tensor can have negative eigenvalues. Then, the acceleration of the electron has an opposite to the force (F~ = eE~ ) direction. Bloch oscillations!!

G. Concept of holes

Consider an almost full band. As an example take the sc lattice.

~ (k) = const. − 2W (cos(akx) + cos(aky) + cos(akz)) , (140)

~ π Near the top of the band, where the unoccupied states are, at k0 = a (1, 1, 1) the eﬀective mass is negative h¯2 m∗ = − . (141) 2W a2 Instead one can introduce holes. The unoccupied states are now called occupied by holes, while occupied states are unoccupied by holes. So a hole is an absence of an electron. This is useful since the fully occupied by electrons band does not contribute to the current. The current provided by the band n (we do not use the band index as we limit ourselves here to a single band) is given by e Z d3k ~j = X X n(~k, s)~v(~k, s) = e X n(~k, s)~v(~k, s) , (142) V 1.B.Z (2π)3 s ~k∈1.B.Z s where n(~k, s) is the occupation number of the state with crystal momentum ~k and spin s = ±1/2 (in the band n). The velocity’s components are given by ~ ~ 1 ∂(k, s) vα(k, s) = . (143) h¯ ∂kα Here we generalized the band energy (~k) to depend on the spin index (~k, s). We did not yet consider spin-orbit coupling. Thus (~k) does not depend on the spin s. In general case, when the SO coupling is present, it does. The time reversal symmetry requires in general case (~k, s) = (−~k, −s). Thus v(−~k, −s) = −v(~k, s). In the absence of spin-orbit coupling (~k) = (−~k) and v(−~k) = −v(~k). The occupation numbers are between 0 and 1. For the fully occupied band n(~k, s) = 1. Thus we have Z 3 Z 3 ~ X d k ~ e X d k ∂(k, s) jα = e 3 vα(k, s) = 3 = 0 . (144) s 1.B.Z (2π) h¯ s 1.B.Z (2π) ∂kα

32 The integral vanishes since (~k, s) is periodic with the period given by the vectors of the reciprocal lattice. ~ ~ For the holes we have the occupation numbers nh(k, s) = 1−n(k, s). A hole characterized by ~k, s moves with the same velocity as the electron with ~k, s. This follows just from the fact that Schr¨odinger equation for the state do not depend on whether the state is occupied. Then we obtain for the current density

Z 3 Z 3 ~ X d k ~ ~ X d k h ~ i ~ j = e 3 n(k, s)~v(k, s) = −e 3 1 − n(k, s) ~v(k, s) s 1.B.Z (2π) s 1.B.Z (2π) Z 3 X d k ~ ~ = −e 3 nh(k, s)~v(k, s) . (145) s 1.B.Z (2π) Thus we can say that the charge of the hole is −e (that is positive). Finally we recall the equation of motion

X ∗ (m )ijv˙j = eEi . (146) j

∗ ∗ Deﬁning the hall eﬀective mass as mh ≡ −m we obtain

X ∗ (mh)ijv˙j = −eEi . (147) j

Thus if the electronic eﬀective mass is negative it is more convenient to use the picture of holes. They have positive charge and positive mass. Thus far we characterized holes by the crystal momentum and the spin of the absent electron ~ ~ ~ k, s. It is more logical to say that the hole has a crystal momentum kh = −k and spin sh = −s. We ~ ~ also deﬁne the energy of the hole as h(kh, sh) = const. − (k, s). For the constant it is convenient to chose the upper edge of the band. Then we can have the usual relation

1 ∂ (~k , s ) 1 ∂(~k, s) ~v (~k , s ) = h h h = = ~v(~k, s) . (148) h h h ~ ~ ¯h ∂kh ¯h ∂k ∗ It is now easy to check that the usual relation is satisﬁed for the hole eﬀective mass mh.

IX. BLOCH ELECTRONS IN MAGNETIC FIELD

e ~ For ~pkin = ~p − c A we obtained

d e ~p = ~v × B~ + eE.~ (149) dt kin c

33 Recalling also that the eﬀective Hamiltonian reads ~p e ! H = − A~ + eφ(~r) (150) n h¯ ch¯ ~ we conclude that the Bloch wave vector is related to ~pkin, i.e.,h ¯k = ~pkin. This gives for the case E~ = 0 d e ~k = ~v(~k) × B~ , (151) dt hc¯ where ~ drj ~ 1 ∂n(k) = vj(k) = . (152) dt h¯ ∂kj First, we observe that the energy is conserved: d d~k =h~v ¯ · = 0 . (153) dt dt ~ ~ Second, the vector dk/dt is perpendicular to B, i.e. kz = const. if z is the direction of B~ . Thus, in the k−space the motion is along lines of equal energy which belong to planes perpendicular to B~ . One obtains these lines by cutting the equal-energy surfaces (e.g., the Fermi surface) by planes ⊥ B~ . B

FIG. 5: Fermi surface cut by a plane perpendicular to B~ .

Next we consider the trajectory in the real space. We obtain ˙ e e eB2 B~ × ~k = B~ × ~v(~k) × B~ = ~v(~k)(B~ )2 − B~ (B~ · ~v(~k)) = ~v , (154) hc¯ hc¯ hc¯ ⊥ ~ ~ where ~v⊥ is the component of the velocity perpendicular to B. Dividing by |B| we obtain ˙ eB ~b × ~k = ~v , (155) hc¯ ⊥ where ~b ≡ B/~ |B~ |. Integrating over time from 0 to t we obtain Be ~b × (~k (t) − ~k (0)) = (~r (t) − ~r (0)) , (156) ⊥ ⊥ hc¯ ⊥ ⊥ Thus the trajectory in the ~r−space is obtained from the trajectory in ~k−space by a π/2 rotation around the axis of the magnetic ﬁeld and a rescaling by a factorhc/ ¯ (eB).

34 A. Closed and open orbits

If the orbit in the k-space is closed, then also the orbit in the r-space is closed (see Fig. 6a). However, there exist situations when the orbit in the k-space is open (see Fig. 6b).

B B r(t) k(t)

k(t) r(t)

FIG. 6: a) Closed trajectories;b) Open trajectories.

This happens when the Fermi surface reaches the border of the Brillouin zone.

B. Cyclotron frequency

Assume the direction of the magnetic ﬁeld is z, i.e., ~b = ~z. Also assume that all the orbits are closed. We write the equations of motion

˙ eB ~z × ~k = ~v (157) hc¯ ⊥ in components: eB eB k˙ = v , k˙ = − v . (158) x hc¯ y y hc¯ x From here we obtain e2B2 (dk )2 + (dk )2 = (v2 + v2)(dt)2 (159) x y h¯2c2 x y Introducing the length element along the trajectory in the reciprocal space dk ≡ q 2 2 (dkx) + (dky) we obtain dk |e|B = v , (160) dt hc¯ ⊥ q 2 2 2 ¯hc where v⊥ ≡ vx + vy. It is useful to introduce the so called magnetic length lB ≡ |e|B . We get dk l2 = v , (161) B dt ⊥

35 The inverted relation reads 2 dk dt = lB , (162) v⊥ For the period of the orbit we obtain

I dk I dk T = l2 =hl ¯ 2 , (163) B B ∂ v⊥ dq where dq is the diﬀerential of the wave vector perpendicular to the trajectory (it must be perpendicular because the gradient along the ~k-trajectory vanishes, since = const.) and laying in the plane of the trajectory, i.e., kz = const.. This gives

I I 2 dq 2 dq T =hl ¯ dk =hl ¯ dk . (164) B ∂ B d dq dq We choose the diﬀerential dq so that at any point of the trajectory k(t) we have d = const., i.e., d is independent of k(t). We identify the integral along the trajectory in the k−space

Z dkdq = dS , (165) where S is the area of the closed orbit. Thus

∂S 2 T =hl ¯ B . (166) ∂ kz=const

One can also deﬁne the ”cyclotron mass”

h¯2 ∂S(, k ) m (, k ) ≡ z , (167) c z 2π ∂ so that the cyclotron frequency is equal to

2π h¯ |e|B ωc = = 2 = . (168) T lBmc c mc For a simple parabolic band the cyclotron mass is equal to the eﬀective mass. In more complicated cases they are diﬀerent (exercise).

C. Semiclassical quantization (Bohr-Sommerfeld) of orbits

We have seen that the eﬀective Hamiltonian

~p e ! H = − A~ (169) n h¯ ch¯

36 is obtained by substitution e h¯~k → ~p − A.~ (170) c We obtain I I e ~p · d~r = h¯~k + A~ · d~r (171) c From Be ~z × (~k (t) − ~k (0)) = (~r (t) − ~r (0)) , (172) ⊥ ⊥ hc¯ ⊥ ⊥ multiplying both sides from the left with ~z× we obtain

e h¯~k = ~r × B~ + const. . (173) ⊥ c ⊥

Thus

I e I ~p · d~r = ~r × B~ + A~ · d~r c ⊥ e I e I = − B~ · ~r × d~r + A~ · d~r c ⊥ c e e = (−2Φ + Φ) = − Φ , (174) c c where Φ is the magnetic flux through the closed orbit (in ~r-space). The quantization condition reads H ~p · d~r = 2πhn¯ (quasi-classics justified for n 1). This is Bohr’s quantization condition. More precise is the replacement n → n + γ, where γ = 1/2 is called the phase mismatch. We obtain hc Φ = − (n + γ) = Φ (n + γ) , (175) e 0 where Φ0 ≡ hc/|e| is the flux quantum. The allowed areas of the orbits in the ~r-space read Φ A = 0 (n + γ) . (176) n B and in the ~k−space

1 B2e2 Φ 2πB|e| 2π S = A = 0 (n + γ) = (n + γ) = (n + γ) . (177) n 4 n 2 2 2 lB h¯ c B hc¯ lB Comparing now with h¯2 ∂S(, k ) m (, k ) ≡ z , (178) c z 2π ∂ and 2π |e|B ωc = = , (179) T c mc

37 we observe that the energy diﬀerence between the levels n and n + 1 is 1 h¯2 2π (n + 1, kz) − (n, kz) = (Sn+1 − Sn) = =hω ¯ c (180) ∂S(,kz) 2 2πmc l ∂ B Thus we obtain the Landau levels quasi-classically. In what follows we will obtain Landau levels for a parabolic band exactly.

D. Berry phase and modiﬁcation of Bohr-Sommerfeld

1. Transformation to the instantaneous basis

Consider a slowly changing in time Hamiltonian H(t). It is convenient to introduce a vector of parameters ~χ upon which the Hamiltonian depends and which changes in time. H(t) = H(~χ(t)). Diagonalize the Hamiltonian for each ~χ. Introduce instantaneous eigenstates |n(~χ)i, such that H(~χ) |n(~χ)i = En(~χ) |n(~χ)i. Since H(~χ) changes continuously, it is reasonable to assume that |n(~χ)i do so as well. Introduce a unitary transformation

X X R(t, t0) ≡ |n(~χ(t0))i hn(~χ(t))| = |n0i hn(~χ(t))| . (181) n n

Thus the new Hamiltonian is given by

H˜ = RHR−1 + ih¯RR˙ −1 (183)

The ﬁrst term is diagonal. Indeed

X R(t, t0)H(t)R(t0, t) = |n0i hn(~χ(t))| H(~χ(t)) |m(~χ(t))i hm0| nm X = En(~χ(t)) |n0i hn0| . (184) n Thus transitions can happen only due to the second term which is proportional to the time derivative of R, i.e., it is small for slowly changing Hamiltonian. This can be seen as a justiﬁcation of the adiabatic theorem. The adiabatic theorem says that if the initial state was one of the eigenstates |n0i ≡ |n(~χ(t0))i then at time t the system will be with a large probability in the state |n(~χ(t))i.

38 2. Geometric Phase

The operator ih¯RR˙ −1 may have diagonal and oﬀ-diagonal elements. The latter is re- sponsible for transitions and is not discussed later. Here we discuss the role of the diagonal elements, e.g.,

D ˙ −1 ˙ ~ Vnn(t) = hn0| ih¯RR |n0i = ih¯ hn˙ (~χ(t))| n(~χ(t))i = ih¯ ~χ ∇n(~χ) n(~χ)i . (185)

This (real) quantity serves as an addition to energy En(~χ), i.e., δEn = Vnn. Thus, state

|n0i acquires an additional phase Z Z δΦn = − dtδEn = −i dt hn˙ (~χ(t))| n(~χ(t))i . (186)

This phase is well deﬁned only for closed path, i.e., when the Hamiltonian returns to itself. Indeed the choice of the basis |n(~χ)i is arbitrary up to a phase. Instead of |n(~χ)i we could have chosen e−iΛn(~χ) |n(~χ)i. Instead of (185) we would then obtain

˙ Vnn(t) = ih¯ hn˙ (~χ(t))| n(~χ(t))i +h ¯Λn(~χ(t)) . (187)

Thus the extra phase is, in general, not gauge invariant. For closed path we must choose the basis |n(~χ)i so that it returns to itself. That is |n(~χ)i depends only on the parameters

~χ and not on the path along which ~χ has been arrived. This means Λn(t0) = Λn(t0 + T ), ˙ where T is the traverse time of the closed contour. In this case the integral of Λn vanishes and we are left with

Z Z D ~ ΦBerry,n ≡ δΦn = −i dt hn˙ (~χ(t))| n(~χ(t))i = −i d~χ ∇n ni (188)

This is Berry’s phase. It is a geometric phase since it depends only on the path in the parameter space and not on velocity along the path. Physical meaning (thus far) only for superpositions of diﬀerent eigenstates.

3. Example: Spin 1/2

˙ We consider a spin-1/2 in a time-dependent magnetic ﬁeld B~ (t): ~b = Ω~ × ~b, where ~b ≡ B/~ |B~ | and Ω~ is the angular velocity. The instantaneous position of B~ is determined by the angles θ(t) and ϕ(t). The Hamiltonian reads 1 H(t) = − B~ (t) · ~σ (189) 2

39 FIG. 7: Time-dependent magnetic ﬁeld.

We transform to the rotating frame with |Φ(t)i = R(t) |Ψ(t)i such that ~b is the z-axis in the rotating frame. Here |Ψi is the wave function in the laboratory frame while |Φ(t)i is the wave function in the rotating frame. It is easy to ﬁnd R−1 since it transforms a spin along the z-axis (in the rotating frame) into a spin along B~ (in the lab frame), i.e., into the E E E ~ ~ −1 ~ time dependent eigenstate n(B(t)) = ↑ (B(t)) . Namely R |↑zi = ↑ (B(t)) . Thus the simplest option would be −1 − iϕ(t) σ − iθ(θ) σ R (t) = e 2 z e 2 y . (190)

However this option does not satisfy the periodicity condition. Indeed after a full rotation that includes a change of ϕ by 2π has been accomplished θ(t0 + T ) = θ(t0), ϕ(t0 + T ) =

ϕ(t0) + 2π we have

ϕ+2π iθ −1 −i σz − σy −1 R (t0 + T ) = e 2 e 2 = −R (t0) . (191)

Thus we choose instead

−1 − iϕ(t) σ − iθ(θ) σ iϕ(t) σ R (t) = e 2 z e 2 y e 2 z . (192)

This gives 1 RHR−1 = − |B~ | σ , (193) 2 z and −1 1 − iϕ(t) σ 1 1 iϕ(t) σ iRR˙ = ϕ˙(1 − cos θ) σ + e 2 z ϕ˙ sin θσ − θσ˙ e 2 z . (194) 2 z 2 x 2 y We obtain h ˙ −1i ϕ˙ iRR = (1 − cos θ) σz (195) diag 2

40 For the Berry phase this would give 1 Z Φ = ∓ ϕ˙(1 − cos θ)dt , (196) ↑/↓,Berry 2 The correct, gauge invariant Berry’s phase reads 1 Z 1 Z Φ = ± ϕ˙(cos θ − 1)dt = ± dϕ (cos θ − 1) . (197) ↑/↓,Berry 2 2 It is given by the solid angle (see Fig. 8).

FIG. 8: Solid angle.

4. Geometric interpretation

Once again we have a Hamiltonian, which depends on time via a vector of parameters

~χ(t), i.e., H(t) = H(~χ(t)). The Berry’s phase of the eigenvector |Ψni is given by t t Z Z I 0 ~ ˙ 0 ~ Φn,Berry = i hΨn|∂tΨnidt = i hΨn|∇~χΨni~χdt = i hΨn|∇~χΨnid~χ. (198) C t0 t0 We introduce a ”vector potential” in the parameter space ~ ~ An(~χ) ≡ hΨn|i∇~χ|Ψni , (199) which gives I ~ Φn,Berry = An(~χ)d~χ. (200) C Further Z Z ~ ~ ~ ~ ~ Φn,Berry = ∇~χ × An · dS = Bn · dS. (201) ~ ~ ~ Thus the Berry’s phase is given by the flux of an effective ”magnetic field” Bn ≡ ∇~χ × An. Clearly, one needs at least two-dimensional parameter space to be able to traverse a contour with a flux through it (see Fig. 9).

41 FIG. 9: Contour in the parameter space.

5. Modiﬁcation of the Bohr-Sommerfeld quantization of cyclotron orbits

Consider again the eﬀective Schr¨odinger equation for the periodic part of the Bloch function un,~k(~r)   2 ~ ~ 2 ~ h¯ (k − i∇) En(k) −  u ~ (~r) = U(~r)u ~ (~r) . (202) 2m n,k n,k

Assume ~k(t) performs a cyclotron motion along a closed trajectory and also assume that ~ the band energy En(k) is well separated from the neighbouring band. Then an adiabatic approximation can be justiﬁed. We consider then ~χ(t) ≡ ~k(t) to be a slowly changing parameter and the system follows adiabatically being all the time in the state un,~k(t). It acquires then the Berry phase

Z D ~ ~ ΦBerry,n = −i dk ∇kun,~k un,~ki (203)

The quantization condition should be now modiﬁed as follows

1 I ~p · d~r + Φ = 2π(N + γ) . (204) h¯ Berry,n

(since n denotes here the band index, the Landau level index was changed to N). This can also be seen as a change of the constant

Φ γ → γ − Berry,n . (205) 2π

Example: graphene.

E. Magnetic susceptibility

Magnetization: M~

42 B~ = H~ + 4πM~ (206)

Susceptibility ∂M ! χ = (207) ∂H H~ =0 For small H~ we have M~ = χH~ . Then B~ = (1 + 4πχ)H~ = µH~ . Finally χ χ M~ = B~ = B.~ (208) µ 1 + 4πχ

If χ 1 we have M~ ≈ χB. Internal energy dU = dQ + dA = T dS − Md~ H~ (209)

Free energy F = U − T S , dF = −SdT − Md~ H~ (210)

From here ∂F ! M = − (211) ∂H T and ∂2F ! χ = − 2 (212) ∂H T,H=0

P −Ei/(kBT ) Thus we need the free energy F = −kBT ln Z, where Z = i e .

1. Grand canonical ensemble

In the grand canonical ensemble we have instead Ω = −kBT ln ZΩ, where ZΩ = h i P µN P −Ei,N /(kBT ) N e i e . For free Fermions this gives

X µ − k Ω = −kBT ln 1 + exp . (213) k kBT

F. Bohr-van-Leeuven Theorem

No magnetization due to classical motion of electrons. Classical partition function: Z H N Z = d3rd3p exp − (214) kBT

43 and 2 ~p − e A~(~r) H = c + U(~r) (215) 2m Integration over ~p eliminates the eﬀect of A~.

X. PARAMAGNETISM PAULI AND DIAMAGNETISM LANDAU

A. Pauli paramagnetism

H = H0 + HZeeman . (216) gµ B~σ~ H = − B , (217) Zeeman 2 e¯h where g = 2, µB = 2mc . The energy enters into the expression for Ω as µ − . Thus for spin ”up” we can say that the chemical potential is µ + µBB while for spin ”down” it is µ − µBB. For the potential Ω we thus obtain

1 1 ∂2Ω Ω = (Ω (µ + µ H) + Ω (µ − µ H)) = Ω + µ2 H2 0 . (218) 2 0 B 0 B 0 2 B ∂µ2

We obtain 2 ! 2 1 ∂ Ω 1 2 ∂ Ω0 χpara = − 2 = − µB 2 . (219) V ∂H T,H=0,µ V ∂µ (Ω should be divided by the volume to get the magnetization as density).

At T = 0 all states up to E = EF are occupied where E = ±µBB. In other words for electrons with σz = +1 we have all states up to = EF + µBB occupied. For σz = −1 we have all states up to = EF − µBB occupied. The diﬀerence of total densities is given by

Z EF +µBB Z EF −µBB n+ − n− = ν()d − ν()d ≈ 2µBBν(EF ) , (220) 0 0 where ν() is the density of states (pro spin). The magnetization is given by

2 M = µB(n+ − n−) = 2µBBν . (221)

Finally

2 2 χ = 2µBν = µBνs , (222) where νs ≡ 2ν is the total density of states at the Fermi energy (including spin).

44 B. Landau levels

Consider now free electrons or equivalently a parabolic band. The magnetic ﬁeld B~ k ~z. The vector potential is, e.g., A~ = (0, Bx, 0). The Hamiltonian reads   h¯2 ∂2 ∂ ieB !2 ∂2 H = −  + − x +  (223) 2m∗ ∂x2 ∂y hc¯ ∂z2 Ansatz ψ = φ(x)eikyyeikzz (224)

This gives h¯2 "∂2φ ieB 2 # Hψ = − + ik − x φ − k2 φ eikyyeikzz = Eφ eikyyeikzz (225) 2m∗ ∂x2 y hc¯ z Thus h¯2 "∂2φ ieB 2 # h¯2k2 ! − + ik − x φ = E − z φ (226) 2m∗ ∂x2 y hc¯ 2m∗

h¯2 ∂2φ h¯2 eB 2 h¯2k2 ! − + k − x φ = E − z φ (227) 2m∗ ∂x2 2m y hc¯ 2m∗

h¯2 ∂2φ e2B2 hc¯ !2 h¯2k2 ! − + x − k φ = E − z φ (228) 2m∗ ∂x2 2m∗c2 eB y 2m∗

h¯2 ∂2φ m∗ω2 h¯2k2 ! − + c (x − x )2 φ = E − z φ , (229) 2m∗ ∂x2 2 0 2m∗ ¯hc |e|B 2 ¯hc 2 where x0 ≡ eB ky and ωc = mc . We can also introduce lB ≡ |e|B . Thus x0 = lBky. The energy levels h¯2k2 1 E = z +hω ¯ n + . (230) 2m c 2

C. Degeneracy of the Landau Level

The degeneracy is obtained by introducing the volume V = LxLyLz and introducing the periodic boundary conditions in the y and z directions. Then k = 2πnz and k = 2πny . The z Lz y Ly ¯hc ky values are also limited by the fact that x0 ≡ eB ky ∈ [0,Lx]. Thus the number of states Ly max min LxLy |e|B for a given kz is given by 2π (ky − ky ) = 2π ¯hc . The number of states in an interval dkz for each Landau level is thus given by L L eB L |e|BV dn = x y z dk = dk (231) n 2π hc¯ 2π z (2π)2hc¯ z

45 D. Landau diamagnetism

The energy levels h¯2k2 1 E = z +hω ¯ n + . (232) 2m∗ c 2 |e|B with ωc = m∗c . LxLy eB The number of states for a given kz is given by 2π ¯hc . The number of states in dkz at the level n L L |e|B L |e|BV dn = x y z dk = dk . (233) n 2π hc¯ 2π z (2π)2hc¯ z We obtain

2 2   ¯h kz 1  LxLy |e|B X µ − 2m∗ − hω¯ c n + 2 Ω = −kBT ln 1 + exp   2π hc¯ k T n,kz B 2 2 ∞   ¯h kz 1  |e|BV X Z µ − 2m∗ − hω¯ c n + 2 = −kBT 2 dkz ln 1 + exp   (2π) hc¯ kBT n −∞ kBT |e|BV X 1 = − 2 F n + (234) (2π) hc¯ n=0 2 where ∞   2 2  Z ¯h kz µ − 2m∗ − hω¯ cx F (x) ≡ dkz ln 1 + exp   (235) kBT −∞ We also introduce ∞   2 2  Z ¯h kz y − 2m∗ f(y) ≡ dkz ln 1 + exp   . (236) kBT −∞ We use the variant of the Euler-Maclaurin formula

∞ ∞ 1 Z 1 X F n + = dxF (x) + F 0(0) + ..., (237) 2 24 n=0 0 We obtain ∞ Z 1 Z µ dxF (x) = dyf(y) . (238) hω¯ c −∞ 0 Thus the integral contribution to Ω reads

µ ∗ µ kBT |e|BV 1 Z kBT m V Z Ωint = − 2 dyf(y) = − 2 dyf(y) = Ω0 . (239) (2π) hc¯ hω¯ c −∞ (2πh¯) −∞ We also obtain 0 0 F (0) = −hω¯ cf (y = µ) (240)

46 and the correction

µ 1 k T |e|BV 1 k T |e|BV ∂2 Z δΩ = B hω¯ f 0(µ) = B hω¯ f(y)dy 24 (2π)2hc¯ c 24 (2π)2hc¯ c ∂µ2 −∞ 1 ∂2 1 heH¯ !2 ∂2 = − (¯hω )2 Ω (µ) = − Ω (µ) . (241) 24 c ∂µ2 0 24 m∗c ∂µ2 0

Thus

2 ! !2 2 2 2 2 1 ∂ Ω 1 1 eh¯ ∂ µB m ∂ χdia = − 2 = ∗ 2 Ω0(µ) = ∗ 2 Ω0(µ) . (242) V ∂H T,H=0,µ V 12 m c ∂µ 3V m ∂µ Comparing with 1 ∂2Ω χ = − µ2 0 . (243) para V B ∂µ2 e¯h and with µB = 2mc we obtain 1 m 2 χ = − χ (244) dia 3 m∗ para

E. van Alphen - de Haas eﬀect

The correction to δΩ in Eq. (241) is proportional to H2. The next correction is an oscillating function of H with amplitude ∝ H5/2. The magnetization M = −∂Ω/∂H is also oscillating and the amplitude of oscillations at low temperatures may become bigger than the non-oscillating contribution. This is the van Alphen - de Haas eﬀect (predicted theoretically by Landau). Qualitative explanation of oscillations due to changing occupation of Landau levels as function of H.

XI. BOLTZMANN EQUATION, ELASTIC SCATTERING ON IMPURITIES.

A. Kinematics

For quasiclassical description of electrons we introduce the Boltzmann distribution func- ~ ~ tion fn(k, ~r, t). This is the probability to find an electron in state n, k at point ~r at time t. More precisely is f/V the probability density to find an electron in state n,~k in point ~r. This means the probability to find it in a volume element dV is given by fdV/V .

47 We consider both ~k and ~r deﬁned. This means that we consider wave packets with both ~k and ~r (approximately) deﬁned. The uncertainty relation ∆k∆r ∼ 1 allows us to choose both ∆k and ∆r small enough. The electron density and the current density are given by 1 n(~r, t) = X f (~k, ~r, t) (245) V n n,~k,σ e ~j(~r, t) = − X ~v f (~k, ~r, t) (246) V ~k n n,~k,σ The equations of motion  ~  d 1 ∂n(k) ~r = ~v~ =   , (247) dt k h¯ ∂~k

d~k e h¯ = −eE~ − ~v × B~ . (248) dt c determine the evolution of the individual ~k(t) and ~r(t) of each wave packet. If the electrons would only obey the equations of motion the distribution function would satisfy ~ ~ fn(k(t), ~r(t), t) = fn(k(0), ~r(0), 0) (249)

Thus, the full time derivative would vanish df ∂f ˙ = + ~k · ∇~ f + ~r˙ · ∇~ f = 0 (250) dt ∂t k r However, there are processes which change the distribution function. These are collisions with impurities, phonons, other electrons The new equation reads ! df ∂f ~˙ ~ ˙ ~ ∂f = + k · ∇kf + ~r · ∇rf = , (251) dt ∂t ∂t Coll where ∂f = I[f] is called the collision integral. ∂t Coll Using the equations of motion we obtain the celebrated Boltzmann equation

∂f e 1 − E~ + ~v × B~ · ∇~ f + ~v · ∇~ f = I[f] . (252) ∂t h¯ c k k r

B. Collision integral for scattering on impurities

The collision integral describes processes that bring about change of the state of the electrons, i.e., transitions. There are several reasons for the transitions: phonons, electron- electron collisions, impurities. Here we consider only one: scattering on impurities.

48 ~ Scattering in general causes transitions in which electron which was in the state n1, k1 ~ is transferred to the state n2, k2. We will suppress the band index as in most cases we consider scattering within a band. The collision integral has two contribution: ”in” and

”out”: I = Iin + Iout. The ”in” part describes transitions from all the states to the state ~k:

X ~ ~ ~ ~ Iin[f] = W (k1, k)f(k1, ~r)[1 − f(k, ~r)] , (253)

~k1 ~ ~ ~ ~ where W (k1, k) is the transition probability per unit of time (rate) from state k1 to state k ~ ~ ~ given the state k1 is initially occupied and the state k is initially empty. The factors f(k1) and 1 − f(~k) take care for the Pauli principle. The ”out” part describes transitions from the state ~k to all other states:

X ~ ~ ~ ~ Iout[f] = − W (k, k1)f(k, ~r)[1 − f(k1, ~r)] , (254)

~k1 The collision integral should vanish for the equilibrium state in which 1 f(~k) = f = . (255) 0 ~ exp (k)−µ + 1 kBT This can be rewritten as   (~k) − µ exp   f0 = 1 − f0 . (256) kBT

The requirement Iin[f0] + Iout[f0] is satisﬁed if  ~   ~  ~ ~ (k1) ~ ~ (k) W (k, k1) exp   = W (k1, k) exp   . (257) kBT kBT We only show here that this is suﬃcient but not necessary. The principle that it is always so is ~ ~ called ”detailed balance principle”. In particular, for elastic processes, in which (k) = (k1), we have ~ ~ ~ ~ W (k, k1) = W (k1, k) . (258)

In this case (when only elastic processes are present we obtain)

X ~ ~ ~ ~ X ~ ~ ~ ~ I[f] = W (k1, k)f(k1)[1 − f(k)] − W (k, k1)f(k)[1 − f(k1)]

~k1 ~k1 X ~ ~ ~ ~ = W (k1, k) f(k1) − f(k) . (259)

~k1 Thus, Pauli principle does not play a role in this case.

49 C. Relaxation time approximation

We introduce f = f0 + δf. Since I[f0] = 0 we obtain

X ~ ~ ~ ~ I[f] = W (k1, k) δf(k1) − δf(k) . (260)

~k1

Assume the rates W are all equal and P δf(~k ) = 0 (no change in total density), then ~k1 1 I[f] ∼ −δf(~k). We introduce the relaxation time τ such that δf I[f] = − . (261) τ This form of the collision integral is more general. That is it can hold not only for the case assumed above. Even if this form does not hold exactly, it serves as a simple tool to make estimates. ~ More generally, one can assume τ is k-dependent, τ~k. Then δf(~k) I[f(~k)] = − . (262) τ~k

We will keep writing τ even if we mean that it is actually τ~k.

D. Condutivity

Within τ-approximation we determine the electrical conductivity. Assume oscillating electric ﬁeld is applied E~ (t) = Ee~ −iωt. The Boltzmann equation reads ∂f e f − f − E~ · ∇~ f + ~v · ∇~ f = − 0 . (263) ∂t h¯ k k r τ Since the ﬁeld is homogeneous we expect homogeneous response δf(t) = δfe−iωt. This gives e 1 − E~ · ∇~ f = iω − δf . (264) h¯ k τ

In the l.h.s. we replace f with f0. This gives e ∂f 1 − 0 h~v¯ · E~ = iω − δf . (265) h¯ ∂ k τ Thus we obtain eτ ∂f δf = 0 ~v · E~ (266) 1 − iωτ ∂ ~k

50 For the current density we obtain ~j(t) = ~je−iωt, where

e ~j = − X ~v δf(~k) V k ~k,σ 2e2 τ ∂f = − X 0 (~v · E~ )~v V 1 − iωτ ∂ ~k ~k ~k Z d3k τ ∂f = −2e2 0 (~v · E~ )~v . (267) (2π)3 1 − iωτ ∂ ~k ~k P We deﬁne the conductivity tensor σ via jα = α σα,βEβ. Thus Z d3k τ ∂f σ = −2e2 0 v v . (268) α,β (2π)3 1 − iωτ ∂ α β

At low enough temperatures, i.e., for kBT µ, ∂f π2 0 ≈ −δ( − µ) − (k T )2δ00( − µ) , (269) ∂ 6 B Assuming τ is constant and the band energy is isotropic (eﬀective mass is simple) we obtain

2e2τ Z dΩ ∂f σ = − ν()d 0 v v α,β 1 − iωτ 4π ∂ α β 2e2τν(µ) Z dΩ 2e2τν(µ) v2 = v v = F δ . (270) 1 − iωτ 4π α β (1 − iωτ) 3 α,β For dc-conductivity, i.e., for ω = 0 we obtain

2e2τν(µ)v2 e2τν (µ)v2 σ = F δ = s F δ , (271) α,β 3 α,β 3 α,β where νs = 2ν is the total density of states.

E. Determining the transition rates

Impurities are described by an extra potential acting on electrons

X Uimp(~r) = v(~r − ~aj) , (272) j where ~aj are locations of the impurities. In the Born approximation (Golden Rule) the rates are given by

~ ~ 2π 2 ~ ~ W (k1, k) = U ~ ~ δ((k1) − (k)) , (273) h¯ imp,k1,k

51 ~ where the delta function is meaningful since we use W in a sum over k1. Thus far we ∗ normalized the Bloch wave function to the volume V . That is hψk| ψki = V (this also means that the Wannier functions were not normalized to unity but to v = V/N). For the matrix element in the Golden Rule we need state normalized to 1. Thus we have

1 Z X ∗ i(~k−~k1)~r U ~ ~ = dV v(~r − ~aj)u (~r)u~ (~r)e (274) imp,k1,k ~k1 k V j

We assume all impurities are equivalent. Moreover we assume that they all have the same position within the primitive cell. That is the only random thing is in which cell there is ~ ~ an impurity. Then ~aj = Rj + δ~a. Shifting by Rj in each term of the sum and using the periodicity of the functions u we obtain

1 Z X i(~k−~k1)R~ j ∗ i(~k−~k1)~r U ~ ~ = e dV v(~r − δ~a)u (~r)u~ (~r)e imp,k1,k ~k1 k V j 1 X ~ ~ ~ = v ei(k−k1)Rj (275) ~k1,~k V j

This gives 2 1 X ~ ~ ~ ~ U = |v |2 ei(k−k1)(Rj −Rl) . (276) imp,~k1,~k 2 ~k1,~k V j,l ~ This result will be put into the sum over k1 in the expression for the collision integral I. ~ The locations Rj are random. Thus the exponents will average out. What remains are only diagonal terms. Thus we replace

2 1 2 U ~ ~ → |v~ ~ | Nimp , (277) imp,k1,k V 2 k1,k where Nimp is the total number of impurities. In other terms what we perform is the averaging over positions of the impurities. This gives for the collision integral

X ~ ~ ~ ~ I[f] = W (k1, k) f(k1) − f(k)

~k1 2π Nimp X 2 ~ ~ ~ ~ = |v~ ~ | δ((k1) − (k)) f(k1) − f(k) h¯ V 2 k1,k ~k1 Z 3 2π d k1 2 ~ ~ ~ ~ = nimp |v~ ~ | δ((k1) − (k)) f(k1) − f(k) , (278) h¯ (2π)3 k1,k where nimp ≡ Nimp/V .

52 ~ ~ We introduce the surface S deﬁned by (k1) = (k) and then

3 d k dSdk⊥ dSd = = (279) (2π)3 (2π)3 (2π)3 ∂ ∂k⊥ Now we can integrate over energy and we obtain

Z ~ ~ 2π dS(k1) 2 ~ ~ I[f(k)] = nimp |v~ ~ | f(k1) − f(k) . (280) ~ k1,k h¯ ∂(k1) (2π)3 ∂k⊥ Note that the density of states is given by

Z dS ν = (281) (2π)3 ∂ ∂k⊥

F. Transport relaxation time

As we have seen the correction to the distribution function due to application of the ~ electric ﬁeld was of the form δf ∼ E · ~vk. In a parabolic band (isotropic spectrum) this would be δf ∼ E~ · ~k. So we make an ansatz

δf = −~nk · ~g() , (282)

~ ~ ~ ~ where ~nk ≡ k/|k|. For isotropic spectrum conservation of energy means |k| = |k1|, the matrix element v depends on the angle between ~k and ~k only, the surface S is a sphere. ~k1,~k 1 Then we obtain Z dS Z dΩ ν = = ν . (283) (2π)3 ∂ 4π ∂k⊥ Thus dS dΩ = ν (284) (2π)3 ∂ 4π ∂k⊥ and

Z 2π dΩ1 2 ~ ~ I[δf] = nimp ν |v~ ~ | δf(k1) − δf(k) h¯ 4π k1,k Z 2π dΩ1 2 = nimpν|~g| |v(θ~ ~ )| cos θ~ − cos θ~ . (285) h¯ 4π k,k1 k,~g k1,~g ~ ~ We choose direction k as z. Then the vector k1 is described in spherical coordinates by θ ≡ θ and φ . Analogously the vector ~g is described by θ = θ and φ . Then ~k1 ~k,~k1 ~k1 ~g ~k,~g ~g dΩ = sin θ dθ dφ . 1 ~k1 ~k1 ~k1

53 From simple vector analysis we obtain

cosθ = cos θ cos θ + sin θ sin θ cos(φ − φ ) . (286) ~g,~k1 ~g ~k1 ~g ~k1 ~g ~k1

The integration then gives

Z nimpν 2 I[δf] = |~g| sin θ~ dθ~ dφ~ |v(θ~ )| × 2¯h k1 k1 k1 k1 × cos θ − cos θ cos θ − sin θ sin θ cos(φ − φ ) ~g ~g ~k1 ~g ~k1 ~g ~k1 Z πnimpν 2 = |~g| cos θ~g sin θ~ dθ~ |v(θ~ )| (1 − cosθ~ ) . (287) h¯ k1 k1 k1 k1

Noting that |~g| cos θ~g = ~g · ~nk = −δf we obtain δf I[δf] = − , (288) τtr where 1 πn ν Z = imp dθ |v(θ)|2 sin θ(1 − cosθ) (289) τtr h¯ Note that our previous ”relaxation time approximation” was based on total omission of the ”in” term. That is in the τ-approximation we had

X ~ ~ ~ ~ ~ X ~ ~ I[f] = W (k1, k) δf(k1) − δf(k) ≈ −δf(k) W (k1, k) . (290)

~k1 ~k1 Thus 1 πn ν Z = X W (~k ,~k) = imp dθ |v(θ)|2 sin θ . (291) τ 1 h¯ ~k1

The diﬀerence between τtr (transport time) and τ (momentum relaxation time) is the 2 factor (1 − cosθ) which emphasizes backscattering. If |v(θ)| = const. we obtain τtr = τ.

G. Local equilibrium, Chapman-Enskog Expansion

Instead of global equilibrium with given temperature T and chemical potential µ in the whole sample, consider a distribution function f(~r,~k) corresponding to space dependent T (~r) and µ(~r): 1 f0 = h i . (292) exp k−µ(~r) + 1 kBT (~r) This state is called local equilibrium because also for this distribution function the collision integral vanishes: I[f0] = 0. However this state is not static. Due to the kinematic terms in

54 ~ the Bolzmann equation (in particular ~vk · ∇rf) the state will change. Thus we consider the state f = f0 + δf and substitute it into the Bolzmann equation. This gives (we drop the magnetic ﬁeld)

∂δf e − E~ · ∇~ (f + δf) + ~v · ∇~ (f + δf) = I[δf] . (293) ∂t h¯ k 0 k r 0

We collect all the δf terms in the r.h.s.:

e ∂δf e − E~ · ∇~ f + ~v · ∇~ f = I[δf] + + E~ · ∇~ δf − ~v · ∇~ δf . (294) h¯ k 0 k r 0 ∂t h¯ k k r

We obtain ! ~ ∂f0 ~ (k − µ) ~ ∇rf0 = − ∇rµ + ∇rT (295) ∂k T and

∂f0 ~ ~ k − µ ~ ∂δf e ~ ~ ~ − ~vk (∇rµ + eE) + ∇rT = I[δf] + + E · ∇kδf − ~vk · ∇rδf . (296) ∂k T ∂t h¯ In the stationary state, relaxation time approximation, and neglecting the last two terms (they are small) we obtain

∂f0 ~ ~ k − µ ~ δf − ~vk (∇rµ + eE) + ∇rT = − , (297) ∂k T τtr ∂f0 ~ ~ k − µ ~ δf = ~vk τtr (∇rµ + eE) + ∇rT . (298) ∂k T Thus we see that there are two ”forces” getting the system out of equilibrium: the ~ ~ ~ ~ electrochemical ﬁeld: Eel.ch. ≡ E + (1/e)∇µ and the gradient of the temperature ∇T . ~ ~ More precisely one introduces the electrochemical potential φel.ch. such that Eel.ch. = E + ~ ~ ~ ~ (1/e)∇µ = −∇φel.ch. = −∇φ + (1/e)∇µ. Thus φel.ch. = φ − (1/e)µ. One also introduces

µel.ch. ≡ −eφel.ch. = µ − eφ. On top of the electric current

e ~j (~r, t) = − X ~v δf(~k, ~r, t) (299) E V ~k ~k,σ we deﬁne the heat current

1 ~j (~r, t) = X( − µ)~v δf(~k, ~r, t) (300) Q V k ~k ~k,σ

This expression for the heat current follows from the deﬁnition of heat dQ = dU − µdN.

55 This gives       ~ ~jE K11 K12 Eel.ch.   =     (301)      ~  ~jQ K21 K22 ∇T/T For the electrical current density we obtain e ~j = − X ~v δf(~k) E V k ~k,σ e ∂f − µ = − X τ 0 ~v · eE~ + k ∇~ T ~v . (302) V tr ∂ ~k el.ch. T ~k ~k,σ

Thus for K11 we obtain e2 ∂f K = − X τ 0 v v . (303) 11αβ V tr ∂ k,α k,β ~k,σ

For K12 this gives e ∂f K = − X τ 0 ( − µ)v v . (304) 12αβ V tr ∂ k k,α k,β ~k,σ For the heat current density we obtain 1 ~j = X( − µ)~v δf(~k) Q V k k ~k,σ 1 ∂f − µ = X τ ( − µ) 0 ~v · eE~ + k ∇~ T ~v . (305) V tr k ∂ ~k el.ch. T ~k ~k,σ

Thus for K21 we obtain e ∂f K = X τ ( − µ) 0 v v . (306) 21αβ V tr k ∂ k,α k,β ~k,σ

For K22 this gives 1 ∂f K = X τ 0 ( − µ)2 v v . (307) 22αβ V tr ∂ k k,α k,β ~k,σ

1) K11 is just the conductivity calculated earlier.

2) K12 = −K21. This is one of the consequences of Onsager relations. Thermo-power etc. K12 6= 0 only if density of states asymmetric around µ (no particle-hole symmetry).

3) For K22 we use ∂f π2 0 ≈ −δ( − µ) − (k T )2δ00( − µ) , (308) ∂ 6 B This gives 1 ∂f K = X τ 0 ( − µ)2 v v 22αβ V tr ∂ k k,α k,β ~k,σ

56 Z dΩ ∂f = 2τ ν()d 0 ( − µ)2v v tr 4π ∂ α β π2 Z dΩ 2π2 = −2τ (k T )2ν v v = − (k T )2νv2 τ δ . (309) tr 3 B 4π α β 9 B F tr α,β ~ Thus, for thermal conductivity κ deﬁned via ~jQ = −κ∇T we obtain 2 K22 2π 2 2 κ = − = kBT νvF τtr (310) kBT 9 Comparing with the electrical conductivity 2 σ = νv2 τ (311) 3 F tr We obtain the Wiedemann-Franz law: κ k2 T π2 = B . (312) σ e2 3

H. Onsager relations

The relation       ~ ~jE K11 K12 Eel.ch.   =     (313)      ~  ~jQ K21 K22 ∇T/T can be slightly rewritten to ﬁt Onsager’s logic. The entropy production is given by Z ~j · E~ Z ∇~ · ~j S˙ = dV E − dV Q (314) T T The last term expresses the heat brought to dV by the heat currents. We perform partial integration in the last term and obtain Z ~j · E~ Z 1 S˙ = dV E + dV ~j · ∇~ (315) T Q T Thus Z ˙ ~ ~ S = dV ~jE · XE + ~jQ · XQ (316)

~ E~ ~ ∇~ T with XE = T and XQ = − T 2 . The linear response relations read           ~ E~ ~jE Q11 Q12 XE Q11 Q12   =     =    T  (317)  ~     ~     ∇~ T  jQ Q21 Q22 XQ Q21 Q22 − T 2

The Onsager theorem states that the matrix Qij is symmetric.

57 XII. MAGNETO-CONDUCTANCE, HALL EFFECT

A. Hall eﬀect

We consider a situation when a relatively strong magnetic ﬁeld B~ is applied and, on top of that, a weak electric ﬁeld E~ . For simplicity we consider =h ¯2k2/(2m), where m can be the band mass. The Boltzmann equation reads

∂f e 1 − E~ + ~v × B~ · ∇~ f + ~v · ∇~ f = I[f] . (318) ∂t h¯ c k k r

δf We assume I[f] = − τ (by τ we mean τtr). As long as we do not consider very high magnetic ﬁelds where Landau quantization is important we still have the Fermi distribution ~ function f0 for E = 0. However we no longer can neglect the magnetic force (Lorentz) acting on the δf function. Thus the stationary Boltzmann equation for δf reads:

e e − E~ · ∇~ f = I[δf] + ~v × B~ · ∇~ δf , (319) h¯ k 0 hc¯ k k or ! ∂f δf e −e E~ · ~v 0 = − + ~v × B~ · ∇~ δf , (320) k ∂ τ hc¯ k k We look for a solution in the form (in analogy to the calculation of conductivity)

∂f ! δf = τe X~ · ~v 0 (321) k ∂

We obtain (using ~v =h ¯~k/m)

hτe¯ ∂f ! ∂2f ! ∇~ δf = 0 X~ + τe (X~ · ~v ) 0 h~v¯ (322) k m ∂ k ∂2 k

The second term multiplied in the Boltzmann equation by ~v × B~ gives zero. Thus we obtain ! ! ! ∂f ∂f e hτe¯ ∂f −e E~ · ~v 0 = −e X~ · ~v 0 + ~v × B~ · 0 X,~ (323) k ∂ k ∂ hc¯ k m ∂ or τe E~ · ~v = X~ · ~v − ~v × B~ · X,~ (324) k k mc k or ~ ~ ~ ~ E · ~vk = X · ~vk − ωcτ ~vk × b · X, (325)

58 where ~b ≡ B/~ |B|. Note that now we assumed both e and m positive. However we should also be ready to change the sign of one of them if we have holes. We make an ansatz for X~ :

X~ = |E~ |(α~e + β~b + γ(~e ×~b)) , (326) where ~e ≡ E/~ |E|. This gives

~ ~ ~e · ~vk = α~e · ~vk + βb · ~vk + γ(~e × b) · ~vk ~ ~ ~ − ωcτ(~vk × b)(α~e + βb + γ(~e × b)) ~ ~ = α~e · ~vk + βb · ~vk + γ(~e × b) · ~vk h ~ ~ ~ i − ωcτ α(b × ~e) · ~vk + γ(~e × b) · (~vk × b) ~ ~ = α~e · ~vk + βb · ~vk + γ(~e × b) · ~vk h ~ ~ ~ i − ωcτ −α(~e × b) · ~vk + γ~vk~e − γ(b~e)(~vkb) . (327)

~ ~ We collect coeﬃcients in front of ~e · ~vk, b · ~vk, and (~e × b) · ~vk. This gives

α − 1 − ωcτγ = 0 ~ β + ωcτγ(~e · b) = 0

γ + ωcτα = 0 (328)

We thus obtain 1 α = 2 2 1 + ωc τ ωcτ γ = − 2 2 1 + ωc τ 2 2 ωc τ ~ β = 2 2 (~e · b) (329) 1 + ωc τ Thus ~ 1 ~ ~ ~ 2 2 ~ ~ ~ X = 2 2 E − ωcτ(E × b) + ωc τ (E · b)b (330) 1 + ωc τ The current density is given by ~ ~j = σ0X, (331)

2 2 2 √ where σ0 = 3 e νF τvF (with ν being density of states per spin). With ν() ∝ we obtain n = 2(2/3)F νF we obtain 2 2 2 ne τvF ne τ σ0 = 2 = ∗ . (332) 2F m

59 This gives for the conductivity tensor (choosing ~b = ~z)

  1 −ωcτ 0   σ0   σ = 2 2  ωcτ 1 0  . (333) 1 + ωc τ    2 2  0 0 1 + ωc τ

The inverse tensor of resistivity reads

  1 ωcτ 0   −1   ρ = σ = ρ0  −ω τ 1 0  , (334)  c    0 0 1

1 where ρ0 ≡ . σ0 P Thus from Eα = β ραβjβ follows

~ ~ E = ρ0(~j − ωcτb × ~j) (335)

Hall eﬀect. Hall coeﬃcient E ω τ R = ⊥ = ρ c (336) Bj 0 B eB With ωc = m∗c we obtain m∗ eB τ 1 R = = . (337) e2nτ m∗c B enc

Note that for the hall coeﬃcient R the sign of the charge is important.

B. Magnetoresistance

We obtained ρxx = ρyy = ρzz = ρ0 = const.. Thus no magnetoresistance. In general one should distinguish two cases

1. Closed orbits

We have obtained for the parabolic band (with only closed orbits)

  1 −ωcτ 0   σ0   σ = 2 2  ωcτ 1 0  . (338) 1 + ωc τ    2 2  0 0 1 + ωc τ

60 At high magnetic ﬁelds ωcτ 1 we obtain

 1 1  2 2 − 0  ωc τ ωcτ   1 1  σ ∼ σ0  2 2 0  . (339)  ωcτ ωc τ    0 0 1

2. Open orbits

For open orbits the situation is different. Imagine the open orbit is in direction ky, Then there is a finite average velocity vx. Thus change of vx is possible exactly as in the case of no magnetic field. This gives

  1 − 1 0  ωcτ   1 1  σ ∼ σ0  2 2 0  . (340)  ωcτ ωc τ    0 0 1

For the resistivity tensor we obtain

  1/2 ωcτ/2 0    2 2  ρ ∼ ρ0  −ω τ/2 ω τ /2 0  . (341)  c c    0 0 1

Thus strong magnetoresistance. Namely 1) The Hall coeﬃcient R is 2 times smaller; 2)

ρyy is greatly enhanced. Since open orbits appear usually only for certain directions of the magnetic ﬁeld, one can expect strong dependence ofρ ˆ or R on the direction of B~ .

C. Quantum Hall Eﬀect (QHE)

Qualitative discussion: Fig. 10. The simple and somewhat naive explanation is provided in Fig. 11. As above we use the gauge A~ = (0, Bx, 0). The Hamiltonian (now in 2D) reads

  h¯2 ∂2 ∂ ieB !2 H = −  + − x  (342) 2m∗ ∂x2 ∂y hc¯

Ansatz ψ = φ(x)eikyy (343)

61 FIG. 10: Quantum Hall Eﬀect.

FIG. 11: Simple explatation. Black circles - occupied state, empty circles - unoccupied states.

This gives h¯2 ∂2φ m∗ω2 − + c (x − x )2 φ = Eφ , (344) 2m∗ ∂x2 2 0 ¯hc |e|B 2 ¯hc 2 where x0 ≡ eB ky and ωc = m∗c . We can also introduce lB ≡ |e|B . Thus x0 = lBky. The energy levels 1 E =hω ¯ n + . (345) c 2 Geometric quantization gives k = 2πny . The k values are also limited by the fact that y Ly y ¯hc 2 x0 ≡ eB ky ∈ [0,Lx]. Thus we obtain a set of ”stripes” localized at xny = 2πlBny/Ly in the x-direction and being plane waves in the y-direction. Each such stripe is denoted by a small circle in Fig. 11. Due to the potential walls limiting the sample, the energy of the ”stripe”- states is lifted near the walls. Thus these states acquire a dispersion E(xny ) = E(ky). This in turn means the states at the edge have a non-vanishing group velocity vy(ky) = (1/h¯)∂E/∂ky.

62 These velocities are opposite at the opposite sides. Thus we obtain chiral 1D modes at the edges.

Now assume the chemical potential at the right edge µR is diﬀerent (higher) from that at the left edge µL. We would now have more electrons on the right edge, and they would contribute more to the current in the positive y-direction. Let us calculate this extra current.

e X e X I = vy(ky) = ∂E/∂ky . (346) Ly hL¯ y µL

E(k )=µ y R 2 e Z dky ∂E(ky) e e V I = = (µR − µL) = V = . (347) h¯ 2π ∂ky 2πh¯ 2πh¯ RK E(ky)=µL This caclulation was done for the edge modes of a single Landau level. If N Landau levels are occupied, and the chemical potential is between the N-th and the N + 1-th levels, we obtain

I = NV/RK . (348)

Thus the Hall resistance is given by RK/N.

1. Relation to the Berry phase, Berry curvature, anomalous velocity, top. insulators

Here we follow Ref. [1]. We provide here only the results without proving them. We have already introduced (see Eq. 203) the Berry phase acquired by a band electrons on a closed contour in the k-space: I D E ~ ~ ΦBerry,n = dk un,~k i∇k un,~k . (349) Here the quasi-momentum ~k ∈ 1.B.Z. is considered to be a parameter. We can thus introduce the vector potential: D E ~ ~ ~ An(k) ≡ un,~k i∇k un,~k . (350) Using the Stockes theorem we can rewrite

I Z ~ ~ ~ ~ ~ ΦBerry,n = dk · An = dS ∇k × An . (351) S We introduce thus the Berry curvature (quasi magnetic ﬁeld)

D E ~ ~ ~ ~ ~ Ωn ≡ ∇k × An = i ∇kun,~k × ∇kun,~k . (352)

63 it turns out that the Berry curvature gives rise to the anomalous velocity, so that   1 ∂(~k) d~k ~v = +   × Ω~ . (353) h¯ ∂~k dt

(if a magnetic ﬁeld is present the band energy gets also an extra contribution of the type −~m(k)B~ ). If only the electric ﬁeld is applied we get

1 ∂(~k) e ~v = + E~ × Ω~ . (354) h¯ ∂~k h¯ Assuming the band is fully occupied we get for the current density

2 Z d ~ e X ~ e ~ d k ~ ~ j = ~v(k) = E × d Ω(k) . (355) V k h¯ 1.B.Z. (2π)

For example in 2D (xy) there is only one component of the Berry curvature, namely Ωz:

Ωz = i h∂kx u| ∂ky ui − ∂ky u ∂kx ui . (356)

One obtains a Hall-like response with

e2 Z d2k σxy = Ωz . (357) h¯ 1.B.Z. (2π)2

It turns out that the quantity

Z 1 2 nC = d k Ωz , (358) 2π 1.B.Z. is a topological invariant called the Chern number. It is integer. Thus

e2 σ = n . (359) xy 2πh¯ C

In most cases Ωz = 0. This is so, e.g., if both time reversal and inversion symmetry are present. However, of the time-reversal is violated (example: anomalous Hall effect), the Berry curvature may appear. To describe the quantum Hall effect (strong magnetic field) one should consider special values of magnetic field such that the flux through p elementary cells is equal to qΦ0 (q is integer). Then the magnetic field introduces the magnetic lattice (roughly unit cell consisting of p elementary cells) and the magnetic Brillouin zone. Every band splits into multiple topological ones with quantized Hall conductivity. See Ref. [2].

64 XIII. FERMI GAS

We consider a collection (gas) of N free electrons (fermions with spin 1/2). The ground state is given by Y † |φ0i = ak,σ |0i (360) |k|

The value of kF is determined by the number of particles N. Namely     Z 3 X X 2V 3 kF N =  1 = 2  1 = d k = V · (361) (2π)3 3π2 |k|

2 1/3 Thus kF = (3π n) , where n ≡ N/V is the density of electrons.

A. One-particle correlation function

0 ˆ † ˆ 0 Gσ(r − r ) = hφ0| Ψσ(r)Ψσ(r ) |φ0i (363)

Meaning: amplitude to remove an electron with spin σ at r0 and insert it back at r. 0 0 0 Clearly Gσ(0) = n/2. So we deﬁne gσ(r − r ) such that Gσ(r − r ) = (n/2)gσ(r − r ) and gσ(0) = 1.

0 1 X −ikr+ik0r0 † Gσ(r − r ) = e hφ0| aˆk,σaˆk0,σ |φ0i V k,k0 Z 3 d k −ik(r−r0) n 3(sin x − x cos x) = 3 e = · 3 , (364) |k|

0 where x ≡ kF |r − r |.

65 B. Two-particle correlation function

Probability to ﬁnd a particle at r0 with spin σ0 if at r there is already a particle with spin σ.

2 0 2 † † 0 0 g 0 (r − r ) = hφ | Ψˆ (r)Ψˆ 0 (r )Ψˆ 0 (r )Ψˆ (r) |φ i σ,σ n 0 σ σ σ σ 0 2 2 0 0 = [hφ | ρˆ (r)ˆρ 0 (r ) |φ i − δ(r − r )δ 0 · n)] (365) n 0 σ σ 0 σ,σ We obtain 2 0 2 1 X −i(k−k0)r −i(q−q0)r0 † † 0 gσ,σ (r − r ) = 2 e · e hφ0| aˆk,σaˆq,σ0 aˆq0,σ0 aˆk0,σ |φ0i (366) n V k,k0,q,q0 If σ 6= σ0 the calculation is simple 2 0 2 X gσ,σ0 (r − r ) = hφ0| nˆk,σnˆq,σ0 |φ0i = 1 (367) V n k,q If, however, σ = σ0, then we use Wick’s theorem (in this case it is not diﬃcult to prove)

† † hφ0| aˆk,σaˆq,σaˆq0,σaˆk0,σ |φ0i = (δk,k0 δq,q0 − δk,q0 δq,k0 )nk,σnq,σ (368)

Thus

2 2 2 0 9(sin x − x cos x) g (r − r0) = X 1 − e−i(k−q)(r−r ) = 1 − σ,σ V n x6 |k|

0 where again x = kF |r − r |. There is a ”hole” due to the Pauli principle. One can check that n Z d3r(g (r) − 1) = −1 (370) 2 σ,σ

−1 Exactly one electron is missing. The radius of the hole ∼ kF . From the density we obtain the volume taken by one electron δV = 1/n. Radius of a sphere corresponding to δV is 1/3 1/3 3 3 9π −1 obtained from 1/n = δV = (4π/3)rF . Thus, indeed, rF = 4πn = 4 kF . In what follows it will be useful to introduce a dimensionless parameter by dividing rF by the Bohr 2 2 radius a0 =h ¯ /(mee ) and obtain

1/3 2 rF 9π mee rs ≡ = 2 (371) a0 4 h¯ kF

66 1.0

0.8

0.6

0.4

0.2

2 4 6 8 10

FIG. 12: Function gσ,σ(x).

C. Jellium model, energy of the ground state

We consider now the gas of interacting electrons. There are N electrons in volume V . The positively charged N ions are distributed homogeneously over the volume V . One neglects the crystalline structure and considers ions as an absolutely homogeneous charge density n = N/V . Thus the Hamiltonian of the electrons reads

Z ( 2 ) ˆ X 3 h¯ ˆ † ˆ ˆ † (1) ˆ H = d r − Ψσ(r)∆Ψσ(r) + Ψσ(r)U (r)Ψσ(r) σ 2m ZZ X 1 3 3 ˆ † ˆ † (2) ˆ ˆ + d r1d r2 Ψσ1 (r1)Ψσ2 (r2)U (r1 − r2)Ψσ2 (r2)Ψσ1 (r1) , (372) σ1,σ2 2 where 2 Z 2 (2) e (1) 3 0 0 e U (r1 − r2) = and U (r) = − d r n(r ) 0 (373) |r1 − r2| |r − r | Since the density of ions n(r0) does not depend on r0, we obtain, that U (1)(r) is also r- (1) independent (except for boundary eﬀects) and is given by U (r) = −nU0, where U0 = R 3 e2 d r |r| . This integral would diverge if V → ∞. We want to calculate the average value of the Hamiltonian in the free electrons’ ground state

E = hφ0| H |φ0i . (374)

That is we do not look for the real ground state of H, but take the simple one (|φ0i) and calculate the expectation value of the energy.

67 1. Kinetic energy

Z ( 2 ) X 3 h¯ ˆ † ˆ Ekin = hφ0| d r − Ψσ(r)∆Ψσ(r) |φ0i σ 2m h¯2k2 3 3 h¯2k2 = X = NE = N F = N , (375) 2m 5 F 5 2m kin |k|

2. Potential energy

Z X 3 n ˆ † (1) ˆ o 2 Epot = hφ0| d r Ψσ(r)U (r)Ψσ(r) |φ0i = −n VU0 = −nNU0 . (376) σ

3. Interaction energy

ZZ X 1 3 3 ˆ † ˆ † (2) ˆ ˆ Eint = hφ0| d r1d r2 Ψσ1 (r1)Ψσ2 (r2)U (r1 − r2)Ψσ2 (r2)Ψσ1 (r1) |φ0i σ1,σ2 2 ZZ X 1 3 3 (2) ˆ † ˆ † ˆ ˆ = d r1d r2 U (r1 − r2) hφ0| Ψσ1 (r1)Ψσ2 (r2)Ψσ2 (r2)Ψσ1 (r1) |φ0i σ1,σ2 2 ZZ 2 1 3 3 (2) X n = d r1d r2 U (r1 − r2) gσ1,σ2 (r1 − r2) 2 σ1,σ2 4 ZZ ! 1 3 3 (2) 2 X 2 = d r1d r2 U (r1 − r2) n − Gσ(r1 − r2) (377) 2 σ The ﬁrst term in called the Hartree term. It gives

2 2 EHartree = n VU0/2 = nNU0/2 = N U0/(2V ) . (378)

It does not exactly cancel Epot as sometimes (wrongly) stated. To get a full cancellation one has to consider the energy of the ion charges interacting with themselves (not included in 2 the model). This one is also given by Eion−ion = n VU0/2. More logical within the current 2 model would be to say that Epot + EHartree = −nelnionVU0 + nelVU0/2 is minimized by 2 nel = nion = n. The minimal value of Epot + EHartree is given by −n VU0/2.

68 The second term is called Fock or exchange contribution

ZZ 1 3 3 (2) X 2 EFock = − d r1d r2 U (r1 − r2) Gσ(r1 − r2) 2 σ Z V 3 (2) X 2 = − d r U (r) Gσ(r) 2 σ Z 2 " #2 9n 3 e sin kF |r| − kF |r| cos kF |r| = −N d r 3 4 |r| (kF |r|)

= −Nexch , (379) where ∞ 9πne2 Z (sin x − x cos x)2 3e2 = − dx = − k . (380) exch k2 x5 4π F F 0 Thus the total energy balance reads

E "2.21 0.916# e2 = kin + exch = 2 − (381) N rs rs 2a0

Minimum is reached for rs ≈ 4.83. This value corresponds to Alkali metals. Around these values of rs the ground state of the Fermi see type is a good approximation.

For much bigger values, rs → ∞, which corresponds to a dilute limit, the energy per electron approaches zero. It turns out one can ﬁnd a better ground state with a smaller energy per electron: the Wigner crystal. In this state the electron avoid each other and. thus, minimize the interaction energy.

For rs → 0 the energy per electron becomes positive, thus it seems the system is unstable, i.e., the electrons would be ”better oﬀ” out of the system. This is however not so. In this limit the Fermi gas is a very good approximation. The positive kinetic energy is still compensated by the (inﬁnite) negative energy (work function) [Epot + EHartree]/nV = −nU0/2.

D. Hartree-Fock equations as a variational problem

One gets the same if one ﬁrst tries to minimize the functional

E[Φ] = hΦ|H|Φi (382) under the condition hΦ|Φi = 1. As Hamiltonian we have again (372). As variational wave functions we take the Slater determinants built out of the unknown single particle wave

69 functions φ1,σ(r), φ2,σ(r),.... Using the book-keeping machinery of the second quantization we get

Z ( 2 ) X 3 ∗ h¯ (1) E[Φ] = d r φn,σ(r) − ∆ + U (r) φn,σ(r) n,σ 2m ZZ X 1 3 3 ∗ ∗ (2) + d r1d r2 φn1,σ1 (r1)φn2,σ2 (r2)U (r1 − r2)φn2,σ2 (r2)φn1,σ1 (r1) , n1,σ1;n2,σ2 2 ZZ X 1 3 3 ∗ ∗ (2) − d r1d r2 φn1,σ(r1)φn2,σ(r2)U (r1 − r2)φn2,σ(r1)φn1,σ(r2) . (383) n1,n2,σ 2 Since we have to keep the wave functions normalized we have to add Lagrange multipliers:

Z ˜ X 3 ∗ E[Φ] = E[Φ] − En d r φn,σ(r)φn,σ(r) . (384) n,σ

˜ ∗ Taking the variation δE/δφn,σ(r) = 0 we obtain the system of Hartee-Fock equations:

( h¯2 ) E φ (r) = − ∆ + U (1)(r) φ (r) n n,σ 2m n,σ ZZ X 3 ∗ (2) + d r1 φn1,σ1 (r1)U (r1 − r)φn,σ(r)φn1,σ1 (r1) n1,σ1 ZZ X 3 ∗ (2) − d r1 φn1,σ(r1)U (r1 − r)φn,σ(r1)φn1,σ(r) . (385) n1

XIV. FERMI LIQUID

A. Spectrum of excitations of the ideal Fermi gas

We begin again with the ideal Fermi gas (no interactions). The Fermi distribution function (the average occupation of a level with energy k) reads 1 n = (386) F k−µ e kBT + 1

¯h2k2 Here k = 2m . In general the density n = N/V is ﬁxed. Thus the chemical potential is 2 2 ¯h kF 2 1/3 temperature dependent, µ(T ). We deﬁne F = 2m = µ(0). One can obtain kF = (3π n) .

At T = 0 the Fermi function is a step function nF = θ(F − k). At T > 0 the step gets smeared. This can be thought of as a result of excitation of ”quasiparticles”. Namely, assume we transfer one electron with energy 1 < F to a state with energy 2 > F . The energy we have to pay is equal to 2 −1. We say that we create two quasiparticles: 1) one of the particle (electron) type with energy ξ1 = 2 − F and the other of the antiparticle (hole)

70 type with energy ξ2 = F − 1. (One uses here the name ”hole” again, although these holes have little to do with the holes introduced earlier for band electrons with negative eﬀective mass and/or almost ﬁlled bands). If both 1 and 2 are close to F we obtain p2 p2 ξ = 1 − F ≈ (p − p )p /m = v (p − p ) 1 2m 2m 1 F F F 1 F p2 p2 ξ = F − 2 ≈ (p − p )p /m = v (p − p ) (387) 2 2m 2m F 2 F F F 2

It is instructive to study the dependence of the total energyhω ¯ ≡ ξ1 + ξ2 on the total momentumh~q ¯ ≡ ~p1 − ~p2. It is easy to show that h¯2q(q + 2k ) for q < 2k 0 < hω¯ < F F 2m h¯2q(q − 2k ) h¯2q(q + 2k ) for q > 2k F < hω¯ < F . (388) F 2m 2m

Provide picture. Exercise: structure factor, reveals the spectrum of quasiparticle - hole continuum.

B. Landau hypothesis

Fermi liquid is a state (one of the possible states) of fermions (electrons or He3 particles) with interactions. The interactions can be strong. Nevertheless Landau proposed the following: At low temperatures (when there are not many excitations) the excitation spectrum of the Fermi liquid has the same form as that of the free Fermi gas. Namely, the excitations are characterized by their momentum. There exists a special momentum pF related to the density of the liquid. The energy of quasiparticles reads ξ1 = vF (p1 − pF ) and of quasiholes ξ2 = vF (pF − p2). Now vF is just a parameter ∗ with dimensions of velocity. One can also introduce an effective mass via m vF = pF . This effective mass has nothing to do with the band effective mass. The Landau Hypothesis can be proved by the diagrammatic technique (course TKM 2). Here we just provide a motivating argument about weakness of relaxation in a Fermi gas with weak interaction. We consider an initial state with a filled Fermi see and in addition we have an electron (quasiparticle) with momentum ~p1 and energy 1 (ξ1 = 1 −F > 0). We assume p1 ∼ pF and ξ1 → 0. More precisely ξ1 F . The only possible scattering process should take an electron below the Fermi level with energy 2 < F . Two ”new” electrons will

71 0 0 0 0 0 0 be created with energies 1 > F and 2 > F . We have 1 + 2 = 1 + 2 or ξ1 + ξ2 = ξ1 + ξ2 0 0 0 with ξ1 > 0, ξ2 < 0, ξ1 > 0, and ξ2 > 0. The energy ξ2 is given by the energy conservation. 0 0 Two energies ξ2 and ξ1 are ”free”. We obtain |ξ2| < ξ1 and ξ1 < ξ1. Thus, the volume of the phase space available for the scattering process can be estimated from above to be smaller 2 2 than ν ξ1 , where ν is the density of states at the Fermi level. Of course one also has to take into account the conservation of momentum. In 3D this does not change the result and one 2 arrives at the scattering rate: γ ∼ ξ1 /F ξ1. Thus, it is the ﬁlled Fermi sphere which prevents particle from scattering. This is why quasiparticles with a given momentum have long lifetimes. In a Fermi liquid the quasiparticles will have energies of order kBT . Thus they are ”good” quasiparticles and the Fermi liquid description holds as long as kBT F .

1. Implications

One of the important implications of the Landau hypothesis is the fact that the speciﬁc heat is still given by the formula similar to that of the free gas: π2 C = k2 T ν (µ) , (389) V 3 B s where νs is the (total including spin) density of states at the Fermi energy. However in a Fermi liquid it is given by p m∗ ν = F . (390) s π2h¯3 Here instead of the free mass we have the eﬀective mass.

C. Gas model

Once having postulated the quasiparticles near the Fermi momentum pF with the eﬀective mass m∗ we can ”go back” and postulate a gas of (quasi) particles with mass m∗ which ﬁll the whole Fermi see. We describe it by the distribution function n~p,σ such that at T = 0 all states with p < pF are occupied. At low temperatures this description is equivalent to the one with quasiparticles. We postulate that the energy eigenstates are product states characterized just by the occupation numbersn ˆp,σ = 0, 1, with np,σ ≡ hnˆp,σi. Then the entropy of the system is given by

X S = −kB [np,σ ln np,σ + (1 − np,σ) ln(1 − np,σ)] . (391) p,σ

72 P This expression follows from the deﬁnition S = −kB s ws ln ws, where s denote the microscopic states characterized by which one-particle states are occupied and which are not.

The probability of a particular microscopic state ws is given by the product of occupation probabilities np for occupied one-particle states and 1 − np for unoccupied ones.

Proof: Assume we have already computed the entropy S{p} for a set of momenta {p} =

(p1, p2, . . . , pN ). That is only the states from {p} can be occupied by particles. We have P S{p} = −kB s ws ln ws, where s is restricted to the appropriate occupation states of {p}. We want to add to the set another momentum q. The new macroscopic states have the probabilities wsnq if the state q is occupied and ws(1 − nq) if it is not. Then X S{p},q = −kB [wsnq ln [wsnq] + ws(1 − nq) ln [ws(1 − nq)] ] s = S{p} − kB [nq ln nq + (1 − nq) ln(1 − nq)] . (392)

Thus, the expression for the entropy follows from the combinatorics and has nothing to do with the energy of the states characterized by the occupation numbers. P In a free Fermi gas the (average) energy would read E = p,σ pnp,σ. From this follows the variation of energy for a variation of the occupation probabilities δn, namely δE = P p,σ pδnp,σ. In the Fermi liquid theory one uses a slightly more general relation. One considers the distribution function as a matrix in the spin spacen ˆp = np,αβ. While in the free gas it was sufficient to fix a spin quantization axis and to consider only diagonal matrices np,σ ≡ np,σσ, in the Fermi liquid theory the spin-spin interaction between the quasiparticles makes a more general consideration necessary. Thus the basic relation reads X X ˆ δE = p,αβ δnp,βα = Tr ˆp δnp . (393) p,αβ p The trace is over spin variables. The relation (393) serves in the Fermi liquid theory as the definition of quasiparticle energy ˆp. That is creating a particle with momentum p would cause δnˆp and a respective change in E. However in the Fermi liquid the energy is not P given by a sum of single particle energies: E 6= Tr p ˆpnˆp. Rather ˆp is a functional of the occupation numbers for all momenta {p}: ˆp = ˆp(δnˆp1 , δnˆp2 ,...). Here by δnˆp we mean the deviation from the T = 0 step function. We rewrite the entropy in the matrix form

X S = −kBTr [ˆnp lnn ˆp + (1 − nˆp) ln(1 − nˆp)] . (394) p

73 P Using δN = Tr p δnˆp we look for a maximum of entropy for a given average energy and average number of particles. Thus we maximize S0 = S + αE + βN. From the condition δS0 = 0 we obtain " ˆ − µ! #−1 nˆ(p) = exp p + 1 , (395) kBT where µ and T are related to α and β. (To prove it is better to diagonalizen ˆp for each p.)

Have we obtained the usual Fermi occupation probability? Not really. Since ˆp depends on occupation numbers of all other states, we cannot independently determinen ˆp.

At T = 0 we can deﬁne F ≡ µ(T = 0). Note: It is a remarkable fact that the relation 2 1/3 kF = (3π n) holds also in the Fermi liquid, i.e., the value of pF =hk ¯ F is unchanged (proof of this is not simple).

D. Landau function f

If one varies the occupation probabilities one obtains for the energy functional p,σ

1 X ˆ 0 δˆp = f(p, p ) δnˆp0 . (396) V p0

The same with indexes reads

1 X 0 δp,αβ = f(p, p )αβ;γδ δnp0,γδ . (397) V p0

ˆ 0 In the Landau theory one postulates the function f(p, p ) to be independent of δnˆp. Thus one postulates that the ”quasiparticle energy” ˆp is a linear functional, while the total energy

E is the quadratic functional of δnˆp. Here, again, δnˆp are the deviations from the T = 0 step function. Thus we obtain

1 X ˆ 0 ˆp − F = vF (p − pF ) 1ˆ + f(p, p )δnˆp0 . (398) V p0

ˆ 0 0 0 Obviously the function f(p, p ) is symmetric: f(p, p )αβ,γδ = f(p , p)γδ,αβ. Only the momenta at the Fermi surface are important. Thus it is convenient to use the ˆ 0 0 function νF f(p, p ). Then in an integration over p only the angular dependence will remain. Usually the function fˆ(p, p0) has spin-independent part and a spin-dependent part:

ˆ 0 νF f = F (θ)1ˆ + G(θ) ~σ~σ . (399)

74 Here θ is the angle between p and p0. The form with explicit spin indexes reads

νF fαβ,γδ = F (θ)δαβδγδ + G(θ)~σαβ~σγδ . (400)

The second term has the usual form of exchange interaction.

E. Zero sound

Here we disregard the spin degree of freedom. The Landau function f modiﬁes the Boltzmann equation. Namely, assume we have an r-dependent deviation from equilibrium:

n(p, r, t) = n0(p) + δn(p, r, t) . (401)

We use p as a Hamilton function of the particle and conclude that the equation of motion ~ ~ (with the external force Fext, e.g., −eE) reads d ∂ ~p = F~ − p . (402) dt ext ∂~r

We obtain ∂ ∂δ 1 ∂δn 0 p = p = X f(p, p0) p (403) ∂~r ∂~r V p0 ∂~r (we omit the spin indexes). Analogously

d ∂ ~r = p . (404) dt ∂~p

The Boltzmann equation reads

∂n ∂n ∂n + ~r˙ + ~p˙ = I[n] (405) ∂t ∂~r ∂~p ˙ We retain only ﬁrst order terms (omitting thus a correction to ~r ≈ vF ~n) and we assume

Fext = 0: ∂δn ∂δn ∂δ ∂n + v ~n − p 0 = I[n] (406) ∂t F ∂~r ∂~r ∂~p Here ~n is a unity vector in the direction of ~p. We consider oscillations with high frequency ωτ 1. Then one can neglect the collision 0 integral. Taking into account that ∂n0/∂~p = −vF ~nδ(p − F ) we obtain

∂δn ∂δn 1 X 0 ∂δnp0 + vF ~n + vF ~nδ( − F ) f(p, p ) = 0 . (407) ∂t ∂~r V p0 ∂~r

75 0 i(~k~r−ωt) We look for a solution of the type δn = x(~n)δ(p − F )e , where x(~n) is an unknown function of the direction ~n. We obtain Z dΩn0 0 (ω − v ~n~k)x(~n) = v ~n~k ν f(θ 0 )x(~n ) . (408) F F 4π F ~n,~n ~ We chose direction of k as ~z and introduce s ≡ ω/(kvF ). This gives Z 0 0 dΩn0 (s − cos θ)x(θ, φ) = cos θ F (θ 0 )x(θ , φ ) . (409) ~n,~n 4π

Simplest case: F = F0 = const. If s > 1, i.e., the sound velocity ω/k bigger than vF we have a solution of the form cos θ x(θ) = const · . (410) s − cos θ Discuss: shape of the deformation. To ﬁnd s substitute (410) into (409). The condition reads π Z 2π sin θdθ cos θ F = 1 . (411) 0 4π s − cos θ 0 With y = −cosθ we obtain 1 Z dy y s s + 1 −F = −F + F ln = 1 . (412) 0 2 s + y 0 0 2 s − 1 −1

Solutions exist for F0 > 0. Discuss the case F0 → 0, s → 1. The deviation from equilibrium only for small θ.

XV. PHONONS

We come back to the potential energy Vion (see Eq. (12)) describing the interaction energy of slow ions due to their direct Coulomb repulsion and due to the electronic ground state energy. The positions of ions are given by

k ~ k ~rn = Rn + ~a , (413) ~ where Rn are the Bravais lattice vectors. The superscript k stands for the ion number k in k k the unit cell. The deviations are denoted by ~un or in components by un,α, where α = x, y, z. k As un,α = 0 is the absolute minimum of the potential energy we can expand and obtain

1 X k,k0 k k0 Vion = V0 + Φn,n0;α,β un,α un0,β (414) 2 n,n0,α,β,k,k0

76 The equations of motion:

k X k,k0 k0 mku¨n,α = − Φn,n0;α,βun0,β (415) n0,β,k0

k,k0 The coeﬃcients Φn,n0;α,β have certain symmetries: 1) Translational symmetry

k,k0 ~ ~ k,k0 Φn,n0;α,β = Φ(Rn − Rn0 )α,β (416)

2) Symmetry (partial derivatives are symmetric)

k,k0 k0,k Φn,n0;α,β = Φn0,n;β,α (417)

3) Homogeneous shift should not produce any force

X k,k0 Φn,n0;α,β = 0 (418) n0,k0

k √ k It is very useful to introduce amplitudes An,α ≡ mk un,α. We look for solutions of the following form k k i ~qR~ n−ωt An,α = Aα(q) e (419) or equivalently k k Aα(q) i ~qR~ n−ωt un,α = √ e (420) mk The eigenmodes are found from

det(ω2 1ˆ − Dˆ) = 0 , (421) where the matrix Dˆ is given by

0 0 ~ ~ k,k X k,k −i~q Rn−Rn0 Dα,β (~q) = Dn,n0;α,β e n0 1 0 ~ ~ 1 0 X k,k −i~q Rn−Rn0 k,k = √ Φn,n0;α,β e = √ Φ(~q)α,β . (422) n0 mkmk0 mkmk0

P −iqRn We have introduced the Fourier transform, which is deﬁned via B(q) = n Bne and

P iqRn Bn = (1/N) q B(q)e . Here N is the total number of the Bravais cells in the lattice

(N = N1N2N3). The wave vector q belongs to the ﬁrst Brillouin zone. It is quantized due to the ﬁnite size of the crystal.

77 k,k0 k0,k k,k0 k0,k ∗ From the symmetry Φn,n0;α,β = Φn0,n;β,α follows that Dα,β (~q) = [Dβ,α (~q)] . This means that the matrix Dˆ is hermitian and that 3M solutions exist, where M is the number of ions k k in a unit cell. We denote solutions by the subscript j: ωj(~q) and ej,α. The eigenvectors ej,α should be normalized (see below). k,k0 k,k0 k,k0 ∗ Since Φn,n0;α,β are all real, it is easy to show that Dα,β (−~q) = [Dα,β (~q)] . This means that k k ∗ ωj(−~q) = ωj(~q) and ej,α(−~q) = [ej,α(~q)] . (423) P k,k0 Now consider the limit ~q → 0. From the property n0,k0 Φn,n0;α,β = 0 follows P √ k,k0 P √ k,k0 n0,k0 mkmk0 Dn,n0;α,β = 0. This in turn means k0 mkmk0 Dα,β (~q = 0) = 0. There k √ are, therefore, 3 modes for which ωj(0) = 0. In these modes ej,α(0)/ mk is independent of k k. That is all ions are shifted exactly the same, i.g., un,α is independent of k. All other 3(M − 1) modes are optical. We obtain from the equation of motion for ~q = 0:

k k0 2 Aα(0) X k,k0 Aβ (0) ω (0) mk √ = Φn,n0;α,β √ . (424) mk n0,β,k0 mk0

P k,k0 Using again n,k Φn,n0;α,β = 0 and assuming ω(0) 6= 0 we sum over n, k and obtain

k X Aα(0) X k mk √ = mk uα(0) = 0 (425) k mk k Thus in optical modes the center of mass is constant. Acoustic modes are divided into 1 longitudinal and 2 transversal.

A. Quantization of phonon modes.

The kinetic energy of vibrations reads

1 X k 2 1 X ˙ k 2 T = mk(u ˙ n,α) = (An,α) . (426) 2 n,k,α 2 n,k,α The potential energy reads

1 X k,k0 k k0 1 X k,k0 k k0 U = Φn,n0;α,β un,α un0,β = Dn,n0;α,β An,α An0,β . (427) 2 n,n0,α,β,k,k0 2 n,n0,α,β,k,k0

k,k0 The Fourier transform of the D matrix Dα,β (~q) is a Hermitian matrix. Thus, it has 3M k 2 orthonormal eigenvectors ej,α(~q) with real eigenvalues ωj (~q):

X k k ∗ ej,α[ej0,α] = δj,j0 . (428) α,k

78 Another property is k k ∗ ej,α(−~q) = [ej,α(~q)] (429)

k k We expand the amplitudes An,α(~q) using the eigenvectors ej,α(~q): √ k X k Aα(~q, t) = NQj(~q, t)ej,α(~q) . (430) j √ The normalizing factor N is chosen to simplify the later expressions. We obtain

k 1 X k i~qR~ n 1 X k i~qR~ n An,α(t) = Aα(~q, t) e = √ ej,α(~q) Qj(~q, t) e , (431) N ~q N j,~q

k where N is the total number of unit cells (N = N1N2N3). Since An,α(t) is real we must have ∗ Qj(−~q, t) = [Qj(~q, t)] . k Using othonormality of the vectors ej,α(~q) we obtain 1 2 1 X ˙ X ˙ ˙ T = Qj(~q, t) = Qj(~q)Qj(−~q) , (432) 2 j,~q 2 j,~q and for the potential energy we obtain

1 X 2 2 1 X 2 U = ωj (~q) |Qj(~q, t)| = ωj (~q) Qj(~q)Qj(−~q) . (433) 2 j,~q 2 j,~q To formulate a Lagrangian theory it would be better to have real coordinates instead of complex Qj(~q). Alternatively one can use Qj(~q) and Qj(−~q) as independent variables.

To simplify we will suppress the index j and use Q(~q) = Qq. The conjugated variables: ∂T P = = Q˙ . (434) q ˙ −q ∂Qq The Hamiltonian: 1 X 1 X 2 H = PqP−q + ωq QqQ−q . (435) 2 q 2 j,~q We introduce the creation and annihilation operators:

† 1 aq = √ (ωqQ−q − iPq) , 2ωq 1 aq = √ (ωqQq + iP−q) . (436) 2ωq The inverse relations † aq + a−q Qq = √ , 2ωq rω P = i q (a† − a ) . (437) q 2 q −q

79 This gives X † 1 H = ωq aqaq + . (438) q 2 k It is important to express the physical ﬁeld un,α

Ak k n,α 1 X k i~qR~ n un,α = √ = √ ej,α(~q) Qj(~q) e mk Nmk j,~q

1 X h k i~qR~ n k −i~qR~ n i = √ Qj(~q) ej,α(~q) e + Qj(−~q) ej,α(−~q) e 2 Nmk j,~q

1 X 1 h k i~qR~ n † k ∗ −i~qR~ n i = √ √ aj,~q ej,α(~q) e + aj,~q [ej,α(~q)] e . (439) 2Nmk j,~q ωq

B. Phonon density of states

Z d3q Z dS 1 D(ω) = X δ(ω − ω (q)) = X (440) 3 j 3 ~ j (2π) j (2π) |∇q ωj| ~ Van-Hove singularities: ∇q ωj = 0. For ~q → 0 we consider only acoustic phonons. We obtain

ωs = cs(~nq)|~q| , (441) where s = 1, 2, 3 counts the acoustic modes. Then Z 2 Z 2 * + X dΩq q (ω) 2 X dΩq 1 3ω 1 D(ω) = 3 = ω 3 3 = 2 3 , (442) s (2π) cs(~nq) s (2π) cs(~nq) 2π cs where * + 1 1 X Z dΩq 1 3 ≡ 3 . (443) cs 3 s 4π cs(~nq)

C. Speciﬁc heat

Bose function: 1 nj,q = , (444) eβ¯hωj,q − 1 −1 where β ≡ (kBT ) . The chemical potential µ = 0 since the number of phonons is not ﬁxed and in the ground state there are no phonons.

80 The internal energy

X 1 X U = hω¯ j,q nj,k + = U0 + hω¯ j,q nj,k . (445) j,q 2 j,q

Speciﬁc heat 1 ∂U CV = . (446) V ∂T V Two universal properties: 1) Maximal frequency of phonons 2) Linear dependence of ω(q) at small q and ω (sound waves) allow for high- and low-temperature expansions.

1. High temperatures

kBT hω¯ max (only for acoustic phonons).

1 1 x x2 ! = 1 − + + ... , x 1 . (447) ex − 1 x 2 12

 !2  1 ∂ X kBT 1 hω¯ s,q 1 hω¯ s,q CV = hω¯ s,q 1 − + + ... V ∂T s,q hω¯ s,q 2 kBT 12 kBT 2 2 ! N 1 h¯ hωs,qi = 3kB 1 − 2 + ... , (448) V 12 (kBT )

(the ﬁrst term is Dulong-Petit law) where

2 1 X 2 hωs,qi = ωs,q (449) 3N s,q

If the temperature is also higher than the maximum frequency of he optical phonons, then 3 → 3M. All phonons contribute.

2. Low temperatures

kBT hω¯ max (only acoustic phonons relevant).

1 ≈ e−x , x 1 . (450) ex − 1

81 ∞ 1 ∂ ∂ Z 1 C = X hω¯ n = X dωD(ω)¯hω V V ∂T s,q s,q ∂T eβ¯hω − 1 s,q s 0 ∞ ∞ 3 * 1 + ∂ Z hω¯ 3 3 * 1 + ∂ 1 Z x3 = X dω = X dx 2π2 c3 ∂T eβ¯hω − 1 2π2 c3 ∂T h¯3β4 ex − 1 s s 0 s s 0 ∞ 3 * 1 + 4k4 T 3 Z x3 3 * 1 + 4k4 T 3 π4 = B dx = B . (451) 2π2 c3 h¯3 ex − 1 2π2 c3 h¯3 15 s 0 s

D. Debye and Einstein approximations

How to interpolate between low and high temperatures. Simpliﬁed model. For acoustic phonons - Debye model. For optical phonons - Einstein model.

1. Debye

The dispersion law ω = cq postulated for all q. Instead of the ﬁrst Brillouin zone one takes a sphere so that the number of q’s in this sphere is equal to N. That is the radius of the sphere qD is given by 4π V q3 · = N (452) 3 D (2π)3

Debye frequency ωD = cqD

Debye temperature kBΘD =hω ¯ D This gives q 1 ∂ ∂ ZD 4πq2dq hcq¯ C = X hω¯ n = 3 V V ∂T s,q s,q ∂T (2π)3 eβ¯hcq − 1 s,q 0 q ZD 4πq2dq (¯hcq)2eβ¯hcq ∂β ! = 3 − (2π)3 (eβ¯hcq − 1)2 ∂T 0 Θ /T 1 4π(¯hc)2 1 ZD x4ex = 3 2 3 5 dx x 2 kBT (2π) (βhc¯ ) (e − 1) 0 N T 3 = 9kB f(T/ΘD) , (453) V ΘD where 1/y Z x4ex f(y) = dx . (454) (ex − 1)2 0

82 FIG. 13: Function f.

2. Einstein

Optical phonons. Neglect dispersion

ω(~q) = ω0 . (455)

hω¯ U = U + (3M − 3)N 0 . (456) 0 hω¯ 0 e kBT − 1

2 hω¯ 0 ¯hω 0 e kBT 1 ∂U N kBT CV = = (3M − 3) kB 2 . (457) V ∂T V hω¯ 0 e kBT − 1

E. Neutron scattering

The spectrum of phonons can be measured by scattering of neutrons on the material. One measures the diﬀerential cross-section:

d2σ 1 δN (Ω, ω) = f , (458) dΩdω dΩdω Ni

83 where Ni is the ﬂux density of the incident neutrons (number of particles per area and time),

δNf (Ω, ω) is the ﬂux (number per time) of neutrons scattered into the solid angle interval dΩ around Ω with energy transfer in the intervalhdω ¯ aroundhω ¯ . The energy transfer is given by 2 h¯ 2 2 hω¯ = (kf − ki ) = Ei − Ef , (459) 2Mn

where ki and kf are the initial and ﬁnal wave vectors of the neutrons. Ei and Ef are the initial and ﬁnal energies of the crystal. Golden rule:

−βEi ~ ~ X e ~ ~ V 3 δN(Ω, ω) = δN(ki → kf ) = W (ki, i → kf , f) V ni 3 d kf , (460) i,f Z (2π)

where n is the density of incident neutrons, N = n v = n ¯hki . With i i i i i Mn M d3k = k2dk dΩ = k n (¯hdω)dΩ (461) f f f f h¯2

we obtain 2 2 2 −βEi d σ Mn V kf X e ~ ~ = W (ki, i → kf , f) (462) dΩdω 2πh¯ 2π ki i,f Z To calculate the rate W we take the interaction potential of the neutrons with the ions

to be u0δ(~rneutron − ~rion) (neutrons interact mostly with nuclei). With ~r = ~rneutron and ~ ~rion = ~rn = Rn + ~un (for simplicity we assume one ion per elementary cell, i.e., only acoustic phonons) X U(~r) = u0δ(~r − ~rn) . (463) n,k 3 R 3 Here u0 has dimensions E · L , u0 = d rU. Golden rule:

2π W (~k , i → ~k , f) = |hi, k |U|f, k i|2 δ(E − E +hω ¯ ) (464) i f h¯ i f f i

84 This gives

2 2 2 −βEi 2 d σ Mn V kf e 2π u X 0 X i~q·~rn −i~q·~rn0 = 2 hi|e |fihf|e |ii δ(Ef − Ei +hω ¯ ) dΩdω 2πh¯ 2π ki i,f Z h¯ V n,n0

2 2 −βEi Mn u kf e Z dt 0 X X i~q·~rn −i~q·~rn0 (t) iωt = 2 hi|e e |iie 2πh¯ h¯ ki i Z n,n0 2π 2 2 Mn u kf Z dt 0 X i~q·~rn −i~q·~rn0 (t) iωt = 2 he e ie 2πh¯ h¯ ki n,n0 2π (466)

P P i~q·~rn R 3 i~q·~r Using ρ(~r) = n,k δ(~r − ~rn) we obtain n,k e = d r ρ(~r)e = ρ−~q. Thus

2 2 2 Z d σ Mn u0 kf dt iωt = 2 hρ−~q(0)ρ~q(t)ie dΩdω 2πh¯ 2πh¯ ki 2π 2 2 Mn u0 kf = 2 S(~q, ω) , (467) 2πh¯ 2πh¯ ki where Z Z dt iωt X dt i~q·~rn −i~q·~r 0 (t) iωt S(~q, ω) ≡ hρ−~q(0)ρ~q(t)ie = he e n ie . (468) 2π n,n0 2π We obtain

X Z dt ~ ~ S(~q, ω) = hei~q·(Rn+~un) e−i~q·(Rn0 +~un0 (t))ieiωt n,n0 2π X ~ ~ Z dt = ei~q·(Rn−Rn0 ) hei~q·~un e−i~q·~un0 (t)ieiωt n,n0 2π X ~ Z dt = N e−i~q·Rn hei~q·~u0 e−i~q·~un(t)ieiωt . (469) n 2π Next we consider the average hei~q·~u0 e−i~q·~un(t)i , (470)

We use the formula obtained earlier √

h¯ X 1 h i~qR~ n † ∗ −i~qR~ n i un,α = √ √ aj,~q ej,α(~q) e + aj,~q [ej,α(~q)] e . (471) 2Nm j,~q ωq

85 We also use a relation heiφ1 e−iφ2 i = e−2W ehφ1φ2i , (472) where 4W = hφ1φ1 + φ2φ2i. This relation holds if the operators φ1 and φ2 satisfy the Wick † theorem. They do if φ1/2 is linear in a and a .

Proof Thus we obtain hei~q·~u0 e−i~q·~un(t)i = e−2W eh[~q·~u0(0)][~q·~un(t)]i , (473)

2 2 where 2W = h[~q · ~u0(0)] i = h[~q · ~un(t)] i. The last equation follows from stationarity and translational symmetry. Thus we obtain

X ~ Z dt S(~q, ω) = Ne−2W e−i~q·Rn eh[~q·~u0(0)][~q·~un(t)]i eiωt (474) n 2π

F. Results

One can expand the exponent

1 eh[~q·~u0(0)][~q·~un(t)]i = 1 + h[~q · ~u (0)][~q · ~u (t)]i + h[~q · ~u (0)][~q · ~u (t)]i2 + ... (475) 0 n 2 0 n and obtain

S = S0 + S1 + ..., (476) where 2 −2W X S0 = N e δ(ω) δ~q,K~ , (477) K~ where K~ are the vectors of the reciprocal lattice. Physical meaning: 1) ω = 0 elastic ~ ~ ~ processes, no energy transfer. 2) ~q = K - von Laue condition. Indeed ~q = kf − ki. 3) The factor e−2W is called Debye-Waller factor. It shows that the motion of ions reduces (smears) the scattering probability. Also zero point motion contributes: W 6= 0 for T = 0. One obtains ~ 2 1 2 h¯ X |~q · ~ej(k)| W = h[~q · ~u0(0)] i = [2nB(ωk) + 1] (478) 2 4Nm ωk j,~k β¯hω −1 where nB(ω) = (e − 1) is the Bose function.

86 Consider now 1-Phonon processes. We obtain

2 −2W X h¯|~q · ~es(~q)| S1(~q, ω) = Ne [n~q,sδ(ω − ωs(~q)) + (n−~q,s + 1)δ(ω + ωs(~q))] (479) s 2mωs(~q) The two terms correspond to emission and absorption of phonons by a neutron.

XVI. PLASMA OSCILLATIONS, THOMAS-FERMI SCREENING

A. Plasma oscillations

Consider a gas of charged particles (charge e, density n0) in an oppositely charged static background. The Maxwell equation that governs the dynamics reads

∇~ · E~ = 4πρ . (480)

It is equivalent to 1 ∂E~ 4π ∇~ × B~ − = ~j (481) c ∂t c for B~ = 0 together with the continuity equation:ρ ˙ + ∇~ · ~j = 0. In Fourier space this gives i~q · E~ (~q, ω) = 4πρ and, with the continuity equation ~q ·~j = ωρ this gives (all ﬁelds are longitudinal) iωE~ (~q, ω) = 4π~j(~q, ω). Equation of motion d2~r m = eE~ (482) dt2 2 ~ and ~j = en0~v lead to −iωm~j = n0e E. This gives 4πn e2 ω2 = 0 (483) p m One can also associate the dielectric constant via ρ = −∇·~ P~ = −i~q ·P~ . Thus (everything longitudinal) iρ j n e2 1 ω2 P = = i = − 0 E = − p E (484) q ω ω2m 4π ω2 Thus ω2 ! D = E = E + 4πP = E 1 − p (485) ω2 and ω2 ! (ω) = 1 − p . (486) ω2 In general this is the q → 0 limit of the Lindhard function (~q, ω).

87 B. Thomas-Fermi screening

One studies reaction of the electrons on an external charge/potential. The Poisson equation for the external charge/potential reads

q2φext(~q) = 4πρext(~q) (487)

For the full charge/potential q2φ = 4πρ (488)

The dielectric constant is deﬁned by

4πρext = ∇~ · D~ = i~q · D~ = i~q(q)E~ = i~q(q)(−i~qφ) = (q)q2φ (489)

Thus (q)φ = φext (490)

We introduce susceptibility χ(q) by

ρind = ρ − ρext = χ(q)φ(q) (491)

That is the charge density is induced by the full potential. Then

q2 4π (φ − φext) = χφ and = 1 − χ (492) 4π q2

The idea how to calculate χ is as follows. If the potential φ is a slowly changing function of coordinate, the local electrochemical potential is given by µe.c(r) = µ(r)−eφ(r) (charge of electron is −e). The system is in local equilibrium if µe.c(r) = const. = µ0. The local electron density is determined by the chemical potential µ(r) = µ0 +eφ(r), i.e., n(~r) = n0(µ0 +eφ(~r)), where 2 X 1 Z 1 n0(µ) = = 2 d ν() (493) β(k−µ) β(−µ) V k e − 1 e − 1 In linear response

∂n ρind = −e(n (µ + eφ) − n (µ )) = −e2φ 0 ≈ −e2φ[2ν(µ )] (494) 0 0 0 0 ∂µ 0

Thus we obtain ∂n 4πe2 ∂n χ = −e2 0 and = 1 + 0 (495) ∂µ q2 ∂µ

88 We deﬁne ∂n k2 ≡ 4πe2 0 = 4πe2[2ν] = 4πe2ν (496) TF ∂µ s Thus k2 = 1 + TF (497) q2 This is the static limit ω → 0 of the Lindhard function (~q, ω). In real space this means the following. If a point charge Q is introduces as an external charge. Then Q 4πQ φext(~r) = → φext(q) = (498) r q2 and φext 4πQ φ = = 2 2 (499) q + kTF The inverse Fourier transform gives

Q φ(~r) = e−r kTF (500) r

It is important to notice (although we did not prove it) that is independent of ω up to hω¯ ∼ EF = µ. Thus, electrons screen instantaneously.

C. Dielectric constant of a metal

We have now ions and electrons and want to calculate their total dielectric constant deﬁned via (ω, ~q)φtotal(ω, ~q) = φext(ω, ~q) (501)

One can apply the following logic. Consider the potential/charge of ions as part of the external one. Then elφtotal = φext + φion (502)

Another logic is to consider electrons as external. Then

ion total ext el 0 φ = φ + φ (503)

Since φtotal = φext + φion + φel we add the two equations and obtain

el ion total total ext ( + 0 )φ = φ + φ (504)

89 or el ion total ext ( + 0 − 1)φ = φ (505) or el ion = + 0 − 1 . (506) For electrons one takes Thomas-Fermi k2 el(ω, q) = 1 + TF (507) q2 For ions - dielectric constant of plasma oscillations Ω2 ion(ω, q) = 1 − p , (508) 0 ω2 2 4πnion(Ze)2 where Ωp = M ion . Thus Ω2 k2 = 1 − p + TF . (509) ω2 q2 ion This way we took into account the bare ions. Thus the notation 0 . Alternatively we can obtain the same if we consider ions ”dressed” by screening electrons. That is ”dressed” ions respond to a potential already screened by electrons:

total 1 ext 1 1 ext φ = ion φscreened = ion el φ (510) dr dr Thus we obtain ion el el ion = dr = + 0 − 1 (511) This gives Ω2 k2 p TF 2 2 1 − 2 + 2 Ω ω (q) ion = ω q = 1 − p = 1 − , (512) dr k2 k2 2 1 + TF 2 TF ω q2 ω 1 + q2 where Ω2 Ω2 ω2(q) ≡ p = p q2 (513) k2 2 2 TF kTF + q 1 + q2 2 For q kTF this gives sound with Ω c = p (514) kTF We return back to the total dielectric constant and obtain k2 ! ω2(q)! = ionel = 1 + TF 1 − (515) dr q2 ω2 or 1 q2 ω2 = 2 2 2 2 (516) (ω, q) (q + kTF ) (ω − ω (q))

90 D. Eﬀective electron-electron interaction

Unscreened Coulomb interaction gets screened

4πe2 4πe2 4πe2 ω2(q) ! 2 → 2 = 2 2 1 + 2 2 (517) q q (q + kTF ) ω − ω (q)

Eﬀective interaction between electrons V eff is obtained by the following substitution ~k,~k0

~ ~ 0 ~q = k − k hω¯ = k − k0 (518)

At ω ω(q) - overscreening, attraction with retardation.

XVII. ELECTRON-PHONON INTERACTION, FROLICH-HAMILTONIAN¨

A. Derivation without taking into account screening

For simplicity we consider one ion per Primitive Unit (Primitive Cell) of the Bravais Lattice. Thus only acoustic phonons. The potential felt by an electron (neglecting screening by other electrons) is given by

X ion ~ X ion ~ X ~ ion ~ U(~r) = V (~r − Rn − ~un) ≈ V (~r − Rn) − ∇V (r − Rn)~un , (519) n n n ~ where ~un is the deviation of the ion from the equilibrium position Rn. In the second quantization this becomes

Z X † X h~ ion ~ i δH = − dV Ψσ(~r) ∇V (r − Rn) ~unΨσ(~r) (520) σ n

With Ψ (~r) = √1 P ψ (~r)c and Ψ† (~r) = √1 P ψ∗ (~r)c† and assuming the Bloch σ V k k,σ σ σ V k k,σ σ functions are spin independent we obtain

X † δH = − hk1| el.ph. |k2i ck1,σck2,σ (521) k1,k2,σ with Z X 1 ∗ h~ ion ~ i hk1| el.ph. |k2i = ~un dV ψk2 (~r) ∇V (~r − Rn) ψk1 (~r) (522) n V We expand 1 ~ ~ ion ~ X ion i(~r−Rn)~p ∇V (~r − Rn) = (i~p)Vp e (523) V ~p

91 Then Z 1 X X ion −i~pR~ n 1 ∗ i~p~r hk1| el.ph. |k2i = (i~p~un)Vp e dV ψk1 (~r)ψk2 (~r)e (524) V n p V

We now use the second quantized form for ~un: √

X h¯ h † i i~qR~ n ~un = q aj,~q + aj,−~q ~ej(~q) e . (525) j,~q 2NMωj,q Substituting and using

X i(~q−~p)R~ n X e = N δ~p,~q+G~ , (526) n G~ where ~q ∈ 1.B.Z and G~ are vectors of the reciprocal lattice, we obtain √ h i i Nh¯ (~q + G~ )~e 1 X ion j,q h † i hk1| el.ph. |k2i = Vq+G q aj,~q + aj,−~q V j,q,G 2Mωj,q 1 Z × dV ψ∗ (~r)ψ (~r)ei(~q+G~ )~r (527) V k1 k2 For the matrix element we obtain

1 Z ~ 1 Z ~ ~ ~ dV ψ∗ (~r)ψ (~r)ei(~q+G)~r = dV u∗ (~r)u (~r)ei(k2−k1+~q+G)~r V k1 k2 V k1 k2 1 1 Z X i(~k2−~k1+~q+G~ )R~ n ∗ i(~k2−~k1+~q+G~ )~r = e dV uk1 (~r)uk2 (~r)e (528) N n VP.U. P.U. We use again the relation (526) which can be written as 1 X i(~k2−~k1+~q+G~ )R~ n X e = δ~ ~ ~ ~ 0 (529) N k1,k2+~q+G−G n G~ 0 Here, however, the possible choices for G~ 0 are severely limited. For each G~ only one term in ~ 0 ~ ~ the sum remains (one value of G ) such that k2, k1, ~q ∈ 1.B.Z. It is more convenient, thus, to write 1 X i(~k2−~k1+~q+G~ )R~ n e = δq,k1−k2+K , (530) N n where K~ ∈ R.L. is chosen so that ~q = ~k − ~k + K~ ∈ 1.B.Z. Obviously, this choice is ~k1,~k2 1 2 unique. This gives Z Z 1 ∗ i(~q+G~ )~r 1 ∗ iG~ +K~ ·~r dV ψk1 (~r)ψk2 (~r)e = δq,k1−k2+K dV uk1 (~r)uk2 (~r)e . (531) V VP.U. P.U. Thus we obtain

X ~ ~ † h † i δH = M(k1, k2, ~q, j) ck1,σck2,σ aj,~q + aj,−~q , (532) k1,k2,q,j,σ

92 where √ h ~ i 1 i Nh¯ (~q + G)~ej,q M(~k ,~k , ~q, j) = − X V ion δ 1 2 V q+G q q,k1−k2+K G~ 2Mωj,q Z 1 ∗ i(G~ +K~ )·~r × dV uk1 (~r)uk2 (~r)e . (533) VP.U. P.U. Once again, K~ is uniquely chosen so that ~q ∈ 1.B.Z.

B. Including screening

ion 4πe2 Our naive derivation assumed no screaning and thus Vp = p2 . Less naively we ion 4πe2 4πe2 ion should include screening by substituting Vp = 2 2 ≈ 2 . This means V (~r) = (p +kTF ) kTF 2 2 3 4πe aTF δ (~r).

C. Direct derivation with screening

The polarization P~ (~r) = en~u(~r) (assume Z = 1) creates a charge density ρion = −∇~ · P~ . The interaction of an electron with this charge density is given by Z Z 3 ion 2 3 ~ U(~r) = −e d r1Q(~r − ~r1)ρ (~r1) = e n d r1Q(~r − ~r1)∇~u(~r1) , (534) where ~ 4π Q(k) = 2 . (535) kTF In the continuous limit √ X h¯ h † i i~q~r ~u(~r) = q aj,~q + aj,−~q ~ej(~q) e . (536) j,~q 2NMωj,q Thus X † δH = hk1| el.ph. |k2i ck1,σck2,σ (537) k1,k2,σ with 1 Z Z hk | el.ph. |k i = e2n d3r d3r ψ∗ (~r)Q(~r − ~r )∇~ ~u(~r )ψ (~r) (538) 1 2 V 1 k2 1 1 k1

√ 1 2 X i h¯ [~q · ~ej,q] h † i hk1| el.ph. |k2i = e n Q(q) q aj,~q + aj,−~q V j,q 2NMωj,q Z ∗ i~q·~r × dV ψk1 (~r)ψk2 (~r)e . (539)

93 With n = N/V we reproduce the previous result. We only lost the ”umklapp” processes due to the continuous approximation for ~u(~r).

D. Phonon induced interaction between electrons

We simplify somewhat. We ”forget” about ”umklapp” processes, and also use plane waves

instead of Bloch functions. For each ~q only one phonon mode (longitudinal with ~ej,q k ~q) contributes. Then X † h † i Hel−ph = M(~q) ck+q,σck,σ a~q + a−~q , (540) k,q,σ where √ ion 1 Nh¯ M(~q) = −iVq q q . (541) V 2Mωq

ion 2 2 with Vq = 4πe aTF . ~ Consider a process in which an electron with momentum k1 emits virtually a phonon ~ with momentum ~q, so that its new momentum is k1 − ~q. Then an electron with momentum ~ ~ k2 absorbs the photon and its momentum becomes k2 + ~q. In the initial state the energy is E = + . In the virtual state the energy is 0 ~k1 ~k2 E = + +hω ¯ . 1 ~k1−~q ~k2 q The second order amplitude of this process reads |M(~q)|2 |M(~q)|2 = (542) E − E − − hω¯ 0 1 ~k1 ~k1−~q q Another process which interferes with the ﬁrst one is as follows. Electron with momentum ~ ~ k2 emits a phonon with momentum −~q. Then electron with momentum k1 absorbs the phonon. The amplitude reads |M(~q)|2 (543) − − hω¯ ~k2 ~k2+~q q Conservation of energy requires + = + . ~k1 ~k2 ~k1−~q ~k2+~q The total amplitude reads |M(~q)|2 |M(~q)|2 + − − hω¯ − − hω¯ ~k1 ~k1−~q q ~k2 ~k2+~q q 2|M(~q)|2hω¯ = q (544) ( − )2 − (¯hω )2 ~k1 ~k1−~q q We observe that if | − | ω the sign of the interaction matrix element is negative, ~k1 ~k1−~q q i.e., we obtain attraction.

94 1. Comparison

We should compare the phonon mediated interaction 2|M(~q)|2hω¯ q (545) ( − )2 − (¯hω )2 ~k1 ~k1−~q q where √ ion 1 Nh¯ M(~q) = −iVq q q . (546) V 2Mωq

ion 2 2 with Vq = 4πe aTF , with the one obtained earlier (a factor 1/V needed)

1 4πe2 1 4πe2 1 4πe2 ω2(q) ! 2 → 2 = 2 2 1 + 2 2 (547) V q V q V (q + kTF ) ω − ω (q)

2 2 Neglecting q in comparison with kTF we see that we have to compare

2 ωq 1 2 2 2 Nh¯ 2 ωq ∗ = 2|M(~q)| = 2 2(4πe aTF ) q (548) h¯ V 2Mωq h¯ with 1 ∗∗ = 4πe2a2 ω2 (549) V TF q We obtain

2 2 1 2 2 2 N 2 1 2 2 2 N ωq 1 2 2 2 N ωq ∗ = 2 (4πe aTF ) q = 2 (4πe aTF ) 2 = 2 (4πe aTF ) 2 2 V M V M cs V M ΩpaTF 2 1 2 2 2 N ωq M = 2 (4πe aTF ) 2 2 = ∗∗ (550) V M aTF 4πnie

XVIII. BCS THEORY OF SUPERCONDUCTIVITY

A. Phonon induced interaction between electrons

X † h † i Hel−ph = M(~q) ck+q,σck,σ a~q + a−~q , (551) k,q,σ where √ ion 1 Nh¯ M(~q) = −iVq q q . (552) V 2Mωq

ion 2 2 with Vq = 4πe aTF .

95 The total interaction amplitude reads 2|M(~q)|2hω¯ g V = q = k1,k2,q (553) k1,k2,q ( − )2 − (¯hω )2 V ~k1 ~k1−~q q (Introduction of g is convenient since g does not contain extensive quantities like V or N - check this. g has dimensions energy × volume). This amplitude is only taken on-shell as far as electrons are concerned. Thus

( − )2 = ( − )2 ~k1 ~k1−~q ~k2 ~k2+~q

. That is the eﬀective second quantized interaction between electrons due to phonons reads 1 H = X g c† c† c c (554) el−el−ph 2V k1,k2,q k1+q,σ1 k2−q,σ2 k2,σ2 k1,σ1 k1,σ1,k2,σ2,q The noninteracting Hamiltonian reads

X † H0 = kck,σ ck,σ (555) k,σ

B. Cooper problem (L. Cooper 1955)

The interaction is attractive and considerable as long as the energy transfer | − | ~k1 ~k1−~q hω¯ q ≤ hω¯ D. We simplify the model as follows:   −g if | − | ≤ hω¯  ~k1 ~k1−~q D gk1,k2,q = (556)  0 if | − | > hω¯  ~k1 ~k1−~q D Cooper considered a pair of electrons above the ﬁlled Fermi sphere. That is the Fermi sphere is given by Y † |Φ0i = ck,σ |0i , (557) k≤kF ,σ Cooper explored the following state

X † † |Φi = ψ(k1, σ1, k2, σ2)ck1,σ1 ck2,σ2 |Φ0i (558) k1>kF ,σ1,k2>kF ,σ2

The wave function ψ(k1, σ1, k2, σ2) is antisymmetric, i.e, ψ(k1, σ1, k2, σ2) = −ψ(k2, σ2, k1, σ1) (indeed the second quantization is organized so that even if we use here not an antisym-

metric function, only the antisymmetric part will be important). We use ψ(k1, σ1, k2, σ2) =

96 α(k1, k2)χ(σ1, σ2). Further we restrict ourselves to the states with zero total momentum, ~ ~ k1 + k2 = 0. We also restrict ourselves to the layer of states with energies [EF ,EF +hω ¯ D]. Any pair out of this layer interacts with any other pair. Thus

X ~ † † |Φi = α(k)χ(σ1, σ2)ck,σ1 c−k,σ2 |Φ0i (559) EF <k

E |Φi = (H0 + Hel−el−ph) |Φi (560)

We count the energy from the energy of the ﬁlled Fermi sphere. Then

X ~ † † E |Φi = 2kα(k)χ(σ1, σ2)ck,σ1 c−k,σ2 |Φ0i k,σ1,σ2 g − X α(~k)χ(σ , σ )c† c† |Φ i (561) V 1 2 k+q,σ1 −k−q,σ2 0 k,σ1,σ2,q This gives g (2 − E)α(k) = X α(k ) (562) k V 1 EF <k1

1 gC C = X (565) V (2k − E) EF <k1

Approximating the density of states by a constant ν() = ν0 (this is density of states per spin) we obtain 1 1 E +hω ¯ − E/2 = ln F D (567) gν0 2 EF − E/2 Thus 2EF + 2¯hωD − E 2 = e gν0 (568) 2EF − E

97 2 gν (2EF − E)(e 0 − 1) = 2¯hωD (569)

For weak coupling gν0 1 we obtain

2 − gν 2EF − E = 2¯hωDe 0 (570)

2 − gν E = 2EF − 2¯hωDe 0 (571)

The binding energy per electron is then found from E = 2EF − 2∆

2 − gν ∆ =hω ¯ De 0 (572)

1. Symmetry

Since α(k) = α(−k), i.e, symmetric, the spin part of the wave function χ must be antisymmetric - singlet.

C. BCS state (J. Bardeen, L. Cooper, and R. Schrieﬀer (BCS), 1957)

1) Everything done in the grand canonical ensemble. The grand canonical partition function

X −β(En,N −µN) ZΩ = e (573) n,N shows that at T = 0 one has to minimize HG = H − µN. We obtain

1 g H = X( − µ)c† c − X c† c† c c (574) G k k,σ k,σ 2 V k1+q,σ1 k2−q,σ2 k2,σ2 k1,σ2 k,σ k1,σ1,k2,σ2,q

where the interaction term works only if the energy transfer k1+q − k1 is smaller than the

Debye energyhω ¯ D. Although the Hamiltonian conserves the number of particles, BCS constructed a trial wave function which is a superposition of diﬀerent numbers of particles:

Y † † |BCSi = (uk + vkck,↑c−k,↓) |0i . (575) k with the purpose to use uk and vk as variational parameters and minimize hBCS| HG |BCSi.

98 For this purpose one can introduce a reduced BSC Hamiltonian. Only terms of this Hamiltonian will contribute to the average with BCS trial functions. The reduced Hamilto- nian is the one in which k1 = −k2 and σ1 = −σ2:

X † 1 g X † † HBCS = (k − µ)ck,σ ck,σ − ck+q,σ c−k−q,−σ c−k,−σ ck,σ . (576) k,σ 2 V k,q,σ Renaming k0 = k + q we obtain

X † 1 g X † † HBCS = (k − µ)ck,σ ck,σ − ck0,σ c−k0,−σ c−k,−σ ck,σ , (577) k,σ 2 V k,k0,σ or X † g X † † HBCS = (k − µ)ck,σ ck,σ − ck0,↑ c−k0,↓ c−k,↓ ck,↑ , (578) k,σ V k,k0 Also the condition on k and k0 gets simpliﬁed. We just demand that

µ − hω¯ D < k, k0 < µ +hω ¯ D . (579)

1. Averages

Normalization:

Y ∗ ∗ Y † † 1 = hBCS| |BCSi = h0| (uk2 + vk2 c−k2,↓ck2,↑) (uk1 + vk1 ck1,↑c−k1,↓) |0i k2 k1 Y 2 2 = (|uk| + |vk| ) . (580) k

2 2 We further restrict ourselves to real uk and vk such that uk + vk = 1. Thus only one of them is independent. The following parametrization is helpful: uk = cos φk, vk = sin φk. We obtain

† hBCS| ck,↑ ck,↑ |BCSi Y † Y † † = h0| (uk2 + vk2 c−k2,↓ck2,↑)ck,↑ ck,↑ (uk1 + vk1 ck1,↑c−k1,↓) |0i k2 k1 2 = vk (581)

† hBCS| ck,↓ ck,↓ |BCSi Y † Y † † = h0| (uk2 + vk2 c−k2,↓ck2,↑)ck,↓ ck,↓ (uk1 + vk1 ck1,↑c−k1,↓) |0i k2 k1 2 = v−k (582)

99 † † hBCS| ck0,↑ c−k0,↓ c−k,↓ ck,↑ |BCSi Y † † Y † † 0 0 = h0| (uk2 + vk2 c−k2,↓ck2,↑) ck ,↑ c−k ,↓ c−k,↓ ck,↑ (uk1 + vk1 ck1,↑c−k1,↓) |0i k2 k1

= ukvkuk0 vk0 (583)

This gives X 2 g X hBCS| HBCS |BCSi = 2 (k − µ)vk − ukvkuk0 vk0 (584) k V k,k0

We vary with respect to φk

∂ g 2 2 X hBCS| HBCS |BCSi = 4(k − µ)vkuk − 2 (uk − vk) uk0 vk0 = 0 . (585) ∂φk V k0 g P 0 0 We introduce ∆ = V k0 uk vk and obtain

2 2 2(k − µ)vkuk = ∆(uk − vk) . (586)

Trivial solution: ∆ = 0. E.g., the Fermi sea: uk = 0 and vk = 1 for k < µ and uk = 1 and vk = 0 for k > µ. We look for nontrivial solutions: ∆ 6= 0. Then from

(k − µ) sin 2φk = ∆ cos 2φk (587) we obtain ∆ sin 2φk = 2ukvk = q (588) 2 2 ∆ + (k − µ)

2 2 k − µ cos 2φk = uk − vk = q (589) 2 2 ∆ + (k − µ) g P Then from deﬁnition of ∆ = V k ukvk we obtain the self-consistency equation g X ∆ ∆ = q (590) 2V 2 2 k ∆ + (k − µ) or

¯hωD g X 1 gν0 Z 1 1 = q = dξ √ 2V 2 2 2 2 2 k ∆ + (k − µ) ∆ + ξ −¯hωD

¯hωD/∆ Z √ ¯hωD/∆ 1 2 2¯hωD = gν0 dx√ = gν0 ln( 1 + x + x) ≈ gν0 ln (591) 1 + x2 0 ∆ 0

We have assumed ∆ hω¯ D. This gives 1 − ν g ∆ = 2¯hωDe 0 (592)

100 2. Total energy

We want to convince ourselves that the total energy of the new state is smaller that the energy of the trivial solution (fully ﬁlled Fermi sphere).

X 2 g X EBCS = hBCS| HBCS |BCSi = 2 (k − µ)vk − ukvkuk0 vk0 k V k,k0 X 2 X = 2 (k − µ)vk − ∆ ukvk , (593) k k whereas X ENorm = hNorm| HBCS |Normi = 2 (k − µ)θ(µ − k) . (594) k We obtain

X 2 X ∆E = EBCS − ENorm = 2 (k − µ)(vk − θ(µ − k)) − ∆ ukvk , (595) k k

With ξk = k − µ,

2 2 1 − cos 2φk 1 ξk vk = sin φk = = − q (596) 2 2 2 2 2 ∆ + ξk and ∆ ukvk = q (597) 2 2 2 ∆ + ξk we obtain    2  X 1 ξk ∆ ∆E = 2ξk  − q − θ(−ξk) − q  (598) 2 2 2 2 2 k 2 ∆ + ξk 2 ∆ + ξk

¯hω Z D "1 ξ # ∆2 ! ∆E = 2V ν dξ ξ − √ − θ(−ξ) − √ 0 2 2 ∆2 + ξ2 2 ∆2 + ξ2 0 ¯hω   Z D ξ2 ∆2 = 2V ν0dξ ξ − √ − q  ∆2 + ξ2 2 2 0 2 ∆ + ξk ¯hω /∆ ZD √ ! 2 2 1 = 2V ν0∆ dx x − 1 + x + √ (599) 2 1 + x2 0

The last integral is convergent and forhω ¯ D ∆ can be taken to ∞. The integral gives −1/4. Thus ν ∆2 ∆E = −V 0 . (600) 2 Roughly energy ∆ per electron in window of energies of order ∆.

101 D. Excitations

We want to consider the BCS ground state as vacuum and ﬁnd the quasiparticle excitations above it. Let us start with the normal state, i.e., vk = θ(−ξk) and uk = θ(ξk). For

ξk > 0 we have

ck,σ |Normi = 0 (601) while for ξk < 0 † ck,σ |Normi = 0 (602) we introduce   ck,σ if ξk < 0 αk,σ ≡ (603)  †  ±c−k,−σ if ξk > 0 or equivalently † αk,σ = ukck,σ ± vkc−k,−σ (604)

(the sign to be chosen).

We see, thus, that αk,σ |Normi = 0, whereas

† † αk,σ = ukck,σ ± vkc−k,−σ (605) creates an excitation of energy |ξk|. For the BCS state we obtain

† Y † † αk,σ |BCSi = (ukck,σ ± vkc−k,−σ) (uq + vqcq,↑c−q,↓) |0i (606) q We see that the proper choice of sign is

† αk,σ = ukck,σ − σvkc−k,−σ (607) and

αk,σ |BCSi = 0 . (608)

The conjugated (creation) operator reads

† † αk,σ = ukck,σ − σvkc−k,−σ (609)

One can check the commutation relations

n † o αk,σ, α 0 0 = δk,k0 δσ,σ0 (610) k ,σ +

102 n † † o {αk,σ, αk0,σ0 } = 0 α , α 0 0 = 0 (611) + k,σ k ,σ + The inverse relations read:

† † † ck,σ = ukαk,σ + σvkα−k,−σ , ck,σ = ukαk,σ + σvkα−k,−σ (612)

1. Mean ﬁeld

We adopt the mean ﬁeld approximation for the BCS Hamiltonian.

X † g X † † HBCS = (k − µ)ck,σ ck,σ − ck0,↑ c−k0,↓ c−k,↓ ck,↑ k,σ V k,k0 (613)

Note that in the interaction the terms with k = k0 are absent, since the matrix element of the electron-phonon interaction is proportional to the momentum transfer q = k − k0. Thus † † the only averages we can extract in the interaction term are hc−k,↓ ck,↑i and hck,↑ c−k,↓i. We use 1)

† † † † ck,↑ c−k,↓ = (ukαk,↑ + vkα−k,↓)(ukα−k,↓ − vkαk,↑)

2 † † 2 † † = ukαk,↑α−k,↓ − vkα−k,↓αk,↑ + ukvk(1 − αk,↑αk,↑ − α−k,↓α−k,↓) (614)

2 2 † † † † c−k,↓ck,↑ = ukα−k,↓αk,↑ − vkαk,↑α−k,↓ + ukvk(1 − αk,↑αk,↑ − α−k,↓α−k,↓) (615)

† † In the BCS ground state we obtain hc−k,↓ ck,↑i = vkuk and hck,↑ c−k,↓i = vkuk. We use

AB = hAi hBi + hAi (B − hBi) + (A − hAi) hBi + (A − hAi)(B − hAi) and neglect the last term. The mean ﬁeld Hamiltonian reads

MF X † g X † † HBCS = (k − µ)ck,σ ck,σ + hck0,↑ c−k0,↓i hc−k,↓ ck,↑i k,σ V k,k0 g X † † g X † † − hck0,↑ c−k0,↓i c−k,↓ ck,↑ − ck0,↑ c−k0,↓ hc−k,↓ ck,↑i V k,k0 V k,k0 2 X † X X † † ∆ = ξkck,σ ck,σ − ∆c−k,↓ ck,↑ − ∆ck,↑ c−k,↓ + V (616) k,σ k k g

103 Substituting the expressions for c operators in terms of α operators we obtain a diagonal Hamiltonian (exercise) X † H = Ekαk,σ αk,σ + const. , (617) k,σ q 2 2 where Ek = ∆ + ξk. For proof one needs 3)

† † † † ck,↑ ck,↑ + c−k,↓ c−k,↓ = (ukαk,↑ + vkα−k,↓)(ukαk,↑ + vkα−k,↓) † † +(ukα−k,↓ − vkαk,↑)(ukα−k,↓ − vkαk,↑)

2 2 † † 2 † † = (uk − vk)(αk,↑αk,↑ + α−k,↓α−k,↓) + 2vk + 2ukvk(αk,↑α−k,↓ + α−k,↓αk,↑) (618)

2. Nambu formalism

Another way to get the same is to use the Nambu spinors. First we obtain     ξ −∆ c 2 MF X † k k,↑ X † ∆ H = c c     + ξkc ck,↓ + V (619) BCS k,↑ −k,↓    †  k,↓ g k −∆ 0 c−k,↓ k

P † P † P † Next we rewrite k ξkck,↓ ck,↓ = k ξk(1 − ck,↓ ck,↓) = k ξk(1 − c−k,↓ c−k,↓). This gives     ξ −∆ c 2 MF X † k k,↑ X ∆ H = c c     + ξk + V (620) BCS k,↑ −k,↓    †  g k −∆ −ξk c−k,↓ k   ξk −∆ q   2 2 The eigenvalues of the matrix   read ±Ek, where Ek = ∆ + ξk. For the −∆ −ξk eigenvectors we get       ξ −∆ u u  k   k   k      = Ek   (621) −∆ −ξk −vk −vk and       ξ −∆ v v  k   k   k      = −Ek   (622) −∆ −ξk uk uk Thus     ξ −∆ E 0 †  k   k  U   U =   , (623) −∆ −ξk 0 −Ek

104 where   u v  k k  U ≡   (624) −vk uk

We obtain     ξ −∆ c 2 MF X † † k † k,↑ X ∆ H = c c UU   UU   + ξk + V (625) BCS k,↑ −k,↓    †  g k −∆ −ξk c−k,↓ k

Diagonalizing the 2 × 2 matrix for each k we obtain     E 0 α 2 MF X † k k,↑ X ∆ H = α α     + ξk + V (626) BCS k,↑ −k,↓    †  g k 0 −Ek α−k,↓ k

Using again the commutation relations for the α operators we obtain

2 MF X † X ∆ HBCS = Ekαk,σ αk,σ + (ξk − Ek) + V . (627) k,σ k g

E. Finite temperature

q 2 2 We obtained the energy spectrum Ek = ∆ + ξk in the mean-ﬁeld approximation assuming that hc−k,↓ck,↑ i = vkuk, where the averaging is in the ground state, i.e., there are no quasi-particles excited. For T > 0 some quasi-particles get excited and the value of hc−k,↓ck,↑ i changes. Namely, we obtain

hc−k,↓ck,↑ i = vkuk(1 − 2nk) , (628) where n = f(E ) = 1 . k k eβEk +1 If we still want to have the Hamiltonian diagonalized by the Bogolyubov transformation, we have to redeﬁne ∆ as

g X g X ∆ = hc−k,↓ck,↑ i = ukvk(1 − 2nk) (629) V k V k q 2 2 Then, however, ∆ is temperature dependent and thus Ek = ∆ + ξk is also temperature dependent. We must do everything self-consistently. From ∆ ukvk = q (630) 2 2 2 ∆ + ξk

105 we obtain the new self-consistency equation

g X ∆ βEk ∆ = q tanh (631) 2V 2 2 2 k ∆ + ξk

To ﬁnd the critical temperature Tc we assume that ∆(Tc) = 0. This gives

¯hω β¯hω /2 Z D βξ Z D g X 1 β|ξk| tanh 2 tanh x 1 = tanh = gν0 dξ = gν0 dx (632) 2V |ξk| 2 ξ x k 0 0

Assuminghω ¯ D/(2kBTc) 1 we obtain

hω¯ D 1 ≈ gν0 ln (633) kBTc or − 1 ∆(T = 0) k T =hω ¯ e gν0 = (634) B c D 2 More precise calculation gives

− 1 ∆(T = 0) k T = 1.14¯hω e gν0 = (635) B c D 1.76

For T ∼ Tc and T < Tc one can obtain s T ∆(T ) ≈ 3.06kBTc 1 − (636) Tc

1. More precise derivation

We have to minimize the grand canonical potential Ω = U − µN − TS = hHBCSi − TS. For the density matrix we take (the variational ansatz)

1 −β P E n ρ = e k,σ k k,σ , (637) Z † where nk,σ = αk,σαk,σ are the occupation number operators of the quasi-particles while Ek are the energies of the quasiparticles (to be determined). Here

† αk,σ = ukck,σ − σvkc−k,−σ (638) with vk = sin φk and uk = cos φk and φk is another variational parameter. We thus obtain

MF X † g X † † hHBCSi = ξkck,σ ck,σ − hck0,↑ c−k0,↓i hc−k,↓ ck,↑i k,σ V k,k0 !2 X h 2 2 2i g X = 2ξk (uk − vk)f(Ek) + vk − ukvk(1 − 2f(Ek)) (639) k V k

106 For the entropy we have

X S = −2kB [f(Ek) ln f(Ek) + (1 − f(Ek)) ln(1 − f(Ek))] (640) k

We vary with respect to φk and with respect to Ek independently. This gives ∂Ω = 4ξkukvk(1 − 2f(Ek)) ∂φk ! 2g X 2 2 − ukvk(1 − 2f(Ek)) (1 − 2f(Ek)) (uk − vk) = 0 (641) V k Introducing g X g X ∆ = hc−k,↓ck,↑ i = ukvk(1 − 2nk) (642) V k V k we obtain the old equation

ξk sin 2φk = ∆ cos 2φk (643)

Thus all the formula remain but with new ∆.

∂Ω ∂hHMF i ∂S = BCS − T ∂Ek ∂Ek ∂Ek 2 2 ∂f ∂f ∂S = 2ξk(uk − vk) + 4∆ukvk − T ∂Ek ∂Ek ∂Ek q 2 2 ∂f ∂f = 2 ξk + ∆ − 2Ek = 0. (644) ∂Ek ∂Ek Thus we obtain q 2 2 Ek = ξk + ∆ (645)

F. Heat capacity

∂S ! CV = T . (646) ∂T V Using for S Eq. (640) we obtain

X ∂f X ∂f CV = −2kBT (−βEk) = 2 Ek (647) k ∂T k ∂T

1 Let’s introduce g(x) = ex+1 . Then f(Ek) = g(βEk). ∂f = βg0 (648) ∂Ek

107 ! ! ! ∂f 0 ∂β ∂Ek 0 β ∂Ek ∂f Ek ∆ ∂∆ = g · Ek + β = g · −Ek + β = − + (649) ∂T ∂T ∂T T ∂T ∂Ek T Ek ∂T Thus ! X Ek ∆ ∂∆ ∂f CV = 2 Ek − + (650) k T Ek ∂T ∂Ek

First, we analyze at T → Tc. There Ek ≈ ξk. With ∂f π2 ≈ −δ(E) − (k T )2δ00(E) , (651) ∂E 6 B and s T ∆(T ) ≈ 3.06kBTc 1 − (652) Tc

We obtain for T = Tc − 0 Z ξ2 ! ∂f Z ∂∆2 ∂f C (T − 0) = 2ν dξ − + ν dξ V c 0 T ∂ξ 0 ∂T ∂ξ 2π2ν k2 = 0 B T + (3.06)2ν k2 T = C (T + 0) + ∆C (653) 3 c 0 B c V c V

Thus one obtains ∆C V ≈ 1.43 (654) CV (Tc + 0) ∂∆ Jump in ∂T leads to jump in CV (see Fig. 14).

FIG. 14: Heat capacitance of vanadium.

∆ − k T For kBT kBTc ∼ ∆(0) one obtains CV ∝ e B .

108 G. Isotope eﬀect.

−1/2 We see that Tc ∝ ∆(T = 0) ∝ ωD ∝ M , where M is the ion mass. This dependence can be observed by using materials with diﬀerent isotope content. It was one of the major motivations for the phonon mechanism of attraction.

XIX. ELECTRODYNAMICS OF SUPERCONDUCTORS.

A. London equations

The zero resistivity and the Meissner eﬀect are closely related. Assume the electrons are accelerated without resistance:

m~v˙ = eE~ (655)

With ~j = ne~v we obtain ˙ Λ~j = E,~ (656)

m where Λ = ne2 . The Maxwell equation reads: 1 ˙ ∇~ × E~ = − B~ (657) c Thus we obtain ∂ 1 Λ∇~ × ~j + B~ = 0 (658) ∂t c But inside the superconductor both B~ = 0 and ~j = 0. Thus F. London and H. London postulated: 1 Λ∇~ × ~j + B~ = 0 (659) c

1. Time-independent situation

An external magnetic ﬁeld is applied. We consider magnetization currents explicitly, thus we use microscopic Maxwell equation:

4π ∇~ × B~ = ~j (660) c

109 This gives 4π ∇~ × (∇~ × B~ ) = ∇~ · (∇~ · B~ ) − (∇~ 2)B~ = −(∇~ 2)B~ = ∇~ × ~j (661) c Substituting the London equation we obtain 4π (∇~ 2)B~ = B~ (662) c2Λ

q c2Λ q c2m One introduces the London penetration depth λL = 4π = 4πne2 .

B. Another form of London equations

B~ = ∇~ × A~ (663)

With this the London equation 1 Λ∇~ × ~j + B~ = 0 (664) c reads 1 Λ∇~ × ~j + ∇~ × A~ = 0 (665) c If both ∇~ · ~j = 0 and ∇~ · A~ = 0 (Coulomb gauge) this gives 1 ~j = − A~ (666) Λc In this form the London equation is convenient to connect to the microscopic theory.

C. Microscopic derivation of London equation

2 ~p − e A~ H = c (667) kin 2m with ~p = −ih¯∇~ . In second quantized form 2 Z −ih¯∇~ − e A~ X † c 2 Hkin = dV Ψσ(r) Ψσ(r) = H0,kin + H1 + O(A ) , (668) σ 2m where Z e X † ~ H1 = − dV Ψσ(r) A ~p Ψσ(r) (669) mc σ

110 (the order of operators A~ and ~p unimportant since ∇~ · A~ = 0). The ﬁrst order correction to the BCS ground state |0i = |BCSi is

X hl| H1 |0i |Φ1i = |li (670) l6=0 E0 − El

Current. Velocity ~p − e A~ ~v = c (671) m Current density ~p − e A~ 2 ~ X † c X † ~p e ~ X † j = e Ψσ(r) Ψσ(r) = e Ψσ(r) Ψσ(r) − A Ψσ(r)Ψσ(r) σ m σ m cm σ

= ~jp + ~jd (672)

(In ~jp one has to symmetrize: half ~p works to the right and half to the left). The ~jd contribution immediately gives the London equation

e2n ~j = − A~ (673) d mc ~ Another contribution linear in A could come from ~jp:

h~jpi = hΦ1|~jp |0i + h0|~jp |Φ1i (674)

To calculate |Φ1i we need hl| H1 |0i, where |li is an excited state. We assume i~q~r A = ~aqe (675) and ~q · ~aq = 0. Using Ψ = √1 P c eikr we obtain σ V k k

he¯ X † ~ H1 = − ck+q,σck,σ(k~aq) (676) mc k,σ

We use † † † ck,σ = ukαk,σ + σvkα−k,−σ , ck,σ = ukαk,σ + σvkα−k,−σ (677) and conclude that † † † hl| ck+q,σck,σ |0i = σuk+qvk hl| αk+q,σα−k,−σ |0i (678)

111 but also

Thus in both cases |li the same, i.e., the same two quasiparticles created. For this particular |li we obtain

he¯ hl| H |0i = − (~k~a )σu v + ((−~k − ~q)~a )σu v 1 mc q k+q k q k k+q he¯ = − (~k~a )σ (u v − u v ) (680) mc q k+q k k k+q

For ~q → 0 we see that the matrix element vanishes. Together with the fact that |E0−El| > 2∆ this gives ”rigidity” and

h~jpi = 0 (681)

D. Pippard vs. London, coherence length.

The matrix element (680) vanishes for q → 0. Let us analyze it more precisely. We have

v v u s s u u1 ξk+q 1 ξk 1 ξk u1 ξk+q uk+qvk − ukvk+q = t + − − + t − (682) 2 2Ek+q 2 2Ek 2 2Ek 2 2Ek+q

For ξk Ek ∼ ∆ we obtain 1 hv¯ q u v − u v ≈ (ξ − ξ ) ≈ F (683) k+q k k k+q 2∆ k+q k 2∆

This introduces the coherence length:

hv¯ ξ ≡ F (684) ∆

¯hvF (one usually deﬁnes ξ0 = π∆ ). The more general than London relation is called Pippard relation:

Z X 3 0 0 0 jα(~r) = − d r Qα,β(~r − ~r )Aβ(~r ) (685) β where the kernel Q decays on the distance of order ξ. ξ is the size of a Cooper pair.

Two limits: ξ < λL - London limit, ξ > λL - Pippard limit.

112 E. Superconducting density

At T = 0 we obtained e2n ~j = − A~ (686) mc Here n is the total electron density. Note that transition to pairs does not change the result. Namely the substitution n → n/2, m → 2m, and e → 2e leaves the result unchanged. At T > 0 not all the electrons participate in the super current. One introduces the superconducting density ns(T ) and the normal density nn(T ), such that ns + nn = n. Thus e2n ~j = − s A~ (687) s mc

Calculations show that near the critical temperature, i.e., for Tc − T Tc n T s ≈ 2 1 − (688) n Tc (arguments with moving liquid) The new penetration depth is deﬁned as s c2m λ (T = 0) T −1/2 L √ λL(T ) = 2 ≈ 1 − (689) 4πnse 2 Tc

F. Critical ﬁeld

One applies external magnetic field H. It is known that the field is expelled from the superconductor (Meissner effect). That is inside the superconductor B = 0. When the field reaches the critical field Hc the superconductivity is destroyed and the field penetrates the metal. Naive (but correct) argument: The total (free) energy of a cylindrical superconductor consists of the bulk free energy Fs and the energy of the induced currents screening the external magnetic field. We have B = 0 = Bext +Binduced (recall that H = Bext). The energy 2 of the induced currents is given by Binduced/(8π). Thus the total energy of a superconductor 2 reads Fs + H /(8π). For H = Hc the free energy of a superconductor and of a normal metal should be equal H2 F + c = F . (690) s 8π n

The less naive thermodynamic argument involves the free enthalpy G = F − HB/(4π) (see the book by Abrikosov).

113 At zero temperature (F = U − TS) we have ν ∆2 F − F = 0 (691) n s 2 Thus we ﬁnd √ Hc(T = 0) = 2 πν0∆(T = 0) (692)

−1/2 In particular also for Hc we have the isotope eﬀect, Hc ∝ M .

For T → Tc − 0 one obtains (no proof) T Hc(T ) = 1.735Hc(0) 1 − (693) Tc

G. Order parameter, phase

Thus far ∆ was real. We could however introduce a diﬀerent BCS groundstate:

Y iφ † † |BCS(φ)i = (uk + e vkck,↑c−k,↓) |0i . (694) k Exercise: check that

2π Z dφ |BCS(N)i = |BCS(φ)i e−iNφ (695) 2π 0 gives a state with a ﬁxed number of electrons N. We obtain for ∆

g X g X iφ iφ ∆ = hc−k,↓ck,↑ i = ukvke = |∆|e (696) V k V k Usual gauge transformation: A~ → A~ + ∇~ χ (697)

ie χ Ψ → Ψe hc¯ (698)

We identify φ e = − χ (699) 2 hc¯ Thus hc¯ A~ → A~ − ∇~ φ (700) 2e and e2n hc¯ ! ~j = − s A~ − ∇~ φ (701) s mc 2e

114 H. Ginsburg-Landau Theory

Theory works for T ≈ Tc. One introduces the order parameter

rn Ψ = s eiφ (702) 2

1. Landau Theory

One postulates for the free energy

Z Z ( b ) dV F = dV F + a|Ψ|2 + |Ψ|4 (703) n 2

In order to describe the phase transition one postulates a = ατ, where

T − T τ = c (704) Tc and α > 0, b > 0. By varying we obtain |Ψ|2: a + b|Ψ|2 = 0 (705)

For τ < 0 this gives ατ α T − T |Ψ|2 = − = c (706) b b Tc For τ > 0 we have |Ψ|2 = 0. Phase transition. We deﬁne a Ψ2 ≡ − . (707) 0 b

2. Ginsburg-Landau Theory, equations

Theory for inhomogeneous situations, currents and magnetic ﬁelds. One postulates for the free energy   Z Z 2 ~ 2  2 b 4 1 ~ 2e ~ B  dV F = dV Fn + a|Ψ| + |Ψ| + −ih¯∇ − A Ψ + (708)  2 4m c 8π 

Here, for a while, we consider the superconductor on its own. Thus B~ is the ﬁeld induced by the currents in the superconductor itself. Below we will include the external ﬁeld.

115 Here we have to vary with respect to Ψ regarding Ψ∗ as independent. This gives

1 2e 2 −ih¯∇~ − A~ Ψ + aΨ + b|Ψ|2Ψ = 0 (709) 4m hc¯

Varying with respect to A~ (using div[a · b] = b · rota − a · rotb) gives

4π ∇ × B~ = ~j (710) c with 2ieh¯ (2e)2 ~j = − Ψ∗∇~ Ψ − Ψ∇~ Ψ∗ − |Ψ|2A~ (711) 4m 2mc iφ(~r) For Ψ = Ψ0e we obtain again the London equation 4e2 hc¯ ! e2n hc¯ ! ~j = − Ψ2 A~ − ∇~ φ = − s A~ − ∇~ φ (712) s 2mc 0 2e mc 2e

Thus we obtain the London penetration depth s c2m s c2m λL = 2 = 2 2 . (713) 4πnse 8πΨ0e

3. Coherence length

Coherence length is obtained by considering small ﬂuctuations of the amplitude of Ψ. So ~ 2 we assume A = 0, and Ψ = Ψ0 + δΨ (both real), and Ψ0 = −a/b. Then we obtain h¯2 − ∇2δΨ + δΨ(a + 3bΨ2) = 0 . (714) 4m 0

±x/ξ In the normal state Ψ0 = 0 and a > 0 we obtain solutions of the type e , where h¯ ξ = √ (715) 4ma

2 In the superconducting state Ψ0 = −a/b, a < 0 h¯2 h¯2 − ∇2δΨ + δΨ(a + 3bΨ2) = − ∇2δΨ − 2aδΨ = 0 . (716) 4m 0 4m

We still deﬁne the coherence length as in the normal case

h¯ ξ = q . (717) 4m|a|

√ However the solutions look like e± 2x/ξ.

116 4. External ﬁeld

~ If a superconductor is placed in an external magnetic field H0 the proper free energy reads Z Z 1 Z dV F = dV F − H~ dV B.~ (718) H 4π 0 ~ ~ ~ ~ ~ Here B is the total magnetic field, B = H0 + Bi. Here Bi is the field induced by currents in the superconductor. Thus   Z Z 2 ~ 2 ~ ~  2 b 4 1 ~ 2e ~ B H0 · B  dV FH = dV Fn + a|Ψ| + |Ψ| + −ih¯∇ − A Ψ + − .  2 4m c 8π 4π  (719) 2 Note, that this gives the same Ginsburg-Landau equations. Indeed B /(8π) − BH0/(4π) = 2 Bi /8π + const. and we vary, actually, the field Bi. 2 In the normal state we have B = H0 and FH = Fn −H0 /(8π). Deep in the superconductor 2 b 4 a2 (ατ)2 B = 0 and FH = Fn + a|Ψ| + 2 |Ψ| = Fn − 2b = Fn − 2b . Thus we obtain the critical field

Hc, i.e., the value of H0 above which the normal state has a lower free energy. We obtain 2 (ατ)2 Hc /(8π) = 2b and s 4πa2 H = . (720) c b

5. Reduced Ginsburg-Landau equations

We deﬁne √ √ 0 0 0 0 Ψ = Ψ/Ψ0 , r = r/λL ,B = B/(Hc 2) ,A = A/(λLHc 2) (721)

We obtain the Ginsburg-Landau equations in the reduced form (omitting the primes)

2 −iκ−1∇~ − A~ Ψ − Ψ + |Ψ|2Ψ = 0 , (722)

i ∇~ × (∇~ × A~) = − Ψ∗∇~ Ψ − Ψ∇~ Ψ∗ − |Ψ|2A,~ (723) 2κ where λ κ = L . (724) ξ Thus, everything depends on κ.

117 The free energy in these units reads    2  Z Z 2 Z  ~  Hc  2 1 4 i∇ ~ ~ 2 ~ ~  dV FH = dV Fn + dV −|Ψ| + |Ψ| + − − A Ψ + B − 2H0 · B .

4π  2 κ  (725) Integrating by part, disregarding the boundary, using the Ginsburg-Landau equations, and adding an unimportant constant we obtain Z Z H2 Z 1 dV F = dV F + c dV − |Ψ|4 + (B~ − H~ )2 . (726) H n 4π 2 0

I. Surface energy

Let us estimate the surface energy of an interface between superconducting and normal phases. We assume H0 = Hc, i.e., both phases are possible. In the normal phase we have √ 0 the critical magnetic ﬁeld B = Hc (B = 1/ 2). In the superconducting phase B = 0 and 0 −1 Ψ = Ψ0 (Ψ = 1). The order parameter varies on the scale ξ (κ ). The magnetic ﬁeld varies on the scale λL (1). We consider a quasi-one dimensional situation. All the quantities depend only on x. A~ is along y (A~ = A(x)~y) and, thus, B~ is along z. We can take Ψ to be real. Then

κ−2∇2Ψ + (1 − A2)Ψ − Ψ3 = 0 , (727)

∇2A − Ψ2A = 0 . (728)

Consider 2 cases:

a) ξ λL (κ 1)(superconductor of the 1-st type). In this case there is a layer on the interface of thickness ξ where the magnetic ﬁeld vanishes and the order parameter vanishes, 2 Hc i.e., the state is normal. We see that there is an additional cost of ∼ ξ 8π per unit of area. The logic: the work of expelling the magnetic ﬁeld has been performed but no energy reduction through the order parameter appearance. Thus the surface energy is positive in this case and the system avoids interfaces.

b) ξ λL (κ 1)(superconductor of the 2-nd type). In this case there is a layer of thickness λL where the magnetic ﬁeld is present and also the order parameter has its 2 Hc bulk value. The surface energy is then negative and equal ∼ −λL 8π . The logic: magnetic ﬁeld not expelled in the layer, thus no energy cost. The energy is reduced by having the superconducting order parameter. Thus the system likes to have interfaces.

118 √ The critical value of κ at which the surface energy vanishes is given by κc = 1/ 2.

J. Higgs mechanismus

We consider again the GL free energy density (action):

2 ~ 2 2 b 4 1 2e B F = a|Ψ| + |Ψ| + −ih¯∇~ − A~ Ψ + 2 4m c 8π b 1 2e 2e (∇~ × A~)2 = a|Ψ|2 + |Ψ|4 + −ih¯∇~ − A~ Ψ ih¯∇~ − A~ Ψ∗ + .(729) 2 4m c c 8π q Consider small ﬂuctuations around the real solution Ψ0 = −a/b.

Ψ(~r) = Ψ0 + φ1(~r) + iφ2(~r) , (730)

~ where φ1 and φ2 are real. Considering also A(~r) to be small we expand the action to second ~ order in φ1, φ2 and A: " # 1 2e2 2 2 2 2e δF (2) = Ψ2 A~ +h ¯2 ∇~ φ +h ¯2 ∇~ φ − 2¯h Ψ A~∇~ φ 4m c 0 1 2 c 0 2 (∇~ × A~)2 − 2aφ2 + + higher orders . (731) 1 8π We still have the gauge freedom: 2ie A~0 = A~ + ∇~ χ , Ψ0 = Ψ exp χ . (732) hc¯ To keep A~0 small we perform an inﬁnitesimal gauge transformation, which then reduces to 0 2ie Ψ ≈ Ψ(1 + iχ˜) = (Ψ0 + φ1 + iφ2)(1 + iχ˜), whereχ ˜ ≡ ¯hc χ. In terms of the deviations we obtain 0 0 φ1 = φ1 − φ2χ˜ , φ2 = φ2 + φ1χ˜ + Ψ0χ˜ . (733)

It is actually suﬃcient to keep only the term Ψ0χ˜. Thus, we can always ﬁnd a gauge 0 transformation such that φ2 = 0. Dropping the primes we obtain

2 h¯ 2 δF (2) = ∇~ φ − 2aφ2 4m 1 1 (∇~ × A~)2 1 2e2 2 + + Ψ2 A~ + higher orders . (734) 8π 4m c 0

Thus we obtain two modes. The ﬁrst mode, φ1, called also Higgs mode, has a characteristic length, which coincides with the coherence length ξ. The second mode is described by ﬁeld

119 A~. The transversal components of A~ are characterized by the London penetration depth (cf. Eq. (713)), i.e., 8π 2e2 4πe2n λ−2 = Ψ2 = s . (735) L 4m c 0 mc2 This can also be seen as the photon mass. Our theory has no time-dependence, but is otherwise complete with respect to the transversal components of the ﬁeld A~. This means, in the relativistic dispersion relation E2 = µ2c4 + c2p2 we should take E =hω ¯ = 0. Then p2 = −µ2c2. Since p2 < 0, we obtain spatial decay, i.e., penetration depth. Identifying 2 2 −2 p = −h¯ λL , we obtain the photon mass 4πe2n (µc2)2 =h ¯2 s = (¯hω )2 . (736) m ps

Here 4πe2n ω2 = s (737) ps m is the plasma frequency of the superconducting electronic liquid. At T = 0 it coincides with the usual plasma frequency. The variation of (734) with respect to the longitudinal component of A~ results simply in ~ Ak = 0. Thus, unlike in full Higgs case, no longitudinal photon appears at ω = 0. In order to treat the longitudinal modes (plasmons) properly we have to introduce time-dependence and the scalar potential. This is beyond the scope of this text.

K. Flux quantization

In the bulk of a superconductor, where ~js = 0, we obtain hc¯ A~ − ∇~ φ = 0 (738) 2e

I hc¯ I hc¯ hc Ad~ ~l = ∇~ φd~l = 2πn = n = nΦ (739) 2e 2e 2e 0 This quantization is very important for, e.g., a ring geometry. If the ring is thick enough

(thicker than λL) the total magnetic ﬂux threading the ring is quantized.

120 L. Josephson eﬀect

We consider now a tunnel junction between two superconductors with diﬀerent phases

φL and φR. The Hamiltonian reads

H = HBCS,L + HBCS,R + HT , (740) where the tunneling Hamiltinoan reads

X h † † i HT = T Rk1,σLk2,σ + Lk2,σRk1,σ . (741) k1,k2,σ

(R) Here Rk,σ ≡ ck,σ is the annihilation operator of an electron in the left superconductor. Two important things: 1) microscopically the electrons and not the quasiparticles tunnel; 2) tunneling conserves spin.

iφL/2 iφR/2 A gauge transformation Lk,σ → e Lk,σ and Rk,σ → e Rk,σ ”removes” the phases from the respective BCS wave functions (making vk, uk, and ∆ real) and renders the tunneling Hamiltonian

X h † −iφ/2 † iφ/2i HT = T Rk1,σLk2,σe + Lk2,σRk1,σe , (742) k1,k2,σ where φ ≡ φR − φL. Josephson [3] used (742) and calculated the tunneling current. We do so here for a time- independent phase diﬀerence φ. The current operator is given by time derivative of the P † number of particles in the right lead NR = k,σ Rk,σRk,σ

ie ie h i I = −eN˙ = − [H ,N ] = X T R† L e−iφ/2 − L† R eiφ/2 . (743) R h¯ T R h¯ k1,σ k2,σ k2,σ k1,σ k1,k2,σ The ﬁrst order time-dependent perturbation theory gives for the density matrix of the system in the interaction representation

t t t −i R dt0H (t0) i R dt0H (t0) Z −∞ T ˜ −∞ T 0 0 ρ(t) = T e ρ0T e ≈ ρ0 − i dt [HT (t ), ρ0] . (744) −∞

For the expectation value of the current this gives

Z t Z t 0 0 0 0 hI(t)i = Tr{ρ(t)I(t)} = −i dt Tr {[HT (t ), ρ0]I(t)} = −i dt Tr {[I(t),HT (t )] ρ0} . −∞ −∞ (745)

121 We use

† † hBCS| ck,↑(t1) c−k,↓(t2) |BCSi † † = hBCS| ukαk,↑(t1) + vkα−k,↓(t1) ukα−k,↓(t2) − vkαk,↑(t2) |BCSi

−iEk(t1−t2) = vkuke , (746) and

hBCS| ck,↑(t1) c−k,↓(t2) |BCSi

† † = hBCS| ukαk,↑(t1) + vkα−k,↓(t1) ukα−k,↓(t2) − vkαk,↑(t2) |BCSi

−iEk(t1−t2) = −vkuke , (747)

After some algebra we obtain (from the anomalous correlators, the rest gives zero)

t Z X h 0 0 i 2 −iφ 0 −i(Ek1 +Ek2 )(t−t ) i(Ek1 +Ek2 )(t−t ) hI(t)i = − 2eT e dt vk1 uk1 vk2 uk2 e − e −∞ k1,k2 t Z X h 0 0 i 2 iφ 0 −i(Ek1 +Ek2 )(t−t ) i(Ek1 +Ek2 )(t−t ) + 2eT e dt vk1 uk1 vk2 uk2 e − e −∞ k1,k2 v u v u ∆2 = 8eT 2 sin(φ) X k1 k1 k2 k2 = 2eT 2 sin(φ) X E + E E E (E + E ) k1,k2 k1 k2 k1,k2 k1 k2 k1 k2 2 2 2 −1 = 2π T ν e∆¯h sin(φ) = Ic sin(φ) , (748) where the Josephson critical current is given by

gT e∆ π∆ Ic = = , (749) 4¯h 2eRT

2 2 2 where gT = 2 × 4π T ν is the dimensionless conductance of the tunnel junction (factor 2 h 1 accounts for spin), while the tunnel resistance is given by RT = 2 . This is the famous e gT Ambegaokar-Baratoﬀ relation [4] (see also erratum [5]).

Thus we have obtained the first Josephson relation I = Ic sin φ. We have introduced the variable φ as the difference of two phases φ = φR − φL. The gauge invariant definition reads

Z R 2e ~ ~ φ = φR − φL − Adl . (750) hc¯ L As a shortest way to the second Josephson relation we assume that an electric ﬁeld exists in the junction and that it is only due to the time-dependence of A~. Then we obtain

2e Z R " ∂ # 2e Z R 2e φ˙ = − A~ d~l = Ed~ ~l = − V, (751) hc¯ L ∂t h¯ L h¯

122 where V is the voltage. Here we all the time treated e as the charge of the electron, i.e., e < 0. Usually one uses e as a positive quantity. Then

2eV φ˙ = . (752) h¯

An alternative way to derive this is to start with a diﬀerence of (time-dependent) chemical potentials

X † X † H = HL + HR − eVL(t) Lk,σLk,σ − eVR(t) Rk,σRk,σ + HT , (753) k,σ k,σ where VL/R are the applied electro-chemical potentials (in addition to the constant chemical potential µ, which is included in HL and HR). A transformation with

t t e Nˆ R V (t0)dt0 e Nˆ R V (t0)dt0 U = e h¯ L L e h¯ R R (754)

In the new Hamiltonian H˜ = iUU˙ −1 + UHU −1 . (755) the terms with VL and VR are cancelled and instead the electronic operators are replaced by, e.g, L → ULU −1 = LeiφL/2 , (756)

t 2e R 0 0 ˙ ˙ ˙ 2e where φL = const. − ¯h VL(t )dt and, thus, φ = φR − φL = − ¯h V .

M. Macroscopic quantum phenomena

1. Resistively shunted Josephson junction (RSJ) circuit

Consider a circuit of parallelly connected Josephson junction and a shunt resistor R.A

Josephson junction is simultaneously a capacitor. An external current Iex is applied. The Kirchhoﬀ rules lead to the ecquation

V I sin φ + + Q˙ = I . (757) c R ex

¯h ˙ As Q = CV and V = 2e φ. Thus we obtain h¯ hC¯ I sin φ + φ˙ + φ¨ = I . (758) c 2eR 2e ex

123 C

Iex Ic

FIG. 15: RSJ Circuit.

1 ˙ It is very convenient to measure the phase in units of magnetic ﬂux, so that V = c Φ (in SI units V = Φ):˙ ch¯ Φ Φ Φ = φ = 0 φ , φ = 2π . (759) 2e 2π Φ0 Then the Kirchhoﬀ equation reads

Φ Φ˙ CΦ¨ Ic sin 2π + + = Iex , (760) Φ0 cR c or in SI units ˙ Φ Φ ¨ Ic sin 2π + + CΦ = Iex . (761) Φ0 R

There are two regimes. In case Iex < Ic there exists a stationary solution φ = arcsin(Iex/Ic). All the current ﬂows through the Josephson contact as a super-current. Indeed V ∝ φ˙ = 0.

At Iex > Ic at least part of the current must ﬂow through the resistor. Thus a voltage develops and the phase starts to ”run”.

2. Particle in a washboard potential

The equation of motion (761) can be considered as an equation of motion of a particle with the coordinate x = Φ. We must identify the capacitance with the mass, m = C, the inverse resistance with the friction coeﬃcient γ = R−1. Then we have

∂U mx¨ = −γx˙ − , (762) ∂x

124 U(Φ)

FIG. 16: Washboard potential.

FIG. 17: Macroscopic Quantum Tunneling (MQT). where for the potential we obtain

Φ U(Φ) = −EJ cos 2π − IexΦ , (763) Φ0 where I Φ hI¯ E ≡ c 0 = c (764) J 2π 2e is called the Josephson energy. The potential energy U(Φ) has a form of a washboard and is called a washboard potential. In Fig. 16 the case Iex < Ic is shown. In this case the potential has minima and, thus, classically stationary solutions are possible. When the external current is close to the critical value a situation shown in Fig. 17 emerges. If we allow ourselves to think of this situation quantum mechanically, then we would conclude that only a few quantum levels should remain in the potential well. Moreover a tunneling process out of the well should become possible. This tunneling process was named

125 Macroscopic Quantum Tunneling because in the 80-s and the 90-s many researchers doubted the fact one can apply quantum mechanics to the dynamics of the ”macroscopic” variable Φ. It was also argued that a macroscopic variable is necessarily coupled to a dissipative bath which would hinder the tunneling. Out these discussions the famous Caldeira-Leggett model emerged [6, 7].

3. Quantization

We write down the Lagrangian that would give the equation of motion (762 or 761). Clearly we cannot include the dissipative part in the Lagrange formalism. Thus we start from the limit R → ∞. The Lagrangian reads

CΦ˙ 2 CΦ˙ 2 Φ L = − U(Φ) = + EJ cos 2π + IexΦ . (765) 2 2 Φ0 We transform to the Hamiltonian formalism and introduce the canonical momentum

∂L Q ≡ = CΦ˙ . (766) ∂Φ˙ The Hamiltonian reads

Q2 Q2 Φ H = + U(Φ) = − EJ cos 2π − IexΦ . (767) 2C 2C Φ0 The canonical momentum corresponds to the charge on the capacitor (junction). The usual commutation relations should be applied

[Φ,Q] = ih¯ . (768)

In the Hamilton formalism it is inconvenient to have an unbounded from below potential.

Thus we try to transform the term −IexΦ away. This can be achieved by the following canonical transformation i R = exp − Q (t)Φ , (769) h¯ ex t R 0 0 where Qex(t) ≡ Iex(t )dt . Indeed the new Hamiltonian reads

2 ˜ −1 ˙ −1 (Q − Qex(t)) Φ H = RHR + ih¯RR = − EJ cos 2π . (770) 2C Φ0

126 The price we pay is that the new Hamiltonian is time-dependent. The Hamiltonian (770) is very interesting. Let us investigate the operator Φ 2e 1 i cos 2π = cos Φ = exp 2e Φ + h.c. (771) Φ0 h¯ 2 h¯ We have i i exp 2e Φ |Qi = |Q + 2ei , exp − 2e Φ |Qi = |Q − 2ei . (772) h¯ h¯ Thus in this Hamiltonian only the states diﬀering by an integer number of Cooper pairs get connected. The constant oﬀset charge remains undetermined. This, however, can be absorbed into the bias charge Qex. Thus, we can restrict ourselves to the Hilbert space |Q = 2emi.

4. Phase and Number of particles (Cooper pairs)

2π Z dφ A† |BCS(N)i = |BCS(φ)i e−i(N+1)φ = |BCS(N + 1)i . (775) 2π 0 We have seen that the excitations above the BCS ground state have an energy gap ∆. Thus, if T ∆ no excitations are possible. The only degree of freedom left is the pair of conjugate variables N, φ with commutation relations [N, e−iφ] = e−iφ. Indeed the ground state energy is independent of φ. This degree of freedom is, of course, non-existent if the number of particles is ﬁxed. Thus a phase of an isolated piece of a superconductor is quantum mechanically smeared between 0 and 2π and no dynamics of the degree of freedom N, φ is possible. However in a bulk superconductor the phase can be space dependent φ(~r). One can still add a constant phase to φ(~r) + φ0 without changing the state. More precisely the phase

φ0 is smeared if the total number of particles is fixed. However the difference of phases, i.e., the phase gradient can be well defined and corresponds to a super-current.

127 5. Josephson energy dominated regime

(2e)2 In this regime EJ EC , where EC = 2C is the Cooper pair charging energy. Let us ﬁrst neglect EC completely, i.e., put C = ∞. Recall that C plays the role of the mass. Then the Φ Hamiltonian reads H = −EJ cos 2π . On one hand it is clear that the relevant state are Φ0 those with a given phase, i.e., |Φi. On the other hand, in the discrete charge representation the Hamiltonian reads

E H = − J X (|m + 1i hm| + |mi hm + 1|) . (776) 2 m

P ikm The eigenstates of this tight-binding Hamiltonian are the Bloch waves |ki = m e |mi with the wave vector k belonging to the ﬁrst Brillouin zone −π ≤ k ≤ π. The eigenenergy 2πΦ reads Ek = −EJ cos(k). Thus we identify k = φ = . Φ0

6. Charging energy dominated regime

In this regime EJ EC . The main term in the Hamiltonian is the charging energy term (Q − Q (t))2 (2em − Q )2 H = ex = ex . (777) C 2C 2C

The eigenenergies corresponding to diﬀerent values of m form parabolas as functions of

Qex (see Fig. 18). The minima of the parabolas are at Qex = 0, 2e, 4e, . . .. The Josephson Φ tunneling term serves now as a perturbation HJ = −EJ cos 2π . It lifts the degeneracies, Φ0 e.g., at Qex = e, 3e, 5e, . . ..

If a small enough external current is applied, Qex = Iext the adiabatic theorem holds and the system remains in the ground state. Yet, one can see that between the degeneracies at Qex = e, 3e, 5e, . . . the capacitance is charged and discharged and oscillating voltage

V = ∂E0/∂Qex appears. Here E0(Qex) is the energy of the ground state. The Cooper pairs tunnel only at the degeneracy points. In between the Coulomb blockade prevents the Cooper pairs from tunneling because this would cost energy.

128 eigen energies

Q=0 Q=2e Q=4e 0 e 2e 3e 4e Qex

FIG. 18: Eigen levels in the coulomb blockade regime. Diﬀerent parabolas correspond to diﬀerent values of Q = 2em. The red lines represent the eigenlevels with the Josephson energy taken into account. The Josephson tunneling lifts the degeneracy between the charge states.

[1] D. Xiao, M.-C. Chang, and Q. Niu, Rev. Mod. Phys. 82, 1959 (2010), URL https://link. aps.org/doi/10.1103/RevModPhys.82.1959. [2] D. J. Thouless, M. Kohmoto, M. P. Nightingale, and M. den Nijs, Phys. Rev. Lett. 49, 405

(1982), URL https://link.aps.org/doi/10.1103/PhysRevLett.49.405. [3] B. D. Josephson, Physics Letters 1, 251 (1962). [4] V. Ambegaokar and A. Baratoﬀ, Phys. Rev. Lett. 10, 486 (1963). [5] V. Ambegaokar and A. Baratoﬀ, Phys. Rev. Lett. 11, 104 (1963). [6] A. O. Caldeira and A. J. Leggett, Phys. Rev. Lett. 46, 211 (1981). [7] A. O. Caldeira and A. J. Leggett, Annals of Physics 149, 374 (1983).

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