Lecture Notes on Many-Body Theory

Home , Bose gas, Fermi point, Fock state, Quantization (physics)

Lecture notes on many-body theory

Michele Fabrizio

updated to 2013, but still work in progress!! All comments are most welcome! Contents

1 Landau-Fermi-liquid theory: Phenomenology 5 1.1 The Landau energy functional and the concept of quasiparticles ...... 5 1.2 Quasiparticle thermodynamics ...... 9 1.2.1 Specific heat ...... 13 1.2.2 Compressibility and magnetic susceptibility ...... 13 1.3 Quasiparticle transport equation ...... 15 1.3.1 Quasiparticle current density ...... 18 1.4 Collective excitations in the collisionless regime ...... 20 1.4.1 Zero sound ...... 22 1.5 Collective excitations in the collision regime: first sound ...... 25 1.6 Linear response functions ...... 27 1.6.1 Formal solution ...... 28 1.6.2 Low frequency limit of the response functions ...... 31 1.6.3 High frequency limit of the response functions ...... 32 1.7 Charged Fermi-liquids: the Landau-Silin theory ...... 34 1.7.1 Formal solution of the transport equation ...... 37 1.8 Dirty quasiparticle gas ...... 39 1.8.1 Conductivity ...... 41 1.8.2 Diffusive behavior ...... 43

2 Second Quantization 48 2.1 Fock states and space ...... 48 2.2 Fermionic operators ...... 50 2.2.1 Second quantization of multifermion-operators ...... 52 2.2.2 Fermi ﬁelds ...... 55 2.3 Bosonic operators ...... 57 2.3.1 Bose ﬁelds and multiparticle operators ...... 57 2.4 Canonical transformations ...... 59 2.4.1 More general canonical transformations ...... 60

1 2.5 Examples and Exercises ...... 62 2.6 Application: fermionic lattice models and the emergence of magnetism ...... 65 2.6.1 Hubbard models ...... 69 2.6.2 The Mott insulator within the Hubbard model ...... 71 2.7 Spin wave theory in the Heisenberg model ...... 73 2.7.1 More rigorous derivation: the Holstein-Primakoﬀ transformation . . . . . 81

3 Linear Response Theory 83 3.1 Linear Response Functions ...... 83 3.2 Kramers-Kronig relations ...... 86 3.2.1 Symmetries ...... 87 3.3 Fluctuation-Dissipation Theorem ...... 88 3.4 Spectral Representation ...... 90 3.5 Power dissipation ...... 91 3.5.1 Absorption/Emission Processes ...... 93 3.5.2 Thermodynamic Susceptibilities ...... 94

4 Hartree-Fock Approximation 98 4.1 Hartree-Fock Approximation for Fermions at Zero Temperature ...... 98 4.1.1 Alternative approach ...... 102 4.2 Hartree-Fock approximation for fermions at finite temperature ...... 106 4.2.1 Preliminaries ...... 106 4.2.2 Variational approach at T 6=0 ...... 110 4.3 Mean-Field approximation for bosons and superfluidity ...... 114 4.3.1 Superfluid properties of the gauge symmetry breaking wavefunction . . . 119 4.4 Time-dependent Hartree-Fock approximation for fermions ...... 124 4.4.1 Bosonic representation of the low-energy excitations ...... 126 4.5 Application: antiferromagnetism in the half-filled Hubbard model ...... 130 4.5.1 Spin-wave spectrum by time dependent Hartree-Fock ...... 134 4.6 Linear response of an electron gas ...... 139 4.6.1 Response to an external charge ...... 142 4.6.2 Response to a transverse field ...... 147 4.6.3 Limiting values of the response functions ...... 148 4.6.4 Power dissipated by the electromagnetic field ...... 156 4.6.5 Reflectivity ...... 158 4.7 Random Phase Approximation for the electron gas ...... 160

5 Feynman diagram technique 163 5.1 Preliminaries ...... 163 5.1.1 Imaginary-time ordered products ...... 163

2 5.1.2 Matsubara frequencies ...... 164 5.1.3 Single-particle Green’s functions ...... 168 5.2 Perturbation expansion in imaginary time ...... 172 5.2.1 Wick’s theorem ...... 173 5.3 Perturbation theory for the single-particle Green’s function and Feynman diagrams174 5.3.1 Diagram technique in momentum and frequency space ...... 177 5.3.2 The Dyson equation ...... 180 5.4 Other kinds of perturbations ...... 185 5.4.1 Scalar potential ...... 185 5.4.2 Coupling to bosonic modes ...... 186 5.5 Two-particle Green’s functions and correlation functions ...... 189 5.5.1 Diagrammatic representation of the two-particle Green’s function . . . . . 190 5.5.2 Correlation functions ...... 191 5.6 Coulomb interaction and proper and improper response functions ...... 194 5.6.1 Screened interaction and corresponding Dyson equation ...... 195 5.7 Irreducible vertices and the Bethe-Salpeter equations ...... 197 5.7.1 Bethe-Salpeter equation for the vertex functions ...... 198 5.8 The Ward identities ...... 200 5.9 Consistent approximation schemes ...... 204 5.9.1 Example: the Hartree-Fock approximation ...... 206 5.10 Some additional properties and useful results ...... 208 5.10.1 The occupation number and the Luttinger-Ward functional ...... 208 5.10.2 The thermodynamic potential ...... 211 5.10.3 The Luttinger theorem ...... 214

6 Landau-Fermi liquid theory: a microscopic justiﬁcation 217 6.1 Preliminaries ...... 217 6.1.1 Vertex and Ward identities ...... 220 6.2 Correlation functions ...... 223 6.2.1 Conserved quantities ...... 225 6.3 Coulomb interaction ...... 226

7 Kondo eﬀect and the physics of the Anderson impurity model 228 7.1 Brief introduction to scattering theory ...... 229 7.1.1 General analysis of the phase-shifts ...... 234 7.2 The Anderson Impurity Model ...... 235 7.2.1 Variation of the electron number ...... 239 7.2.2 Energy variation ...... 240 7.2.3 Mean-ﬁeld analysis of the interaction ...... 242 7.3 From the Anderson to the Kondo model ...... 244

3 7.3.1 The emergence of logarithmic singularities and the Kondo temperature . 245 7.3.2 Anderson’s scaling theory ...... 248

8 Introduction to abelian Bosonization 252 8.1 Interacting spinless fermions ...... 252 8.1.1 Spin wave theory ...... 255 8.1.2 Construction of the eﬀective Hamiltonian ...... 256 8.2 Interactions ...... 263 8.2.1 Umklapp processes ...... 266 8.3 Bosonization of the Heisenberg model ...... 268 8.4 The Hubbard model ...... 272

4 Chapter 1

Landau-Fermi-liquid theory: Phenomenology

One of the ﬁrst thing that one learns in a Solid State Physics course is that the thermodynamic and transport properties of metals are well described in terms of non-interacting electrons; the Drude-Sommerfeld-Boltzmann theory of metals. Yet, this evidence is somehow surprising in view of the fact that actual electrons interact mutually via Coulomb repulsion, which is not weak at all. This puzzle was solved brilliantly in the end of the 50’s by Landau, as we are going to discuss in what follows. We will start by analyzing the case of a neutral Fermi system, like 3He. Later we shall discuss charged Fermi systems, relevant to metals.

1.1 The Landau energy functional and the concept of quasiparticles

Let us start from a non-interacting Fermi gas at very low temperature T . Right at T = 0 we know that the ground state is the Fermi sea that is obtained by ﬁlling with two opposite-spin electrons all momentum states with energy smaller than the chemical potential µ (the Fermi energy). In other words, if nkσ is the occupation number of each momentum state – nkσ being zero or one because of Pauli principle – then the Fermi-sea occupation numbers are

( (0) (0) 1 if k ≤ µ nkσ = (0) . 0 if k > µ At ﬁnite temperature, the equilibrium values of the occupation numbers are given by the Fermi- Dirac distribution −1 (0) (0) (0) β (k −µ) nkσ = f k − µ = 1 + e ,

5 with β = 1/KBT the inverse temperature. Any excited state can be uniquely identiﬁed by the variation of the occupation numbers with respect to equilibrium, namely through

(0) δnkσ = nkσ − nkσ , (1.1) and costs an energy (0) X (0) δE [{δnpα}] = k − µ δnkσ, (1.2) kσ that is a simple functional of the δnkσ’s. Now suppose that we switch on smoothly the interaction. Each non-interacting excited state will evolve smoothly into a fully-interacting one. The Landau hypothesis was that the non-interacting and the fully interacting states are adiabatically connected; in other words they are in one-to-one correspondence. This implies in particular that each fully-interacting excited state can be uniquely identiﬁed, like its non-interacting partner, by the deviation with respect to equilibrium of the occupation numbers δnkσ. Consequently, its excitation energy δE must be a functional of the δnkσ’s, i.e. δE [{δnpα}] . (1.3)

——— Remark This hypothesis may look simple but it is actually deeply counter-intuitive. Suppose we have a “non-interacting” Hamiltonian H0 and a fully-interacting one H = H0 +H1, and suppose to follow the spectrum, that we assume discrete, within a speciﬁc symmetry-subspace of the Hamiltonian H(λ) = H0 + λ H1, with λ that increases smoothly from 0 to 1. The evolution of the spectrum must resemble that one drawn schematically in Fig. 1.1 with a series of avoided crossings that allow to follow adiabatically the n-th excited level at λ = 0 into the n-th excited level at λ = 1. Note that the crossing are avoided because the states have the same symmetry. If we were to consider levels of diﬀerent symmetries nothing would prevent crossings. However, while the adiabatic evolution does occur and it is a well accepted phenomenon in models with discrete spectra, it is equally common wisdom that it can not happen when the spectrum is continuous, as in a bulk metal. Landau’s revolutionary hypothesis that the adiabatic evolution also occurs in a bulk system, at least for the low energy excited states, and the fact that, by this simple assumption, one can justify and predict a lot of physical properties is one of the major achievements of the contemporary Condensed Matter Theory. ———

Let us therefore assume that the excitation energy of the fully-interacting Fermi system is indeed a functional of the same δnkσ’s as its non-interacting partner, and further suppose

6 E

0 1 λ

Figure 1.1: Adiabatic evolution of a discrete spectrum within a speciﬁc symmetry-subspace. that the temperature is very small. In this case, any weak deviation from equilibrium must correspond to δnkσ 1, that justiﬁes a Taylor expansion of (1.3), namely

X 1 X X 3 δE [{δn }] = ( − µ) δn + f 0 0 δn δn 0 0 + O δn . (1.4) pα kσ kσ 2 kσ k σ kσ k σ kσ kk0 σσ0 This is the famous Landau’s energy functional. Note that

• even though the two terms are apparently of diﬀerent orders, in reality they are of the same order since the excitation energies kσ − µ are of the same order as the deviations of the occupation numbers;

(0) • kσ should not be confused with the non-interacting energies kσ .

Rigorously speaking, the δnkσ’s that appear in (1.4) only serve as labels to identify excited states and they do not correspond to deviations of the occupation numbers of the real particles, as in the non-interacting case (1.2). Just for this reason, Landau coined for δnkσ the deﬁnition “quasiparticle” occupation number, as if the real-particle excitations were substituted in the presence of interaction by “quasiparticle” excitations. The Fermi gas of these quasiparticles is the so-called “Landau-Fermi liquid”.

——— Remark This idea is actually ubiquitous in all Solid State Physics. Let us for instance recall brieﬂy how phonons arise. One starts from a model of interacting ions and electrons. Because of their larger mass with respect to the electrons and of the Coulomb interaction among them and with the electrons, the ions localize, thus forming a lattice, and the low energy excitations become the small ﬂuctuations around the equilibrium positions, whose quantization gives rise to phonons, which are bosons. In other words, one begins with real particles, the ions, yet their low-energy

7 excitations in the presence of interaction are new bosonic “quasiparticles”, the phonons, that turns out to be generally weakly interacting among each other and with the electrons. This allows to treat such an interaction within perturbation theory, unlike the original Coulomb interaction. In a more general context, since the only many-body systems that can be solved exactly in any dimensions are free bosons or fermions, it is a very natural, and fortunately successful, approach to attempt a description of the low-energy excitations of, even strongly, interacting models in terms of eﬀective “quasiparticles”, bosons or fermions, whose interaction is weak enough to be treated perturbatively. ———

Going back to the Landau energy functional (1.4), in the case in which spin isotropy is preserved the following equivalences hold:

fk↑ k0↑ = fk↓ k0↓, fk↑ k0↓ = fk↓ k0↑.

In this case, the choice of a spin quantization axis must not be inﬂuential, hence it is convenient to rewrite (1.4) in a form that is explicitly spin isotropic. For that purpose, let us introduce charge deviations from equilibrium

δρk = δnk↑ + δnk↓, (1.5) and spin ones x y z δσk = δσk, δσk, δσk , (1.6) where the z-component is simply z δσk = δnk↑ − δnk↓. With these deﬁnitions the second term of the energy functional (1.4) can be written as

1 X S A f 0 δρ δρ 0 + f 0 δσ · δσ 0 , (1.7) 2 kk k k kk k k kk0 where 1 1 f S = (f + f ) , f A = (f − f ) , (1.8) 2 ↑↑ ↑↓ 2 ↑↑ ↑↓ as can be readily demonstrated by equating (1.4) and (1.7) assuming a deviation from equilibrium of the form z δσk = (0, 0, δσk) .

8 1.2 Quasiparticle thermodynamics

Let us derive, starting from the Landau’s energy functional, the thermodynamical properties of the quasiparticles. We note that, if we consider a deviation from equilibrium that consists, in the absence of interaction, to add δN particles in particular momentum-states, i.e. an excited state identiﬁed by δnkσ such that X δnkσ = δ N, (1.9) kσ such a state must evolve in the Landau’s hypothesis into a state with the same “quasiparticle” excitations δnkσ and furthermore, since the total number of particles is conserved along the adiabatic evolution, with the same deviation δN from equilibrium. In other words, although quasiparticles have not to be confused with real particles, still the sum of all quasiparticle deviations of the occupation numbers gives the variation of the total number of real particles, i.e. Eq. (1.9).

——— Remark It is important to note that the adiabatic evolution hypothesis guarantees that only true conserved quantities, like the total number of particles or the total spin, keep the same expressions in terms of quasiparticle occupation numbers as for the real particles. Conserved quantities refer, rigorously speaking, only to the fully interacting Hamiltonian, whereas the non-interacting one has generally much larger symmetry. In other words, the Landau’s theory of Fermi liquid has only to do with conserved quantities but its use is not at all justiﬁed for non-conserved ones, even if they are conserved in the non-interacting limit. This fact is often underestimated and may lead easily to wrong conclusions. ———

Since the labeling of the states is identical to non-interacting electrons, the phase-space volume counting is the same, hence also the formal expression of the entropy-change δS: " # X δS = δ −KB nkσ ln nkσ + (1 − nkσ) ln (1 − nkσ) . (1.10) kσ Thermodynamic equilibrium implies an extremum of the free-energy F , namely that

δF = δE − T δS − µ δN = 0. (1.11)

Let us solve this equation by assuming that the deviation from equilibrium is induced by a deviation of the quasiparticle occupation numbers δnkσ, that amounts to impose δF = 0. δnkσ δn=0

9 By Eq. (1.4) we ﬁnd that

δE X = k + fkσ k0σ0 δnk0σ0 ≡ ¯kσ, (1.12) δnkσ k0σ is the quasiparticle energy in the presence of excited quasiparticles, including itself, while what it is needed is δE = k, δnkσ δn=0 that is obviously independent of δn. Since all terms in (1.11) have exactly the same expression as for non-interacting particles, one readily realizes that the quasiparticle equilibrium distribution is the Fermi-Dirac distribution −1 0 0 β (k−µ) nkσ = nk = f (k − µ) = 1 + e , (1.13) the only difference with respect to non-interacting real particles being the renormalized band dispersion k. We must point out that f (k − µ) corresponds to global equilibrium – all quasiparticles are in equilibrium among themselves. Analogously, we could define a distribution function for the “local” equilibrium that minimizes the free-energy of a single excited quasiparticle in the presence of other excited quasiparticles. It is clear from Eq. (1.12) that the local-equilibrium distribution function is the Fermi-Dirac distribution with argument the quasiparticle energy ¯kσ, i.e. −1 0 β (¯kσ−µ) n¯kσ = f (¯kσ − µ) = 1 + e , which is a functional of the occupation number deviations. In accordance, we can define a deviation with respect to local equilibrium as

δn¯kσ = nkσ − f (¯kσ − µ) , (1.14) as opposed to that one with respect to global equilibrium

0 δnkσ = nkσ − nk 0 0 = δn¯kσ +n ¯kσ − nk 0 ∂nk = δn¯kσ + (¯kσ − kσ) ∂k ∂f (k − µ) = δn¯kσ + (¯kσ − kσ) ∂k X ' δn¯kσ − δ (k − µ) fkσ k0σ0 δnk0σ0 , (1.15) k0σ

10 where we have assumed to be close to equilibrium and at very low temperature.

Now let us consider, as conventionally done, a space-isotropic system, where k = (|k|) and use the parametrization

0 ∂nk X X δnkσ = − Ylm (Ωk) δnlmσ ' δ (k − µ) Ylm (Ωk) δnlmσ, (1.16) ∂k lm lm where Ylm (Ωk) are spherical harmonics identified by the Euler angles of the unit vector in the direction of k, being its modulus fixed by the δ-function to be right on the Fermi sphere. (The δ-function in the right hand side of (1.16) derives from the fact that both sides of that equation must consistently be of first order in the deviation from equilibrium.) In addition, we introduce the Legendre decomposition of fkk0 assuming that it depends only on the relative angle θkk0 between the two momenta, their modula being on the Fermi sphere. Therefore S(A) X S(A) fkk0 = fl Pl (θkk0 ) , (1.17) l 1 where Pl (θkk0 ) are Legendre polynomials. In the non-abelian representation in which the energy functional is explicitly spin-isotropic, one readily finds through (1.15) the following relations between the charge/spin deviations at global and local equilibrium:

0 ∂nk X S δρk = δρ¯k + 2 fkk0 δρk0 , (1.18) ∂k k0 0 ∂nk X A δσk = δσ¯ k + 2 fkk0 δσk0 . (1.19) ∂k k0

One realizes that, because of k = (|k|) and of the identity Z dΩk0 1 P (θ 0 ) Y 0 0 (Ω 0 ) = δ 0 Y 0 0 (Ω ) , 4π l kk l m k ll 2l0 + 1 l m k it follows that

0 X X S ∂nk − fl Pl (θkk0 ) Yl0m0 (Ωk0 ) δρl0m0 ∂k k0 l X X S X ' fl Pl (θkk0 ) δ (k0 − µ) Yl0m0 (Ωk0 ) δρl0m0 k0 l l0m0

1Conventionally, the argument of the Legendre polynomials is indicated as cos θ ∈ [−1 : 1]. Here, in order to simplify notations, we have decided to use as argument θ

11 Z V X S dΩk0 = N f δρ 0 0 P (θ 0 ) Y 0 0 (Ω 0 ) 2 l l m 4π l kk l m k ll0m0 V X S 1 = N f 0 δρ 0 0 Y 0 0 (Ω ) , 2 l l m 2l0 + 1 l m k l0m0

S A and analogously for δσk with f → f , where N = N (µ) is the quasiparticle density of states

1 X N () = δ ( − ) , V k kσ at the chemical potential. Therefore Eqs. (1.18) and (1.19) have the simple solution

δρ¯lm δρlm = , (1.20) F S 1 + l 2l + 1 δσ¯ lm δσlm = ,, (1.21) F A 1 + l 2l + 1 where the Landau F -parameters are deﬁned through

S(A) S(A) Fl = V N fl . (1.22) It also common to deﬁne A-parameters through

F S(A) AS(A) = l , (1.23) l F A 1 + l 2l + 1 so that

S S Fl δρlm = Al δρ¯lm, (1.24) A A Fl δσlm = Al δσ¯ lm., (1.25)

We note that the above formulas are valid only for an isotropic system, like 3He, or for a metal with an approximately spherical Fermi surface. In general, one must use, instead of spherical harmonics and Legendre polynomials, other basis functions appropriate to the symmetry of the lattice. However, in what follows we just take into account the isotropic case, unless otherwise stated.

12 1.2.1 Speciﬁc heat Let us start the calculation of the quasiparticle thermodynamic properties at low temperature from the simplest one: the speciﬁc heat. It is easy to show through (1.10) (we set KB = 1 as well as ~ = 1) that T ∂S T ∂S 2 cV = ' + O T V ∂T N,V V ∂T µ,V 0 0 0 T X ∂nk nkσ 1 X 2 ∂nk = − ln 0 = − (k − µ) V ∂T 1 − n TV ∂k kσ kσ kσ 1 Z ∂f() π2 = − d N ( + µ) 2 = N T. (1.26) T ∂ 3 This is the same expression as in the absence of interaction with the quasiparticle density of states N instead of the free-particle one N (0)

(0) (0) X (0) N = N (µ) = δ k − µ . kσ

(0) Therefore, if cV is the speciﬁc heat of the free particles, then c N V = . (1.27) (0) N (0) cV

1.2.2 Compressibility and magnetic susceptibility Suppose that we perturb the system with a static homogeneous ﬁeld that may modify the chemical potentials, µσ = µ → µ+δµσ, for spin up and spin down real particles. At equilibrium, the quasiparticle occupation numbers will change accordingly as

0 ∂nk δnkσ = (δkσ − δµσ) , (1.28) ∂k where X δkσ = fkσ k0σ0 δnk0σ0 , k0σ0 is the change in the quasiparticle energy due to the fact that the external ﬁeld changes all quasiparticle occupation numbers. It is important to note that δµσ that appears in (1.28) coincides with the ﬁeld that acts on the real particles because the latter couples to conserved quantities, otherwise nothing could guarantee that the two are equal. Once again this demonstrates that this theory only addresses the response of conserved quantities.

13 By means of Eq. (1.15) we ﬁnd that

0 0 ∂nk X ∂nk δn¯kσ = δnkσ − fkσ k0σ0 δnk0σ0 = − δµσ, (1.29) ∂k ∂k k0σ0 showing that the deviation from local equilibrium is the same as for free particles with dispersion k in the presence of the ﬁeld.

Compressibility The compressibility κ is defined by 1 ∂V 1 ∂n κ = − = , V ∂P n2 ∂µ where n is the total density. Since a variation in the total density of particles coincides with the variation in the quasiparticle one, we find through (1.20) that 2 1 X 1 X δn = δn = δρ V kσ V k kσ k 1 1 δρ¯ 1 √ √ 00 = N δρ00 = N S = S δn.¯ 2 4π 2 4π 1 + F0 1 + F0 The variation δn¯ with respect to a variation of chemical potential is the same as for free particles with density of states N , namely δn¯ = N . δµ Therefore we finally obtain that 1 N 1 N κ = = κ(0), (1.30) 2 S S (0) n 1 + F0 1 + F0 N where κ(0) is the compressibility of the original free particles which have density of states N (0).

Magnetic susceptibility A magnetic ﬁeld B, e.g. in the z-direction, introduces a Zeeman term in the Hamiltonian 1 δH = −g µ B (N − N ) , B 2 ↑ ↓ 2Note that r 1 Y (Ω) = . 00 4π

14 where Nσ is the total number of spin-σ electrons, that acts as a diﬀerence of chemical potential for spin up and down electrons: δµ↑ −δµ↓ = g µB B. Since the total magnetization is also conserved, the quasiparticles acquire the same diﬀerence in chemical potentials. The magnetization per unit volume δm can therefore be calculated by the quasiparticles through

1 X 1 X z 1 z δm = g µB δnk↑ − δnk↑ = g µB δσ = g µB √ N δσ 2V 2V k 4 4π 00 k k 1 δσ¯z δm¯ √ 00 = g µB N A = A . 4 4π 1 + F0 1 + F0 Once again, since the local equilibrium magnetization δm¯ is equivalent to the response of free particles with density of states N , instead of N0 as for the original particles, we obtain that the magnetic susceptibility χ is δm 1 N χ = = χ(0), (1.31) A (0) δB 1 + F0 N where χ(0) is the susceptibility the original free-particles.

1.3 Quasiparticle transport equation

Let us suppose to perturb the Landau-Fermi liquid by an external probe that varies in space on a wavelength 1/q and in time on a period 1/ω. If

q kF , ω µ, where kF is the Fermi momentum ((kF ) = µ), the probe will aﬀect only quasiparticles extremely close to the Fermi-sea. In this limit, we can safely adopt a semi-classical approach and introduce a space- and time-dependent density of quasiparticle occupation numbers at momentum |k| ∼ kF

nkσ(x, t), and its space and time Fourier transform

nkσ(q, ω).

Fixing both coordinate x and momentum k is allowed in spite of the Heisenberg principle ∆x ∆k ' 1, because 1 ∆k ' ' q k , ∆x F so that the quantum indeterminacy of momentum ∆k is much smaller than its typical value kF .

15 Since the quasiparticle occupation number is not a conserved quantity, its density should follow a Liouville equation

∂n (x, t) ∂n (x, t) ∂x ∂n (x, t) ∂k kσ + kσ + kσ = I (x, t) , (1.32) ∂t ∂x ∂t ∂k ∂t kσ where Ikσ(x, t) is a collision integral and corresponds to the transition rate within a volume dx around x of all processes in which quasiparticles have a transition from any other state into (k, σ) minus the transition rate of the inverse processes. In the absence of the external probe, quasiparticles are at equilibrium and their density is uniform. Once the probe is applied, the density deviates from its uniform equilibrium value and acquires space and time dependence. In general the external force that acts on the real particles can not be identified with the force felt by the quasiparticles unless the external field couples to a conserved quantity. This is the case of a scalar potential coupled to the charge density, or a magnetic field that couples to the spin density. In those cases, the deviation from equilibrium of the energy becomes non-uniform hence can be generalized into

Z X Z δE(t) = dx δE(x, t) = dx k + Vσ(x, t) δnkσ(x, t) kσ 1 X X Z + dxdy δn (x, t) f 0 0 (x − y) δn 0 0 (y, t), (1.33) 2 kσ kσ k σ k σ kk0 σσ0 where Vσ(x, t) represents the action of the external probe, V↑ = V↓ for a scalar potential and V↑ = −V↓ for a magnetic ﬁeld, and we assume that the quasiparticle interaction is instantaneous and only depends on the distance. The above expression provides also the proper deﬁnition of the quasiparticle excitation energy as

X Z ¯kσ(x, t) = k + Vσ(x, t) + dy fkσ k0σ0 (x − y) δnk0σ0 (y, t). (1.34) k0σ0 For neutral particles we can assume that

fkσ k0σ0 (x − y) = fkσ k0σ0 δ(x − y), in which case X ¯kσ(x, t) = k + Vσ(x, t) + fkσ k0σ0 δnk0σ0 (x, t). (1.35) k0σ0 By means of the Hamiltonian equations for conjugate variables ∂x ∂H ∂k ∂H = , = − , ∂t ∂k ∂t ∂x

16 Eq. (1.32) can be ﬁnally written as

∂n (x, t) ∂n (x, t) ∂¯ (x, t) ∂n (x, t) ∂¯ (x, t) kσ + kσ kσ − kσ kσ = I (x, t) , (1.36) ∂t ∂x ∂k ∂k ∂x kσ

At linear order in the external probe Vσ(x, t), the occupation density

0 nkσ(x, t) = nkσ + δnkσ(x, t), and, consistently, ¯kσ(x, t) = k + δkσ(x, t), where δE X δkσ(x, t) = − k = Vσ(x, t) + fkσ k0σ0 δnk0σ0 (x, t). δnkσ(x, t) k0σ0 Consequently, the linearized transport equation (1.36) reads

∂δn (x, t) ∂δn (x, t) ∂ ∂n0 ∂δ (x, t) kσ + kσ k − k kσ ∂t ∂x ∂k ∂k ∂x 0 " # ∂δnkσ(x, t) ∂δnkσ(x, t) ∂k ∂nk ∂Vσ(x, t) X ∂δnk0σ0 (x, t) = + − + f 0 0 ∂t ∂x ∂k ∂k ∂x kσ k σ ∂x k0σ0 ∂δn (x, t) ∂δn (x, t) ∂ = kσ + kσ k ∂t ∂x ∂k 0 " # ∂nk ∂k ∂Vσ(x, t) X ∂δnk0σ0 (x, t) − + fkσ k0σ0 ∂k ∂k ∂x ∂x k0σ0 ∂δn (x, t) ∂δn (x, t) = kσ + v · kσ ∂t k ∂x 0 " # ∂nk ∂Vσ(x, t) X ∂δnk0σ0 (x, t) − vk · + fkσ k0σ0 ∂k ∂x ∂x k0σ0 0 ∂δnkσ(x, t) ∂δnkσ(x, t) ∂nk ≡ + vk · + vk · Fkσ(x, t) ∂t ∂x ∂k = Ikσ (x, t) , (1.37) where ∂ v = k k ∂k is the quasiparticle group velocity. The Landau transport equation (1.37) resembles the conventional Boltzmann equation apart from the fact that the eﬀective force Fkσ(x, t) depends self-consistently on the occupation density.

17 1.3.1 Quasiparticle current density Let us set V (x, t) = 0 in (1.37). We note that 1 X δn (x, t) = δρ(x, t), V kσ kσ where δρ(x, t) is the deviation of the density with respect to the uniform equilibrium value, and moreover that 1 X I (x, t) = 0, V kσ kσ by particle conservation (recall the meaning of the collision integral Ikσ). Through Eq. (1.37) and using the relation (1.29) between the deviations from global and local equilibrium, we ﬁnd the following equation for the evolution of δρ(x, t) in the absence of external probes:

" 0 # ∂δρ(x, t) ∂ 1 X ∂nk X + · vk δnkσ(x, t) − fkσ k0σ0 δnk0σ0 (x, t) ∂t ∂x V ∂k kσ k0σ0 ∂δρ(x, t) ∂ 1 X = + · v δn¯ (x, t) = 0. (1.38) ∂t ∂x V k kσ kσ Since the integral over the whole space of δρ(x, t) is the variation of the total number of real particles, which is conserved, then δρ(x, t) should also satisfy a continuity equation ∂δρ(x, t) ∂J(x, t) + = 0, ∂t ∂x that allows to identify the current density J(x, t) (note that, at equilibrium, the current is zero hence J(x, t) = δJ(x, t)) 1 X J(x, t) = v δn¯ (x, t) V k kσ kσ " 0 # 1 X ∂nk X = vk δnkσ(x, t) − fkσ k0σ0 δnk0σ0 (x, t) V ∂k kσ k0σ0 " 0 # 1 X X ∂nk0σ0 = δnkσ(x, t) vk − fkσ k0σ0 vk0 V ∂k0 kσ k0σ0 1 X ≡ δn (x, t) J . (1.39) V kσ kσ kσ

This equation shows that the actual current matrix element Jkσ is not the group velocity vk – the moving quasiparticle induces a ﬂow of other quasiparticles because of their mutual interaction.

18 We observe that at very low temperature

0 ∂nk0σ0 ∂f (k0 − µ) = ' −δ (k0 − µ) , ∂k0 ∂k0 namely k0 lies on the Fermi surface. If the system is isotropic, then on the Fermi sphere

∂(|k|) k kF k vk = = vF = , ∂k |k| m∗ |k| with vF and m∗ the Fermi velocity and eﬀective mass, respectively, of the quasiparticles. Us- ing an expansion like (1.16) in terms of spherical harmonics we ﬁnd that, for instance the z-component of the current density, Jz, is 1 X J (x, t) = (v ) δn¯ (x, t) z V k z kσ kσ 1 X kz X = v δ ( − µ) Y (Ω ) δn¯ (x, t) V F |k| k lm k lmσ kσ lm Z vF X dΩ = N δρ¯ (x, t) cos θ Y (Ω) 2 lm 4π lm lm S ! vF vF F1 = √ N δρ¯10(x, t) = √ N 1 + δρ10(x, t) 2 12π 2 12π 3 S !" (0) # vF F1 vF = 1 + √ N δρ10(x, t) (0) 3 2 12π vF S ! vF F1 1 X (0) = 1 + vk δnkσ(x, t), (1.40) (0) 3 V z vF kσ

(0) where vk is the group velocity of the free particles on the Fermi surface ∂(0)(|k|) k k k v(0) = = v(0) = F . (1.41) k ∂k F |k| m |k|

The x and y component can be found analogously and involve combinations of δn1 +1(x, t) and δn1 −1(x, t). Therefore in general it holds that

S ! vF F 1 X (0) J(x, t) = 1 + 1 v δn (x, t) (0) 3 V k kσ vF kσ S ! m F1 1 X (0) = 1 + vk δnkσ(x, t). (1.42) m∗ 3 V kσ

19 As a concluding remark we notice that we could also follow the same analysis in connection with S A the spin-current density, the only diﬀerence being that F1 would be substituted by F1 .

Translationally invariant systems In a system that is not only isotropic but also translationally invariant, the current matrix element is k J = , k m with m the mass of the real particles. Since the current is conserved, the total current per unit volume must be given in terms of quasiparticles by

1 X k kF Jz = δnkσ = √ N δρ10. (1.43) V m 2m 12π kσ Comparing (1.43) with (1.40) in the case in which the current density, hence the occupation density, are constant in time and space, i.e. Jz(x, t) = Jz and δρ10(x, t) = δρ10, we conclude that, when translational invariance is unbroken, like in 3He, the following relation holds

m F S ∗ = 1 + 1 , (1.44) m 3

S namely the eﬀective mass and the Landau parameter F1 are not independent.

1.4 Collective excitations in the collisionless regime

The collision integral in the transport equation determines a typical collision time τ. If one is interested in phenomena that occur on a time scale smaller than τ, namely on frequencies ω 1/τ, then the collision integral can be safely neglected and the transport equation (1.37) in frequency and momentum space becomes

0 " # ∂nk X (ω − q · vk) δnkσ(q, ω) + q · vk Vσ(q, ω) + fkσ k0σ0 δnk0σ0 (q, ω) = 0. (1.45) ∂k k0σ0 The solution of this equation in the absence of external ﬁeld gives information about the collective excitations of the quasiparticle gas. These excitations can also be thought as collective vibrations of the Fermi sphere. From this point of view, let us suppose that, along the direction identiﬁed by the Euler angles (θ, φ), the z-axis being directed along the momentum q, the Fermi momentum kF = kF (sin θ cos φ, sin θ sin φ, cos θ) for a spin σ quasiparticle changes into kF σ = kF + uσ(θ, φ) (sin θ cos φ, sin θ sin φ, cos θ) ,

20 as if kF σ(θ, φ) = kF → kF σ(θ, φ) = kF + δkF (θ, σ) = kF + uσ(θ, φ). The occupation number for a spin σ quasiparticle with momentum k = k (sin θ cos φ, sin θ sin φ, cos θ) changes accordingly as

∂n0 n = n0 ' θ (k − k (θ, φ)) → n = n0 − k δk = n0 + δn . kσ kσ F σ kσ kσ ∂k F kσ kσ We note that ∂n0 δn = − k δk kσ ∂k F 0 0 ∂nk ∂k ∂nk = − δkF = − vF uσ(θ, φ). (1.46) ∂k ∂k ∂k Inserting (1.46) into (1.45) with V = 0 we get

0 X ∂nk0σ0 0 0 (ω − vF q cos θ) uσ(θ, φ) + vF q cos θ fkσ k0σ0 uσ0 (θ , φ ) = 0. (1.47) ∂k0 k0σ0

For small vibrations of the Fermi sphere, we can assume that fkσ k0σ0 only depends on the angle 0 between k and k , fkσ k0σ0 = fσσ0 (θkk0 ), their modula being equal to the equilibrium value of the Fermi momentum kF . We note that, since

0 ∂n 0 0 k σ ' −δ (|k0|) − µ , ∂k0 it follows that 0 Z 0 2 0 X ∂nk0σ0 0 0 4π |k | d|k | 0 fkσ k0σ0 uσ0 (θ , φ ) = −2 V 3 δ (|k |) − µ ∂k0 (2π) k0σ0 Z 1 dΩk0 × f 0 (θ 0 ) u 0 (Ω 0 ) 2 4π σσ kk σ k Z Z 1 dΩk0 1 dΩk0 = −V N f 0 (θ 0 ) u 0 (Ω 0 ) = − F 0 (θ 0 ) u 0 (Ω 0 ) , 2 4π σσ kk σ k 2 4π σσ kk σ k where, generalizing Eq. (1.22), the F -parameters are deﬁned through

Fσσ0 (θkk0 ) = V N fσσ0 (θkk0 ) .

If furthermore we deﬁne, for spin-isotropic models,

S A u = u↑ + u↓, u = u↑ − u↓,

21 S A S A F↑↑ = F↓↓ = F + F ,F↑↓ = F↓↑ = F − F , we ﬁnd, deﬁning λ = ω/vF q, the following equation Z S(A) dΩk0 S(A) S(A) (λ − cos θ ) u (θ , φ ) − cos θ F (θ 0 ) u (θ 0 , φ 0 ) = 0. (1.48) k k k 4π kk k k We write, dropping the indices S and A, X u(θ, φ) = Ylm(θ, φ) ulm, lm X F (θkk0 ) = Pl (θkk0 ) Fl, l so that Eq. (1.48) can be re-written as X cos θk Y (θ , φ ) u (λ − cos θ ) − F = 0. (1.49) lm k k lm k 2l + 1 l lm Since r4π cos θ = Y (θ, φ), 3 10 l is not a good quantum number of Eq. (1.49), while m is. In particular m = 0 corresponds to longitudinal vibrations of the Fermi sphere, m = 1 to dipolar vibrations and m = 2 to quadrupolar ones. We observe that a solution of (1.49) at ω = 0 would imply that a deformation of the Fermi sphere would be cost-free, namely that the Fermi sphere is unstable. At λ = 0, l is a good quantum number, so that a λ = 0 solution of (1.49) occurs whenever there is a value of l such that, see Eq. (1.23), F S(A) F S(A) 1 + l = 0 = l , (1.50) 2l + 1 S(A) Al S(A) which is the condition for the instability of the Landau-Fermi liquid. We note that, since Fl is generally non zero, the instability implies a singularity of the Landau A-parameters.

1.4.1 Zero sound S(A) (S(A) Let us assume, as commonly done, that only F0 and F1 are not negligible, and study Eq. (1.49) for longitudinal vibrations, m = 0. In this case, dropping for simplicity the labels S and A, we obtain

X cos θ (λ − cos θ) u(θ) = (λ − cos θ) Y (θ , φ ) u = cos θ F Y u + F Y u . (1.51) lm k k lm 0 00 00 3 1 10 10 lm

22 We deﬁne,

r 1 C = F u , (1.52) 4π 0 00 1r 3 D = F u , (1.53) 3 4π 1 10 and, by noting that r 1 r4π Y = , cos θ = Y , 00 4π 3 10 we get the formal solution C cos θ + D cos2 θ u(θ) = . (1.54) λ − cos θ In terms of this solution

r 1 Z dΩ C cos θ + D cos2 θ u = 4π 00 4π λ − cos θ = CI1 + DI2, (1.55) 1r 3 Z dΩ C cos θ + D cos2 θ u = cos θ 3 4π 10 4π λ − cos θ = CI2 + DI3, (1.56) where Z 1 dz zn In = . (1.57) −1 2 λ − z Therefore the self-consistency condition that has to be satisﬁed is

C = F0 (CI1 + DI2) , (1.58)

D = F1 (CI2 + DI3) , (1.59) which have a solution if 2 I2 1 = F0 I1 + F1 (1.60) 1 − F1 I3 By realizing that

λ λ + 1 I = −1 + ln , 1 2 λ − 1 I2 = λ I1, 1 I = − + λ2 I , 3 3 1

23 we ﬁnally get that (1.60) is satisﬁed if

2 1 = F0 + λ A1 I1, (1.61) where, by deﬁnition, F1 A1 = . 1 + F1/3 We note that, for λ ≤ 1, π I ' −1 + λ2 − i λ, 1 2 has a ﬁnite imaginary part. This implies that the frequency ω that solves (1.61) is necessarily complex, hence that it does not correspond to a stable collective mode but rather to a damped one. Therefore the only stable solutions of (1.61) must correspond to λ ≥ 1.

For λ ' 1 + δλ, with 0 < δλ 1,

1 e2 δλ I ' − ln , 1 2 2 leading to 1 δλ ' 2 exp −2 1 + . (1.62) F0 + A1 which is consistent with the assumption δλ 1 only if

0 ≤ F0 + A1 1.

In other words, when the Landau parameters are positive and very small, the stable collective modes propagate like sound modes with a velocity

1 v ' vF 1 + 2 exp −2 1 + , F0 + A1 slightly larger than the Fermi velocity. These modes correspond to a deformation of the Fermi sphere of the form C cos θ + D cos2 θ u(θ) ' , 1 − cos θ + δλ that are strongly peaked to θ = 0, namely in the direction of the propagating wave-vector q. In the opposite case of λ 1, 1 1 I ' + , 1 3λ2 5λ4

24 and the solution of (1.60) is readily found to be

2 2 ω F0 A1 F1 λ = 2 2 = + 1 + , (1.63) vF q 3 5 3 that requires the right hand side to be positive and much greater than one. Again this solution describes collective modes that propagate acoustically with a velocity this time much bigger than vF and that correspond to a deformation of the form

C cos θ + D cos2 θ u(θ) ' , λ that involves the whole Fermi sphere.

These collective modes that may emerge in the collisionless regime and propagate like sound modes were called by Landau “zero sounds”, as opposed to the ﬁrst sound that exists in the collision regime. The existence of these new modes of a Landau-Fermi liquid depends on the quasiparticle interaction parameters F , that act as a restoring force absent for non-interacting S A S(A) electrons. Usually, for repulsive interactions, F0 is positive and F0 negative, while A1 is negligible.3 In this case only charge zero-sound, i.e. a Fermi-sphere deformation in the channel

S u = u↑ + u↓, is a stable collective mode, while a spin zero-sound in the channel

A u = u↑ − u↓, is not.

1.5 Collective excitations in the collision regime: ﬁrst sound

For frequencies much smaller than the typical collision rate, the collision integral can not be neglected anymore and, as a result, any deformation of the Fermi sphere is destined to decay. The only exceptions are those excitations that correspond in the limit q → 0 to conserved

3 For repulsive interaction, the effective mass m∗ increases with respect to the non-interacting system – the quasiparticles move more slowly than free ones. Furthermore, the magnetic susceptibility increases even more, since the slowed down quasiparticles with opposite spin repel each other, hence can react quite efficiently to a magnetic field. On the contrary, the compressibility κ decreases, as adding particles costs more energy. Since

κ m∗ 1 χ m∗ 1 (0) = S , (0) = A , κ m 1 + F0 χ m 1 + F0

S A if follows that F0 > −1 + m∗/m > 0 and F0 ≤ 0.

25 quantities, in particular to charge and momentum densities, in which case it is guaranteed P P that kσ Ikσ = kσ k Ikσ = 0. Charge and momentum density excitations correspond to deformations of the Fermi sphere in the symmetric S-channel and with angular momenta l = 0 and l = 1, respectively, and m = 0. Through Eq. (1.49), the excitations in this regime satisfy the equation

0 = (cos θ − λ) Y00(Ω) u00 + (cos θ − λ) Y10(Ω) u10 F S +F S cos θ Y (Ω) u + 1 cos θ Y (Ω) u . (1.64) 0 00 00 3 10 10

Note that this equation diﬀers from Eq. (1.51) for the zero sound because in that case F0 and F1 are assumed to be non-zero, but the u(θ) solving the equation has all l-components, unlike in this case. To solve (1.64), we multiply both sides once by Y00(Ω) and integrate over the solid angle Ω, and do the same multiplying by Y10(Ω), thus obtaining the following set of equations:

r1 F S −λ u + 1 + 1 u = 0, 00 3 3 10 r1 1 + F S u − λ u = 0. 3 0 00 10 This set of equations has solution if

2 S 2 ω 1 S F1 λ = 2 2 = 1 + F0 1 + , (1.65) vF q 3 3 which corresponds to acoustic waves propagating with velocity s, where v2 F S s2 = F 1 + F S 1 + 1 . (1.66) 3 0 3 We note that, for an isotropic system as that we are assuming,

3 kF m∗kF kF vF = , N = 2 , n = 2 , m∗ π 3π so that v2 n F = 3 m∗N hence, through Eq. (1.30), 1 F S s2 = 1 + 1 . (1.67) nm∗κ 3 One can readily realize that this is the ordinary sound velocity compatible with thermodynamics. When local thermodynamic equilibrium is enforced by collisions – the hydrodynamic

26 regime – the equations of motion for the charge density, ρ(x, t), through the continuity equation, and for the velocity ﬁeld through the pressure P (x, t),4

∂ρ(x, t) m F S = −n 1 + 1 ∇ · v(x, t) ∂t m∗ 3 ∂v(x, t) m n = −∇ P (x, t), ∂t where n is the average density, can be solved using the local thermodynamic relation ∂P 1 δP (x, t) = δρ(x, t) = δρ(x, t), ∂ρ n κ where κ is the thermodynamic compressibility. The solution gives ordinary sound with a velocity that coincides with (1.67).

1.6 Linear response functions

An external ﬁeld induces deviations of the quasiparticle occupation numbers, and the linear relation between these deviations and the ﬁeld responsible for them are the so-called “linear response functions”. Let us go back to the transport equation (1.37), and assume in what follows that the collision integral is negligible. After space and time Fourier transform, Eq. (1.37) is5

0 ∂nk X (q · vk − ω) δnkσ(q, ω) − q · vk fkσ k0σ0 δnk0σ0 (q, ω) ∂k k0σ0

4We assume that the particle velocity ﬁeld is related to the current as in Eq. (1.42), where by deﬁnition

1 X (0) n v(x, t) = v δn (x, t). V k kσ kσ

5 In deriving the transport equation we assume implicitly the system to be initially, for instance at time t → −∞, at equilibrium. This implies that the external field that moves away from equilibrium must initially be absent, i.e. V (x, t → −∞) → 0. A simple and convenient choice is to represent, for all times before the measure, performed e.g. at t = 0, an external field that oscillates with frequency ω as V (x, t) = V (x, ω) e−i ω t+η t, with η an infinitesimal positive number. Consequently, the response of the system must follow the same time behavior, i.e. −i ω t+η t δnkσ(x, t) = δnkσ(x, ω) e , so that the time dependence cancels out from (1.37). When taking the time derivative, ∂δn (x, t) kσ = −i (ω + iη) δn (x, ω) e−i ω t+η t, ∂t kσ which implies that the frequency ω has to be interpreted in whatever follows as ω + iη.

27 0 ∂nk = q · vk Vσ(q, ω). (1.68) ∂k We stress once more that the external field that enters this equation can be identified with the actual field acting on the true particles only if it refers to conserved quantities, hence the charge, in which case Vσ(q, ω) = V (q, ω), (1.69) and the spin, V↑(q, ω) = −V↓(q, ω). (1.70) Let us write 0 q · vk ∂nk δnkσ(q, ω) = − Xkσ(q, ω), (1.71) q · vk − ω ∂k through which Eq. (1.68) transforms into

0 X q · vk0 ∂nk0 Xkσ(q, ω) − fkσ k0σ0 Xk0σ0 q · vk0 − ω ∂k0 k0σ0 = −Vσ(q, ω). (1.72)

1.6.1 Formal solution Let us write Eq. (1.72) in matricial way like h i Iˆ− fˆKˆ X~ = −V,~ (1.73) where the matrices have matrix elements

Ikσ k0σ0 = δkk0 δσσ0 ,

fkσ k0σ0 = fkσ k0σ0 , 0 q · vk ∂nk Kkσ k0σ0 = δkk0 δσσ0 , q · vk − ω ∂k and the vectors X~ and V~ X~ = Xkσ(q, ω), kσ V~ = Vσ(q, ω). kσ The formal solution of (1.73) is h i−1 X~ = − Iˆ− fˆKˆ V.~ (1.74)

28 Let us suppose to introduce another matrix Aˆ, with elements

ˆ A = Akσ k0σ0 (q, ω), (1.75) kσ k0σ0 that satisﬁes the set of equations h i Iˆ− fˆKˆ Aˆ = f.ˆ (1.76)

One readily derives from this equation that

h i−1 Aˆ = Iˆ− fˆKˆ f,ˆ hence that h i−1 h i−1 h i Aˆ Kˆ = Iˆ− fˆKˆ fˆKˆ = Iˆ− fˆKˆ fˆKˆ − Iˆ+ Iˆ h i−1 = −Iˆ+ Iˆ− fˆKˆ .

In other words h i−1 Iˆ− fˆKˆ = Aˆ Kˆ + I,ˆ (1.77) so that h i X~ = − Iˆ+ Aˆ Kˆ V,~ that explicitly means

0 X q · vk0 ∂nk0 Xkσ(q, ω) = − δkk0 δσσ0 + Akσ k0σ0 (q, ω) Vσ0 (q, ω), (1.78) q · vk0 − ω ∂k0 k0σ0 hence that 0 0 q · vk ∂nk X X q · vk0 ∂nk0 δnkσ(q, ω) = δσσ0 + Akσ k0σ0 (q, ω) Vσ0 (q, ω), (1.79) q · vk − ω ∂k q · vk0 − ω ∂k0 σ0 k0 which is the formal solution of (1.68). The functions Abkσ k0σ0 (q, ω) are the so-called quasiparticle scattering amplitudes. We deﬁne as usual

S A Akσ k0σ(q, ω) = Akk0 (q, ω) + Akk0 (q, ω), S A Akσ k0σ0 (q, ω) = Akk0 (q, ω) − Akk0 (q, ω),

0 with σ 6= σ . We also introduce a vertex λσ that, in case of a scalar potential Vσ(q, ω) = V (q, ω), is deﬁned as λσ = 1,

29 while it is λσ = δσ↑ − δσ↓, for a Zeeman-splitting magnetic ﬁeld Vσ(q, ω) = V (q, ω) δσ↑ − δσ↓ = V (q, ω) λσ.

Upon multiplying both sides of (1.79) by the corresponding vertex and summing over k and σ and dividing by the volume, we ﬁnd that the charge, δρ(q, ω), and spin, δσ(q, ω), density deviations at linear order in the corresponding ﬁelds are given by

δρ(q, ω) = χS(q, ω) V (q, ω) δσ(q, ω) = χA(q, ω) V (q, ω), where 0 0 S(A) 2 X q · vk ∂nk X (S(A) q · vk0 ∂nk0 χ (q, ω) = 1 + 2 Akk0 (q, ω) . (1.80) V q · vk − ω ∂k q · vk0 − ω ∂k0 k k0

Thus −χS(q, ω) is n2 times the dynamical compressibility κ(q, ω), while −χA(q, ω) is the dynamical spin-susceptibility.

Limiting cases We readily realize that, if q = 0 and ω 6= 0, then the matrix Kˆ = 0. In this case Aˆ = fˆ, namely

Akσ k0σ0 (0, ω) = fkσ k0σ0 . (1.81)

In the opposite limit, q 6= 0 and ω = 0, Eq. (1.76) reads explicitly

0 X ∂np A 0 0 (q, 0) − f A 0 0 (q, 0) = f 0 0 , kσ k σ kσ pα ∂ pα k σ pα k σ pα p showing that Akσ k0σ0 (q, 0) is independent on q. By writing

X S A V N Akσ k0σ0 (q, 0) = Pl (θkk0 ) Al ± Al , l

0 S where the + sign refers to σ = σ and the − one otherwise, one readily recognize that Al and A Al coincide with the Landau A-parameters of Eq. (1.23).

30 1.6.2 Low frequency limit of the response functions

For frequencies ω vF q, we can expand Akσ k0σ0 (q, ω) as

Akσ k0σ0 (q, ω) = Akσ k0σ0 (q, 0) + δAkσ k0σ0 (q, ω), (1.82) where δA vanishes linearly at ω = 0. We note that q · v ω k = 1 + , q · vk − ω q · vk so that the linear variation δA has to satisfy the equation 0 0 X ∂np X ∂np ω δA 0 0 (q, ω) − f δA 0 0 (q, ω) = f A 0 0 (q, 0). (1.83) kσ k σ kσ pα ∂ pα k σ kσ pα ∂ q · v pα k σ pα p pα p p

−1 The integral over Ωp of the function (q · vp) is singular, hence we need some regularization. Indeed, as discussed in the footnote [5], the frequency ω above is actually ω + iη, with η an inﬁnitesimally small positive number. Therefore

q · vk q · vk ω ω → = 1 + ' 1 + i π ω δ (q · vk) + P , q · vk − ω q · vk − ω − iη q · vk − iη q · vk where P indicates the principal value. Therefore, within the S and A channels, Eq. (1.83) becomes (the principal value averages to zero hence does not contribute)

0 0 S(A) X S(A) ∂np S(A) X S(A) ∂np S(A) δA (q, ω) − 2 f δA (q, ω) = 2 π i ω f δ (q · v ) A (q, 0). kk0 kp ∂ pk0 kp ∂ p pk0 p p p p (1.84) At ﬁrst order in ω, Eq. (1.80) becomes (we drop the indices S and A) 0 0 2 X ∂nk X ∂nk0 χ(q, ω) ' 1 + 2 Akk0 (q, 0) V ∂k ∂k0 k k0 0 0 1 X ∂nk X ∂nk0 +2 π i ω δ (q · vk) 1 + 2 Akk0 (q, 0) V ∂k ∂k0 k k0 0 0 1 X ∂nk ∂nk0 +4 π i ω Akk0 (q, 0) δ (q · vk0 ) V ∂k ∂k0 kk0 0 0 1 X ∂nk ∂nk0 +4 δAkk0 (q, ω). (1.85) V ∂k ∂k0 kk0 Since Z dΩ P (θ) = δ , 4π l l0

31 Z dΩ 1 δ (q vF cos θ) = 4π 2 q vF it follows from Eq. (1.84) that Z Z dΩk dΩk0 dΩp dΩk0 1 1 δAkk0 (q, ω) + F0 δApk0 (q, ω) = −π i ω F0 A0, 4π 4π 4π 4π V N 2 q vF namely Z dΩk dΩk0 1 1 F0 1 1 2 δAkk0 (q, ω) = −π i ω A0 = −π i ω A0. 4π 4π V N 2 q vF 1 + F0 V N 2 q vF Therefore Eq. (1.85) is found to be

χ(q, ω) ' −N (1 − A0) ω −i π N (1 − A0) 2 q vF ω +i π N A0 2 q vF ω 2 −i π N A0 2 q vF N ω 2 = − − i π N (1 − A0) 1 + F0 2 q vF N ω −2 = − − i π N (1 + F0) . (1.86) 1 + F0 2 q vF

1.6.3 High frequency limit of the response functions

Let us now consider the case ω vF q → 0 when (again the S and A indices are not explicitly shown) Apk0 (q, ω) = fkk0 . In this case 2 q · vk q · vk (q · vk) ' − − 2 , q · vk − ω ω ω and, since the angular average of q · vk vanishes, the response functions are

0 2 X ∂nk 2 χ(q → 0, ω) = − 2 (q · vk) ω V ∂k k 0 0 4 X ∂nk ∂nk0 0 0 + 2 (q · vk)(q · vk ) fkk . (1.87) ω V ∂k ∂k0 kk0

32 We note that 0 Z 2 X ∂nk 2 dΩ 2 (q · vk) = −N (vF q cos θ) V ∂k 4π k 1 = − N v2 q2, 3 F 0 0 4 X ∂nk ∂nk0 2 2 fkk0 (q · vk)(q · vk0 ) = N vF q V ∂k ∂k0 kk0 Z X dΩk dΩk0 × F cos θ cos θ 0 P (θ 0 ) l 4π 4π k k l kk l Z X dΩk dΩk0 = N v2 q2, F F l 4π 4π lm r r 4π 4π 4π ∗ × Y (θ ) Y (θ 0 ) Y (θ ) Y (θ 0 ) 3 10 k 3 10 k 2l + 1 lm k lm k 1 = N v2 q2 F . F 1 9 Therefore Eq. (1.87) is v2 q2 F χ(q → 0, ω) = N F 1 + 1 . (1.88) 3ω2 3 In a system that is also translationally invariant, besides being isotropic, in the S channel the following relations hold

m k mk F N = ∗ F = F 1 + 1 , π2 π2 3 −1 kF kF F1 vF = = 1 + , m∗ m 3 k3 n = F , 3π2 hence n q2 κ(q → 0, ω) = , (1.89) m ω2 in agreement with the f-sum rule.

33 1.7 Charged Fermi-liquids: the Landau-Silin theory

Now, instead of a neutral Fermi system, let us consider a charged one and discuss how all previous results change. The first thing to note is that the quasiparticle interaction fkσ k0σ0 (x − y) in Eq. (1.33) will now contain, besides a short range contribution, also a long range Coulomb one, so that can be assumed to be of the form e2 f 0 0 (x − y) = f 0 0 δ(x − y) + . (1.90) kσ k σ kσ k σ |x − y| Another novel feature that we have to take into account is the possible presence of a transverse electromagnetic field. If the system is translationally invariant or if inter-band matrix elements of the current density are negligible, the role of a transverse field is that the conjugate momentum is not anymore k but e K = k − A(x, t), c where A is the transverse vector potential felt by the real electrons. Thus the quasiparticle occupation density in the semi-classical limit can be still parametrized by nKσ(x, t). Nevertheless, it is more convenient to define the occupation density in the equivalent way

nKσ(x, t) ≡ nkσ(x, t) = n e (x, t), (1.91) K+ c A(x,t) σ since, in the presence of A, the quasiparticle excitation energy, Eq. (1.34), changes simply into

¯kσ(x, t) → ¯K+eA/c σ(x, t) = ¯kσ(x, t). From the above equation it follows that ∂¯ (x, t) ∂¯ (x, t) K+eA/c σ = kσ = v (x, t) ∂K ∂k kσ is the proper quasiparticle group velocity, and

∂¯K+eA/c(x, t) X e ∂Ai(x, t) = v (x, t) ∂x c kσ i ∂x i Z ∂Vσ(x, t) X ∂nk0σ0 (y, t) + + dy f 0 0 (x − y) . ∂x kσ k σ ∂y k0σ0 Through the deﬁnition Eq. (1.91), it follows that ∂n (x, t) ∂n (x, t) e ∂n (x, t) ∂A(x, t) Kσ = kσ + kσ · , ∂t K x ∂t k x c ∂k ∂t ∂nKσ(x, t) ∂nkσ(x, t) X e ∂nkσ(x, t) ∂Ai(x, t) = + , ∂x ∂x c ∂k ∂x K k i i ∂n (x, t) ∂n (x, t) Kσ = kσ . ∂K x ∂k x

34 Putting everything together we ﬁnd the following transport equation for charged electrons: ∂n (x, t) e ∂n (x, t) ∂A(x, t) I (x, t) = kσ + kσ · kσ ∂t c ∂k ∂t ∂nkσ(x, t) X e ∂nkσ(x, t) ∂Ai(x, t) +v (x, t) · + v (x, t) kσ ∂x kσ j c ∂k ∂x ij i j Z ∂nkσ(x, t) h∂Vσ(x, t) X ∂nk0σ0 (y, t)i − · + dy f 0 0 (x − y) ∂k ∂x kσ k σ ∂y k0σ0 X e ∂nkσ(x, t) ∂Ai(x, t) − v (x, t) c ∂k kσ i ∂x ij j j P We note that the two terms with the ij can be written as

X e ∂nkσ(x, t) ∂Ai(x, t) ∂Aj(x, t) v (x, t) − c ∂k kσ j ∂x ∂x ij i j i e ∂n (x, t) = − v (x, t) × H(x, t) · kσ , c kσ ∂k where H(x, t) = ∇ × A(x, t) is the magnetic field. Since the external transverse electric field is 1 ∂A(x, t) E(x, t) = − , c ∂t the final expression of the transport equation reads ∂n (x, t) ∂n (x, t) I (x, t) = kσ + v (x, t) · kσ kσ ∂t kσ ∂x Z ∂nkσ(x, t) h∂Vσ(x, t) X ∂nk0σ0 (y, t)i − · + dy f 0 0 (x − y) ∂k ∂x kσ k σ ∂y k0σ0 ∂n (x, t) e ∂n (x, t) −e kσ · E(x, t) − v (x, t) × H(x, t) · kσ . (1.92) ∂k c kσ ∂k

Let us now expand the transport equation (1.92) at linear order in the deviation from equilibrium 0 nkσ(x, t) = nkσ + δnkσ(x, t). First we assume an ac electromagnetic ﬁeld acting as a perturbation, in which case e ∂n (x, t) e ∂n0 − v (x, t) × H(x, t) · kσ ' − v × H(x, t) · k c kσ ∂k c k ∂k

35 0 e ∂nk = − vk × H(x, t) · vk = 0. c ∂k This shows that an ac magnetic field does not contribute to the linearized transport equation, that becomes ∂δn (x, t) ∂δn (x, t) I (x, t) = kσ + v · kσ kσ ∂t k ∂x 0 Z ∂nk h ∂Vσ(x, t) X ∂δnk0σ0 (y, t)i + vk · − − e E(x, t) − dy fkσ k0σ0 (x − y) ∂k ∂x ∂y k0σ0 ∂δn (x, t) ∂δn¯ (x, t) = kσ + v · kσ ∂t k ∂x 0 ∂nk h ∂Vσ(x, t) i + vk · − − e E(x, t) , (1.93) ∂k ∂x where we recall that 0 Z ∂nk X δn¯kσ(x, t) = δnkσ(x, t) − dy fkσ k0σ0 (x − y) δnk0σ0 (y, t), ∂k k0σ0 is the deviation from local equilibrium. In the presence of a dc magnetic field H(x), a subtle issue arises in connection with the Lorenz’s force term e ∂n (x, t) − v (x, t) × H(x, t) · kσ . c kσ ∂k Indeed, a dc field, unlike an ac one, can be assumed as integral part of the unperturbed Hamilto- nian and, if large, can be taken as a zeroth order term, H(x) = H0(x). This requires to expand at linear order vkσ(x, t) and ∂nkσ(x, t)/∂k, X Z ∂ h i v (x, t) ' v + dy f 0 0 (x − y) δn 0 0 (y, t) , kσ kσ ∂k kσ k σ k σ k0σ0 ∂n (x, t) ∂n0 ∂δn (x, t) kσ ' k + kσ , ∂k ∂k ∂k leading to the transport equation 0 ∂δnkσ(x, t) ∂δn¯kσ(x, t) ∂nk h ∂Vσ(x, t) i Ikσ(x, t) = + vk · + vk · − − e E(x, t) ∂t ∂x ∂k ∂x e ∂δn¯ (x, t) − v × H (x, t) · kσ , (1.94) c k 0 ∂k which is the appropriate one to discuss properties like magnetoresistance, cyclotron resonances, etc. However, in what follows we will not consider such a physical situation of a large dc field, hence we will just focus on Eq. (1.93).

36 1.7.1 Formal solution of the transport equation We note that, because of (1.90), 2 S S e A A f 0 (x − y) = f 0 δ (x − y) + , f 0 (x − y) = f 0 δ (x − y) , (1.95) kk kk |x − y| kk kk showing that only the response in the S charge-channel is going to be modiﬁed by the Coulomb repulsion with respect to the previously studied case of neutral particles. Hence let us assume that Vσ(x, t) = −e φ(x, t), is the conventional scalar potential and, neglecting the collision integral, let us write the transport equation for the charge deviation

δnk(x, t) = δnk↑(x, t) + δnk↓(x, t), which is found to be 0 ∂δnk(x, t) ∂δn¯k(x, t) ∂nk h ∂φ(x, t) i + vk · − 2 vk · − e − e E(x, t) = 0, (1.96) ∂t ∂x ∂k ∂x with 0 Z ∂nk X S δn¯k(x, t) = δnk(x, t) − 2 dy fkk0 (x − y) δnk0 (y, t). ∂k k0 In the Fourier space 0 0 2 ∂nk X S ∂nk 4πe 0 δn¯k(q, ω) = δnk(q, ω) − 2 fkk0 δnk (q, ω) − 2 2 δρ(q, ω), ∂k ∂k q k0 where 1 X δρ(q, ω) = δn 0 (q, ω), V k k0 is the deviation of the total charge density. We then find 0 ∂nk X S (ω − vk · q) δnk(q, ω) + 2 vk · q fkk0 δnk0 (q, ω) ∂k k0 0 ∂nk −i e 2 vk · E(q, ω) = 0, (1.97) ∂k where E(q, ω) is the internal electric field felt by the quasiparticles, q E(q, ω) = E(q, ω) − i q φ(q, ω) + i 4 π e δρ(q, ω), (1.98) q2 that contains the external transverse and longitudinal fields as well as the longitudinal field created by the same quasiparticles.

37 Longitudinal response Let us assume a zero transverse ﬁeld, and deﬁne as

4 π e2 V (q, ω) = −e φ(q, ω) + δρ(q, ω), (1.99) q2 the internal scalar potential. One easily recognizes that the response to the internal scalar potential is formally the same as for neutral particles, so that, by means of Eq. (1.80), we find that S δρ(q, ω) = χ (q, ω) V (q, ω) ≡ χ∗(q, ω) V (q, ω), (1.100) that also define the so-called “proper” charge response-function χ∗. Inserting the expression of V (q, ω), Eq. (1.99), and solving for δρ(q, ω) we finally obtain χ (q, ω) δρ(q, ω) = ∗ V (q, ω) ≡ χ(q, ω) V (q, ω),, (1.101) 4 π e2 1 − χ (q, ω) q2 ∗ that defines the “improper” response function χ, which is the response to the external field, as opposed to the proper one, i.e. the response to the internal field. Eq.(1.101) provides the definition of the longitudinal dielectric constant ||(q, ω) as

4 π e2 (q, ω) = 1 − χ (q, ω). (1.102) || q2 ∗ In the limit ω → 0 ﬁrst and then q → 0, we can make use of Eq. (1.86) and ﬁnd that

4 π e2 N lim lim ||(q, ω) = lim 1 + → ∞, (1.103) q→0 ω→0 q→0 2 S q 1 + F0 which deﬁnes the eﬀective Thomas-Fermi screening-length s S π 1 + F0 λTF = . e2 N In the limit q → 0, we can use Eq. (1.88) to get

2 2 2 2 4 π e vF q F1 ωp lim ||(q, ω) = 1 − N 1 + ≡ 1 − , (1.104) q→0 q2 3ω2 3 ω2 that provides the deﬁnition of the conduction-band plasma-frequency

4 π e2N v2 F ω2 = F 1 + 1 , (1.105) p 3 3

38 as the pole of 1/||(0, ω). Note that, for a translationally invariant system

4 π n e2 ω2 = , p m with n the electron density and m the real electron mass, that coincides with the conventional expression for the plasma oscillations.

1.8 Dirty quasiparticle gas

So far we have not taken into account the collision integral, that is however important at frequencies smaller than the typical collision rate. At very low temperature T , the main contribution to the collision rate comes from impurity scattering. Indeed, collisions due to phonons vanish at low temperature, as phonon occupations go exponentially to zero as T → 0, as well as collisions due to scattering oﬀ other quasiparticles. 6 Taking into account only the scattering oﬀ (non- magnetic) impurities, and moreover assuming a short-range impurity potential, we can write the collision integral as X h i Ikσ(x, t) = −2π Wkp nkσ(x, t) (1 − npσ(x, t)) − npσ(x, t) (1 − nkσ(x, t)) p ×δ ¯kσ(x, t) − ¯pσ(x, t) , (1.106)

6 0 An excited quasiparticle at ¯k > µ can in principle decay into a quasiparticle at momentum k with ¯k0 > µ 0 plus a quasiparticle-quasihole pair, namely a quasiparticle with momentum p with ¯p0 > µ and a quasihole with 0 0 momentum p and energy ¯p < µ. Because of momentum conservation k = k + p − p, and of energy conservation

¯k − µ = (¯k0 − µ) + (¯p0 − µ) − (¯p − µ) .

The collision integral expressed through the Fermi golden-rule will have the general form (we discard for simplicity the spin label)

1 X 0 0 I = W 0 0 δ k + p − k − p δ ( + − 0 − 0 ) n n 0 n 0 (1 − n ) , k V 2 kp p k k p k p k k p p k0pp0 where W is the modulus square of the scattering amplitude. Since one momentum is ﬁxed by the momentum conservation, say p, the sum becomes a double sum over k0 and p0 with the condition that

|¯k0 − µ| + |¯p0 − µ| + |¯k0+p0−k − µ| = |¯k − µ| ' T, the last almost equivalence deriving from the fact that the energy variation with respect to the chemical potential of a quasiparticle excitation at temperature T is of the order of T itself. This implies that both (¯k0 − µ) and (¯p0 − µ) are positive but smaller that T , so that, by simple phase space arguments, the collision integral is found to decay approximately as T 2. Extending this analysis, one readily realizes that any decay into a quasiparticle plus m quasiparticle-quasihole pairs occurs with probability T 2m. Therefore the collision integral due to scattering oﬀ quasiparticles is negligible at low temperature.

39 where the energy conservation involves, as it should, the true quasiparticle excitation energy

δE X Z ¯kσ(x, t) = = k + dy fkσ k0σ0 (x − y) δnk0σ0 (y, t). δnkσ(x, t) k0σ0 We note that h i nkσ(x, t) (1 − npσ(x, t)) − npσ(x, t) (1 − nkσ(x, t)) δ ¯kσ(x, t) − ¯pσ(x, t) h i = nkσ(x, t) − npσ(x, t) δ ¯kσ(x, t) − ¯pσ(x, t)

h 0 0 i = n¯kσ(x, t) − n¯pσ(x, t) + δn¯kσ(x, t) − δn¯pσ(x, t) δ ¯kσ(x, t) − ¯pσ(x, t) h i = δn¯kσ(x, t) − δn¯pσ(x, t) δ ¯kσ(x, t) − ¯pσ(x, t) , so that X h i Ikσ(x, t) = −2π Wkp δn¯kσ(x, t) − δn¯pσ(x, t) δ ¯kσ(x, t) − ¯pσ(x, t) , (1.107) p showing that the collision integral only depends upon deviations from local equilibrium. Within linear approximation X h i Ikσ(x, t) ' −2π Wkp δn¯kσ(x, t) − δn¯pσ(x, t) δ (k − p) . (1.108) p

Let us write 1 X W = W P (θ ) , kp V l l kp l 0 ∂nk X δn¯kσ(x, t) = − δn¯lm σ(x, t) Ylm (Ωk) ∂k lm so that Eq. (1.108) becomes

0 ∂nk X Ikσ(x, t) = π N W0 δn¯lm σ(x, t) Ylm (Ωk) ∂k lm 0 ∂nk X Wl −π N δn¯lm σ(x, t) Ylm (Ωk) ∂k 2l + 1 lm 0 ∂nk X 1 = δn¯lm σ(x, t) Ylm (Ωk) , (1.109) ∂k τl lm

40 where the inverse relaxation time is deﬁned through 1 Wl = π N W0 − τl 2l + 1 2 π X h i = W δ ( − ) 1 − P (θ ) . (1.110) V kp k p l kp p

We note that 1/τ0 = 0. Therefore the transport equation in Fourier space and for the charge S component reads

0 ∂nk X S −ω δn (q, ω) + q · v δn (q, ω) − 2 q · v f 0 δn 0 (q, ω) k k k ∂k k kk k k0 0 0 ∂nk ∂nk X 1 +i e 2 vk · E(q, ω) = −i δn¯lm(q, ω) Ylm (Ωk) ∂k ∂k τl lm 0 ∂nk = −ω δnk(q, ω) + q · vk δn¯k(q, ω) + i e 2 vk · E(q, ω), (1.111) ∂k where we recall that q E(q, ω) = E(q, ω) + i 4 π e δρ(q, ω), (1.112) q2 is the internal ﬁeld felt by quasiparticles, E including both transverse and longitudinal components, and we have deﬁned

0 ∂nk X S δn¯ (q, ω) = δn (q, ω) − 2 f 0 δn 0 (q, ω), k k ∂k kk k k0 as the deviation from local equilibrium excluding the long-range interaction.

1.8.1 Conductivity

We observe that Z −e δρ(q = 0, ω) = dt ei ω t δN(t), is the frequency Fourier-transform of the change in the total electron charge. If we assume that the system is maintained neutral at any time by a compensating positive ionic charge-density, then the q = 0 component of the field generated by the electrons must be canceled exactly by the ionic field. In other words, δρ(q, ω) must not include any q = 0 component, i.e. δρ(0, ω) = 0. Thus, taking q = 0 in (1.111) we find

0 0 ∂nk ∂nk X 1 − ω δnk(ω) + i e 2 vk · E(ω) = −i δn¯lm(ω) Ylm (Ωk) . (1.113) ∂k ∂k τl lm

41 We observe that

1 k r4π vk · E(ω) = k · E(ω) = Y10 (Ωk) Ez(ω) m∗ m∗ 3 1 1 +Y (Ω ) E (ω) − iE (ω) + Y (Ω ) E (ω) + iE (ω) , 1+1 k 2 x y 1−1 k 2 x y namely that only the l = 1 components are coupled to the electric ﬁeld, so that the actual transport equation reads [dropping the label (l,m)=(1,m)]

0 ∂nk 1 −ω δnk(ω) + i e 2 vk · E(ω) = i δn¯k(ω). ∂k τ1 Since, for l = 1 deviations δn¯ δn = , F S 1 + 1 3 we ﬁnally get ! ∂n0 F S 1 δn¯ (ω) = 2 e k 1 + 1 v · E(ω), (1.114) k ∂ 3 1 k k −iω + τ where τ τ = 1 . (1.115) F S 1 + 1 3 The electric current is therefore e X J(ω) = − v δn¯ (ω) V k k k 0 S ! e X ∂nk F1 1 = − vk 2 e 1 + vk · E(ω) (1.116) V ∂k 3 1 k −iω + τ ! v2 F S 1 = e2 N F 1 + 1 E(ω) ≡ σ(ω) E(ω), (1.117) 3 3 1 −iω + τ which deﬁnes the conductivity σ(ω). We note that

2 2 vF m∗ kF kF n N = 2 2 = , 3 π 3m∗ m∗

42 with n the average electron density. Thus the conductivity is ! n e2 F S 1 σ(ω) = 1 + 1 , (1.118) m 3 1 ∗ −iω + τ with a dc value ! n e2 τ F S σ = σ(0) = 1 + 1 . (1.119) m∗ 3

This expression shows that the eﬀective mass that controls dc transport is not m∗ but m ∗ , F S 1 + 1 3 which, for a system that is translationally invariant before adding the impurities, is just the real-electron mass m.

1.8.2 Diﬀusive behavior

Let us now consider a ﬁnite q such that vF q τ 1 as well as ωτ 1, and assume a longitudinal electric ﬁeld E||q, q taken to be along the z-axis. If we write

0 ∂nk X δn¯k(q, ω) = − δn¯lm(q, ω) Ylm (Ωk) , ∂k lm and analogously for δnk, we note that only the m = 0 component is coupled to the external field. If we substitute these expressions for δn and δn¯ into Eq. (1.111), multiply both sides of ∗ this equation by Yl0 (Ωk) and integrate over the solid angle, by means of s Z (2l + 1) (2l + 1) 2 dΩY ∗(Ω) Y (Ω) Y (Ω) = 1 2 Cl0 , l0 l1,0 l2,0 4π (2l + 1) l10 l20 where Cl0 are Clebsch-Gordan coefficients, we find the following set of equations (we discard l10 l20 the label m = 0) s 0 2 r X 2l + 1 l0 4π δn¯l(q, ω) − ω δnl(q, ω) + q vF Cl00 10 δn¯l0 (q, ω) − 2 i e vF E(q, ω) δl1 = i . 2l + 1 3 τl l0 (1.120)

43 We observe that only l = 1 is directly coupled to the ﬁeld, so that, for any l > 1, we ﬁnd   s 0 2  τlω  X 2l + 1 l0 − − i δn¯l(q, ω) + τl q vF C 0 δn¯l0 (q, ω)  F S  2l + 1 l 0 10 1 + l l06=1 2l + 1 r 3 2 = −τ q v Cl0 δn¯ (q, ω). l F 2l + 1 10 10 1

This implies that all deviations with l > 1 are driven by q vF τl δn¯1 δn¯1, therefore are negligible. What remains are just the components with l = 0, 1, satisfying the coupled equations

ω r1 − S δn¯0(q, ω) + q vF δn¯1(q, ω) = 0, 1 + F0 3 r r ω 1 4π δn¯1(q, ω) − S δn¯1(q, ω) + q vF δn¯0(q, ω) − 2 i e vF E(q, ω) = i , F 3 3 τ1 1 + 1 3 whose solutions are readily obtainable:

r S ! 4π F1 −i ω τ δn¯1(q, ω) = 2 e vF 1 + ! E(q, ω), 3 3 q2 v2 τ F S iω − F 1 + F S 1 + 1 + ω2 τ 3 0 3 r1 v q δn¯ (q, ω) = F 1 + F S δn¯ (q, ω) 0 3 ω 0 1 √ ! 4π F S = 2 e v 1 + 1 1 + F S F 3 3 0

−i q vF τ × ! E(q, ω). q2 v2 τ F S iω − F 1 + F S 1 + 1 + ω2 τ 3 0 3

Therefore the electric current is found to be

vF J(q, ω) = −e √ N δn¯1(q, ω) 2 12π ! r4π F S −i ω τ 2 e v 1 + 1 " F 3 3 # vF = −e √ N ! E(q, ω) 2 12π q2 v2 τ F S iω − F 1 + F S 1 + 1 + ω2 τ 3 0 3

44 2 S ! n e τ F1 iω = 1 + 2 2 S E(q, ω) m∗ 3 q vF τ S F1 2 iω − 3 1 + F0 1 + 3 + ω τ

2 S ! n e τ F1 iω = 1 + 2 2 E(q, ω) m∗ 3 iω − D q + ω τ ≡ σ(q, ω) E(q, ω), (1.121) where ! v2 τ F S D = F 1 + F S 1 + 1 , (1.122) 3 0 3

45 is the diffusion coefficient. 7 Seemingly, the charge density is 1 δρ(q, ω) = √ N δn0(q, ω) 2 4π 1 −i D q = S N 2 2 e E(q, ω). (1.123) 1 + F0 iω − D q + ω τ We recall that, for a longitudinal field,

4π e2 −e E(q, ω) = e iq φ(q, ω) − i q δρ(q, ω) ≡ −iq V (q, ω), q2

7Indeed this behavior is characteristic of diffusive modes. If E(q, ω) = 0, the two coupled equations reduce to r ω 1 − S δn¯0(q, ω) + q vF δn¯1(q, ω) = 0, 1 + F0 3 r ω 1 δn¯1(q, ω) − S δn¯1(q, ω) + q vF δn¯0(q, ω) = i , F 3 τ1 1 + 1 3 √ S √ F1 If we multiply the first by N /2 4π , and the second by N vF 1 + 3 /2 12π, we find

J(q, ω) −ω δρ(q, ω) + q = 0 −e J(q, ω) D 1 J(q, ω) −ω + q δρ(q, ω) = i . −e τ τ −e The ﬁrst equation is nothing but the continuity equation (the electric current is −e times the charge current j):

∂ρ(x, t) + ∇ · j(x, t) = 0. ∂t The second equation for ωτ 1, gives j(q, ω) = −i q D δρ(q, ω), namely that the charge current, j(x, t) = −D ∇ ρ(x, t), is proportional to minus the space-gradient of the charge density, the proportionality constant being the diffusion coefficient. Inserting this expression in the continuity equation, we find ∂ρ(x, t) − D ∇2 ρ(x, t) = 0, ∂t showing that charge diffuses with equation of motion

−iω + D q2 δρ(q, ω) = 0.

46 where V is the internal scalar potential felt by the electrons. Therefore,

D q2 δρ(q, ω) = κ V (q, ω), (1.124) iω − D q2 + ω2 τ with κ the compressibility. This equation provides an expression of the “proper” response function in the presence of disorder

D q2 χ∗(q, ω) = κ , (1.125) iω − D q2 + ω2 τ hence of the “improper” one χ (q, ω) χ q, ω) = ∗ . (1.126) ( 4πe2 1 − χ∗(q, ω) q2

47 Chapter 2

Second Quantization

The first difficulty encountered in a many-body problem is how to deal with a many-body wavefunction. The reason is that a many-body wavefunction has to take into account both the indistinguishability of the particles and their statistics, while any operator, including the Hamiltonian, does not have in first quantization such properties. Hence it would be desirable to have at disposal an alternative scheme where the indistinguishability principle as well as the statistics of the particles were already built in the expression of the operators. This is actually the scope of second quantization.

2.1 Fock states and space

Let us take a system of N particles, either fermions or bosons. The Hilbert space spans a basis of N-body orthonormal wavefunctions which should satisfy both the indistinguishability principle as well as the appropriate statistics of the particles. The simplest way to construct this space is as follows. We start by choosing an orthonormal basis of single particle wavefunctions:

φa(x), a = 1, 2, ··· . (2.1) Here x is a generalized coordinate which include both the space coordinate x as well as e.g. the z-component, σ, of the spin, which is half-integer for fermions and integer for bosons. The suﬃx a is a quantum label and Z ∗ dx φa(x) φb(x) = δab

X ∗ φa(x) φa(y) = δ(x − y). a R P R 0 [The dx . . . means σ dx ... , while δ(x − y) ≡ δ(x − y)δσσ0 with x = (x, σ) and y = (y, σ )]. A generic N-body wavefunction with the appropriate symmetry properties can be constructed

48 through the above single-particle states. Since the particles are not distinguishable, we do not need to know which particle occupies a speciﬁc state. Instead, what we need to know are just the occupation numbers na’s, i.e. the number of particles occupying each single-particle state φa. This number is either na = 0, 1 for fermions, because of Pauli principle, or an arbitrary integer na ≥ 0 for bosons. Apart from that, the occupation numbers should satisfy the trivial particle-conservation constraint X na = N. a Since the occupation numbers are the only ingredients we need in order to build up the N-body wave-function, we can formally denote the latter as the ket

|n1, n2,... i, (2.2) which is called a Fock state, while the space spanned by the Fock states is called Fock space. Within the Fock space, the state with no particles, the vacuum, will be denoted by |0i. For instance, if the N fermions with coordinates xi, i = 1, ··· ,N, occupy the states aj, j = 1, ··· ,N, with a1 < a2 < ··· < aN , namely na = 1 for a ∈ {aj}, otherwise na = 0, then the appropriate wave-function is the Slater determinant

φa1 (x1) ··· φa1 (xN ) r 1 φa2 (x1) ··· φa2 (xN ) Ψ (x , . . . , x ) = , (2.3) {na} 1 N . . . . . N! . . . . .

φaN (x1) ··· φaN (xN ) which is therefore the ﬁrst quantization expression of the Fock state |{na}i with the same occupation numbers. The above wave-function satisﬁes the condition of being anti-symmetric if two coordinates are interchanged, namely two columns in the determinant. Analogously, it is antisymmetric by interchanging two rows, i.e. two quantum labels. On the contrary, if we have N bosons with coordinates xi, i = 1, ··· ,N, which occupy the states aj, j = 1,...,M, M ≤ N and a1 < a2 < ··· < aM , with occupation numbers naj , then the appropriate wave-function is the permanent sQ j nj! X Φ{n }(x1, . . . , xN ) = φa (xp ) . . . φa (xp ) a N! 1 1 1 na1 p

φa (xp ) . . . φa (xp ) ... 2 na1 +1 2 na1 +na2

. . . φaM (xpN−n +1 ) . . . φaM (xpN ), (2.4) aM where the sum is over all non-equivalent permutations p’s of the N coordinates 1. Indeed the wave-function is even by interchanging two coordinates or two quantum labels.

1 Non-equivalent means for instance that φi(x)φi(y) is equivalent to φi(y)φi(x)

49 In conclusion, the space spanned by all possible Slater determinants built with the same basis set of single-particle wavefunctions constitues an appropriate Hilbert (Fock) space for many-body fermionic wavefunctions. Analogously the space spanned by all possible permanents is an appropriate Hilbert space for many-body bosonic wavefunctions. In the following we will introduce operators acting in the Fock space. We will consider separately the fermionic and bosonic cases.

2.2 Fermionic operators

† Let us introduce the creation, ca, and annihilation, ca, operators which add or remove, respec- † tively, one fermion in state a. The operator caca ﬁrst annihilates then creates a particle in a, which can be done as many times as many particles occupy that state. Therefore

† caca|nai = na|nai, (2.5)

† so it acts like the occupation number operator caca ≡ nˆa. Since by Pauli principle na = 0, 1, then † † caca|0i = 0, caca|na = 1i = |na = 1i. (2.6) † Analogously, the operator caca ﬁrst creates then destroys a fermion in state a, which can not do if a is occupied, while it can do once if empty. Therefore

† † caca|0i = |0i, caca|na = 1i = 0. (2.7) Thus, either a is empty or occupied, the following equation holds

† † caca + caca |na = 0, 1i = |na = 0, 1i, which leads to the operator identity

† † n † o caca + caca = ca, ca = 1, (2.8) where the symbol {... } means the anti-commutator. Moreover, since one can not create nor destroy two fermions in the same state, it also holds

n o n † † o ca, ca = ca, ca = 0. (2.9)

Eqs. (2.8) and (2.9) are the anti-commutation relations satisﬁed by the fermion operators with the same quantum label. Going back to (2.6) and (2.7), we readily see that they are all satiﬁed if

† † ca|0i = |na = 1i, ca|na = 1i = 0

50 ca|0i = 0, ca|na = 1i = |0i, † also showing that ca is the hermitean conjugate of ca. Let us consider now the action of the above operators on a Fock state. First of all we need to provide a prescription to build a Fock state by the creation operators. We shall assume that, if † | na = 1i = ca | 0i, then by deﬁnition and for b 6= a † † † cb | na = 1i = cbca | 0i ≡| nb = 1, na = 1i. (2.10) Since the Slater determinant, hence the corresponding Fock state, is odd by interchanging two rows it follows that † † | nb = 1, na = 1i = − | na = 1, nb = 1i ≡ −cacb | 0i. (2.11) Comparing (2.11) with (2.10) we conclude that

† † † † n † †o cacb = −cbca → ca, cb = 0, (2.12) The hermitean conjugate thus implies that also n o ca, cb = 0. (2.13) We note that because of (2.9), both Eqs. (2.12) and (2.13) remain valid even if a = b. † Finally we need to extract the reciprocal properties of cb and ca for a 6= b. We assume the following result: † † † ca | na = 1, nb = 1i = cacacb | 0i ≡| nb = 1i = cb | 0i. (2.14) † † † † † Since | 0i = caca | 0i, and cacb = −cbca, it follows that † † † † −cacbca | 0i = cbcaca | 0i, namely that, for a 6= b, n †o ca, cb = 0. (2.15) All the above anti-commutation relations can be cast in the general formulas

n †o n † †o n o ca, cb = δab, ca, cb = 0, ca, cb = 0. (2.16) If we extend our prescription (2.10) to more than two electrons, we ﬁnd that any Fock state has the simple expression n Y † i | n1, n2,... i = ci | 0i, (2.17) i≥1 where the occupation numbers ni = 0, 1.

51 2.2.1 Second quantization of multifermion-operators Let us consider the single-particle operator in ﬁrst quantization X Vˆ = V (xi), (2.18) i where the sum runs over all particles and V (xi) is an operator acting both on space coordinates and spins. We want to calculate the matrix element

0 0 ˆ hn1, n2,... |V |n1, n2,... i, (2.19)

0 among two Slater determinats, equivalently Fock states, with occupation numbers ni’s and ni’s, respectively. Since the operator Vˆ conserves the number of particles, then

X X 0 ni = nj = N. i j Being the particles not distinguishable, it derives that

0 0 ˆ 0 0 hn1, n2,... |V |n1, n2,... i = Nhn1, n2,... |V (x1)|n1, n2,... i Z ∗ = dx1 dx2 . . . dxN Ψ{n0}(x1, x2, . . . , xN ) V (x1)Ψ{n}(x1, x2, . . . , xN )

We assume that in the Fock ket |n1, n2,... i, the states ai, i = 1,...,N are occupied, with 0 0 a1 < a2 < ··· < aN , while in |n1, n2,... i are occupied the bj’s, j = 1,...,N. We can expand the Slater determinant in the ﬁrst column, corresponding to particle x1 and get

N 1 X i−1 Ψ (x1, x2, . . . , xN ) = √ (−1) φa (x1)Ψ (x2, . . . , xN ), {n} i {n:nai =0} N i=1 where Ψ (x2, . . . , xN ) is the Slater determinant of the N − 1 particles x2, x3, . . . , xN {n:nai =0} with the same occupation numbers as Ψ{n}(x1, x2, . . . , xN ) apart from state ai which is empty. Analogously

N ∗ 1 X j−1 ∗ ∗ 0 0 Ψ{n }(x1, x2, . . . , xN ) = √ (−1) φbj (x1) Ψ{n0:n =0}(x2, . . . , xN ) . bj N j=1 Therefore

0 0 ˆ hn1, n2,... |V |n1, n2,... i = Z X i+j ∗ 0 0 (−1) dx1φbj (x1) V (x1)φai (x1) hn1, n2, . . . , nbj − 1,... |n1, n2, . . . , nai − 1,... i i,j

52 X = V hn0 , n0 , · · · | c† c | n , n ,... i, (2.20) bj ,ai 1 2 bj ai 1 2 i,j the last expression following from (2.17) and (2.16). Therefore the second quantization expression of the single particle operator (2.18) is

ˆ X † V = Vi,j ci cj, (2.21) i,j where Z ∗ 2 Vi,j = dx φi(x) V (x) φj(x).

Let us continue and consider a two particle operator

1 X Uˆ = U(x , x ). (2.22) 2 i j i6=j

The matrix element between two Fock states is now N(N − 1) hn0 , n0 ,... |Uˆ|n , n ,... i = hn0 , n0 ,... |U(x , x )|n , n ,... i (2.23) 1 2 1 2 2 1 2 1 2 1 2 In this case we need to expand the Slater determinants in the ﬁrst two columns:

N 1 X Ψ (x , x , . . . , x ) = (−1)i−1(−1)j−1−θj,i φ (x ) φ (x ) {n} 1 2 N p aj 2 ai 1 N(N − 1) i6=j=1

Ψ (x3, . . . , xN ), (2.24) {n:nai =0,naj =0} where now Ψ (x3, . . . , xN ) is the Slater determinant of the N−2 particles x3, x4, . . . , xN {n:nai =0,naj =0} in which, diﬀerently from Ψ{n}(x1, x2, . . . , xN ), the states ai and aj are empty. The θj,i function is = 1 if j > i and = 0 if j < i. Analogously

1 N ∗ X i−1 j−1−θj,i ∗ ∗ Ψ 0 (x , x , . . . , x ) = (−1) (−1) φ (x ) φ (x ) {n } 1 2 N p bi 1 bj 2 N(N − 1) i6=j=1 ∗ Ψ{n0:n0 =0,n0 =0}(x3, . . . , xN ) . (2.25) bi bj

2 Notice that if the operator V (x) acts on the spins, namely V (x) → Vσσ0 (x), then Z X ∗ 0 Vi,j = dx φi(x, σ) Vσσ0 (x) φj (x, σ ). σσ0

53 Therefore

N N N(N − 1) 1 X X hn0 , n0 ,... |U(x , x )|n , n ,... i = (−1)i+j+θj,i (−1)m+n+θn,m 2 1 2 1 2 1 2 2 i6=j=1 m6=n=1 Z ∗ ∗ dx1 dx2 φbm (x1) φbn (x2) U(x1, x2) φaj (x2) φai (x1) Z ∗ dx3 . . . dxN Ψ{n0:n0 =0,n0 =0}(x3, . . . , xN ) Ψ{n:n =0,n =0}(x3, . . . , xN ) bm bn ai aj

1 X 0 0 † † = Ub ,b ;a ,a hn , n , · · · | c c c c | n1, n2,... i, (2.26) 2 m n j i 1 2 bm bn aj ai i,j,m,n hence the second quantization expression of the two-body interaction 1 X Uˆ = U c†c†c c , (2.27) 2 i,j;k,l i j k l ijkl where Z ∗ ∗ Ui,j;k,l = dx dy φi(x) φj(y) U(x, y)φk(y)φl(x).

Analogously, one can introduce m-particle operators 1 X Uˆ = U(x , x , . . . , x ), m m! 1 2 m i16=i26=···6=im which translate in second quantization language into 1 X Uˆ = U m m! i1,i2,...,im;jm,jm−1,...,j1 i1,j1,i2,j2,...,im,jm c† c† . . . c† c c . . . c , (2.28) i1 i2 im jm jm−1 j1 with Z ∗ ∗ ∗ Ui1,i2,...,im;jm,jm−1,...,j1 = dx1 dx2 . . . dxm φi1 (x1) φi2 (x2) . . . φim (xm)

U(x1, x2, . . . , xm)φjm (xm)φjm−1 (xm−1) . . . φj1 (x1).

We conclude by emphasizing the advantages of second quantization with respect to ﬁrst quantization. In the former, any multiparticle operator depends explicitly on the particle coordinates. It is only the wave-function which contains information about either the indistiguishability of the particles as well as their statistics. On the contrary, in second quantization those properties are hidden in the deﬁnition of creation and annihilation operators, hence the multiparticle

54 operators do not depend anymore on the particle coordinates. Moreover in second quantization we can also introduce operators which have no first-quantization counterpart. For instance we can define particle-non-conserving operators which connect subspaces with different numbers of particles of the whole Hilbert space. For instance

X † † ∗ ∆i,j ci cj + ∆i,j cjci , i,j is an operator which connects Fock states with particle numbers diﬀering by two. As we shall see such operators are useful for discussing superconductivity.

2.2.2 Fermi fields Till now we have defined fermionic operators for a given set of single-particle wavefunctions. Let us now introduce new operators which are independent of this choice. We define annihilation and creation Fermi fields by

X † X ∗ † Ψσ(x) ≡ Ψ(x) = φi(x) ci , Ψ(x) = φi(x) ci . (2.29) i i They satisfy the following properties

† X ∗ † {Ψ(x), Ψ(y) } = φi(x) φj(y) {ci , cj} ij X ∗ = φi(x) φi(y) = δ(x − y), (2.30) i as well as {Ψ(x), Ψ(y)} = {Ψ(x)†, Ψ(y)†} = 0. (2.31) which are indeed independent upon the basis. If we change the basis via the unitary transformation Uˆ acting on the basis set X φi(x) = Uα,i φα(x), α with UˆUˆ † = I, the identity matrix, then X X Ψ(x) = φi(x) ci = φα(x) Uα,i ci i i,α X = φα(x) cα, (2.32) α

55 showing the proper transformation properties of the fermionic operators X cα = Uα,i ci . (2.33) i The multiparticle operators in second quantization have a very simple expression in terms of the Fermi ﬁelds. For instance it is easy to show that Eq. (2.21) can also be written as Z ˆ X † † V = Vi,j ci cj = dx Ψ(x) V (x) Ψ(x), (2.34) ij and analogously (2.27) as

1 X 1 Z Uˆ = U c†c†c c = dx dy Ψ(x)†Ψ(y)† U(x, y) Ψ(y)Ψ(x), (2.35) 2 i,j;k,l i j k l 2 ijkl and ﬁnally (2.28) as

1 X Uˆ = U m m! i1,i2,...,im;jm,jm−1,...,j1 i1,j1,i2,j2,...,im,jm c† c† . . . c† c c . . . c i1 i2 im jm jm−1 j1 1 Z = dx . . . dx Ψ(x )† ... Ψ(x )† U(x , . . . , x ) Ψ(x ) ... Ψ(x ). (2.36) m! 1 m 1 m 1 m m 1 Let us for instance consider the density operator in ﬁrst quantization

N X ρˆσ(x) = δ(x − xi) δσ,σi . i=1 In second quantization, through Eq. (2.34), it reads Z X † ρˆσ(x) = dy Ψσ0 (y) δ(x − y) δσ,σ0 Ψσ0 (y) σ0 † † = Ψσ(x) Ψσ(x) = Ψ(x) Ψ(x), (2.37) which also ahows that Ψ(x)† is nothing but the operator which creates a particle in coordinate x, while Ψ(x) destroys it.

56 2.3 Bosonic operators

† As previoulsy done for the fermionic case, we introduce the creation, da, and its hermitean conjugate, the annihilation da, operators which respectively create and destroy a boson in state † a. Again the operators dada counts how many times we can destroy and create back a boson in state a, hence it is just the occupation number na. However, since Pauli principle does not hold † for bosons, the operator dada ﬁrst adds one boson in state a, hence increases na → na + 1, then destroys one in the same state. This latter process can be done na + 1 times, being na + 1 the actual occupation number once one more boson has been added. Therefore

† † dada|nai = na|nai, dada|nai = (1 + na)|nai, (2.38) hence the following commutation relation holds

h † i † † da, da = dada − dada = 1, (2.39) where [... ] denotes the commutator. Eq. (2.38) is trivially satisﬁed by

√ † √ da|nai = na|na − 1i, da|nai = na + 1|na + 1i. (2.40)

The permanent, contrary to the Slater determinant, is invariant upon interchanging two quantum labels. As a consequence, bosonic operators corresponding to diﬀerent states commute instead of anti-commuting as the fermionic ones. Namely, for a 6= b,

h † i h † †i da, db = da, db = 0.

Therefore, in general, h †i h † †i h i da, db = δab, da, db = da, db = 0. (2.41) Moreover, through (2.40) and (2.41) we can write a generic Fock state as

ni d† Y i |n1, n2,... i = √ |0i. (2.42) n ! i i

2.3.1 Bose fields and multiparticle operators The analogous role of the Fermi fields is now played by the Bose fields defined through

X † X ∗ † Φ(x) = φa(x) da, Φ(x) = φa(x) da, (2.43) a a

57 which satisfy the commutation relations h i h i Φ(x), Φ(y)† = δ(x − y), [Φ(x), Φ(y)] = Φ(x)†, Φ(y)† = 0, (2.44)

Let us consider the expansion of the permanent (2.4) over e.g. the coordinate x1. One can easily show that it is given by r X na |n , n ,... i = Φ (x , x , . . . , x ) = φ (x )Φ (x , . . . , x ) 1 2 {n} 1 2 N N a 1 {n:na→na−1} 2 N a r X na = φ (x ) |n , n , . . . , n − 1,... i, (2.45) N a 1 1 2 a a √ where the na also enforces that the sum does not include empty states. By means of (2.45) we can write the matrix element of a single-particle operator like (2.18) as 0 0 ˆ 0 0 hn1, n2,... |V |n1, n2,... i = Nhn1, n2,... |V (x1)|n1, n2,... i Z X q 0 ∗ 0 0 0 = na nb dx1 φb(x1) V (x1) φa(x1) hn1, n2, . . . , nb − 1,... |n1, n2, . . . , na − 1,... i ab Z 0 0 † = hn1, n2,... | dxΦ(x) V (x) Φ(x) |n1, n2,... i, thus showing that Z Vˆ = dx Φ(x)† V (x) Φ(x). (2.46)

Analogously we can expand the permanent in two coordinates, e.g x1 and x2,

|n1, n2,... i = Φ{n}(x1, x2, . . . , xN ) = r r X na nb − δab φ (x )φ (x )Φ (x , . . . , x ), N N − 1 b 2 a 1 {n:na→na−1:nb→nb−1} 3 N ab and show that the average of a two-body interaction like (2.22) can be written as N(N − 1) hn0 , n0 ,... |Uˆ|n , n ,... i = hn0 , n0 ,... |U(x , x )|n , n ,... i 1 2 1 2 2 1 2 1 2 1 2 1 X q Z = n0 (n0 − δ )n (n − δ ) dx dx φ (x )∗φ (x )∗ U(x , x ) φ (x )φ (x ) 2 a b ab d c cd 1 2 a 1 b 2 1 1 c 2 d 1 abcd 0 0 0 0 hn1, n2, . . . , na − 1, . . . , nb − 1,... |n1, n2, . . . , nc − 1, . . . , nd − 1,... i. Therefore, like in the fermionic case, we can formally identify 1 Z Uˆ = dx dy Φ(x)†Φ(y)† U(x, y) Φ(y)Φ(x), (2.47) 2

58 as well as for the m-particle operators 1 Z Uˆ = dx . . . dx Φ(x )† ... Φ(x )† U(x , . . . , x ) Φ(x ) ... Φ(x ). (2.48) m m! 1 m 1 m 1 m m 1

2.4 Canonical transformations

In general an interacting Hamiltonian, which contains besides bilinear also quartic and higher order terms in creation and annihilation operators, can not be diagonalized. On the contrary, a bilinear Hamiltonian is diagonalizable by a canonical transformation.

A canonical transformation preserves the commutation/anti-commutation properties of the operators.

Since the interaction is commonly analysed perturbatively starting from an appropriate non- interacting theory, it is useful to begin with bilinear Hamiltonians and introduce the canonical transformations which diagonalize them. The simplest bilinear Hamiltonian is the second quantized expression of a non-interacting ﬁrst-quantization Hamiltonian, which has the general form

ˆ X † H = tab cacb, (2.49) ab both for fermions and bosons. Since Hˆ is hermitean, then

∗ tab = tba.

† If et is the hermitean matrix with elements tab, there exists a unitary transformation Ue, Ue Ue = Ie, with Ie the identity matrix, such that

X † Uαa tab Ubβ = α δαβ. (2.50) ab Therefore, if we deﬁne X ca = Uaα cα, (2.51) α then through (2.50) we ﬁnd

ˆ X X † † X † H = cαUαa tab Ubβcβ = αcαcα. (2.52) αβ ab α

59 We have now to check whether the above is a canonical transformation. If the cα’s are fermionic/bosonic operators then

† X † † X † {ca, cb}± = Uaα Uβb {cα, cβ}± = Uaα Uαb = δab, αβ α where {... }± stands for the anticommutator (+) and commutator (-), respectively. Therefore also the ca’s are fermions/bosons, hence Uˆ is indeed canonical. Once the Hamiltonian has been transformed into the diagonal form (2.52), the problem is solved. Indeed any Fock state constructed through the new basis set with operators cα, namely a wave-function |{nα}i where each state α is occupied by nα particles, is an eigenstate of the Hamiltonian " # X Hˆ |{nα}i = α nα |{nα}i. α

2.4.1 More general canonical transformations In second quantization we have the opportunity to introduce bilinear Hamiltonians which does not conserve the number of particles:

ˆ X † ∗ † † H = tab cacb + ∆ab cacb + ∆ab cbca. (2.53) ab

∗ These Hamiltonians are relevant for a wide class of physical problems. Again tab = tba while

∆ab = −∆ba for fermions, and for bosons ∆ab = ∆ba. The most general transformation which may diagonalize the Hamiltonian is of the form

X † ca = Uaαcα + Vaαcα, (2.54) α † X ∗ † ∗ ca = Uaαcα + Vaαcα. (2.55) α Let us derive the conditions under which the above is a canonical transformation:

† X † † † † {ca, cb}± = Uaα Uβb{cα, cβ}± + Vaα Vβb{cα, cβ}± αβ X † † = Uaα Uαb ± Vaα Vαb ≡ δab α

60 X † † {ca, cb}± = Uaα Vbβ{cα, cβ}± + Vaα Ubβ{cα, cβ}± αβ X = Uaα Vbα ± Vaα Ubα ≡ 0. α This implies that, for fermions

Ue Ue† + Ve Ve † = I,e Ue Ve T + Ve UeT = 0, (2.56) while, for bosons, Ue Ue† − Ve Ve † = I,e Ue Ve T − Ve UeT = 0. (2.57) † In addition, Ue and Ve should diagonalize (2.53), namely, once we write the original ca’s and ca’s in terms of the new operators, we should obtain

ˆ X † H = α cαcα. α We conclude by pointing out that, due to the particle-non-conserving terms, the vacuum of the original particles, |0i, is nomore the vacuum of the new ones, |e0i. Indeed

X † ca|e0i = Vaα cα|e0i= 6 0. α

61 2.5 Examples and Exercises

Let us consider electrons in a square box of linear length L with periodic boundary conditions. As a basis of single-particle wave-functions we use simple plane waves and spinors, namely the quantum label a = (k, s) with 2π k = (n , n , n ) , L x y z being ni’s integers, s = ±1/2 =↑, ↓. Hence the single-particle wavefunctions of the basis set are

1 ik·x φa(x) = √ e δs σ. L3 The annihilation and creation operators are deﬁned as

† ck σ, ck σ, hence the Fermi ﬁelds are 1 X Ψ (x) = √ eik·x c , σ 3 k σ L k 1 X Ψ† (x) = √ e−ik·x c† . σ 3 k σ L k Let us start from the kinetic energy, which is in ﬁrst quantization

N X 2 H = − ~ ∇2, kin 2m i i=1 diagonal in the spin coordinate. Since

1 Z 2 2 k2 dx e−ik·x − ~ ∇2 eip·x = ~ δ , L3 2m 2m k p then, in the plane-wave basis, the kinetic energy in second quantization becomes

X 2 k2 X H = ~ c† c ≡ c† c . (2.58) kin 2m k σ k σ k k σ k σ k σ k σ The ground state with N, assumed to be even, electrons is the Fermi sea |FSi obtained by ﬁlling with a spin-up and spin-down electrons the lowest energy levels up to the Fermi momentum kF deﬁned through X N = 2 ,

|k|≤kF

62 namely Y † † |FSi = ck ↑ck ↓|0i, (2.59) k:|k|≤kF hence the occupation number in momentum space is

nk ↑ = nk ↓ = θ (kF − |k|) , where the θ-function is deﬁned through θ(x) = 1 if x ≥ 0, otherwise θ(x) = 0. Let us add an electron-electron interaction of the general form

1 X H = V (x − x ). int 2 i j i6=j

In second quantization it becomes

1 X Z H = dx dy Ψ† (x)Ψ† (y) V (x − y)Ψ (y)Ψ (x) int 2 σ σ0 σ0 σ σσ0 1 1 Z X X † † −ik1·x −ik2·y ik3·y ik4·x = c c 0 c 0 c dx dy e e e e V (x − y). 3 k1 σ k2 σ k3 σ k4 σ 3 2L 0 L σσ ki,i=1,...,4 We deﬁne the Fourier transform of the potential as Z V (q) = dr e−iq·r V (r), so that 1 X V (x − y) = eiq·(x−y) V (q), L3 q and the interaction is ﬁnally found to be

1 X X H = V (q) c† c† c c . (2.60) int L3 k σ p+q σ0 p σ0 k+q σ σσ0 k p q

We showed that the electron density for spin σ in second quantization is

† ρσ(x) = Ψσ(x)Ψσ(x).

Its Fourier transform is Z −iq·x † X † ρσ(q) = dx e Ψσ(x)Ψσ(x) = ck σck+q σ. k

63 Let us consider interacting electrons also in the presence of a single-particle potential, provided e.g. by the ions, X Z Hpot = dx U(x) ρσ(x), σ so that the total Hamiltonian reads

H = Hkin + Hint + Hpot. Since the Hamiltonian conserves the total number of electrons, the electron density summed over the spins, i.e. ρ = ρ↑ + ρ↓ must satisfy a continuity equation. Let us define the Heisenberg evolution of the density through ρ(x, t) = eiHt ρ(x) e−iHt. Since the integral of the density over the whole volume is the total number of electrons, N, which is conserved, it follows that Z ∂ρ(x, t) ∂ Z ∂N dx = dx ρ(x, t) = ≡ 0. ∂t ∂t ∂t This condition is satisfied if ∂ρ(x, t) = −∇J(x, t), ∂t where J(x, t) is the current density operator. In fact, the integral over the volume of the left hand side is also equal to minus the flux of the current through the surface of the sample. If the number of electrons is conserved, it means that the flux of the current through the surface vanishes, which is the desired result. Exercises: • Calculate the formal expression of the average value of

H = Hkin + Hint, over the Fermi sea wave-function (2.59);

• Using the Heisenberg equation of motion of the operators, according to which ∂ρ(x, t) i = [ρ(x, t), H] , ~ ∂t calculate the expression of the Fourier transform of the current J(q);

• Calculate the following commutator [ρ(q, t), J(−q, t)] , (2.61) which is commonly known as the f-sum rule.

64 2.6 Application: fermionic lattice models and the emergence of magnetism

Let us consider the electron Hamiltonian in the presence of the periodic potential provided by the ions in a lattice:

N X p 2 X X 1 X H = i + V (x − R) + U(x − x ) 2m i 2 i j i=1 i R i6=j

= Hkin + Hel−ion + Hel−el = H0 + Hel−el, (2.62) where R are lattice vectors. We start by rewriting the Hamiltonian in second quantization. For that purpose we need to introduce a basis set of single-particle wavefunctions. Since the Hamilto- nian is spin independent, it is convenient to work with factorized single-particle wavefunctions: φ(x, σ) = φ(x) χσ. As a basis for the space-dependent φ(x) we use Wannier wave-functions φ(x)n,R satisfying the usual conditions Z ∗ X ∗ dx φ(x)n,R1 φ(x)m,R2 = δnm δR1R2 , φ(x)n,Rφ(y)n,R = δ(x − y), n,R as well as

φ(x)n,R+R0 = φ(x − R0)n,R . (2.63) † Consequently we associate to any wavefunction φ(x)n,R χσ creation, cn,R,σ, and annihilation, cn,R,σ, operators, and introduce the Fermi ﬁelds

X † † Ψσ(x) = φ(x)n,R cn,R,σ, Ψσ(x) = (Ψσ(x)) . n,R

Let us start by second quantization of the non interacting part of the Hamiltonian, H0 in Eq. (2.62): " # X Z 2∇2 X H = dx Ψ† (x) −~ + V (x − R) Ψ (x) 0 σ 2m σ σ R X X X = tnm c† c . (2.64) R1,R2 n,R1,σ m,R2,σ σ nm R1,R2

The matrix elements are " # Z 2∇2 X tnm = dx φ(x)∗ −~ + V (x − R) φ(x) , (2.65) R1,R2 n,R1 2m m,R2 R

65 and satisfy nm mn ∗ tR1,R2 = tR2,R1 . (2.66) By the property (2.63) it derives that " # Z 2∇2 X tnm = dx φ(x)∗ −~ + V (x − R) φ(x) R1+R0,R2+R0 n,R1+R0 2m m,R2+R0 R " # Z 2∇2 X = dx φ(x − R )∗ −~ + V (x − R) φ(x − R ) 0 n,R1 2m 0 m,R2 R " # Z 2∇2 X = dx φ(x)∗ −~ + V (x + R − R) φ(x) = tnm , n,R1 2m 0 m,R2 R1,R2 R since X X V (x + R0 − R) = V (x − R). R R Let us now introduce the operators in the reciprocal lattice through the canonical transformation

1 X −ik·R cn,k,σ = √ e cn,R,σ, (2.67) V R and its inverse 1 X ik·R cn,R,σ = √ e cn,k,σ, (2.68) V k where V is the number of lattice sites and k belongs to the reciprocal lattice, namely

X ik·R X −ik·R e = V δk0, e = V δR0. R k One can check that the above transformation is indeed canonical: 1 X {c , c† } = e−ik1·R1 eik2·R2 {c , c† } n,k1,σ1 m,k2,σ2 V n,R1,σ1 m,R2,σ2 R1,R2 1 X = δ δ e−i(k1−k2)·R1 = δ δ δ . nm σ1σ2 V nm σ1σ2 k1k2 R1

Let us substitute (2.68) into (2.64): 1 X X X X −ik1·R1 ik2·R2 nm † H0 = e e t c c . V R1,R2 n,k1,σ m,k2,σ σ nm R1,R2 k1,k2

66 We notice that

X −ik1·R1 ik2·R2 nm e e tR1,R2 R1,R2 1 X = e−ik1·R1 eik2·R2 tnm V R1−R0,R2−R0 R1,R2,R0 1 X = e−ik1·(R1+R0)eik2·(R2+R0) tnm V R1,R2 R1,R2,R0 X 1 X = e−ik1·R1 eik2·R2 tnm e−i(k1−k2)·R0 R1,R2 V R1,R2 R0

X −ik1·(R1−R2) nm nm = δk1k2 e tR1,R2 = V δk1k2 tk1 , R1,R2 where we deﬁne 1 X X tnm ≡ e−ik·(R1−R2) tnm = e−ik·R tnm . (2.69) k V R1,R2 R,0 R1,R2 R Since (2.66) holds it also follows that

nm ∗ X ik·R nm ∗ (tk ) = e tR,0 R X ik·R mn X ik·R mn = e t0,R = e t−R,0 R R X −ik·R mn mn = e tR,0 = tk , R

mn namely the matrix etk, with elements tk , is hermitean. The Hamiltonian is therefore

X X X nm † H0 = tk cn,k,σcm,k,σ. (2.70) σ nm k

As we said etk is hermitean, hence it is possible to write

† nm † etk = Ue(k) ek Ue(k) → tk = U (k)ni i,k U(k)im, with Ue(k)† Ue(k) = Ie. Therefore, upon the canonical transformation X ci,k,σ = U(k)incn,k,σ, n

67 the non interacting Hamiltonian gets a diagonal form

X X X † H0 = i,k ci,k,σci,k,σ. (2.71) σ i k

The index i identiﬁes the bands, and i,k is the energy dispersion in the reciprocal lattice: we have thus obtained the band structure. We can formally write 1 X −iR·k ci,k,σ = √ e ci,R,σ, V R thus introducing a new basis of Wannier functions. Since the Fermi ﬁeld is invariant upon the basis choice, then

X 1 X ik·R Ψσ(x) = φ(x)n,R χσ cn,R,σ = √ φ(x)n,R χσ e cn,k,σ R,n V R,n,k

1 X ik·R † = √ φ(x)n,R χσ e U (k)ni ci,k,σ V R,n,i,k

1 X ik·(R−R0) † X = φ(x) χ e U (k) c ≡ φ(x) 0 χ c , V n,R σ ni i,R0,σ i,R σ i,R0,σ R,R0,n,i,k R0 thus implying the following expression of the new Wannier functions

1 X ik·(R−R0) † φ(x) 0 = e U (k) φ(x) . (2.72) i,R V ni n,R n,R,k

Going back to the Hamiltonian in the diagonal basis, we can also rewrite it as

ˆ X X X † H0 = i,k ci,k,σci,k,σ σ i k 1 X X X X ik·(R1−R2) † = i,k e c c V i,R1,σ i,R2,σ σ i k R1,R2 X X X ≡ ti c† c , (2.73) R1,R2 i,R1,σ i,R2,σ σ i R1,R2 namely like a tight-binding Hamiltonian, diagonal in the band index. Once we know the band structure, the ground state of the non interacting Hamiltonian is simply obtained by ﬁlling all the lowest bands with the available electrons. If the highest occupied band is full, the model is a band insulator, otherwise is a metal. In particular, since each band can accomodate 2V electrons, V of spin up and V of spin down, a necessary condition for a band insulator is to have

68 an even number of available electrons per unit cell. This is not suﬃcient since the bands may overlap.

Exercises: • Calculate the formal expression of the Fourier transform of the density ρ(q) in the basis † of eigenoperators of H0, ci,k,σ and ci,k,σ; • Do the same calculation for the current density J(x), and discuss the meaning of the f-sume rule, Eq. (2.61) in this representation.

2.6.1 Hubbard models

In many physical situations, the Wannier orbitals φ(x)i,R’s in Eq. (2.72) for the highest occupied bands (valence bands) are quite delocalized, hence the lattice vector label R looses its physical meaning. In those cases, although formally exact, the tight-binding way (2.73) of rewriting the non interacting Hamiltonian is of little use. 3 However there is a wide class of materials, commonly called strongly correlated systems, where the tight-binding formalism is meaningful. There the valence bands derive from d or f orbitals of transition metals, rare earth or actinides, and the Wannier orbitals keep noticeable atomic character, thus leading to short range hopping elements ti ’s. Let us consider just the above situation and write down the left-over electron- R1,R2 electron interaction in the Wannier basis. We further assume that there is only one valence band well separated from lower and higher ones, so that we are allowed to neglect interband transition processes induced by the interaction. Since we take into account only one band, we are going to drop the band index i, so that the Wannier orbitals for the valence band are φR(x) and the tight-binding term projected on the same band is X X Hˆ = t c† c . (2.74) 0 R1,R2 R1σ R2σ σ R1,R2 The interaction term within the valence band is given by

1 X X † † Hˆint = UR ,R ;R ,R c c c c , (2.75) 2 1 2 3 4 R1σ1 R2σ2 R3σ2 R4σ1 σ1σ2 R1,R2,R3,R4 where Z ∗ ∗ UR1,R2;R3,R4 = dxdy φR1 (x) φR2 (y) U(x − y) φR3 (y)φR4 (x). (2.76)

Let us make use of our assumption of well localized Wannier orbitals, namely that φR(x) decays suﬃciently fast with |x − R|. Within this assumption, the leading matrix element is when all lattice sites are the same: UR,R;R,R ≡ U. (2.77)

3 i For instance the hopping matrix elements tR1,R2 turns out to be long ranged.

69 This term gives rise to an interaction U X Hˆ = n n , (2.78) U 2 R R R where the operator X † nR = cRσcRσ, σ counts the number of particles at site R. The interaction term (2.78), which simply describes an on-site Coulomb repulsion, plus the hopping (2.74) represent the so-called Hubbard model, which is the prototype of the strongly-correlated lattice models. Let us continue and consider in (2.76) two other cases: either (1) R1 = R4, R2 = R3 with R2 nearest neighbor of R1 or (2) R1 = R3, R2 = R4 still with R2 nearest neighbor of R1. In case (1) we obtain U1 X n n 0 , 2 R R where < RR0 > stands for nearest neighbor sites and

U1 = UR,R0;R0,R. In case (2) we ﬁnd

U2 X X U2 X X c† c† c c = − c† c c† c , 2 Rσ R0σ0 Rσ0 R0σ 2 Rσ Rσ0 R0σ0 R0σ σσ0 σσ0 where U2 = UR,R0;R,R0 . One can easily show that

X † † 1 c c c c = [n n 0 + 4S · S 0 ] , (2.79) Rσ Rσ0 R0σ0 R0σ 2 R R R R σσ0 where the spin density operator 1 X S = c† σ c , R 2 Rα αβ Rβ αβ being σ = (σ1, σ2, σ3), with σi’s the Pauli matrices. Therefore, upon deﬁning V = U1 − U2/2 and Jex = −2U2, the whole nearest neighbor interaction reads

V X Jex X Hˆ = n n 0 + S · S 0 . (2.80) n.n. 2 R R 2 R R

70 Since U1 > U2 > 0, the ﬁrst term describes a nearest-neighbor repulsion, while the second a spin exchange which tends to aligns the spin ferromagnetically since J < 0, so called direct exchange. Therefore, although we have started from a spin independent interaction, in the Wannier basis we have been able to identify a spin interaction, thus showing in a simple way how magnetism emerges out of the Coulomb repulsion.

2.6.2 The Mott insulator within the Hubbard model Let us summarize the approximate Hubbard Hamiltonian which we have so far derived, by further assuming a nearest neighbor hopping:

X X † U X V X Jex X Hˆ = −t c c + n n + n n 0 + S · S 0 Rσ R0σ 2 R R 2 R R 2 R R σ R

= Hˆ0 + Hˆint.

By construction U > V > 0 and Jex < 0. We consider the case in which the number of valence electrons is equal to the number of sites N. In the absence of interaction, the hopping forms a band which can accomodate 2N electrons: therefore the band is half-ﬁlled and the system is metallic.

Exercise: Consider the nearest neighbor tight-binding model

X X † H = −t cRσcR0σ, σ with t > 0 in a two-dimensional square lattice. The number of electrons is equal to the number of sites. Find the shape of the Fermi surface.

Let us analyse the opposite case of a very large U t, V, |Jex|. In this case we should start from the conﬁguration which minimizes the Coulomb repulsion U and treat what is left by perturbation theory. This lowest energy electronic conﬁguration is the one in which each site is singly occupied. Indeed the energy cost in having just an empty site and a doubly occupied one instead of two singly occupied sites is given by U E(2) + E(0) − 2E(1) = 2U + 0 − 2 = U, 2 and is much larger then the energy gain in letting the electrons move, which is of order t. In this situation the model describes an insulator, but of a particular kind. Namely the insulating state is driven by the strong correlation, while the conventional band structure argument would predict always a metal. This correlation-induced insulator is called a Mott insulator. Therefore, as the strength of U increases with respect to the bandwidth proportional to t, an interaction

71 driven metal-to-insulator transition, commonly referred as a Mott transition, should occur at a critical Uc. However, the configuration with one electron per site is hugely degenerate, since the electron can have either spin up or down. There are 2N degenerate states with one electron per site, which are going to be split by the left-over terms in the Hamiltonian. The nearest neighbor interaction V is not effective in splitting the degeneracy. On the contrary, the direct exchange does play a role and tends to prefer the configuration in which all spins are aligned – a ferromagnetic ordering. Yet, this is not the only source of spin correlations. Indeed also the hopping term is able to split the degeneracy within second order in perturbation theory. Let us focus on two states which only differs in the spin configurations of two nearest neighbor sites: |σ , σ i and |σ0 , σ0 i. Within second order in t we obtain a matrix element bewteen these R1 R2 R1 R2 two states given by

2 X X † 1 † 0 0 t hσR1 , σR2 | cR σcR σ + H.c. |ni hn| cR σ0 cR σ0 + H.c. |σR1 , σR2 i, 1 2 E0 − En 1 2 n σσ0 where E0 = NU/2 is the energy of the degenerate ground states with all singly occupied sites, and |ni an excited state with energy En. Since the hopping creates one empty and one doubly occupied site out of two singly occupied ones, E0 − En = −U. Moreover the sum over the intermediate states act as a complete sum, so that we can write also

2 t X † † † † 0 0 − hσR , σR | c c c 0 c 0 + c c c 0 c 0 |σ , σ i. U 1 2 R1σ R2σ R2σ R1σ R2σ R1σ R1σ R2σ R1 R2 σσ0 By means of (2.79) we therefore ﬁnd that the operator responsable of the additional splitting is

2 t X † † † † − c c c 0 c 0 + c c c 0 c 0 U R1σ R2σ R2σ R1σ R2σ R1σ R1σ R2σ σσ0 2 t X † † = 2 c c 0 c 0 c − (nR σ + nR σ) δσσ0 U R1σ R1σ R2σ R2σ 1 2 σσ0 4t2 1 = S · S + (n n − n − n ) . U R1 R2 4 R1 R2 R1 R2

Since the last term is a constant over states where each site is singly occupied, the relevant part is just 2t2 X S · S 0 , (2.81) U R R the missing factor two coming from the fact that in the sum over nearest neighbors each bond is counted twice. Therefore the second order perturbation theory in the hopping gives rise to a novel spin exchange which is antiferromagnetic, usually called super-exchange. All together, the

72 large U eﬀective Hamiltonian describes localized spin-1/2’s (each site occupied by one electron) coupled by the spin exchange J X Hˆ = S · S 0 , (2.82) Heis 2 R R where 4t2 4t2 J = J + = −2U + . ex U 2 U The spin exchange may be either ferromagnetic or antiferromagnetic depending on the strength of the direct- with respect to the super-exchange couplings. The eﬀective Hamiltonian (2.82) describes an Heisenberg model.

2.7 Spin wave theory in the Heisenberg model

We have previoulsy shown how magnetism may emerge in a single band Mott insulator where strong Coulomb repulsion localizes the electrons which become eﬀectively local moments. We have also shown that both the Coulomb repulsion itself, via the direct exchange, and the covalent binding, via the super-exchange proportional to the square of the hopping, induce a coupling among the local moments which has the form

ˆ X HHeis = JR,R0 SR · SR0 . (2.83) R,R0

Notice that, by lattice translational symmetry,

JR,R0 = JR0,R = JR0−R−R0,R−R−R0 = J−R,−R0 , namely the exchange is inversion symmetric. In the single band case we discussed, the spins have magnitude S = 1/2, but, in what follows, we will consider the general case of arbitrary S. As a result, the Mott insulator will likely be magnetically ordered according to the properties of the exchange terms. If the spins were classical vectors, then the classical ground state would show a magnetic order of the form

hSRi = S [u cos(Q · R) + v sin(Q · R)] , (2.84) with u and v two orthogonal unit vectors. Upon inserting this expression into (2.83), we ﬁnd a classical energy

X 0 X N 4 E(Q) = J 0 cos Q · (R − R ) = N J cos (Q · R) = (J + J ) = NJ , R,R R,0 2 Q −Q Q R,R0 R

4 ∗ Since, as we showed, the exchange is inversion symmetric, namely JR,0 = J−R,0, then JQ = JQ = J−Q

73 which is minimized by the value of Q for which the JQ is minimum. The classical ground state energy does not depend on u and v, reﬂecting the global spin-rotational symmetry. In the reciprocal lattice, (2.84) implies an order parameter X SN SN hS i = e−iq·R hS i = (u − iv) δ + (u + iv) δ . (2.85) q R 2 q,Q 2 q,−Q R In reality the spins are quantum operators satisfying the usual commutation relations

h a b i c SR,SR0 = iabc SR δRR0 , which, in the reciprocal lattice, transform into

h a b i c Sq,Sq0 = iabc Sq+q0 . Therefore, through the Heisenberg equation of motion

dSR h i X i = S , Hˆ = −i J 0 S × S 0 , ~ dt R Heis R,R R R R0 we find that dSQ X Jq i = −i S × S . ~ dt N q+Q −q q Since the right hand side is non zero, then the classical order parameter does not correspond quantum mechanically to a conserved quantity. In turns this implies that quantum fluctuations are going to reduce the magnitude of the classical order parameter. There is just one exception when the classical ordering wave-vector Q = 0, which corresponds to a ferromagnetic order. In this case the order parameter is just the total magnetization, which is indeed a conserved quantity. Therefore, if all J’s are negative so that the min(Jq) is at Q = 0, the quantum ground state remains the classical fully polarized ferromagnet. In all other cases when Q 6= 0, we do not know apriori if the classical ordered phase is going to survive quantum fluctuations. A quite simple but effective way to analyse the role of quantum fluctuations is by the so-called spin-wave theory.

We now show how this method works for the simple example of an antiferromagnetic Heisen- berg model with nearest neighbor exchange on an hypercubic lattice. The Hamiltonian is given by ˆ X HHeis = J SR · SR0 , (2.86) with J > 0. The classical ground state is the Ne`elantiferromagnet with order parameter (we choose to break the SU(2) symmetry along the z-axis)

z R hSRi = S cos(Q · R) = S (−1) ,

74 5 where Q = π(1, 1, 1,..., 1) and, if R = (n1, n2, . . . , nd) with d the dimension of the space, then Pd R = i=1 ni. Let us go back to the commutation relation

h a b i c Sq,Sq0 = iabc Sq+q0 , and assume that the left hand side is substituted by the classical average value, i.e.

h a b i Sq,Sq0 = iabz NSδQ,q+q0 . The meaning of this approximation is the following. If we want to analyse the role of quantum ﬂuctuations, we need to study the dynamics of the spin operators which, by the equations of motion, is related to commutators. If the true ground state is a slight modiﬁcation of the classical one, which has to be checked aposteriori, then the commutators can be approximated with their classical averages. Within this approximation

h x y i Sq,S−q+Q ' iNS. Yet we have to further impose that

x z y Sq,SQ = −iSq+Q, y z y Sq,SQ = iSq+Q,

Both conditions can be fulﬁlled if we write

x q SN † Sq = 2 dq + d−q , (2.87) y q SN † Sq = −i 2 dq−Q − d−q+Q ,

z P † SQ = SN − q dqdq.

† The operators dq’s and dq’s are bosonic and satisfy

h † i h i dq, dq0 = δq,q0 , dq, dq0 = 0.

z The expression for the order parameter operator SQ clariﬁes also the meaning of our starting assumption of a quantum ground state slightly diﬀerent from the classical one. Indeed this approximation holds if z hSQi ∼ SN,

5Notice that Q = −Q, being the two connected by a reciprocal lattice vector

75 namely if 1 X hd† d i S. N q q q Therefore the larger is the spin magnitude S the more correct is the approximation, i.e. the weaker are the eﬀects of quantum ﬂuctuations. The Hamiltonian in the reciprocal lattice is 1 X Hˆ = J(q) S · S , Heis N q −q q where d X J(q) = 2J cos(qi) ≡ 2 J γq, i=1 d being the dimension of the space and q = (q1, q2, . . . , qd). Notice that J(Q) = −2 J d.

Since d X J(q + Q) = 2J cos(qi + π) = −J(q), i=1 we can also rewrite the Hamiltonian, compatibly with our assumption of weak quantum ﬂuctu- ations, as 1 X Hˆ = J(q)(S · S − S · S ) Heis 2N q −q q+Q −q+Q q 1 X ' J(q) SxSx + SySy − Sx Sx − Sy Sy 2N q −q q −q q+Q −q+Q q+Q −q+Q q 1 + J(Q) Sz Sz , (2.88) N Q Q At this stage we should substitute (2.87) into (2.88), keep only bilinear terms and diagonalize the Hamiltonian by a canonical transformation. However it is much better to follow an equivalent route where calculations are simpler. We introduce conjugate variables through

r1 x = d + d† , (2.89) q 2 q −q r1 p = i d† − d ,, (2.90) q 2 −q q which satisfy the canonical commutation relation xq, pq0 = iδq,−q0 .

76 In terms of these variables the expressions for the spin operators read √ x Sq = SN xq, √ y Sq = SN pq−Q, 1 1 X Sz = S + N − x x + p p , Q 2 2 q −q q −q q which substituted into (2.88) and keeping up to order S terms, lead to S X Hˆ = J(q)(x x + p p − x x − p p ) Heis 2 q −q q−Q −q+Q q−Q −q+Q q −q q ! 1 X + J(Q) S2N 2 + SN 2 − SN x x + p p N q −q q −q q X = NJ(Q) S (S + 1) + S J(q)(xqx−q − pqp−q) q X −SJ(Q) (xqx−q + pqp−q) q = NJ(Q) S (S + 1) X +S (J(q) − J(Q)) xqx−q + (−J(q) − J(Q)) pqp−q. (2.91) q The constant term NJ(Q) S2 represents the classical Ne`elenergy, the rest being the correction due to quantum mechanics. In order to diagonalize this Hamiltonian, let us consider the following canonical transformation p xq = Kq Xq, (2.92) s 1 pq = Pq, (2.93) Kq which preserves the commutation relations provided Kq = K−q. Upon substitution, the quantum ﬂuctuation term transforms into X S (J(q) − J(Q)) xqx−q + (−J(q) − J(Q)) pqp−q → q X −1 → S Kq (J(q) − J(Q)) XqX−q + Kq (−J(q) − J(Q)) PqP−q. q If we impose that −J(Q) − J(q) K2 = , (2.94) q −J(Q) + J(q)

77 the Hamiltonian becomes that of independent harmonic oscillators, i.e.

1 X Hˆ = NJ(Q) S (S + 1) + ω (X X + P P ) , (2.95) Heis 2 q q −q q −q q with the dispersion relation deﬁned through

p q ωq = 2S (−J(Q) − J(q)) (−J(Q) + J(q)) = 4 SJ (d − γq)(d + γq). (2.96)

Notice that, since d ≥ |γq|, the frequencies are positive real numbers. It is clear that the Hamiltonian has a diagonal form. Indeed if we introduce new bosonic operators through

r1 X = b + b† , q 2 q −q r1 P = −i b − b† , q 2 q −q the Hamiltonian transforms into

X 1 Hˆ = NJ(Q) S (S + 1) + ω b† b + , (2.97) Heis q q q 2 q thus showing that the eigenstates are Fock states in the b-basis |{nq}i with energies " # X 1 Hˆ |{n }i = NJ(Q) S (S + 1) + ω n + |{n }i ≡ E [{n }] |{n }i. Heis q q q 2 q q q q

The elementary bosonic excitations are called spin-waves. The vacuum is the ground state, whose energy per site including quantum ﬂuctuations is

1 X 1 E = −2JdS2 − 2SJd − ω 0 N 2 q q 2JS X q = E − d − (d − γ )(d + γ ) < E , class N q q class q which, as expected, is lower than the classical counterpart Eclass. Let us analyse the energy spectrum of the spin-waves. First we notice that

γq = −γq−Q,

78 hence that ωq = ωq+Q, showing that the true Brillouin zone is half of the original one. This is because the Ne`elstate breaks translational symmetry and the new unit cell contains two sites. Secondly, by expanding (2.96) close to q = 0, which as we said is equivalent to q = Q, and since 1 γ ' d − q · q, q 2 we ﬁnd that q √ ωq = 4 SJ (d − γq)(d + γq) ' 4 SJ d |q| ≡ v |q|, √ namely a linearly vanishing dispersion with velocity v = 4 SJ d. This was actually predictable by the Goldstone theorem which states that when a continuos symmetry is broken, in our case the spin SU(2), and in the absence of long range interaction, there should exist a mode, the Goldstone mode, with vanishing energy as q → 0.

Let us go back to the canonical transformation (2.92) and (2.93). One can prove that this transformation is accomplished by the unitary operator " # i X Uˆ = exp − ln K x p , (2.98) 2 q q −q q namely that † p Uˆ xq Uˆ = Kq xq. (2.99) This result can be obtained through to the following expression ∞ X (−1)n e−A B eA = C , n! n n=0 where C0 = B and Cn = [A, Cn−1]. Exercise: Prove the above statement, namely that (2.98) gives rise to (2.99). Therefore the unitary transformation (2.98) applied to the Hamiltonian (2.91) gives 1 X Uˆ † Hˆ Uˆ = E + ω (x x + p p − 1) , Heis 0 2 q q −q q −q q namely the desired diagonal form. This way of rewriting the canonical transformation is quite useful since it allows to identify the eigenstates in a simple way. Let us consider the state

|{nq}i = Uˆ |{nq}i0, where |{nq}i0 is a Fock state in the old basis with the occupation number conﬁguration {nq}. Then

HˆHeis |{nq}i = HˆHeis Uˆ |{nq}i0

79 " # † X = Uˆ Uˆ HˆHeis Uˆ |{nq}i0 = Uˆ E0 + ωq nˆq |{nq}i0 q " # X = E0 + ωq nq Uˆ |{nq}i0 = E [{nq}] |{nq}i, q namely it is an eigenstate with energy " # X E [{nq}] = E0 + ωq nq . q Analogously, the new vacuum state can be obtained by the old one through

|0i = Uˆ |0i0.

Finally we have to check whether our approximation is valid. Let us evaluate the ground state average value of the order parameter 1 1 1 X M = h0|Sˆz |0i = S + − h0|x x + p p |0i N Q 2 2N q −q q −q q 1 1 X = S + − h0|K X X + K−1 P P |0i 2 2N q q −q q q −q q 1 1 X = S + − K + K−1 2 4N q q q s 1 1 X d + γq = S + − , (2.100) 2 2N d − γ q q where we used the fact that on the new vacuum 1 h0|X X |0i = h0|P P |0i = , q −q q −q 2 −1 and that Kq = Kq+Q. If S −M 1 then the approximation indeed holds. The only dangerous region in the sum is when d ∼ γq, i.e. when the spin-wave energy vanishes. Here the sum behaves as s Z 1 X d + γq dq ∼ . 2N d − γ |q| q q This integral is convergent for d ≥ 2, but is singular in one dimension. Therefore the spin wave approximation is always wrong in one dimension, in agreement with Mermin-Wagner theorem

80 which states that a continuos symmetry can not be broken in 1+1 dimensions (A quantum problem in d dimensions corresponds to a classical one in d + 1). To complete our analysis, let us evaluate the contribution of thermal fluctuations, namely let us move to finite temperature T . Instead of ground state averages, we need the thermal averages 1 1 ω hX X i = hP P i = (1 + 2b(ω )) = coth q , q −q q −q 2 q 2 2T where b(ωq) is the Bose distribution function 1 b(x) = . eβ x + 1 (Exercise: Derive the above result). Therefore, at finite temperature, the order parameter is s 1 1 X d + γq ωq M(T ) = S + − coth . (2.101) 2 2N d − γ 2T q q

We notice that in two-dimensions the contribution of thermal fluctuations is diverging since, for ωq T , ω 2T coth q ∼ , 2T ωq and the sum behaves like T Z dq , v |q|2 which is singular in two-dimensions. This implies that for d = only at T strictly zero there is a true Neèlorder. For dimensions d > 2, Eq. (2.101) allows to estimate the value of the critical temperature Tc above which the order parameter vanishes through the equation M(Tc) = 0. Since J is the only dimensional coupling, one easily realize that the Neèlcritical temperature Tc ∼ J.

2.7.1 More rigorous derivation: the Holstein-Primakoﬀ transformation Let us conclude by showing more rigorously how one can derive the spin wave theory. One can readily demonstrate that the following way of writing spin operators in terms of bosonic one (Holstein-Primakoﬀ transformation) preserves proper spin commutation relations:

z R h † i SR = (−1) S − dRdR , (2.102)

1 † p 1 p S+ = 1 − (−1)R d 2S − nˆ + 1 + (−1)R 2S − nˆ d , (2.103) R 2 R R 2 R R

81 1 † p 1 p S− = 1 + (−1)R d 2S − nˆ + 1 − (−1)R 2S − nˆ d , (2.104) R 2 R R 2 R R

† wheren ˆR = dRdR. Notice that the bosonic vacuum has the properties

z R + − h0|SR|0i = (−1) S, h0|SR|0i = h0|SR|0i = 0, namely it corresponds to the classical Ne`elstate. The square root operators in (2.103) and (2.104) z assure that it is not possible to create more than 2S bosons at any given site, so that hSRi = −S, . . . , S as it should. If one inserts the above expressions in the Heisenberg Hamiltonian and expands it up to order S, the eﬀective Hamiltonian (2.91) is recovered, after moving from bosonic operators to conjugate variables.

Exercises:

• Determine how the speciﬁc heat behaves at low energy for the antiferromagnetic Heisenberg Hamiltonian (2.86) in an hypercubic lattice in d ≥ 3 dimensions.

• Add to the antiferromagnetic Heisenberg Hamiltonian (2.86) with symmetry breaking along the z-direction a magnetic ﬁeld along the x-direction with Fourier components Bq, i.e. a term √ ˆ X x X δH = − B−q Sq ' − SN B−q xq. q q Diagonalize the Hamiltonian in the presence of this term and calculate the ground state energy.

• Calculate the spin wave spectrum of the Heisenberg ferromagnetic Hamiltonian X H = −J SR · SR0 ,

with J > 0 and where < RR0 > means sum over nearest neighbor sites on an hypercubic lattice in d dimensions.

82 Chapter 3

Linear Response Theory

In order to access the physical properties of a system, one has to act on it with some external probe. This amounts to add to the unperturbed Hamiltonian Hˆ0 a time-dependent perturbation of the general form Z Vˆ (t) = dx Aˆ(x) v(x, t), (3.1) where v(x, t) represents the external probe which couples to the hermitean operator Aˆ(x). Our scope is to study the eﬀects of Vˆ (t) on some measurable quantity described e.g. by an operator Bˆ(x), namely to calculate h i B(x, t) ≡ Tr ρˆ(t) Bˆ(x) , beingρ ˆ(t) the time-dependent density matrix in the presence of the perturbation.

3.1 Linear Response Functions

We assume that the perturbation is switched on at time t → −∞. Initially the system is in thermal equilibrium, so that the density matrix

83 where ∂ h i i |φ (t)i = Hˆ + Vˆ (t) |φ (t)i, (3.5) ~∂t n 0 n is the Shrœdinger equation which determines the evolution of the eigenstates of the unperturbed Hamiltonian in the presence of the perturbation. The meaning of (3.4) is that initially the system is described by a statistical ensamble of sub-systems, each in a given eigenstate of Hˆ0 and weigthed by the Boltzmann factor. After we switch on the perturbation, |φni ceases to be an eigenstate of the perturbed Hamiltonian, so it acquires a non-trivial time evolution. Through Eq. (3.5) one readily ﬁnds the equation of motion for the density matrix

∂ h i i ρˆ(t) = Hˆ + Vˆ (t), ρˆ(t) . (3.6) ~∂t 0 We introduce the Dirac, also called interaction, representation of the density matrix as

ˆ ˆ iH0t/~ −iH0t/~ ρˆD(t) = e ρˆ(t) e , which satisﬁes

∂ h i ˆ h i ˆ i ρˆ (t) = − Hˆ , ρˆ (t) + eiH0t/~ Hˆ + Vˆ (t), ρˆ(t) e−iH0t/~ ~∂t D 0 D 0 h i = VˆD(t), ρˆD(t) , (3.7) where Z ˆ ˆ Z iH0t/~ −iH0t/~ VˆD(t) = dx e Aˆ(x) e v(x, t) = dx Aˆ(x, t) v(x, t), being Aˆ(x, t) the Heisenberg evolution of Aˆ(x) with the unperturbed Hamiltonian. We solve (0) (1) (n) (3.7) perturbatively in v, i.e. ρD(t) = ρD (t) + ρD (t) + ... , where ρD (t) contains n-powers of the perturbation. Obviously (0) lim ρˆD(t) =ρ ˆ0 =ρ ˆ . t→−∞ D We will limit our analysis to the linear response, hence we just need the ﬁrst order term which satisﬁes ∂ h i i ρˆ(1)(t) = Vˆ (t), ρˆ(0)(t) , ~∂t D D D with solution Z t (1) i 0 h ˆ 0 i ρˆD (t) = − dt VD(t ), ρˆ0 . (3.8) ~ −∞ Therefore, at linear order,

h i h ˆ ˆ i iH0t/~ iH0t/~ B(x, t) = Tr ρˆ(t) Bˆ(x) = Tr ρˆD(t) e Bˆ(x) e

84 h i h i = Tr ρˆD(t) Bˆ(x, t) = Tr ρˆ0 Bˆ(x, t) h (1) ˆ i +Tr ρˆD (t) B(x, t) h (1) ˆ i = B(x)0 + Tr ρˆD (t) B(x, t) where we used the fact that h i h i Tr ρˆ0 Bˆ(x, t) = Tr ρˆ0 Bˆ(x) = B(x)0, is the unperturbed average value. We then ﬁnd that the variation of the average value is given by

Z t i 0 nh 0 i o B(x, t) − B(x)0 = − dt Tr VˆD(t ), ρˆ0 Bˆ(x, t) ~ −∞ Z t Z i 0 n h 0 io 0 = − dt dy Tr ρˆ0 Bˆ(x, t), Aˆ(y, t ) v(y, t ) ~ −∞ Z ∞ Z i 0 0 n h 0 io 0 = − dt dy θ(t − t )Tr ρˆ0 Bˆ(x, t), Aˆ(y, t ) v(y, t ) ~ −∞ Z ∞ Z 0 0 0 ≡ dt dy χBA(x, y; t − t ) v(y, t ), (3.9) −∞ with the linear response function deﬁned through

0 i 0 h 0 i χBA(x, y; t − t ) = − θ(t − t ) h Bˆ(x, t), Aˆ(y, t ) i, (3.10) ~ where h... i means a thermal and quantum average with the unperturbed density matrix and we recall that the operators evolve in time with Hˆ0. Eq. (3.9) shows that, at linear order, the variation of any measurable quantity is obtained through the linear response function (3.10) which is only related to averages on the unperturbed system. We conclude by noticing that any average of pairs of time-evolved operators in the Heisenberg representation only depends on the time-diﬀerence, since the Schrœdinger equation is time- translationally invariant. In fact

1 0 0 hBˆ(t) Aˆ(t0)i = Tr e−β H ei H t/~ Bˆ e−i H t/~ ei H t /~ Aˆ e−i H t /~ Z 1 0 0 = Tr e−β H ei H (t−t )/~ Bˆ e−i H (t−t )/~ Aˆ = hBˆ(t − t0) Aˆ(0)i Z 1 0 0 = Tr e−β H Bˆ ei H (t −t)/~ Aˆ e−i H (t −t)/~ = hBˆ(0) Aˆ(t0 − t)i. Z

85 3.2 Kramers-Kronig relations

Let us now study the analytical properties of the response function in the frequency domain. In the following we drop the space-coordinate dependence of the response function, which is not relevant for what we are going to demonstrate. The response of an operator Aˆ in the presence of an external probe which couples to Bˆ is therefore i h i χAB(t) = − θ(t) h Aˆ(t), Bˆ i, (3.11) ~ where ˆ ˆ i H0t −i H0t Aˆ(t) = e ~ Aˆ e ~ . The response function (3.11) vanishes for t < 0, which is a consequence of causality. We introduce the Fourier transform through Z ∞ dω −iωt χAB(t) = e χAB(ω), (3.12) −∞ 2π as well as its analytical continuation in the complex frequency plane χAB(z). If we assume, as it is always the case, that χAB(z) does not diverge exponentially for |z| → ∞, we can regard (3.12) as the result of a contour integral

I dz χ (t) = e−iztχ (z), AB 2π AB where the contour is in the upper half plane for t < 0 and in the lower for t > 0. The integral catches all poles lying inside the contour. Since χAB(t) = 0 for t < 0, it follows that

• as consequence of causality χAB(z) is analytic in the upper half plane. • Let us now consider the contour drawn in Fig. 3.1. Since there are no poles enclosed by the contour, it is trivially zero the integral

I χ (z) dz AB = 0. (3.13) C ω − z On the other hand the above integral is also equal to the line integral along the lower edge, hence Z ω− Z ∞ 0 I 0 χAB(ω ) χAB(z) 0 = + dω 0 + dz −∞ ω+ ω − ω z=ω+ exp(iθ);θ∈[π,0] ω − z Z ∞ 0 Z 0 0 χAB(ω ) iθ = P dω 0 − i dθ χAB ω + e , −∞ ω − ω π

86 Im(z)

2ε

Re(z) ω

Figure 3.1: Integration contour in Eq. (3.13). the symbol P denoting the principal value of the integral. In the limit → 0, the above expression simpliﬁes into

Z ∞ 0 0 χAB(ω ) P dω 0 + i π χAB(ω) = 0, (3.14) −∞ ω − ω which implies that

Z ∞ 0 0 dω Re χAB(ω ) Im χAB(ω) = P 0 , (3.15) −∞ π ω − ω Z ∞ 0 0 dω Im χAB(ω ) Re χAB(ω) = −P 0 , (3.16) −∞ π ω − ω known as the Kramers-Kronig relations. Therefore, because of causality, the real and imaginary parts of the response function are not independent. It is possible to rewrite both expressions as

Z ∞ 0 0 dω Im χAB(ω ) χAB(ω) = 0 , (3.17) −∞ π ω − ω − iη with η an inﬁnitesimal positive number.

3.2.1 Symmetries Let us introduce back the space dependence, so that

0 i 0 h 0 i χAB(x, y; t − t ) = − θ(t − t ) h Aˆ(x, t), Bˆ(y, t ) i. (3.18) ~

87 Since both operators are hermitean, it follows that

0 ∗ i 0 h 0 i 0 χAB(x, y; t − t ) = θ(t − t ) h Bˆ(y, t ), Aˆ(x, t) i = χAB(x, y; t − t ). (3.19) ~ By deﬁnition Z iωt χAB(x, y; ω) = dt e χAB(x, y; t), therefore, through (3.19), we ﬁnd that Z ∗ −iωt ∗ χAB(x, y; ω) = dt e χAB(x, y; t) = χAB(x, y; −ω).

This implies that 1 Re χ (x, y; ω) = [χ (x, y; ω) + χ (x, y; −ω)] , AB 2 AB AB is even in frequency, while 1 Im χ (x, y; ω) = [χ (x, y; ω) − χ (x, y; −ω)] , AB 2i AB AB is odd.

3.3 Fluctuation-Dissipation Theorem

Let us introduce other types of correlation functions. The ﬁrst are the so-called structure factors deﬁned through 1 SAB(x, y; t) = hAˆ(x, t) Bˆ(y)i. (3.20) ~ In addition we introduce the dissipation response function

00 1 h ˆ ˆ i 1 χAB(x, y; t) = h A(x, t), B(y) i = [SAB(x, y; t) − SBA(y, x; −t)] , (3.21) 2~ 2 whose meaning will be explained in the following section, as well as the ﬂuctuation one 1 n o F (x, y; t) = h Aˆ(x, t), Bˆ(y) i = ~ [S (x, y; t) + S (y, x; −t)] . (3.22) AB 2 2 AB BA One readily veriﬁes that the former is related to the response function through

00 i χ (x, y; t) = [χ (x, y; t) − χ (y, x; −t)] , AB 2 AB BA

88 which in the frequency domain reads

00 i χ (x, y; ω) = [χ (x, y; ω) − χ (y, x; −ω)] . (3.23) AB 2 AB BA In particular 00 χAA(x, x; ω) = −Im χAA(x, x; ω). (3.24) Through the deﬁnition (3.20) we ﬁnd that

1 ˆ ˆ t ˆ t −βH0 −iH0 ˆ iH0 ˆ SBA(y, x; −t) = Tr e e ~ B(y) e ~ A(x) Z0 1 (t−iβ~) (t−iβ~) −βHˆ0 iHˆ0 −iHˆ0 = Tr e e ~ Aˆ(x) e ~ Bˆ(y) Z0 = SAB(x, y; t − i~β).

Therefore Z Z −iωt iωt SBA(y, x; −ω) = dt e SBA(y, x; t) = dt e SBA(y, x; −t) Z iωt −β~ω = dt e SAB(x, y; t − i~β) = e SAB(x, y; ω), (3.25) namely

00 1 χ (x, y; ω) = S (x, y; ω) 1 − e−β~ω , (3.26) AB 2 AB F (x, y; ω) = ~S (x, y; ω) 1 + e−β~ω . (3.27) AB 2 AB In other words the following relation holds β ω 00 F (x, y; ω) = coth ~ χ (x, y; ω), (3.28) AB ~ 2 AB which is the so-called ﬂuctuation-dissipation theorem. Indeed, if Aˆ = Bˆ and x = y, FAA(x, x; t = 0) is an estimate of the ﬂuctuations of Aˆ. On the other hand

Z dω F (x, x; t = 0) = F (x, x; ω) (3.29) AA 2π AA Z dω β ω 00 = coth ~ χ (x, x; ω), (3.30) ~ 2π 2 AA which relates the ﬂuctuations to the dissipation.

89 3.4 Spectral Representation

The spectral representation of the response functions gives instructive information about their physical meaning. Let us start from the structure factor (3.20) which can be written as (in the following we do not explicitly indicate the space dependence)

1 X ˆ ˆ S (t) = e−βEn hn|eiHt/~ Aˆ e−iHt/~ Bˆ|ni AB Z ~ n 1 X = e−βEn ei(En−Em)t/~hn|Aˆ|mihm|Bˆ|ni. Z ~ nm After Fourier transformation we get 2π X S (ω) = e−βEn hn|Aˆ|mihm|Bˆ|ni δ ( ω + E − E ) . (3.31) AB Z ~ n m nm The meaning is now self-evident. The matrix element hm|Bˆ|ni is the transition amplitude for the excitation of the initial state |ni into the final one |mi induced by the operator Bˆ, while hn|Aˆ|mi decribes the reverse process but now induced by Aˆ. The excitation followed by relaxation process is weigthed by the Boltzmann factor for the initial state and contributes to SAB(ω) only if the energy difference Em − En is ~ω. Thus SAB(ω) is a spectral function which measures the transition amplitude for excitations induced by Bˆ and de-excitation induced by Aˆ with a given energy ~ω. Through (3.25) we also find that

00 π X χ (ω) = e−βEn 1 − e−β~ω hn|Aˆ|mihm|Bˆ|ni δ ( ω + E − E ) AB Z ~ n m nm π X = e−βEn − e−βEm hn|Aˆ|mihm|Bˆ|ni δ ( ω + E − E ) , (3.32) Z ~ n m nm 00 ˆ which means that χAB(ω) is the transition amplitude for |ni → |mi induced by B and |mi → |ni induced by Aˆ weighted by the occupation probability

e−βEn p = , n Z of the initial state |ni minus that one

e−βEm p = , m Z of the ﬁnal state |mi. We notice that

pn − pm = pn(1 − pm) − pm(1 − pn),

90 namely it is the probability of n being occupied and m empty, minus the opposite. In other 00 words, χAB(ω) measures the absorption minus the emission probability of an energy ~ω, namely the total absorption probability. Finally, one can analogously derive the spectral representation of the response function χAB, i 1 X χ (t) = − θ(t) e−βEn AB Z ~ nm h i ei(En−Em)t/~ hn|Aˆ|mihm|Bˆ|ni − e−i(En−Em)t/~ hn|Bˆ|mihm|Aˆ|ni i 1 X = − θ(t) hn|Aˆ|mihm|Bˆ|ni ei(En−Em)t/~ e−βEn − e−βEm . Z ~ nm For the Fourier transform one has to evaluate the integral Z ∞ −i dt eiωt ei(En−Em)t/~. 0 Since the perturbation has been assumed to be switched on at very early times, a meaningful regularization of the above integral is Z ∞ −i dt eiωt ei(En−Em)t/~ e−ηt/~ = ~ , 0 ~ω − (Em − En) + iη where η/~ is the switching rate of the perturbation, and is taken to be an inﬁnitesimal positive number. As a result we ﬁnd that

1 X e−βEn − e−βEm χ (ω) = hn|Aˆ|mihm|Bˆ|ni . (3.33) AB Z ω − (E − E ) + iη nm ~ m n

3.5 Power dissipation

Till now we have formally introduced several response functions. In this section and in the following ones we are going to show how those functions emerge in real experiments. Let us ﬁrst analyse the power dissipated in the presence of the perturbation. Given our starting assumption about the time-evolution of the density matrix (3.4), it is clear that the entropy deﬁned through the phase space occupied by the statistical ensamble remains constant and equal to the thermal equilibrium one, S0. Therefore the system free energy is

F (t) = U(t) − TS0 = (U(t) − U0) + F0, so that ∂F (t) ∂U(t) = . ∂t ∂t

91 On the other hand, since U(t) = Tr ρˆ(t) Hˆ0 , it follows that ∂U(t) i h i = − Tr Hˆ0 + Vˆ (t), ρˆ(t) Hˆ0 ∂t ~ i h i = − Tr ρˆ(t) Hˆ0, Hˆ0 + Vˆ (t) ~ i h i = − Tr ρˆ(t) Hˆ0, Vˆ (t) . (3.34) ~ We assume that the perturbation has the general form X Vˆ (t) = AˆJ vJ (t), J so that ∂U(t) i X h i i X h i = − v (t) Tr ρˆ(t) Hˆ , Aˆ = v (t) h Aˆ , Hˆ i(t), (3.35) ∂t J 0 J J J 0 ~ J ~ J where the last term denotes the average value of the operator within the bracket in the presence of the time-dependent perturbation. We further notice that Z ˆ X 0 0 0 AJ (t) = Tr ρˆ(t) AJ = dt χJJ0 (t − t ) vJ0 (t ), J0 where 0 i 0 h ˆ ˆ 0 i χJJ0 (t − t ) = − θ(t − t ) h AJ (t), AJ0 (t ) i. ~ Therefore, by recalling that the operators evolve with the unperturbed Hamiltonian, we ﬁnd that Z Z X 0 ∂ 0 0 X 0 0 h i 0 i dt χ 0 (t − t ) v 0 (t ) = dt δ(t − t )h Aˆ (t), Aˆ 0 (t) i v 0 (t ) ~ ∂t JJ J J J J J0 J0 Z X 0 i 0 hh ˆ ˆ i ˆ 0 i 0 − dt θ(t − t ) h AJ (t), H0 , AJ0 (t ) i vJ0 (t ) J0 ~ X h ˆ ˆ i h ˆ ˆ i = h AJ (t), AJ0 (t) i vJ0 (t) + h AJ (t), H0 i. J0 After inserting into (3.35) we ﬁnally get Z ∂U(t) X 0 ∂ 0 0 = − dt χ 0 (t − t ) v (t) v 0 (t ) ∂t ∂t JJ J J J,J0

92 i X h ˆ ˆ i − h AJ (t), AJ0 (t) i vJ (t) vJ0 (t). ~ J,J0

0 The last term vanishes since the commutator is odd by interchanging J with J , while vJ (t) vJ0 (t) is even. Hence Z ∂U(t) X 0 ∂ 0 0 = − dt χ 0 (t − t ) v (t) v 0 (t ). (3.36) ∂t ∂t JJ J J J,J0 Let us write 1 v (t) = v e−iωt + v∗ eiωt , (3.37) J 2 J J and deﬁne the power dissipated within a cycle, W , through

ω Z 2π/ω ∂U(t) W = dt . (3.38) 2π 0 ∂t By performing the integral and by means of (3.36) and (3.37) we obtain for W the expression:

ω X ∗ ω X ∗ 00 W = i v v 0 [χ 0 (ω) − χ 0 (−ω)] = v χ 0 (ω) v 0 , (3.39) 4 J J JJ J J 2 J JJ J J,J0 J,J0 where Eq. (3.23) has been used. The power dissipated during a cycle is proportional to what we deﬁned as the dissipation response function, thus explaining its name. Indeed, as we showed in 00 the previous section, χJJ0 (ω) measures the probability of energy absorption during the process, hence its appearance in Eq. (3.39) it is not unexpected. We notice that, since W > 0, it derives 00 that ω χJJ0 (ω) is a positive-deﬁnite quadratic form. In particular

00 ω χJJ (ω) = −ω Im χJJ (ω) > 0, namely the imaginary part of χJJ (ω) is positive for ω < 0 and negative otherwise.

3.5.1 Absorption/Emission Processes The power dissipation is related to the absorption minus emission probability. However, there are other measurements where only absorption or emission is revealed. For instance one can shot on a sample with a beam of particles, either photons, neutrons, electrons etc..., and measure the absorption probability of an energy ~ω. If the coupling between the beam and the sample is represented by an operator Aˆ, the Fermi golden rule tells us that the absorption rate per unit time of an ensamble at thermal equilibrium is 2π X 2 P (ω) = e−βEi hf|Aˆ|ii δ ( ω + E − E ) = S (ω), (3.40) A Z ~ i f AA if which enlights the meaning of the structure factors.

93 3.5.2 Thermodynamic Susceptibilities Let us consider again a perturbation of the form X Z Vˆ (t) = dx AˆJ (x) vJ (x, t). J

Let us further assume that the only time-dependence of the external probes vJ (x, t) comes from a very slow switching rate (adiabatic switching) which just sets the proper regularization of time-integrals as in Eq. (3.33). Hence, for times far away from the time at which the perturbation is switched on, the external probes become constant in time, vJ (x, t) = vJ (x), and the thermodynamic averages lose any time-dependence. In this limit Z ∞ Z X 0 0 AJ (x) − AJ (x)0 = dt dy χJJ0 (x, y; t − t ) vJ0 (y) J0 −∞ X Z = dy χJJ0 (x, y; ω = 0) vJ0 (y). (3.41) J0 Let us now consider a generically perturbed Hamiltonian of the form X Z Hˆ = Hˆ0 + dy AˆJ (y) vJ (y), J which therefore admits stationary eigenstates. The perturbed free-energy turns out to be a functional of the external fields vJ (y), F = F [vJ (y)]. Standard thermodynamics tells us that δF hAˆJ (x)i = , δvJ (x) which is in general different from its average value, hAˆJ (x)i0, in the absence of external fields. For very small external fields one finds at linear order that

Z 2 ˆ ˆ X δ F hAJ (x)i − hAJ (x)i0 = dy vJ0 (y), (3.42) δvJ (x) δvJ0 (y) J0 v=0 where the second derivatives of the unperturbed free-energy are the so-called thermodynamic susceptibilities. Comparing (3.42) with (3.41) one obtains that

δ2F = χJJ0 (x, y; ω = 0), (3.43) δvJ (x) δvJ0 (y) v=0 which relates thermodynamic susceptibilities to the response functions at zero frequency. Notice that thermodynamic stability implies that −χJJ0 (x, y; ω = 0) is positive deﬁnite.

94 Exercises

(1) Consider a free electron Hamiltonian

X † H0 = k ckσckσ, kσ in the presence of a Zeeman splitting due to a slowly varying magnetic ﬁeld oriented along the z-direction B(q, t). The perturbation is 1 V (q, t) = µ g B(q, t) σ , V B −q

where V is the volume, µB the Bohr magneton, g ' 2 the electron gyromagnetic ratio and σq the spin density operator at momentum q deﬁned through

1 X σ = c† c − c† c . q 2 k↑ k+q↑ k↓ k+q↓ k

By linear response theory, the average of the spin-density operator at momentum q is Z 0 0 0 hσqi(t) = dt χ(q, t − t ) gµB B(q, t ) ,

where 0 i 1 0 χ(q, t − t ) = − θ(t − t )h[σq(t), σ−q]i, ~ V is the magnetic response function per unit volume. (Prove that for any p 6= q h[σp(t), σ−q]i = 0, so that only the Fourier component q is aﬀected by the magnetic ﬁeld.)

• Find the formal expression of Z χ(q, ω) = dt eiωt χ(q, t).

• Calculate the magnetic susceptibility, namely

χ = lim χ(q, ω = 0), q→0

at zero temperature in terms of the density of states

1 X N() = δ( − ). V k k

95 T I µ L R

Figure 3.2: Geometry of the tunneling problem (2).

• Prove and discuss that χ(q = 0, ω) = 0.

(2) Consider two disconnected metallic leads, a right one (R) and a left one (L), see Fig. 3.2, described by the unperturbed Hamiltonian

X † † H0 = k cR kσcR kσ + cL kσcL kσ . kσ At some time, a voltage bias V is applied, implying that the two leads are kept at diﬀerent chemical potential, i.e. that the following perturbation is added to the Hamiltonian µ X µ V = c† c − c† c = (N − N ) , µ 2 R kσ R kσ L kσ L kσ 2 R L kσ with µ = eV . At the same time a tunnelling between the two leads is switched on, which is described by the additional perturbation

X † VT = − T (k, p) cR kσcL pσ + H.c., kσ

where we further assume that the tunneling amplitudes T (k, p) depends only on the energy. Therefore the fully perturbed Hamiltonian is

H = H0 + Vµ + VT . (3.44)

• Using the Heisenberg equation of motion with the Hamiltonian (3.44), calculate the expression of the operator of the current ﬂowing from the L to the R lead, which is deﬁned through ∂ i I = (NR − NL) = − [(NR − NL) , H] . ∂t ~

96 • Extending the derivation of the linear response up to second order in the perturbation, calculate the average value of the current I = hIi. (One needs the second order (2) density matrixρ ˆ in the perturbation Vµ + VT , and in particular the mixed term which derives from Vµ VT .) For the explicit calculation it is convenient to introduce the density of states N(), which is by deﬁnition equal for the L and R leads.

97 Chapter 4

Hartree-Fock Approximation

In this chapter we describe the simplest approximate technique to study a model of interacting fermions: the Hartree-Fock approximation. This technique is variational and can be applied both at zero and at finite temperature. Essentially within the Hartree-Fock approximation the effects of the particle-particle interaction are simulated by an effective external field acting on the particles which is self-consistently generated by the same particles. This is also the reason why the Hartree-Fock approximation is a Mean Field theory. The main body is devoted to the Hartree-Fock approximation for interacting fermions. In the last Section the case of bosons is briefly discussed in connection with superfluidity.

4.1 Hartree-Fock Approximation for Fermions at Zero Temper- ature

The Hartree-Fock (HF) approximation at zero temperature consists in

• searching for a Slater determinant which minimizes the total energy.

Since in general the ground state wavefunction is not a single Slater determinant, the HF approach is variational hence the HF energy is an upper bound to the true ground state energy. The trial HF wavefunction is a Slater determinant, namely

φ1(x1) ··· φ1(xN )

1 φ2(x1) ··· φ2(xN ) Φ (x , . . . , x ) = √ , (4.1) HF 1 N . . . . . N! . . . . .

φN (x1) ··· φN (xN ) where N is the electron number and the sigle particle wavefunctions φi(x) are for the meanwhile

98 unknown. They are assumed to belong to a complete set of orthonormal wavefunctions Z ∗ dx φi(x) φj(x) = δij. (4.2)

The Hamiltonian in ﬁrst quantization is given by

N X 1 X Hˆ = T (x , p ) + U(x , x ), i i 2 i j i=1 i6=j where 2 T (x, p) = − ~ ∇2 + V (x) 2m is a single particle contribution including the kinetic energy as well as a potential term. The average value of the Hamiltonian over the wavefunction (4.1) is

hΦHF |Hˆ |ΦHF i EHF = . (4.3) hΦHF |ΦHF i

One has to impose that the variation of EHF vanishes upon varying the trial wavefunction, keeping it still of the form of a Slater determinant. Most generally, this amounts to change one of the single particle wavefunctions, namely

φi(x) → Ni (φi(x) + δφi(x)) . (4.4)

Clearly, if the variation δφi(x) has a ﬁnite overlap with anyone of the φk(x)’s already present in (4.1), the Slater determinant does not change, hence such a variation is irrelevant. Therefore the only meaningful possibility is that

δφi(x) = η φj(x), (4.5) where η is an inﬁnitesimal quantity and φj belongs to the set of wavefunctions (4.2) but does not appear in (4.1), namely j > N. The normalization Ni in Eq. (4.4) is given by Z −2 ∗ ∗ ∗ 2 Ni = dx (φi(x) + η φj(x) )(φi(x) + ηφj(x)) = 1 + |η| .

Therefore Ni ' 1 at linear order in η. This implies that also the normalization of the Slater determinant ΦHF + ηδΦHF remains one at linear order in η, hence

∗ δEHF = η hδΦHF |Hˆ |ΦHF i + ηhΦHF |Hˆ |δΦHF i

= Re η RehδΦHF |Hˆ |ΦHF i + Im η ImhδΦHF |Hˆ |ΦHF i = 0.

99 Since η is an arbitrary inﬁnitesimal number, this implies that

X 1 X Hˆ = t c†c + U c†c†c c , (4.7) ij i j 2 ijkl i j k l ij ijkl where the parameters are the matrix elements over the basis set (4.2). What we need to solve is therefore the equation † ˆ hΦHF |ci cjH|ΦHF i = 0, where i ≤ N and j > N. For that we need the following two equalities, which can be easily derived:

X † † tklhΦHF |ci cjckcl |ΦHF i = tji kl 1 X U hΦ |c†c c† c†c c |Φ i 2 klmn HF i j k l m n HF klmn N 1 X = U + U − U − U 2 jmmi mjim jmim mjmi m=1 N X = Ujmmi − Ujmim, m=1 where the last expression comes from the symmetry relation

Uijkl = Ujilk.

100 Eq. (4.6) implies that N X tji + Ujmmi − Ujmim = 0, (4.8) m=1 if j > N and i ≤ N. In other words, the Slater determinant which minimizes the total energy is constructed by single particle wavefunctions φi’s which have matrix elements obeying (4.8). Let us suppose that we have found instead a set of wavefunctions φi’s satisfying

N X tji + Ujmmi − Ujmim = i δij. (4.9) m=1 This set authomatically satisﬁes also (4.8), hence it does solve our variational problem. In ﬁrst quantization Eq.(4.9) reads

N Z X ∗ ∗ T (x, p) φi(x) + dy U(x, y)(φm(y) φm(y)φi(x) − φm(y) φm(x)φi(y)) = iφi(x), (4.10) m=1 which is the standard Hartree-Fock set of equations. Notice that there might me several Slater determinants built up using N of the wavefunctions solving (4.9) which would satisfy the HF variational principle. Among them one has to ﬁnd the Slater determinant which makes minimum the total energy for N electrons, which can be easily found to be N N X 1 X E (N) = hΦ |Hˆ |Φ i = t + U − U HF HF HF ii 2 immi imim i=1 i,m=1 N N X 1 X = − U − U . (4.11) i 2 immi imim i=1 i,m=1 Notice that, if for N − 1 particles the Hartree-Fock single-particle wavefunctions stay approximately invariant, then

N−1 N−1 X 1 X E (N − 1) ' t + U − U HF ii 2 immi imim i=1 i,m=1 N N X 1 X = t (1 − δ ) + (U − U ) (1 − δ ) (1 − δ ) ii iN 2 immi imim iN mN i=1 i,m=1

= EHF (N) − N , (4.12) showing that the Hartree-Fock single particle energies correspond approximately to the ioniza- tion energies.

101 4.1.1 Alternative approach The Hartree-Fock equations (4.10) are complicated non-linear integral-differential equations. This is especially true if one does not impose any constraint on the form of the variational Slater determinant dictated for instance by the symmetry properties of the Hamiltonian, what is called unrestricted Hartree-Fock approximation. However, very often one expects that the true ground state has well defined properties under symmetry transformations which leave the Hamiltonian invariant. For instance, if the Hamiltonian is translationally and spin-rotationally invariant, one may expect that the true ground state is an eigenstate of the total spin and of the total momentum. For this reason one would like to search for a variational wavefunction within the subspace of Slater determinants which are eigenstates of the total spin and momentum. This amounts to impose symmetry constraints on the general form (4.1) of the variational wavefunction. Yet, this is not a simple task if one keeps working within first quantization. The second quantization approach to the Hartree-Fock approximation that we describe in the following has the big advantage to allow an easy implementation of such symmetry constraints. Let us suppose we have our Hamiltonian written in a basis of single particle wavefunctions {φα(x)} which is more convenient to work with (for instance Block waves) X 1 X Hˆ = t c† c + U c† c† c c . (4.13) αβ α β 2 αβγδ α β γ δ αβ αβγδ

Our scope is to find the basis {φi(x)} which solves the Hartree-Fock equations (4.9). The Hartree-Fock wavefunction (4.1), being a Slater determinant, should be the ground state of a single-particle Hamiltonian, which we define as HˆHF . We write such an Hamiltonian in the following general form ˆ X † HHF = hαβ cαcβ, (4.14) αβ ∗ where we have introduced a set of unknown variational parameters satisfying hαβ = hβα, for the Hamiltonian to be hermitean. The ignorance about the basis set {φi(x)} is reflected in the ignorance about the h’s. The Hartree-Fock wavefunction satisfies:

HˆHF | ΦHF i = E | ΦHF i, (4.15) namely is the lowest energy eigenstate of HˆHF . We deﬁne as

† ∆αβ = hΦHF | cαcβ | ΦHF i, (4.16) the average values of all one-body operators on the wavefunction, which are therefore functional of the variational parameters h. Since X E = hΦHF | HˆHF | ΦHF i = hαβ ∆αβ, (4.17) αβ

102 it follows that ∂E X ∂∆αβ = ∆µν + hαβ . (4.18) ∂hµν ∂hµν αβ

On the other hand, since | ΦHF i is an eigestate of the Hamiltonian HˆHF , the Hellmann-Feynman theorem holds1 that states

∂E ∂HˆHF = hΦHF | =| ΦHF i = ∆µν. (4.19) ∂hµν ∂hµν

By comparing (4.18) with (4.19) we therefore obtain that

X ∂∆αβ hαβ = 0, (4.20) ∂hµν αβ for any hµν. On the other hand, the average value of the original Hamiltonian (4.13) on the Hartee-Fock wavefunction is readily found to be

EHF = hΦHF | Hˆ | ΦHF i X 1 X = t ∆ + ∆ ∆ U − U , (4.21) αβ αβ 2 αβ γδ αγδβ αγβδ αβ αβγδ which is also a functional of the h’s. Minimization of EHF requires as a necessary condition that ∂EHF X ∂∆αβ X = tαβ + ∆γδ Uαγδβ − Uαγβδ = 0. (4.22) ∂hµν ∂hµν αβ γδ

1 The Hellmann-Feynman theorem states that, if | ψi is the normalized eigenstate with eigenvalue E of a Hamiltonian Hˆ that depends on some parameter λ, then

∂E ∂Hˆ = hψ | | ψi. ∂λ ∂λ This theorem can be easily proved. Indeed

103 The Hartree-Fock wavefunction, hence the variational parameters hµν, must therefore satisfy both (4.20) and (4.22), which implies that X hαβ = tαβ + ∆γδ[h] Uαγδβ − Uαγβδ . (4.23) γδ

Since the average values ∆γδ[h] are themselves functional of the h’s, the above set of equations is actually a self-consistency condition for determining the latter variational parameters. These equations are fully equivalent to solving the Hartree-Fock non-linear diﬀerential equations that are found in ﬁrst quantization with one major advantage. Indeed, the symmetry properties of | ΦHF i are completely determined by the parameters hαβ. In turns, this implies that we can implement any desired symmetry operation Oˆ by imposing that h i O,ˆ HˆHF = 0, which leads to simple conditions to be imposed on the hαβ. For instance, if the quantum number α = (a, σ) include an orbital index a and spin σ, and we would like to enforce full spin-rotational symmetry, then we must assume that

h(aσ)(bσ0) = hab, does not depend on the spin indices. If we expect that spin SU(2) is lowered down to U(1), namely the rotation around the magnetization axis, which we can always assume to be also the quantization axis, then we have to impose that, while h(aσ)(bσ) are ﬁnite, h(a↑)(b↓) and h(a↓)(b↑) are identically zero.

We note that (4.23) could be also regarded as the deﬁnition of hαβ in terms of new variational parameters ∆γβ that, only at the end, must be determined in such a way that

† ∆αβ = hΦHF | cαcβ | ΦHF i, (4.24) the average being done on a choosen eigenstate of ˆ X † X HHF = cαcβ tαβ + ∆γδ Uαγδβ − Uαγβδ , (4.25) αβ γδ which depends parametrically on the ∆’s. Moreover it follows that the variational energy h i EHF ∆ = hΦHF | Hˆ | ΦHF i 1 X = hΦ | Hˆ | Φ i − ∆ ∆ U − U HF HF HF 2 αβ γδ αγδβ αγβδ αβγδ

104 h i 1 X = E h∆ − ∆ ∆ U − U , (4.26) 2 αβ γδ αγδβ αγβδ αβγδ

also becomes functional of the ∆’s. Through Eqs. (4.19) and (4.23) it follows that

∂E X ∂EHF ∂hαβ = ∂∆γδ ∂hαβ ∂∆γδ αβ X = ∆αβ Uαγδβ − Uαγβδ . (4.27) αβ

The equations (4.23), (4.25), (4.26) and (4.27) suggest two alternative ways of stating the Hartee-Fock variational problem, which are very convenient to implement:

Given the interacting Hamiltonian

X † 1 X † † Hˆ = tαβ c c + Uαβγδ c c c c α β 2 α β γ δ αβ αβγδ

then the Hartree-Fock variational wavefunction |ΦHF i is the ground state with energy E of the single particle Hamiltonian

ˆ X † X † HHF = tαβ cαcβ + cαcβ∆γδ (Uαγδβ − Uαγβδ) αβ αβγδ

where the parameters ∆αβ have to be determined self-consistently by imposing either of the following two conditions

Condition 1.: † ∆αβ = hΦHF |cαcβ|ΦHF i

Condition 2.: δ 1 X E[∆] − − ∆αβ ∆γδ Uαγδβ − Uαγβδ = 0 δ∆ 2 µν αβγδ

105 4.2 Hartree-Fock approximation for fermions at ﬁnite temperature

A variational approach of the Hartree-Fock type can also be implemented at ﬁnite temperature as a variational minimization of the free energy. Before presenting the method, it is necessary to introduce some preliminary results.

4.2.1 Preliminaries Let us consider a generic Hamiltonian

Hˆ = Hˆ0 + V,ˆ where Hˆ0 is the unperturbed Hamiltonian while Vˆ is the perturbation, e.g. the interaction, which does not commute with Hˆ0. The partition function is deﬁned through

ˆ Z = e−βF = Tr e−βH . (4.28)

We assume that the free energy can be expanded in powers of the perturbation, namely X F = F (n), n=0 where F (n) contains n-powers of Vˆ . Let us deﬁne for real τ’s ˆ ˆ e−Hτ = e−H0τ Sˆ(τ), (4.29) implying that ˆ ˆ Sˆ(τ) = eH0τ e−Hτ . From this expression it follows that

∂Sˆ ˆ ˆ = eH0τ Hˆ − Hˆ e−Hτ ∂τ 0 ˆ ˆ ˆ ˆ = −eH0τ Vˆ e−H0τ eH0τ e−Hτ = −Vˆ (τ)Sˆ(τ), (4.30) where we deﬁne the unperturbed evolution (Dirac or interaction representation) in imaginary time as ˆ ˆ Vˆ (τ) = eH0τ Vˆ e−H0τ . The equation of motion satisﬁed by the S-operator allows a very simple perturbative expan- (0) sion. The 0-th order Sˆ = 1 since at this order Hˆ = Hˆ0. Morover, the boundary condition at

106 τ = 0 is Sˆ(0) = 1, implying that all Sˆ(n)(τ = 0) = 0 for n > 0, since already S(0) = 1 and, for any n and τ, Sˆ(n)(τ) > 0. The ﬁrst order term satisﬁes

∂Sˆ(1) = −Vˆ (τ) Sˆ(0) = −Vˆ (τ). ∂τ The solution with the appropriate boundary condition is Z τ (1) Sˆ (τ) = − dτ1 Vˆ (τ1). (4.31) 0 For a generic n, the equation of motion reads

∂Sˆ(n) = −Vˆ (τ) Sˆ(n−1), ∂τ whose solution can be iteratively found starting from the expression of Sˆ(1), and reads

Z τ Z τn Z τ2 (n) n Sˆ (τ) = (−1) dτn dτn−1 ... dτ1 Vˆ (τn) Vˆ (τn−1) ... Vˆ (τ1). (4.32) 0 0 0 We can rewrite this term as 1/n! times the sum of all the terms which are obtained by permuting the n-indices. This allows us to formally write

n (−1)n Z τ Y Sˆ(n)(τ) = dτ T Vˆ (τ )Vˆ (τ ) ... Vˆ (τ ) , (4.33) n! i τ 1 2 n 0 i=1 where we have introduced the so-called time-ordered product of operators, Tτ , which is deﬁned as follows:

• the time-ordered product Tτ of several operators at diﬀerent (imaginary) times is the product where the operators are ordered in such a way that those at later times appear on the left of those at earlier times. If some of those operators are fermionic-like, namely contain odd number of fermionic operators, the result has to be multiplied by (−1)P , where P is the number of permutations needed to bring the original sequence of only the fermionic operators into the time-ordered one.

For instance the time-ordered product of two operators Aˆ1(τ1) and Aˆ2(τ2) is Tτ Aˆ(τ1)Aˆ(τ2) = θ (τ1 − τ2) Aˆ(τ1)Aˆ(τ2) ± θ (τ2 − τ1) Aˆ(τ2)Aˆ(τ1), where the − sign has to be used if both operators are fermionic-like.

107 Having introduced the time-ordered product, we can write the general expression for the S-operator as − R τ dτ 0 Vˆ (τ 0) Sˆ(τ) = Tτ e 0 . (4.34) Coming back to the partition function, it can be rewritten as

ˆ ˆ Z = Tr e−βH = Tr e−βH0 Sˆ(β) 1 −βHˆ0 = Z0 Tr e Sˆ(β) = Z0hSˆ(β)i0, (4.35) Z0 where (0) −βHˆ0 −βF Z0 = Tr e = e , is the unperturbed partition function and h...i0 means a quantum and thermal average with the uncorrelated Boltzmann weight, namely 1 −βHˆ0 h...i0 ≡ Tr e ... . Z0 In a perturbation expansion

X (n) X (n) (1) (2) hSˆ(β)i0 = hSˆ (β)i0 = S = 1 + S + S + .... n=0 n=0 Therefore (0) 1 2 Z = e−βF 1 − βF (1) + β2 F (1) − βF (2) + ... 2 (0) h i = e−βF 1 + S(1) + S(2) + ... , which leads to 1 1 F (1) = − S(1) = − hSˆ(1)(β)i (4.36) β β 0 1 1 2 1 1 2 F (2) = − S(2) + β F (1) = − hSˆ(2)(β)i + hSˆ(1)(β)i . (4.37) β 2 β 0 2β 0

Let us ﬁrst consider (4.36). We need to calculate

1 Z β (1) −βHˆ0 τHˆ0 −τHˆ0 hSˆ (β)i0 = − Tr e dτ e Vˆ e Z0 0 1 Z β −βHˆ0 = − Tr e dτ Vˆ = −βhVˆ i0. (4.38) Z0 0

108 Hence (1) F = hVˆ i0, (4.39) is the thermal average of the perturbation. From the expression of Sˆ(2)(β) one can easily show that

Z β (2) 1 h i F = − dτ1 dτ2hTτ Vˆ (τ1) − hVˆ i0 Vˆ (τ2) − hVˆ i0 i0. (4.40) 2β 0

Going up in perturbation theory, one ﬁnds that the generic F (n) is obtained by a cumulant expansion in powers of Vˆ , namely an expansion in

Vˆ (τ) − hVˆ i0.

We want now to show that F (2) < 0, which is the ﬁnite temperature analogoue of the known result that second-order perturbation theory always decreases the ground state energy. Let us consider an hermitean operator Aˆ(τ) such that hAˆi0 = 0, a property shared also by Vˆ (τ)−hVˆ i0. Then

Z β Z τ1 h i −βHˆ0 dτ1 dτ2 Tr e Aˆ(τ1)Aˆ(τ2) 0 0 Z β Z τ1 2 X −βn (n−m)(τ1−τ2) ˆ = dτ1 dτ2 e e hn|A|mi n,m 0 0 X 2 e−βm − e−βn e−βn = hn|Aˆ|mi − β . ( − )2 − n,m n m n m

2 ˆ Since the matrix element hn|A|mi is even by interchanging n ↔ m, the ﬁrst term in the square brackets, being odd, vanishes. The second term gives, after symmetrizing the sum P P n,m f(n, m) = 1/2 n,m f(n, m) + f(m, n),

1 X 2 e−βm − e−βn β hn|Aˆ|mi , 2 − n,m n m which is always greater than zero. By this result, one can write

1 X 2 e−βm − e−βn F (2) = − hn| Vˆ − hVˆ i |mi , 2Z 0 − 0 n,m n m hence the desired result F (2) ≤ 0.

109 How do we use this property? Let us consider the perturbed Hamiltonian

Hˆ (λ) ≡ Hˆ0 + λV.ˆ

Then F (λ) = F (0) + λF (1) + λ2F (2) + ..., where, as we showed, the curvature F (2) ≤ 0. Therefore in the interval λ ∈ [0, 1], the function F (λ) has a downwards curvature. This implies that in this interval

F (λ) ≤ F (0) + λF (1), leading to the variational principle

1 ˆ 1 ˆ F = F (λ = 1) ≤ F (0) + F (1) = − ln Tr e−βH0 + Tr e−βH0 Vˆ . (4.41) ˆ β Tr e−βH0

4.2.2 Variational approach at T 6= 0 Let us apply this ﬁnite temperature variational principle to our interacting problem. The interacting Hamiltonian is, as usual,

X 1 X Hˆ = t c† c + U c† c† c c . αβ α β 2 αβγδ α β γ δ αβ αβγδ

We deﬁne an unperturbed Hamiltonian

ˆ X † H0 = hαβ cαcβ, αβ which contains variational parameters hαβ [the similarity with (4.14) is not accidental, as we will show], so that

1 X X Vˆ ≡ Hˆ − Hˆ = U c† c† c c + t − h c† c . (4.42) 0 2 αβγδ α β γ δ αβ αβ α β αβγδ αβ

By the variational principle (4.41), we can get an upper bound to the true free energy by choosing the variational parameters hαβ which minimize the right hand side of (4.41). This is what we intend to do in the following. Since Hˆ0 is quadratic in the fermionic ﬁelds, it can be diagonalized by some unitary transformation X cα = Uαici , i

110 leading to ˆ X † H0 = i ci ci . i

Any many-body eigenstate of Hˆ0, e.g. |ni, with energy En, can be written as a single Slater determinant. The following property then holds 1 † † X † † −βHˆ0 † † hcαcβcγcδi0 = Uiα Ujβ Uγk Uδl Tr e ci cjckcl Z0 ijkl

1 X † † X −βEn † † = Uiα Ujβ Uγk Uδl e hn|ci cjckcl |ni Z0 ijkl n

1 X † † X −βEn † † = Uiα Ujβ Uγk Uδl e (δilδjk − δikδjl) hn|ci ci |nihn|cjcj|ni Z0 ijkl n X † † = Uiα Ujβ Uγk Uδl (δilδjk − δikδjl) fi(T )fj(T ) ijkl † † † † = hcαcδi0hcβcγi0 − hcαcγi0hcβcδi0 ≡ ∆αδ(T )∆βγ(T ) − ∆αγ(T )∆βδ(T ), (4.43) where −βHˆ0 Y −βi Z0 = Tr e = 1 + e , i is the partition function of Hˆ0, 1 fi(T ) = , 1 + eβi is the Fermi distribution function and we have introduced the thermal averages

† ∆αβ(T ) = hcαcβi0. The previous result shows in a particular case the following general result

• the thermal average of a product of n creation and n annihilation operators with a bilinear Hamiltonian Hˆ0, namely

1 ˆ hc† c† . . . c† c c . . . c i = Tr e−βH0 c† c† . . . c† c c . . . c i1 i2 in j1 j2 jn 0 i1 i2 in j1 j2 jn Z0 X P † † † = (−1) hc c i0hc c i0 ... hc c i0, (4.44) i1 jP1 i2 jP2 in jPn P

where (jP1 , jP2 , . . . , jPn ) = P (j1, j2, . . . , jn) is a permutation over the n j-indices and P is the order of the permutation. The most general case where the operators are not ordered in such a way that creation operators appear on the left can be straightforwardly derived

111 from the above result. Essentially one has to consider all possible contractions between a † creation, ci , and an annihilation, cj, operator. If the former is on the left of the latter, the contraction means † hci cji0, otherwise it means † hcjci i0. The sign of a given product of contractions is plus or minus depending on the number of fermionic hops one has to do to bring each pair of operators to be contracted close together.

Coming back to our original scope, we need to solve the equation

∂F (0) ∂F (1) + = 0. ∂hαβ ∂hαβ Let us ﬁrst consider the ﬁrst term on the left-hand side. We note that

∂F (0) T ∂Z = − 0 . (4.45) ∂hαβ Z0 ∂hαβ

On the other hand, if En are the eigenvalues of the many-body eigenstates | ψni of Hˆ0, then

−βEn ∂Z0 X ∂e = ∂h ∂h αβ n αβ ˆ X ∂En X ∂H0 = −β e−βEn = −β e−βEn hψ | | ψ i ∂h n ∂h n n αβ n αβ

X −βEn † = −β e hψn | cαcβ | ψni = −β Z0 ∆αβ(T ), (4.46) n where we made use of the Hellmann-Feynman theorem previously discussed. Therefore, through (4.45) and (4.46) we find that ∂F (0) = ∆αβ(T ), (4.47) ∂hαβ a simple extension of the Hellmann-Feynman theorem at finite temperature. By means of (4.43) we find that F (1) defined through (4.39) and (4.42) is given by

1 X X F (1) = hVˆ i = ∆ (T )∆ (T )(U − U ) + t − h ∆ (T ), 0 2 αβ γδ αγδβ αγβδ αβ αβ αβ αβγδ αβ

112 hence

(1) ∂F X ∂∆γδ X ∂∆γδ = −∆αβ(T ) + tγδ − hγδ + ∆µν (Uγµνδ − Uγµδν) . (4.48) ∂hαβ ∂hαβ ∂hαβ γδ γδµν

The sum of (4.47) and (4.48) gives therefore " # X ∂∆γδ X hγδ − tγδ − ∆µν (Uγµνδ − Uγµδν) = 0, (4.49) ∂hαβ γδ µν which should hold for every pair of indices (αβ). The solution of this equation reads X hγδ = tγδ + ∆µν(T )(Uγµνδ − Uγµδν) µν 1 X ˆ = t + (U − U ) Tr e−βH0 c† c . (4.50) γδ Z γµνδ γµδν µ ν 0 µν

(0) (1) Therefore the single particle Hamiltonian Hˆ0 which minimizes F + F , hence providing an upper bound to the exact free energy, is defined through variational parameters hαβ which have to be determined self-consistently through Eq. (4.50). Notice that this equation is just an extension of the Hartree-Fock Hamiltonian (4.14) at finite temperature, where the variational parameters have to correspond now to thermal averages and nomore to ground state averages. Finally notice that the Hartree-Fock Hamiltonian Hˆ0 represents non-interacting electrons in the presence of a fictitious external field which, through (4.50), is indeed the self-consistent field generated by the same electrons. This is the reason why the Hartree-Fock approximation is also called Mean-Field approximation.

113 4.3 Mean-Field approximation for bosons and superﬂuidity

The na¨ıve Hartree-Fock approximation for interacting bosons, in the spirit of what we did for fermions, might be searching for the permanent of single-particle wavefunctions which has the lowest average value of the Hamiltonian. This is however not so easy in general because a permanent, unlike a Slater determinant, does not vanish if the single particle wavefunctions, out of which it is constructed, overlap each other. In addition, bosonic Hamiltonian fall into two different classes. The first includes models where the number of bosons is not a conserved quantity, e.g. phonons in solids. In this case a permanent, which is by definition a wavefunction for a fixed number of particles, is a very poor approximation for the actual ground state. The second class includes models where the number of bosons is conserved, as for instance models for liquid 4He. Here however one has to face another problem, namely Bose condensation, and a single permanent is not expected to be a good approximation, too. In other words there is not a unique prescription for a simple and accurate mean-field approximation for bosons. In what follows we discuss an Hamiltonian for bosonic particles in the second class, i.e. which conserves the number of bosons, in the context of Bose superfluidity.

Let us consider the Hamiltonian for N bosonic particles X 1 X H = b† b + U(q) b† b† b b , (4.51) q q q 2V p k+q k p+q q q p k where the single particle energy 2 q2 = ~ . q 2m In the absence of interaction, the ground state is simply obtained by putting, i.e. condense, all N particles into the q = 0 state. The interaction is going to mix this state with excited states, by moving particles from q = 0 to q 6= 0. First of all let us rewrite the interaction term as 1 X U(0) X U(q) b† b† b b = b† b† b b 2V p k+q k p+q 2V p k k p q p k p k 1 X X + U(q) b† b† b b 2V p k+q k p+q q6=0 p k U(0) 1 X X = N (N − 1) + U(q) b† b† b b , 2V 2V p k+q k p+q q6=0 p k which shows that the q = 0 component of the interaction is just a constant since the number of bosons is conserved. Therefore, discarding this constant, the Hamiltonian can be written as X 1 X X H = b† b + U(q) b† b† b b , (4.52) q q q 2V p k+q k p+q q q6=0 p k

114 We will assume that the interaction is weak enough that the actual ground state is a slight modiﬁcation of the state where all bosons condense into the state q = 0. Let us consider the following unitary transformation acting on the q = 0 bosons:

hp † i U = exp N0 b0 − b0 . It follows that † p † † † p U b0 U = b0 + N0, U b0 U = b0 + N0. Therefore, if we consider a so-called coherent state

|e0i ≡ U |0i, then

† he0| b0b0 |e0i = N0, † † he0| b0b0 |e0i = he0| b0b0 |e0i = N0.

In other words |e0i contains an average number N0 of bosons at q = 0, but, at the meantime, it † is not eigenstate of b0b0. On the other hand, the number ﬂuctuation behaves as 2 2 † ∆N = he0| b0b0 − N0 |e0i = N0, hence, for large N0, ∆N/N0 → 0. These properties suggest that |e0i might be a good representation of a state in which bosons condense into the zero-momentum state but ﬂuctuations of charge from q = 0 to q 6= 0 are still possible. Therefore, let us assume as a variational wavefunction |Ψi = |e0i |Φq6=0i, where |Φq6=0i includes non-zero momentum bosons, supplemented by the constraint

† X † X † hΨ| b0b0 + bqbq |Ψi = N0 + h bqbq i = N. (4.53) q6=0 q6=0 As we pointed out, such a variational wavefunction should be a faithful representation of the ground state provided N − N0 N, (4.54) which we are going to assume in what follows and check a posteriori. The average of the Hamiltonian over |e0i leads to an eﬀective Hamiltonian for the q 6= 0 bosons which reads

X N0 X N0 X H = b† b + U(q) b† b + U(q) b† b† + b b eff q q q V q q 2V q −q −q q q6=0 q6=0 q6=0

115 √ N0 X 1 XX + U(q) b† b† b + b† b b + U(q) b† b† b b . V p q p+q p+q q p 2V p k+q k p+q q p p+q6=0 q p k6=0

Notice that the ﬁrst two interaction terms are proportional to N0 (N − N0) while the last two 2 can be shown to be of order (N − N0) . Therefore, in view of the assumption (4.54), we are going to start from the Hamiltonian

X N0 X H = b† b + U(q) b† b 0 q q q V q q q6=0 q6=0 N0 X + U(q) b† b† + b b , (4.55) 2V q −q −q q q6=0 and eventually treat as a perturbation √ N0 X H = U(q) b† b† b + b† b b int V p q p+q p+q q p q p p+q6=0 1 XX + U(q) b† b† b b . (4.56) 2V p k+q k p+q q p k6=0 Let us introduce conjugate variables through

r1 x = b + b† , q 2 q −q r1 p = −i b − b† , q 2 q −q which satisfy [xp, p−k] = i δp,k. In terms of these variables (4.55) becomes X 1 1 H = + n U(q) x x + p p − 1 + n U(q) x x − p p , (4.57) 0 2 q 0 q −q q −q 2 0 q −q q −q q6=0 where n0 = N0/V , which can be diagonalized by the canonical transformation s p 1 xq → Kq xq, pq → pq, Kq with 2 q Kq = , (4.58) q + 2n0U(q)

116 leading to 1 X 1 X H = − + n U(q) + ω x x + p p , (4.59) 0 2 q 0 2 q q −q q −q q6=0 q6=0 with the frequency r ωq = q q + 2n0U(q) . (4.60)

For small q, if U(q → 0) is ﬁnite, r n U(0) ω → q 0 = s q, q ~ m thus showing that the long wavelength excitations describe acoustic phonons, which are actually

117 the Goldstone modes since the wave-function explicitly breaks gauge invariance. 2

Let us go back to the Hamiltonian (4.59). We notice that † 1 1 1 hbqbqi = hxqx−q + pqp−q − 1i = Kq + − 2 . 2 4 Kq

Therefore the constraint (4.53) implies the following self-consistency equation for n0

N 1 X 1 = n0 + Kq + − 2 . V 4V Kq q6=0

Since Kq ∼ q for small q’s, this self-consistency equation can be always solved but in one dimension, where the summation diverges. Once more this tells us that a continuos symmetry,

2Indeed the original Hamiltonian has a gauge symmetry, namely is invariant upon the transformation

iφ † −iφ † bq → e bq, bq → e bq, generated by the unitary operator ! X † Tφ = exp iφ bq bq . q

As usual, this implies that, given a generic normalized eigenstate |Ψni with eigenvalue En, the state Tφ |Ψni is 0 also a normalized eigenstate with the same eigenvalue. There are two possible cases: either Tφ |Ψni is, for all φ s, the same |Ψni apart from a phase factor, namely

|hΨn| Tφ |Ψni| = 1, or it is diﬀerent, i.e. |hΨn| Tφ |Ψni| < 1.

The former case means that |Ψni is invariant under a gauge transformation, while the latter case implies that it is not, hence that the eigenvalue En is degenerate. Let us suppose that our variational wave-function is indeed a good approximation of the actual ground state |Ψ0i, which is therefore also characterized by a ﬁnite value of

−iφ hΨ0| b0 |Ψ0i = e hΨ0| b0 |Ψ0i.

This equality, being true for any φ, would necessarily imply that hΨ0| b0 |Ψ0i = 0. Therefore, in order for this average value to be ﬁnite, the ground state has to be degenerate and generically not gauge invariant. Notice that in principle, given one of the degenerate ground states, say |Ψ0i, one can always construct a gauge invariant combination by Z |Ψe 0i = dφ Tφ |Ψ0i. Yet, all other orthogonal combinations will remain not gauge invariant.

118 as the gauge symmetry is, can not be broken spontaneously in one dimension. Notice in addition that the summation vanishes when U(q) → 0, namely when Kq → 1, so that, for suﬃciently small interaction, indeed N − N0 N, thus justifying our approximation.

4.3.1 Superﬂuid properties of the gauge symmetry breaking wavefunction Let us uncover now some interesting properties of the variational wavefunction. By deﬁnition the Fourier transform of the current operator is

X q J = ~ k + b† b . (4.61) q m 2 k k+q k In the spirit of our approximation, the leading term of the current is obtained when either k = 0 or k + q = 0, leading to

p q † p q J ' N ~ b − b = i 2N ~ p . (4.62) q 0 2m q −q 0 2m q

In other words the current is purely longitudinal, i.e. q ∧ Jq = 0. Let us check whether (4.62) is compatible, within the same scheme, with the continuity equation ∂ρ i q = [ρ , H ] = q · J . ~ ∂t q 0 q The density operator at leading order is

X † p † p ρq = bkbk+q ' N0 bq + b−q = 2N0 xq, k and one readily veriﬁes that its commutation with the Hamiltonian H0 indeed reproduces q ·Jq, with the current given by (4.62).

Non-Classical Rotational Inertia What is the consequence of a purely longitudinal current? Let us suppose that our system of N bosons with mass m is enclosed in a cylindrical annulus, initially at rest, with internal radious R and thickness d, with d/R 1. At a given time the cylinder is made moving around its axis, assumed to be the z-axis, with constant angular velocity ω. A normal liquid would be dragged by the walls of the cylinder and rotate at the same angular velocity ω. Neglecting the mass of the cylinder, the energy change at equilibrium due to rotation would simply be at leading order in ω 1 E(ω) − E(0) = I ω2, 2 0

119 2 where the classical moment of inertia I0 ' mNR if d R. Let us consider what happens in our Bose-condensed ﬂuid. The Schrœdinger equation describing the system reads

2 N dΨ(x1, x2,..., xN ; t) − X i = ~ ∇2 Ψ(x , x ,..., x ; t) ~ dt 2m i 1 2 N i=1 1 X + U(x − x ) Ψ(x , x ,..., x ; t) 2 i j 1 2 N i6=j N X + V (xi; t) Ψ(x1, x2,..., xN ; t), i=1 where V (x; t) represents the coupling of a boson with the rotating walls of the cylinder, hence it is explicitly time-dependent. We use cylidrical coordinates, so that the position of particle i is

xi = (r cos θ, r sin θ, z) ≡ ri + zi z, where z is the unit vector in the z-direction. Let us consider the following transformation on the planar coordinate ρi(t) = cos ωt ri + sin ωt ri ∧ z, (4.63) which corresponds to a reference frame in which the cylinder is at rest. In this frame

V (x; t) = V (r, z; t) = V (ρ(t), z) , namely the explicit time-dependence of the interaction with the wall is being transformed into an implicit one via the time-dependence of ρ(t). We rewrite the many-body wavefunction in terms of the new variables, i.e.

Ψ ((r1, z1), (r2, z2),..., (rN , zN ); t) = Ψ ((ρ1, z1), (ρ2, z2),..., (ρN , zN ); t) . In this representation the wavefunction depends both explicitly upon time and also implicitly, via the coordinates ρi’s. Therefore dΨ ((r , z ), (r , z ),..., (r , z ); t) ∂Ψ ((ρ , z ), (ρ , z ),..., (ρ , z ); t) i 1 1 2 2 N N = i 1 1 2 2 N N ~ dt ~ ∂t N X ∂ρ +i i · ∇ Ψ ((ρ , z ), (ρ , z ),..., (ρ , z ); t) ~ ∂t i 1 1 2 2 N N i=1 ∂Ψ ((ρ , z ), (ρ , z ),..., (ρ , z ); t) = i 1 1 2 2 N N ~ ∂t N X +i~ω ρi ∧ z · ∇iΨ ((ρ1, z1), (ρ2, z2),..., (ρN , zN ); t) i=1

120 ∂Ψ ((ρ , z ), (ρ , z ),..., (ρ , z ); t) = i 1 1 2 2 N N ~ ∂t N X −mω ρi ∧ z · J iΨ ((ρ1, z1), (ρ2, z2),..., (ρN , zN ); t) , i=1 where ∇i is assumed to act on the new coordinate basis and by deﬁnition the current operator is i J = − ~ ∇ . i m i One can readily verify that the Laplacian is invariat upon the transformation (4.63), hence the Schrœdinger equation becomes

N ∂Ψ X 2 i = − ~ ∇2 + mωρ ∧ z · J Ψ ~ ∂t 2m i i i i=1 N 1 X X + U(ρ − ρ , z − z ) Ψ + V (ρ , z )Ψ. (4.64) 2 i j i i i i i6j i=1

Unlike in the original representation, this equation is not explicitly time-dependent, hence it admits stationary solutions

iE(w)t/~ Ψ ((ρ1, z1), (ρ2, z2),..., (ρN , zN ); t) = e Ψ ((ρ1, z1), (ρ2, z2),..., (ρN , zN )) .

The minimum energy E(ω) corresponds to the equilibrium condition.3 It is important to recognize that the actual average value of the Hamiltonian is not E(ω) but4 Z Y d E(ω) = dr dz Ψ ((r , z ), (r , z ),..., (r , z ); t)∗ i Ψ ((r , z ), (r , z ),..., (r , z ); t) i i 1 1 2 2 N N ~dt 1 1 2 2 N N i   Z Y ∂ X = dρ dz Ψ ((ρ , z ), (ρ , z ),..., (ρ , z ); t)∗ i − mω ρ ∧ z · J i i 1 1 2 2 N N  ~∂t j j i j

Ψ ((ρ1, z1), (ρ2, z2),..., (ρN , zN ); t)   Z Y ∗ X = dρi dzi Ψ ((ρ1, z1), (ρ2, z2),..., (ρN , zN )) E(ω) − mω ρj ∧ z · Jj i j

3 Notice that the equilibrium condition corresponds to the minimum energy in a reference frame in which the cylinder is at rest. Indeed, it is only in such a reference frame that the walls of the cylinder can stop transfering energy to the liquid. 4The jacobian of the transformation (4.63) is unity.

121 Ψ ((ρ1, z1), (ρ2, z2),..., (ρN , zN )) X = E(ω) − mω hρi ∧ z · Jii. (4.65) i Therefore, in order to evaluate E(ω) we ﬁrst need to study the model described by the Hamiltonian in second quantization Z H = dx Ψ(x)† − ~ ∇2 + A(x) · J + V (x) Ψ(x) 2m 1 Z + dxdy Ψ(x)†Ψ(y)† U(x − y) Ψ(y) Ψ(x) , 2 with A(x) = mω x ∧ z = mω (y, −x, 0), (4.66) playing the role of a vector potential coupled to the current operator

J = −i ~ ∇. m We notice that ∇ · A(x) = 0, namely the vector potential is purely transverse. If we assume that the liquid is homogeneous, then, as we previously proved, the leading component of the current operator is purely longitudinal, hence the net eﬀect of the transverse perturbation to the system is negligible. Therefore, within our approximation, E(ω) ' E(0) does not depend upon ω. Analogoulsy

hJ · Ai ' 0, implying that also E(ω) ' E(0) is ω-independent. In other words, the actual moment of inertia is zero, i.e. the quantum liquid can not be made moving by rotating the cylinder, so called Non-Classical Rotational Inertia (NCRI). In reality the situation may be more complex. Indeed, if the frequency of rotation exceed a critical value ωc1, the lowest energy conﬁguration ceases to be homogeneous, as we assumed so far, because line defects, so-called vortex lines, appear. It is however not our scope here to discuss vortices.

Superfluid flow Let us consider another hypotetical experiment in which the bosonic fluid is made moving inside a capillary with constant velocity v parallel to the walls of the capillary. A normal fluid, due to the friction, will be slowed down by the walls. Let us consider instead what happens with our

122 Bose-condensed fluid. In the reference frame in which the fluid is at rest, hence the capillary moves, we may expect that energy is transferred from the moving walls to the fluid leading to creation of excitations. Let us assume that an excitation of momentum q and energy ωq is present. In the original reference frame in which the capillary is at rest, the actual energy is, by a Galilean tranformation, Nmv2 E(q) = ω + v · q + . q 2 As we said the equilibrium condition corresponds to the minimum of E(q), which is obtained by a momentum antiparallel to v, i.e. q · v = −q v. In this case

Nmv2 E(q) = ω − v q + . q 2 It is clear that it is advantageous to create excitations only if ω v ≥ q . (4.67) q This is the famous Landau criterium for superfluidity. It tells us that the fluid can flow without resistance through a capillary, so-called superfluidity, for velocities below a critical one vc defined by ωq vc = lim . q→0 q

From our previous analysis it derives that vc = s, the velocity of the acoustic phonons in the liquid. We notice that the interaction among the particles is essential for superﬂuidity. Indeed 2 2 an ideal non-interacting Bose gas with dispersion ~ q /2m has

2 2 1 ~ q vc = lim = 0, q→0 q 2m hence it is not superﬂuid.

123 4.4 Time-dependent Hartree-Fock approximation for fermions

Let us move back to the Hartree-Fock approximation for fermions, and suppose to have solved the time-independent Hartree-Fock approximation, namely to have found a set of single particle wave-functions φi(x)’s which diagonalize

N X tij + Uikkj − Uikjk = i δij. (4.68) k=1 The Hartree-Fock wave-function is the Slater determinant

N Y † |ΦHF i = ci |0i, (4.69) i=1

† where ci creates a fermion in state φi. The original Hamiltonian in the Hartree-Fock basis is given by X 1 X Hˆ = t c†c + U c†c†c c . (4.70) ij i j 2 ij,kl i j k l ij ijkl Let us suppose to perturb the Hamiltonian by a time-dependent perturbation

ˆ X † V = Vij(t) ci cj, (4.71) ij and let us study the response of the system to linear order. Therefore the full Hamiltonian Hˆ (t) = Hˆ + Vˆ (t) is time-dependent and the variational principle now concerns the time dependent Shrœdinger equation. Namely we will search within the subspace of time-dependent Slater determinants for a |ΦHF (t)i which satisﬁes

" ∂ ˆ # hΦHF (t)| i~ − H(t) |ΦHF (t)i δ ∂t = 0. (4.72) hΦHF (t)|ΦHF (t)i

We assume that the time-dependent Slater determinant is build up by N single particle normalized wave-functions φα(x, t). By analogy with the conventional derivation of the time- independent Hartree-Fock equations, one ﬁnds that ∂ i φ (x, t) = [T (x, p) + V (x, p, t)] φ (x, t) ~∂t α α N Z X 2 ∗ + dy U(x, y) |φβ(y, t)| φα(x, t) − φβ(y, t) φα(y, t)φβ(x, t) , (4.73) β=1

124 where T (x, p) is the single-particle term of the Hamiltonian, V (x, p, t) the perturbation and U(x, y) the interaction. We assume that the φα(x, t)’s are related to the φi(x)’s by the time- dependent unitary transformation X φα(x, t) = Uiα(t) φi(x), i where Z ∗ Uiα(t) = dx φi(x) φα(x, t).

In terms of the matrix elements, the time-dependent equations (4.73) read

∂ X i U (t) = [t + V (t)] U (t) ~∂t iα ij ij jα j N X X ∗ + Ukβ(t) Ulβ(t) Ujα(t)[Uiklj − Uikjl] . (4.74) β=1 jkl ∂ X i U (t)∗ = − [t + V (t)] U (t)∗ ~∂t iα ji ji jα j N X X ∗ ∗ − Ukβ(t) Ulβ(t) Ujα(t) [Ujlki − Uljki] . (4.75) β=1 jkl

The time-dependent Hartree-Fock wave-function is identiﬁed by the matrix elements

N † X ∗ ∆ij(t) = hΦHF (t)| ci cj |ΦHF (t)i = Uiα(t) Ujα(t) , (4.76) α=1 which, calculated with the time-independent wave-function (4.69) are simply

(0) ∆ij = δij ni, (4.77) with ni = 1 if i ≤ N and ni = 0 otherwise. Through (4.74), (4.75) and (4.76) we obtain the following equations for the ∆ij’s:

∂ X i ∆ (t) = − [t + V (t)] ∆ (t) + [t + V (t)] ∆ (t) ~∂t ij ki ki kj jk jk ik k X + −∆kl(t) ∆mj(t)[Umkli − Ukmli] klm +∆il(t) ∆km(t)[Ujkml − Ujklm] . (4.78)

125 We solve the above equation up to ﬁrst order in the perturbation V . One can easily check that the zero order time-independent term is indeed given by (4.77). The ﬁrst order terms satisfy the equation

" N # ∂ (1) X X (1) i ∆ (t) = − t + (U − U ) ∆ (t) ~∂t ij ki kmmi mkmi kj k m=1 " n # X X (1) + tjk + (Ujmmk − Ujmkm) ∆ik (t) k m=1 X (1) + (ni − nj) ∆ml (t)[Ujmli − Ujmil] + Vji(t)(ni − nj) . (4.79) lm Through Eq. (4.68) we ﬁnally obtain

∂ i − + ∆(1)(t) ~∂t j i ij " # X (1) = (ni − nj) Vji(t) + ∆kl (t)(Ujkli − Ujkil) , (4.80) kl which is our desired result.

4.4.1 Bosonic representation of the low-energy excitations We notice from (4.80) that the only expectation values which are inﬂuenced at linear order by the perturbation are those which either ni = 1 and nj = 0 or viceversa, namely where one of the index refer to an occupied state within (4.69) and the other to an un-occupied one. Let us denote by greek letters α, β, . . . the un-occupied states (> N), hereafter named “particles”, and by roman letters a, b, . . . the occupied ones (≤ N), named “holes”. We denote by

† † bαa = cαca, (4.81) the particle-hole creation operator, which destroys one electron (create one hole) inside the Slater determinant and creates it outside, and its hermitean conjugate

† bαa = cacα. (4.82)

The commutators h i h † † i bαa, bβb = bαa, bβb = 0, vanish, while h † i † † bαa, bβb = δαβ cacb − δab cβcα,

126 has the average value on (4.69) given by

h † i hΦHF | bαa, bβb |ΦHF i = δαβδab (na − nα) = δαβδab, as if the particle-hole creation and annihilation operators were bosonic particles. If we assume that at linear order in V the Hartree-Fock wave-function is sligthly modiﬁed with respect to its time-independent value, in other words we make a similar assumption as we did in deriving spin-wave theory, then we can approximate the commutator by its average value, thus obtaining

h † i h i h † † i bαa, bβb = δαβ δab, bαa, bβb = bαa, bβb = 0, (4.83) which are indeed bosonic commutation relations. We notice that the bosonic vacuum is the time-independent Hartree-Fock wave-function. Since

∆aα(t) = hΦHF (t)|bαa|ΦHF (t)i, Eq. (4.80) implies the following equation of motion for the bosonic operators

∂ i − + b ~∂t α a αa   X † = Vαa(t) + bβb (Uαbβa − Uαbaβ) + bβb (Uαβba − Uαβab) . (4.84) βb

We may now ask the following question: What would be a bosonic Hamiltonian leading to the above equation of motion ? The answer can be readily found and is

X † X † (α − a) bαabαa + bαabβb (Uαbβa − Uαbaβ) αa αβab 1 X + b† b† (U − U ) + b b (U − U ) 2 αa βb αβba αβab βb αa abβα abαβ αβab X † + Vαa(t)bαa + Vaα(t)bαa. (4.85) αa It is now clear that the external ﬁeld simply probes particle-hole excitations which have their own dynamics provided by the Hamiltonian

ˆ X † X † HTDHF = (α − a) bαabαa + bαabβb (Uαbβa − Uαbaβ) αa αβab 1 X + b† b† (U − U ) + b b (U − U ) . (4.86) 2 αa βb αβba αβab βb αa abβα abαβ αβab

127 Due to the presence of the last two terms, the true ground state of (4.86) is not the vacuum, namely the Hartree-Fock wave-function, but another state

A |ΦTDHF i = e |ΦHF i, with A an anti-hermitean operator quadratic in the bosonic operators. The above wave-function is actually an improvement of the Hartree-Fock wave-function that includes the zero-point ﬂuc- tuations of the particle-hole excitations.

The above result can be re-derived in the spirit of the spin-wave theory, that we have already encountered, without even invoking the time-dependent Hartree-Fock. If we assume, to be checked a posteriori, that the Hartree-Fock wave-function is not strongly modiﬁed by the inclusion of quantum ﬂuctuations, we can assume that the commutators

h † † i † † ca cα , cβ cb = δαβ ca cb − δab cβ cα, h † † i † † cα ca , cb cβ = δab cα cβ − δαβ cb ca, h † † i † † ca cb , cc cd = δbc ca cd − δad cc cb, h † † i † † cα cβ , cγ cδ = δβγ cα cδ − δαδ cγ cβ, can be eﬀectively substituted by their average values on the Hartree-Fock wave-function, i.e.

h † † i ca cα , cβ cb = δαβ δab, h † † i cα ca , cb cβ = −δαβ δab, h † † i ca cb , cc cd = 0, h † † i cα cβ , cγ cδ = 0, which justifies the identification with bosonic operators. The only terms of the interaction that may have non vanishing effect when applied either on the right or on the left of the Hartree-Fock wave-functtion are those which contain two particle and two holes, hence they are 1 X H ' U c†c† c c + U c† c†c c int 2 bαβa b α β a αbaβ α b a β abαβ † † † † +Ubαaβ cbcαcacβ + Uαbβa cαcbcβca † † † † +Uαβba cαcβcbca + Uabβα cacbcβcα 1 X = 2 U − U c† c†c c 2 αbβa αbaβ α b β a abαβ

128 † † † † +Uαβba cαcβcbca + Uabβα cacbcβcα. Next we have to express the interaction in terms of bosons, avoiding double counting. The ﬁrst term becomes straightforwardly

X † Uαbβa − Uαbaβ bαa bβb. abαβ

The second term is more delicate to re-express, as one can at will couple α with a or b, thus † † † † getting either bαa bβb or −bαb bβa. Although the two terms correspond to the same fermionic operator, they are independent from the bosonic point of view, hence they both have to be kept, leading to 1 X U − U b† b† + H.c., 2 αβba αβab αa βb abαβ which coincides with the term in (4.86).

129 4.5 Application: antiferromagnetism in the half-ﬁlled Hubbard model

Let us consider again the Hubbard model on an hypercubic lattice in d-dimensions:

X X † X H = −t cR σcR0 σ + H.c. + U nˆR ↑nˆR ↓, (4.87) σ R where † nˆR σ = cR σcR σ, and, as usual, < RR0 > means that R and R0 are nearest neighbor sites. We already showed that for very large U/t and one electron per site, i.e. half-filling, this model is a Mott insulator with a Neèlmagnetic order in d > 1 at zero temperature and in d > 2 at finite temperature below a critical Tc. Let us now try to recover this behavior within the previously described Hartree-Fock theory. The first step is to identify the variational parameters. The interaction, being on-site, leads to local variational parameters which are

† ∆R, σσ0 = hΦHF |cR σcR σ0 |ΦHF i. We search for an Hartree-Fock Slater determinant which describes a Ne`elorder with the antiferromagnetic order parameter along the z-direction, namely we assume that the spin SU(2) symmetry is lowered down to a U(1) symmetry which describes spin-rotations around the z-axis. This implies that we can choose |ΦHF i as an eigenstate of the z-component of the total spin, hence that the only ﬁnite variational parameters are diagonal in the spin index, namely

nR σ = hΦHF |nˆR σ|ΦHF i. P Since the average number of electrons per site is σ nR σ = 1, we use the following parametrization: 1 n = + m (−1)R, (4.88) R ↑ 2 1 n = − m (−1)R,, (4.89) R ↓ 2 Pd where if the vector R = a(n1, n2, . . . , nd), a being the lattice spacing, then R = i=1 ni. In other words the average magnetization is assumed to be 1 Sz = (n − n ) = m (−1)R. R 2 R ↑ R ↓

130 The Hartree-Fock Hamiltonian is therefore

X X † X HHF = −t cR σcR0 σ + H.c. + U (ˆnR ↑ nR ↓ +n ˆR ↓ nR ↑) σ R X X † X R = −t cR σcR0 σ + H.c. − U m (−1) (ˆnR ↑ − nˆR ↓) σ R U X + (ˆn +n ˆ ) . 2 R ↑ R ↓ R The last term is proportional to the total number of electrons, UN/2, which is a conserved quantity. Therefore UN/2 is just a constant which can be dropped leading to the Hartree-Fock Hamiltonian

X X † X R HHF = −t cR σcR0 σ + H.c. − U m (−1) (ˆnR ↑ − nˆR ↓) . (4.90) σ R Let us write this Hamiltonian in momentum space. Since

(−1)R = eiQ·R, where Q = π(1, 1,..., 1)/a, one readily ﬁnds that

X X † † HHF = knˆk σ − U m ck ↑ck+Q ↑ − ck ↓ck+Q ↓ . (4.91) k σ k

If k = (k1, k2, . . . , kd), the bare dispersion is given by

d X k = −2t cos kia. i=1

The original Brillouin zone is an hypercube with linear size 2π/a. Since the Hartree-Fock Hamiltonian breaks translational symmetry – the new unit cell is twice larger – the actual Brillouin zone is half the original one. Let us deﬁne the new Brillouin zone, which we call Magnetic Brillouin zone (MBZ) as the volume which encloses all k-points such that k ≤ 0. Since for any of these points k+Q = −k ≥ 0, the MBZ is indeed twice-smaller than the original one. Then let us deﬁne new fermionic operators with k ∈ MBZ by ak σ = ck σ, bk σ = ck+Q σ, and spinorial operators ak σ ψk σ = . (4.92) bk σ

131 In terms of the latter (4.91) can be written as

X † X † † HHF = k ψk σ τ3 ψk σ − U m ψk ↑ τ1 ψk ↑ − ψk ↓ τ1 ψk ↓, (4.93) k∈MBZ σ k∈MBZ where τi’s, i = 1, 2, 3, are Pauli matrices acting in the spinorial basis. We notice that k Um k τ3 ∓ U m τ1 = −Ek τ3 − ± i τ2 Ek Ek ±2iθk τ2 ∓iθk τ2 ±iθk τ2 = −Ek τ3 e = −Ek e τ3 e , where q 2 2 2 Ek = k + U m , (4.94) and k Um cos 2θk = − , sin 2θk = . Ek Ek Therefore if we deﬁne two new spinors through αk ↑ iθk τ2 αk ↓ −iθk τ2 φk ↑ = = e ψk ↑, φk ↓ = = e ψk ↓, (4.95) βk ↑ βk ↓ the Hamiltonian becomes

X † X † † HHF = − Ek φk σ τ3 φk σ = − Ek αk σαk σ − βk σβk σ , (4.96) k∈MBZ σ k∈MBZ σ namely it acquires a diagonal form. Since Ek > 0 the ground state with a number of electrons equal to the number of sites N is simply obtained by ﬁlling completely the α-band (notice that the MBZ contains N/2 k-points, hence each band can accomodate 2 N/2 = N electrons). Thus the Hartree-Fock Hamiltonian describes a band-insulator with two bands separated by a gap of minimal value 2U m. The Hartree-Fock wave-function satisﬁes:

† hΦHF |φk σ τi φk σ|ΦHF i = δi 3, which becomes at ﬁnite temperature E hφ† τ φ i = δ tanh k . (4.97) k σ i k σ i 3 2T

We still have to impose the self-consistency condition

1 X m = (−1)R hnˆ − nˆ i 2V R ↑ R ↓ R

132 1 X = hψ† τ ψ − ψ† τ ψ i 2V k ↑ 1 k ↑ k ↓ 1 k ↓ k∈MBZ 1 X = hψ† eiθk τ2 τ e−iθk τ2 ψ − ψ† e−iθk τ2 τ eiθk τ2 ψ i 2V k ↑ 1 k ↑ k ↓ 1 k ↓ k∈MBZ 1 X = hφ† (cos 2θ τ + sin 2θ τ ) φ − φ† (cos 2θ τ − sin 2θ τ ) φ i 2V k ↑ k 1 k 3 k ↑ k ↓ k 1 k 3 k ↓ k∈MBZ 1 X Ek = sin 2θ tanh . V k 2T k∈MBZ In other words the self-consistency condition reads

U X 1 Ek tanh = 1. (4.98) 2V Ek 2T k

We solve this self-consistency condition by assuming a constant density of states, namely

1 X Z Z D ··· = d ρ() · · · → ρ d . . . , V 0 k −D where ρ0 = 1/2D, with D ∼ 2dt. At zero temperature we find Z D Uρ0 1 1 = d p , 2 −D 2 + U 2m2 which has solution given by D 1 −1 m = sinh . U Uρ0 This equation has always a solution for any U 6= 0. In particular, for U D, D 1 m ' exp − . (4.99) U Uρ0 This results states that the Hubbard model at half-filling on an hypercubic lattice is always an antiferromagnetic insulator whatever is the value of the Hubbard U. Notice that in the opposite limit of U D, m ∼ D ρ0 = 1/2, which is the expected result since, for large U, electrons localize, one per site, hence behave like local spin-1/2 moments. We can also determine the Neèltemperature, which is the temperature Tc at which the self-consistency condition starts to be verified with m = 0, namely Z D U X 1 k Uρ0 d 1 = tanh ' tanh . (4.100) 2V k 2Tc 2 2Tc k −D

133 One ﬁnds that Tc ∼ U m(T = 0). This implies that

1 Tc ∼ D exp − , Uρ0 for small U, but Tc ∼ U at large U. The latter results contradicts what we previously found in the large U limit, by mapping the Hubbard model onto an Heisenberg model. There, within 2 spin-wave theory, we showed that Tc ∼ J ∼ t /U, hence vanishes for large U. The origin of the disagreement stems from the fact that the Hartree-Fock theory is valid only for an interaction which is weak compared to the electron bandwidth. Essentially the Mott phenomenon, which dominates at large U/t, escapes any description using a single Slater determinant, which is only able to mimick a band-insulator.

4.5.1 Spin-wave spectrum by time dependent Hartree-Fock Rigorously speaking, Hartree-Fock theory only allows to access ground state or, at finite temperature, equilibrium properties, although in an approximate variational manner, but not properties of excited states. Therefore it is not legitimate to interpret the spectrum of the Hartree-Fock Hamiltonian as an approximation of the actual spectrum. For instance we showed that the Heisenberg antiferromagnet, which corresponds to the large U-limit of the Hubbard model at half-filling, has gapless spin-wave excitations. On the contrary the Hartree-Fock Hamiltonian, which describes a band-insulator, has a gap for all excitations. The simplest way to cure this defect, namely to have access to excitations, is by means of the time dependent Hartree-Fock. Here we apply this technique to recover the spin-wave spectrum. Let us derive the bosonic Hamiltonian for the spin-flip particle-hole excitations above the Hartree-Fock ground state. We consider, in terms of the spinors (4.95) which diagonalize the Hartree-Fock Hamiltonian, the following spin-flip operators:

r1 X = φ† τ φ , k;q 2 k ↑ 1 k+q ↓ r1 X† = φ† τ φ , k;q 2 k+q ↓ 1 k ↑ r1 P = φ† τ φ , k;q 2 k ↑ 2 k+q ↓ r1 P † = φ† τ φ . k;q 2 k+q ↓ 2 k ↑ In the framework of time-dependent Hartree-Fock one can assume that

h † i i † † X ,P = δ 0 0 φ τ φ + δ 0 φ τ φ k;q k0;q0 2 k+q,k +q k ↑ 3 k0 ↑ k,k k0+q0 ↓ 3 k+q ↓

134 h † i ' hΦHF | Xk;q,Pk0;q0 |ΦHF i = i δk,k0 δq,q0 , which shows that X and P † are conjugate variables. The energy cost of such spin-ﬂip excitations are described by the free Hamiltonian

X † † H0 = ωk;q Xk;qXk;q + Pk;qPk;q , (4.101) k q where ωk;q = Ek+q + Ek. This simply means that if one destroys an α particle at k and create a β one at k+q, the energy cost is ωk;q. Following the prescriptions of time-dependent Hartree-Fock, we still need to express the Hubbard interaction in terms of these spin-ﬂip excitations. We notice that

† † nˆR ↑nˆR ↓ = cR ↑cR ↑ cR ↓cR ↓ † † + − = −cR ↑cR ↓ cR ↓cR ↑ ≡ SR SR, so that X X U X U nˆ nˆ = U S+ S− = S+ S− , R ↑ R ↓ R R V q −q R R q where + X −iq·R + X † Sq = e SR = ck ↑ck+q ↓. R k We recall that the MBZ is half the original one. For instance, while before the momenta q = 0 and q = Q were diﬀerent, in the MBZ both coincide with momentum zero. Therefore, if we are interested in long wavelength excitations, we need to consider in the original BZ both + q ∼ 0 and q ∼ Q. For q ∼ 0, Sq connects states inside the MBZ, i.e.

+ X † † X † Sq ' ak ↑ak+q ↓ + bk ↑bk+q ↓ = ψk ↑ψk+q ↓. k∈MBZ k∈MBZ

+ Near Q we can consider the operator Sq+Q with small q, which is

+ X † † X † Sq+Q ' ak ↑bk+q ↓ + bk ↑ak+q ↓ = ψk ↑ τ1 ψk+q ↓. k∈MBZ k∈MBZ Therefore in the long wavelength limit it is legitimate to re-write U X U X S+ S− = ψ† ψ ψ† ψ + ψ† τ ψ ψ† τ ψ . V q −q V k ↑ k+q ↓ p+q ↓ p ↑ k ↑ 1 k+q ↓ p+q ↓ 1 p ↑ q k p q ∈MBZ

135 Notice that

† † iθkτ2 iθk+qτ2 ψk ↑ψk+q ↓ = φk ↑ e e φk+q ↓ † √ ' i sin (θk + θk+q) φk ↑ τ2 φk+q ↓ = i 2 sin (θk + θk+q) Pk;q,

† † iθkτ2 iθk+qτ2 ψk ↑ τ1 ψk+q ↓ = φk ↑ e τ1 e φk+q ↓ † √ ' cos (θk − θk+q) φk ↑ τ1 φk+q ↓ = 2 cos (θk − θk+q) Xk;q, so that the interaction Hamiltonian becomes 2U X H = − sin (θ + θ ) sin (θ + θ ) P † P int V k k+q p p+q k;q p;q k p q∈MBZ † + cos (θk − θk+q) cos (θp − θp+q) Xk;q Xp;q. (4.102)

The complete Hamiltonian for the spin-ﬂip excitations is therefore

X † † H = ωk;q Xk;qXk;q + Pk;qPk;q k q∈MBZ 2U X − sin (θ + θ ) sin (θ + θ ) P † P V k k+q p p+q k;q p;q k p q∈MBZ † + cos (θk − θk+q) cos (θp − θp+q) Xk;q Xp;q. (4.103) One might diagonalize this Hamiltonian to obtain the spin-ﬂip excitation spectrum. Here we aim simply to show that (4.103) does contain the spin-waves. First we perform the canonical transformation s 1 Xk;q → Xk;q, ωk;q √ Pk;q → ωk;q Pk;q, after which

X † 2 † H = Xk;qXk;q + ωk;q Pk;qPk;q k q∈MBZ 2U X √ − ω ω sin (θ + θ ) sin (θ + θ ) P † P V k;q p;q k k+q p p+q k;q p;q k p q∈MBZ s 1 † + cos (θk − θk+q) cos (θp − θp+q) Xk;q Xp;q. ωk;qωp;q

136 Next we make a unitary transformation and deﬁne the new operators v u 1 X cos (θp − θp+q) Xeq = u X √ Xp;q, u 2 ωp;q t cos (θk − θk+q) /ωk;q p∈MBZ k∈MBZ v u 1 X cos (θp − θp+q) Peq = u X √ Pp;q, u 2 ωp;q t cos (θk − θk+q) /ωk;q p∈MBZ k∈MBZ as well as all the other orthogonal combinations, Xi;q and Pi;q. The part of the Hamiltonian which involves Xe is simply

2 ! X 2U X cos (θk − θk+q) † 1 − Xeq Xeq. V ωk;q q∈MBZ k∈MBZ Notice that for q = 0 the square bracket vanishes because is nothing but the self-consistency equation (4.98) at zero temperature, namely

U X 1 1 = . V Ek k∈MBZ Therefore for small q 2 2U X cos (θk − θk+q) 1 − ∼ A |q|2, V ωk;q k∈MBZ where A is a constant which can be determined and for large U/t behaves as t2/U 2. All the combinations of X’s, Xi;q’s, which are orthogonal to Xe remain degenerate and described by the term X X † Xi;qXi;q. i q

One realizes that the above |q|2-term is the leading one at small q, hence we can evaluate what is left putting q = 0. Then, through (4.98), we ﬁnd that r r U X 1 Peq ' Pk;q, V Ek k∈MBZ and that the part of the Hamiltonian involving the conjugate variables is

X 2 † 2U X p Um Um † 4Ek Pk;qPk;q − 2 Ek Ep Pk;q Pp;q V Ek Ep k q∈MBZ k p q∈MBZ

137 X 2 † 2 2 X † = 4Ek Pk;qPk;q − 4U m PeqPeq. k q∈MBZ q

This implies that the conjugate variables are governed, at small q, by the term " # X 4U X 2 2 † X † Ek − 4U m Pe Pe = B Pe Pe , V q q q q q k∈MBZ q where B ∼ t2 for large U, and a bunch of other terms involving the orthogonal combinations Pi;q’s which are coupled together and to Peq. Let us neglect for the moment the coupling to Peq. Then Xeq and Peq are governed by the Hamiltonian

X 2 † † A|q| XeqXeq + B PeqPeq, q q p −1 which is diagonalized by the transformation Xeq → Kq Xeq and Peq → Kq Peq, where

B K2 = , q A|q|2 leading to a linear dispersion √ ωq = AB |q| ∼ J |q|. If we diagonalize the Hamiltonian which involves all other orthogonal combinations, we would get for them a gaped spectrum. Notice that q −1 p Peq → Kq Peq ∼ |q| Peq, p therefore the coupling between Pe and all other Pi’s is proportional to |q| hence just renor- malizes the spin-wave velocity. Moreover this coupling term between Pe and all Pi’s actually vanishes for large U since " # X 2 † 2 2 X † X † 4Ek Pk;qPk;q ∼ 4U m PeqPeq + Pi;qPi;q . k q∈MBZ q i so that, for U/t 1, the spin-wave velocity v is, up to subleading terms, √ v = AB, (4.104) with A and B deﬁned previously.

138 4.6 Linear response of an electron gas

Let us consider now the response to an electromagnetic ﬁeld of a solid. We start from the classical macroscopic Maxwell equations of an isolated system

 ∇ · E = 4πρ,     ∇ · B = 0,  (4.105)  1 ∂B  ∇ ∧ E = − c ∂t ,    1 ∂E 4π ∇ ∧ B = c ∂t + c J, where ρ is the total charge density within the system, and J the current density. In general the current receives contributions from the free carriers, from the magnetic moments and from bound charges, so that its expression is ∂P J = J + + c ∇ ∧ M, (4.106) carriers ∂t where P the polarization density of bound charges5 and M the magnetization density. In what follows we will not take into account these two terms, unless explicitly stated. We deﬁne the displacement ﬁeld D through 1 dD 1 ∂E 4π = + J. (4.107) c dt c ∂t c By applying ∇ on both sides we get

1 d 1 ∂ 4π 4π ∂ρ ∇ · D = ∇ · E + ∇ · J = + ∇ · J = 0, (4.108) c dt c ∂t c c ∂t

5The macroscopic charge density is assumed to be the average of the microscopic one over a volume element much bigger then the unit cell and smaller than the coherent volume of the applied electromagnetic ﬁeld. It is reasonable to assume that, at equilibrium, this reference volume can be taken to be neutral, so that the charge density vanishes but eventually for the presence of a ﬁnite polarization P and of higher multipolar components which derive from the shape of the bound-charge density. Away from equilibrium, the charge density can change also because free electrons can move or because external charges are added. Hence, neglecting higher multipoles than the dipole, the generic macroscopic charge density is

ρ(x) = ρfree(x) − ∇ · P(x) + ρext(x), where ρfree is the variation of the free-charge density with respect to equilibrium. The distinction between free and bound charges is however a bit matter of convention. Indeed, at frequencies much bigger then the atomic excitation ones, there is actually no distinction among them. The convention that is usually adopted, which we adopt here as well, is that the frequencies under consideration will always be much smaller then atomic ones.

139 where we made use of the continuity equation ∂ρ + ∇ · J = 0. ∂t We can therefore assume ∇ · D = 0. (4.109) If there is an external charge with density, then (4.108) becomes

1 d 4π dρ ∇ · D = ext , c dt c dt implying ∇ · D = 4πρext. (4.110) In other words the displacement ﬁeld is only sensible to the external charge. In the following we assume that the system is translationally invariant. In this case we can formally write a relationship between the Fourier components of D, D(ω, q), and those of E, E(ω, q): D(ω, q) = (ω, q) E(ω, q), (4.111) where (ω, q) is the dielectric constant, which is actually a matrix acting on the space components, i.e. Di = ij Ej with i, j = 1, 2, 3. If the system is isotropic, we can generally write

q q q q (ω, q) = i j (ω, q) + δ − i j (ω, q), (4.112) ij q2 || ij q2 ⊥ where ||(ω, q) is the longitudinal component of the dielectric constant and ⊥(ω, q) the transverse one. Analogously, for any vector V (ω, q) one can deﬁne the longitudinal component through q · V (ω, q) = q V||(ω, q), as well as the transverse one q V (ω, q) = V (ω, q) − V (ω, q). ⊥ q ||

In the presence of the electromagnetic ﬁeld the Hamiltonian H = H0 + HCoul, where H0 is the kinetic energy while HCoul includes the electron-electron, electron-ion and ion-ion Coulomb interaction, changes according to

X Z 1 e 2 H → dx Ψ (x)† −i ∇ + A(x) Ψ (x) − eφ(x)Ψ (x)†Ψ (x). (4.113) 0 2m σ ~ c σ σ σ σ

140 Since we already includes the Coulomb interaction among the internal charges inside HCoul, the displacement, electric and magnetic ﬁelds are, in terms of the vector and scalar potentials, 1 ∂ D = −∇φ − A , (4.114) || c ∂t || 1 ∂ E = − A , (4.115) ⊥ c ∂t ⊥ B = ∇ ∧ A⊥. (4.116)

In other words the scalar potential and the longitudinal component of the vector potential are related only to the displacement ﬁeld, which is in turn determined by the external charge through

∇ · D|| = 4πρext.

Therefore φ and A|| are actually the external fields φext and A||ext applied on the system. Notice that, since only D|| has physical meaning there is a redumdancy in having both φ and A||. This is a manifestation of the so-called gauge invariance. On the contrary, since we do not include the relativistic current-current interaction among the internal charges, the transverse electric and magnetic fields are those actually felt by the electrons inside the sample, so called internal fields. We conclude by noticing that, in the presence of a vector potential, the charge current operator as obtained through the continuity equation ∂ρ + ∇ · J = 0, ∂t changes according to

X e J(x) = −e −i ~ Ψ (x)† ∇Ψ (x) + Ψ (x)†Ψ (x) A(x, t) m σ σ mc σ σ σ e2 ≡ −e j(x) − ρ(x) A(x, t), (4.117) mc where we deﬁned as −e j(x) the purely electronic current, while the last term is commonly referred to as the diamagnetic term. If we take into account also the Zeeman splitting, the Hamiltonian further contains the term Z Z X † gµB dx B(x, t) · Ψα(x) Sαβ Ψβ(x) = gµB dx A⊥(x, t) · ∇ ∧ σ(x), (4.118) αβ where X † σ(x) = Ψα(x) Sαβ Ψβ(x) , αβ

141 is the spin density operator, being S the spin-1/2 matrices. The Zeeman splitting contributes to the current with the transverse term6

δJ(x) = −c g µB ∇ ∧ σ(x). (4.119)

4.6.1 Response to an external charge

Let us start by studying the response to an external charge density e ρext(x, t). For simplicity we will represent the ions as an inﬁnitely massive uniform background of positive charge which has only the eﬀect of neutralizing on average the electronic charge. In other words, if e n is the charge density of the background andρ ˆ(x) the density operator for the electrons, then Z dxhρˆ(x)i ≡ ρ(q = 0) = V n, where V is the sample volume and ρ(q) the Fourier transform of the density. Notice that the operatorρ ˆ(q = 0) is the total number of electrons, which is conserved, hence we can always writeρ ˆ(q = 0) = V n If we choose a gauge in which A|| = 0, the coupling to the scalar potential φ(x, t) leads to the perturbation

Z e2 δHˆ = − dx dy ρ (x, t) (ˆρ(y) − n) |x − y| ext 1 X 4πe2 = − ρ (q, t) (ˆρ(−q) − V nδ ) V q2 ext q0 q 1 X 4πe2 1 X = − ρ (q, t)ˆρ(−q) ≡ −e φ(q, t)ρ ˆ(−q), (4.120) V q2 ext V q6=0 q6=0 since the Fourier transform of the scalar potential is 4πe φ(q, t) = ρ (q, t). q2 ext In the absence of the external charge, the electron density is uniform hence

hρˆ(y)i = n.

6Notice that the expression of the current can be also determined through the functional derivative δH J(x, t) = −c . δA(x, t)

142 The external charge induces a deviation of the electron density with respect to the unperturbed value n, which we denote as the induced charge

hρˆ(y)i − n = ρind(y, t). (4.121) Its Fourier transform is, for q 6= 0,

ρind(q, t) = hρˆ(q)i, while for q = 0 it vanishes for charge conservation. The induced charge density at frequency ω is given, within linear response theory, by 4πe2 ρ (q, ω) = −eχ(q, ω) φ(q, ω) = − χ(q, ω) ρ (q, ω), (4.122) ind q2 ext where 1 Z ∞ χ(q, ω) = dt eiωt (−iθ(t)h[ˆρ(q, t), ρˆ(−q)]i) , (4.123) V −∞ is the Fourier transform of the density-density linear response function. Commonly it is denoted as the improper density-density linear response function, as opposed to the proper one which is defined through 4πe2 ρ (q, ω) = − χ(q, ω)(ρ (q, ω) − ρ (q, ω)) . (4.124) ind q2 e ext ind The proper function gives the response of the electrons to the internal field, which is generated both by the external charge and by the same electrons. 7 Comparing (4.122) with (4.124) we find that χ(q, ω) χ(q, ω) = e . (4.125) 4πe2 1 − χ(q, ω) q2 e By definition

iq · D||(q, ω) ≡ iq · ||(q, ω) E||(q, ω) = 4πeρext(q, ω),

iq · E||(q, ω) = 4πeρext(q, ω) − 4πeρind(q, ω), hence ||(q, ω)(ρext(q, ω) − ρind(q, ω)) = ρext(q, ω). In other words 1 ρ (q, ω) = 1 − ind , (4.126) ||(q, ω) ρext(q, ω)

7The induced charge refers to electrons, which explains the minus sign in (4.124)

143 ρind(q, ω) ||(q, ω) = 1 + . (4.127) ρext(q, ω) − ρind(q, ω) Through Eqs. (4.122) and (4.126) it derives that

1 4πe2 = 1 + 2 χ(q, ω), (4.128) ||(q, ω) q while by the Eqs. (4.124) and (4.127)

4πe2 (q, ω) = 1 − χ(q, ω). (4.129) || q2 e These two equations relate the longitudinal dielectric constant to the density-density proper and improper response functions. Notice that for the dielectric constant the Kramers-Kr¨onig relations read 1 Z dω0 1 1 Re = 1 − P 0 Im 0 . (4.130) ||(q, ω) π ω − ω ||(q, ω )

Using once more the definition of the displacement field, one finds that d ∇ · D = ∂ ∇ · E + 4π∇ · J, dt t where J is the current flowing inside the system.8 In Fourier space it reads

qω D||(q, ω) = qω E||(q, ω) + 4πi q · J(q, ω).

Since D|| = || E||, we ﬁnd ω J (q, ω) = (q, ω) − 1 E (q, ω) ≡ σ (q, ω) E (q, ω), || 4πi || || || || where σ|| is the longitudinal conductivity which relates the current to the internal electric ﬁeld. Therefore ω σ (q, ω) = (q, ω) − 1 , (4.131) || 4πi || 4πi (q, ω) = 1 + σ (q, ω). (4.132) || ω ||

8 Since we assumed an external probing charge ρext(x, t) which depends on time only explicitly, the total ﬂowing current includes just the induced one.

144 Gauge invariance and its consequences So far we have adopted a gauge where the external scalar field is finite while the external longitudinal vector potential is zero. Now let us do the opposite, keeping the physical displacement field fixed. By gauge invariance the response of the system should be the same. Therefore let us consider an external charge density eρext(x, t) leading to a displacement field e D (q, ω) = ρ (q, ω). || iq ext We assume that iω D (q, ω) = A (q, ω) = (q, ω) E (q, ω). || c || || || A vector potential provides a perturbation

e Z e2 Z δH = dx A(x, t) · j(x) + dx ρ(x) A(x, t) · A(x, t), (4.133) c 2mc2

P † where the electronic current j has been deﬁned by Eq. (4.117) and ρ(x) = σ Ψσ(x) Ψσ(x) is the density operator. The longitudinal value of j is given in linear response by e j (q, ω) = χ (q, ω) A (q, ω), || c || || where, if χlm(x, t) = −iθ(t)h[jl(x, t), jm(0, 0)]i is the current-current response function, then its longitudinal component is deﬁned through Z Z X qlqm χ (q, ω) = dt eiωt dx e−iq·x χ (x, t). || q2 lm lm On the other hand the true charge current operator, see Eq. (4.117), is

e2 J(x, t) = −ej(x, t) − hρ(x)iA (x, t). mc || Since the second term in the right hand side is already linear in the external ﬁeld, the average of the density operator has to be evaluated on the unperturbed state, where hρ(x)i = n. Hence

ne2 J (q, ω) = −ej (q, ω) − A (q, ω) || || mc || e2 n = − χ (q, ω) + A (q, ω) c || m || e2 n = − χ (q, ω) + (q, ω) E (q, ω) ≡ σ (q, ω) E (q, ω). iω || m || || || ||

145 Therefore we ﬁnd that e2 n σ (q, ω) = − χ (q, ω) + (q, ω), || iω || m || which, compared with (4.131) leads to

e2 n ω − χ (q, ω) + (q, ω) = (q, ω) − 1 , iω || m || 4πi || namely

n ω2 1 χ||(q, ω) + = − 2 1 − m 4πe ||(q, ω) ω2 = χ(q, ω). (4.134) q2 This shows that gauge invariance implies that the longitudinal current-current response function can be expressed in terms of the density-density response function. In principle we can also introduce a proper longitudinal current-current response function through

e2 n J (q, ω) = − χ (q, ω) + (q, ω) E (q, ω) || iω || m || || e2 n ≡ − χ (q, ω) + E (q, ω), iω f|| m || which is related by gauge invariance to the proper density-density response function through n q2 χ (q, ω) + = ω2 χ(q, ω). (4.135) f|| m e

We end by noticing that, in the gauge in which the scalar potential is zero, the internal longitudinal vector potential, deﬁned by c A (q, ω) = E (q, ω), || int iω || is related to the external one, c A (q, ω) = A (q, ω) = D (q, ω), || || ext iω || through 1 A|| int(q, ω) = A|| ext(q, ω). (4.136) ||(q, ω)

146 4.6.2 Response to a transverse field By recalling the definition of the displacement field d D = ∂ E + 4πJ, dt t its transverse component is given by

−iω D⊥(q, ω) = −iω ⊥(q, ω) E⊥(q, ω) = −iω E⊥(q, ω) + 4π J⊥(q, ω)

= (−iω + 4πσ⊥(q, ω)) E⊥(q, ω), which provides the relationship between the transverse dielectric constant and conductivity: 4πi (q, ω) = 1 + σ (q, ω). (4.137) ⊥ ω ⊥ We already said that the transverse vector potential is actually the internal one, so the response to it is connected with proper response functions. Following the same reasoning as after Eq. (4.133) but now for the transverse current, we get to the conclusion that

e2 n J (q, ω) = − χ (q, ω) + A (q, ω), (4.138) ⊥ c e⊥ m ⊥ where χe⊥(q, ω) is obtained by the current-current response function through q q q q χ (q, ω) = l m χ (q, ω) + δ − l m χ (q, ω). (4.139) lm q2 || lm q2 e⊥ Notice that, unlike the longitudinal one, the transverse component is proper. Since iω J (q, ω) = σ E (q, ω) = σ A (q, ω), (4.140) ⊥ ⊥ ⊥ c ⊥ ⊥ we ﬁnally obtain e2 n σ (q, ω) = − χ (q, ω) + . (4.141) ⊥ iω e⊥ m By means of (4.115) and (4.116) and the last Maxwell equation we get the usual wave- equation for the internal vector potential

1 ∂2A 4π −∇2 A = − ⊥ + J , ⊥ c2 ∂t2 c ⊥ tot which in momentum and frequency space reads

2 2 2 c q − ω A⊥(q, ω) = 4πc J⊥ tot(q, ω). (4.142)

147 The current appearing in (4.142) is the total one, including the external source of the electromagnetic ﬁeld. The latter satisﬁes a similar equation

2 2 2 c q − ω A⊥ ext(q, ω) = 4πc J⊥ ext(q, ω), in terms of the external current source. Writing J⊥ tot = J⊥ ext + J⊥, with the latter being the induced current as given by (4.140), we ﬁnally get

2 2 2 2 2 2 c q − ω A⊥(q, ω) = c q − ω A⊥ ext(q, ω) + 4πiωσ⊥(q, ω) A⊥(q, ω).

In other words, the internal vector potential, A⊥ = A⊥ int, is related to the applied one by the equation c2 q2 − ω2 A⊥ int(q, ω) = 2 2 2 A⊥ ext(q, ω) c q − ω − 4πiωσ⊥(q, ω) c2 q2 − ω2 = 2 2 2 A⊥ ext(q, ω), (4.143) c q − ω ⊥(q, ω) to be compared with (4.136) valid for the longitudinal component. Let us introduce back the Zeeman contribution to the transverse current given in Eq. (4.119). Within linear response theory and assuming a uniform isotropic system Z Z 0 0 0 0 0 0 σ(x, t) = gµB dy dt χσ(x − y, t − t ) B(y, t ) = gµB dy dt χσ(x − y, t − t ) ∇y ∧ A⊥(y, t ), where χσ is the spin-density response function. Therefore (4.119) is, at linear order, Z 2 0 0 0 δJ(x) = −c (gµB) dy dt ∇x ∧ χσ(x − y, t − t ) ∇y ∧ A⊥(y, t ) Z 2 0 2 0 0 = c (gµB) dy dt ∇ χσ(x − y, t − t ) A⊥(y, t ), (4.144) which, after Fourier transform, is

2 2 δJ⊥(q, ω) = −c q (gµB) χσ(q, ω) A⊥(q, ω). This term modiﬁes (4.143) into

c2 q2 − ω2 A⊥ int(q, ω) = A⊥ ext(q, ω). (4.145) 2 2 2 2 2 2 c q − ω ⊥(q, ω) + 4πc q (gµB) χσ(q, ω)

4.6.3 Limiting values of the response functions The values of the response functions in some particular limits of physical signiﬁcance can be obtained without making any complicated calculation.

148 Consequences of the f-sum rule The charge density operator at q = 0 is the total number of electrons N, which is a conserved quantity, independent upon time. Therefore 1 1 χ(q = 0, t) = −i θ(t) h[ˆρ(0, t), ρˆ(0)]i = −i θ(t) h[N,N]i = 0. V V In other words due to charge conservation it holds that

χ(0, ω) = 0 . (4.146)

The dissipative density-density response function is by deﬁnition

00 1 Z dω 00 χ (q, t) = h[ˆρ(q, t), ρˆ(−q)]i = e−iωt χ (q, ω). 2V 2π By means of the continuity equation

∂χ00 (q, t) 1 ∂ρˆ(q, t) i = h i , ρˆ(−q) i ∂t 2V ∂t 1 = h[q · J(q, t), ρˆ(−q)]i 2V Z dω 00 = e−iωt ω χ (q, ω). 2π On the other hand, for t = 0, the following expression holds

1 nq2 h[q · J(q), ρˆ(−q)]i = , 2V 2m which leads to the so-called f-sum rule

2 Z dω 00 nq ω χ (q, ω) = . (4.147) 2π 2m By the Kramers-Kronig relations

00 Z dω0 χ (q, ω0) Re χ(q, ω) = P . π ω − ω0 We know that the dissipative response function is ﬁnite only for frequencies which correspond to the energies of excitations which may be created by the density operators. In other words χ00 (q, ω0) is bounded from above. Therefore if we consider a frequency much above this upper

149 bound, and we recall that the dissipative response function is odd in frequency, then we ﬁnd that

00 Z dω0 χ (q, ω0) ω0 lim Re χ(q, ω) = 1 + ω→∞ π ω ω 0 2 1 Z dω 00 nq = ω0 χ (q, ω0) = , ω2 π mω2 namely nq2 lim Re χ(q, ω) = . (4.148) ω→∞ mω2 Alternatively, we can re-write the sum-rule (4.147) as

0 0 n 1 Z dω 00 1 Z dω = ω0 χ (q, ω0) = − ω0 Im χ(q, ω0) 2m q2 2π q2 2π ! Z dω0 1 n = − Im χ (q, ω0) + . 2π ω0 || m

In the limit q → 0, there is no distinction between longitudinal and transverse response, so that, through Eq. (4.139), we obtain

Z 0 ! n dω 1 0 n = − lim Im χ||(q, ω ) + 2m q→0 2π ω0 m Z 0 ! dω 1 0 n = − lim Im χ⊥(q, ω ) + q→0 2π ω0 e m ! Z dω0 i = Im σ (0, ω0) . 2π e2 ⊥

In other words, the f-sum rule can also be written as

Z ∞ ne2 dω Re σ⊥(0, ω) = , (4.149) 0 2m which is the so-called optical f-sum rule. By Eq. (4.128) we also ﬁnd that

2 1 ωp lim Re = 1 + 2 , (4.150) ω→∞ ||(q, ω) ω

150 − +

Figure 4.1: Pictorial representation of a plasmon excitation. as well as Z 0 dω 0 1 2 ω Im 0 = −ωp , (4.151) π ||(q, ω ) where r 4πne2 ω = (4.152) p m is the so-called electron plasma frequency.

What is the meaning of the plasma frequency? Let us take a piece of metal and let us displace uniformly, i.e. at q = 0, the valence electrons with respect to the positive background9 by a distance x, as in Fig. 4.1. As a result a surface charge will appear σ = n e x, so that x is subject to a restoring Coulomb force

d2x m = −4πeσ = −4πe2n x, (4.153) dt2 leading to oscillations, so-called plasmons, with frequency ωp. In other words the only longitudinal modes at q = 0 allowed in a metal are plasmons. Taking into account (4.151), which is valid for all q’s, hence also for q → 0, we conclude that 1 π lim Im = − ωp (δ(ω − ωp) − δ(ω + ωp)) , (4.154) q→0 ||(q, ω) 2 which, by Kramer-Kr¨onigtransformation, leads to

1 ω2 Re = 2 2 , ||(0, ω) ω − ωp

9In a real solid we have to imagine that the positive background includes the ions and the bound electrons. In other words the derivation implicitly assumes that we explore excitations at frequency smaller than interband transitions.

151 or, equivalently, 1 ω 1 1 = 1 + p − , (4.155) ||(0, ω) 2 ω − ωp + iη ω + ωp + iη with η an inﬁnitesimal positive number. In addition

ω2 Re (0, ω) = 1 − p . || ω2

Through (4.129) we also ﬁnd that

n q2 lim χ(q, ω) = , (4.156) q→0 e m ω2 which, through the gauge relation (4.135), leads to

lim χ||(q, ω) = 0 . (4.157) q→0 f

Finally, using the deﬁnition of the longitudinal conductivity (4.131), we also ﬁnd that

ne2 1 ne2 Re σ (0, ω) = Re i = π δ(ω) , || m ω + iη m which implies an inﬁnite conductivity of an ideal metal.

Zero frequency response Now suppose we change by the same amount the charge density of the positive background and of the electrons, for instance by changing the volume V . From the point of view of the electrons, this is analogous to changing the chemical potential, hence Z ∂N dx hρˆ(x)i − n = δN = δµ. ∂µ The compressibility is deﬁned by 1 ∂V K = − , V ∂P where P is the pressure. The free-energy F (N,V,T ) can be written as

F (N,V,T ) = V f(n, T ),

152 so that the pressure ∂F ∂f P = − = −f + n , ∂V ∂n hence 1 ∂2f = n2 . K ∂n2 On the other hand ∂F ∂f ∂ν 1 ∂2f µ = = , = , ∂N ∂n ∂N V ∂n2 implying that ∂N = n2 KV. (4.158) ∂µ From the point of view of the Hamiltonian, the action of changing the density of the background and in the meantime changing the electron chemical potential so to maintain charge neutrality corresponds to a time-independent perturbation

1 X δH = V (−q, ω = 0)ρ ˆ(q), V ext q where 4πe2 V (q, ω = 0) = − δρ (q, ω = 0) − V δµ δ . ext q2 back q,0 The internal field is instead 4πe2 V (q, ω = 0) = − (δρ (q, ω = 0) − δρ (q, ω = 0)) − V δµ δ , int q2 back el q,0 and, by definition, δρel(q, ω = 0) = χe(q, ω = 0) Vint(q, ω = 0). Since charge neutrality is maintained, in the limit q → 0 we find

lim Vint(q, ω = 0) = −V δµ, q→0 hence 2 lim δρel(q, ω = 0) = δN = −V δµ lim χ(q, ω = 0) = n K V δµ, q→0 q→0 e leading to the relationship lim χ(q, 0) = −n2 K . (4.159) q→0 e

153 In terms of the dielectric constant it implies that

4πn2e2 lim ||(q, 0) = 1 + K → ∞ . (4.160) q→0 q2

The dielectric constant at zero frequency diverges when the momentum vanishes, provided the compressibility is finite. What is the meaning of a finite or vanishing compressibility? We have shown that 1 ∂N K = . n2V ∂µ It the system is a metal, the number of electrons is a continuos and increasing function of the chemical potential, hence K is finite. On the contrary, if the system is an insulator, the chemical potential lies within an energy gap so that the number of electrons stays constant upon small changes of µ, implying K = 0. In other words a finite or vanishing compressibility discriminates bewteen metals and insulators. Equivalently, a metal has a longitudinal dielectric constant at zero frequency which diverges as q → 0, while for an insulator this limit is finite. To better appreciate this difference, let us consider the response to an external point-like charge, for instance at position x0

−iq·x0 ρext(x, t) = Z δ(x − x0), ρext(q, ω) = Z δ(ω) e .

According to our deﬁnitions, the induced density is

4πe2 4πe2 χ(q, ω) e −iq·x0 ρind(q, ω) = − 2 χ(q, ω) ρext(q, ω) = −Z 2 δ(ω) e q q ||(q, ω) (q, 0) − 1 (q, 0) − 1 −iq·x0 || || = Z δ(ω) e = ρext(q, ω) . ||(q, 0) ||(q, 0)

Therefore the net charge inside the system is 1 e (ρext(q, ω) − ρind(q, ω)) = e ρext(q, ω) . ||(q, 0)

Since ||(q, 0) diverges for small q in a metal, then, in the same limit, the total charge inside the system vanishes, which implies that in a metal the induced charge is always such as to perfectly screen the external one,

Transverse response We already know that lim χ||(q, ω) = 0. q→0

154 Since for q → 0 at fixed frequency there is no difference bewteen transverse and longitudinal fields, this also implies that lim χ⊥(q, ω) = 0 . q→0 f Through Eqs. (4.141) and (4.137) we then find

2 ωp lim σ⊥(q, ω) = − , (4.161) q→0 4πiω

2 ωp lim ⊥(q, ω) = 1 − . (4.162) q→0 ω2

On the other hand we also know that

lim χ(q, 0) = −n2 K, q→0 e so that from the gauge relation we should have n lim χ||(q, 0) = − . q→0 e m

The transverse response at ω = 0 is the response to a finite magnetic field in the absence of electric one, as E⊥ = iω A⊥/c. Since the transverse conductivity is not expected to be singular in the presence of a magnetic field, we have to conclude from (4.141) that also n lim lim χ⊥(q, ω) = − . q→0 ω→0 e m

The approach to this limiting value should be regular, so that we expect that n q2 lim lim χ⊥(q, ω) = − 1 − , (4.163) q→0 ω→0 e 2 m AkF 10 with A a dimensionless constant and kF the Fermi momentum. . Through Eq. (4.141) we then ﬁnd 2 2 q lim −4πiω σ⊥(q, ω) = ωp , q→0 2 AkF which inserted in (4.145) gives

2 2 Ac kF A⊥ int(q, 0) = 2 2 2 2 2 2 A⊥ ext(q, 0), (4.164) Ac kF + ωp + 4π c A kF (gµB) χσ 10A = 4 for free electrons

155 being χσ the magnetic susceptibility, which we know is negative. Since an ω = 0 transverse vector potential corresponds to a constant magnetic field in the absence of electric one, (4.164) implies that the magnetic field can penetrate inside the metal. The Zeeman term would lead to an increased amplitude with respect to the applied one (Landau paramagnetism), while the free-carrier term to a decrease (Landau diamagnetism).11 On the contrary, since ||(q, 0) diverges as q → 0, from (4.136) one realizes that a longitudinal field does not propagate freely inside a metal. Indeed, using Eq. (4.160), we can write 4πn2e2 (q, 0) = 1 + K f(q), || q2 with f(0) = 1. If f(q) were analytic, then through q2 A (q, 0) = A (q, 0), || int q2 + 4πn2e2Kf(q) || ext and upon Fourier transform in real space, we would find a longitudinal vector potential decaying exponentially inside the sample, with a decay length 1 λ = , TF 4πn2e2 which is called the Thomas-Fermi screening length.12 In reality the function f(q) is non-analytic. It has a branch cut starting from q = 2kF due to the Fermi surface. The reason is that it is not possible to create a particle-hole excitation at zero-frequency if the transfered momentum q > 2kF , hence the imaginary part of the density-density response function contains a step-function which turns into a log-singularity in the real part. The final result is that the longitudinal field 3 far away from the external source decays as cos(2kF r)/r , so called Friedel oscillatory behavior.

4.6.4 Power dissipated by the electromagnetic field In the presence of an external longitudinal field, the power dissipated according to the general formula is e2 ω 2 00 e2 ω 2 W = A|| ext(q, ω) χ (q, ω) = − A|| ext(q, ω) Im χ||(q, ω) c 2 || c 2 11For free electrons the diamagnetic term is 1/3 of the paramagnetic one, so the field is actually enhanced within the interior of the system. 12Indeed this is the same screening length one would obtain within the Thomas-Fermi approximation. Namely, if we consider the equation for the internal scalar potential in the presence of a time-independent external charge 2 ∇ φ(x) = −4πρext(x) − 4πρind(x), and we assume a very weak space-dependence so that e ∂N ρ (x) ' − (−eφ(x)) = e2n2K φ(x), ind V ∂µ as if (−eφ(x)) acts like a chemical potential shift, we do find the Thomas-Fermi screening length.

156 e 2 ω 2 − E|| ext(q, ω) Im χ||(q, ω), ω 2

00 where χ||(q, ω) is the dissipative current-current response function which is equal to minus the imaginary part of the linear response function. On the other hand ω σ||(q, ω) Im χ||(q, ω) = Im −i 2 e ||(q, ω) ω σ||(q, ω) ω ω 1 = − 2 Re = − 2 Re ||(q, ω) − 1 e ||(q, ω) e 4πi ||(q, ω) ω2 1 = 2 Im , 4πe ||(q, ω)

ω σ||(q, ω) Im χ||(q, ω) = − 2 Re e ||(q, ω)   ω σ (q, ω) ω 1 = − Re  ||  = − Re σ (q, ω), e2  4πi  e2 | (q, ω)|2 || 1 + σ (q, ω) || ω || which, through |E|| ext| = |||| |E|| int|, implies that

ω 1 2 1 2 W = − Im E|| ext(q, ω) = Re σ||(q, ω) E|| int(q, ω) . (4.165) 8π ||(q, ω) 2

Therefore the power dissipation in the presence of a longitudinal field is controlled either by the imaginary part of the inverse dielectric constant or by the real part of the conductivity, depending whether one has access to the applied field or to the full internal one. In the case of a transverse field e2 ω W = − |A (q, ω)|2 Im χ (q, ω), c 2 ⊥ int f⊥ where ω ω2 Im χ (q, ω) = − Re σ (q, ω) = − Im (q, ω). f⊥ e2 ⊥ 4πe2 ⊥ Therefore ω2 1 W = Re σ (q, ω) |A (q, ω)|2 = Re σ (q, ω) |E (q, ω)|2 2c2 ⊥ ⊥ int 2 ⊥ ⊥ int ω ω2 ω = Im (q, ω) |A (q, ω)|2 = Im (q, ω) |E (q, ω)|2 . (4.166) 8π c2 ⊥ ⊥ int 8π ⊥ ⊥ int

157 z

E i E r

k i k r

x E t k t Figure 4.2: Experimental geometry.

4.6.5 Reflectivity Let us suppose to shoot a beam of light normally to the surface of the sample, which is in contact with the vacuum. The experimental geometry is drawn in Fig. 4.2. The normal direction is the z-direction, so that z > 0 is the vacuum and z < 0 the sample, and the electric field is polarized along the x-direction. By the geometry of the scattering experiment, the incident, reflected and transmitted waves have all the wave-vector along z, ki, kr and kt, respectively. Through Eq. (4.143) it derives that the electromagnetic field propagates inside the sample if ω 2 q2 = (q, ω). c ⊥

Since tha vacuum has ⊥ = 1, the conservation of frequency implies 2 2 2 kt ki = kr = , ⊥(kt, ω) hence kr = −ki > 0. If the sample is non-magnetic, the electric and magnetic ﬁelds, which are both parallel to the surface, should be continuous through the interface vacuum-sample:

x x x y y y Ei + Er = Et ,Bi + Br = Bt . Notice that c By = k Ex, i ω i i y y y and analogously for the reﬂected and transmitted components. Hence Bi + Br = Bt implies, x x x thorugh Ei + Er = Et , that x x x p x x kr (Er − Ei ) = kt Et = − ⊥(kt, ω)kr (Ei + Er ) ,

158 which leads to p x 1 − ⊥(kt, ω) x Er = p Ei . 1 + ⊥(kt, ω) Therefore the reﬂection coeﬃcient is 2 E 2 1 − p (k , ω) r ⊥ t R = = p . (4.167) Ei 1 + ⊥(kt, ω) Since the propagating wave-vector of the light is negligible at the typical excitation frequencies of the sample, we can approximate ⊥(kt, ω) ' ⊥(0, ω). In a metal we showed that

ω2 (0, ω) = 1 − p , ⊥ ω2 which is real and negative for |ω| < ωp. Through Eq. (4.167) it implies that a metal reﬂects perfectly for frequencies below the plasma frequency.

159 4.7 Random Phase Approximation for the electron gas

So far we have introduced the linear response functions of an electron gas and discussed some limiting values which can be obtained without resorting to any explicit calculation. In this section we will evaluate the response functions within the time-dependent Hartree-Fock approximation. As before we consider an electron gas in the presence of a positive background which neu- tralizes its charge. The Hamiltonian is

X 1 X H = c† c + U(q)(ρ(q) − V nδ )(ρ(−q) − V nδ ) , (4.168) k kσ kσ 2V q0 q0 kσ kpq

2 2 2 2 where k = ~ |k| /2m is the kinetic energy, U(q) = 4πe /|q| and ρ(q) the Fourier transforms of the Coulomb interaction and of the electron density, respectively, and n the positive background density. By deﬁnition the Fourier transform of the electron density at q = 0 is equal to V n, thus canceling the positive background charge. Therefore the interaction term is actually

1 X X H = U(q) ρ(q)ρ(−q). (4.169) int V q6=0 kp

Let us start with the Hartree-Fock approximation. If we assume that spin-rotational and space-translational symmetries are preserved in the actual ground state, then we are forced to search for an Hartree-Fock wave-function whose only variational parameters are

† hckσckσi = nkσ, (4.170) with nk↑ = nk↓ = nk. Then, for q 6= 0,

X X † † ρ(q)ρ(−q) = ckσck+qσ cp+qσ0 cpσ0 kp σσ0 X † † → − nk ck+qσck+qσ + nk+q ckσckσ, kσ so that the Hartree-Fock Hamiltonian is   X 1 X H = − U(q) n c† c , (4.171) HF  k V k+q kσ kσ kσ q6=0 which is already diagonal with eigenvalues

1 X = − U(q) n . k HF k V k+q q6=0

160 Let us assume that the Fermi-sea solution, i.e. nk = 1 if |k| < kF and zero otherwise, is still the choice which minimizes the total energy, and let us apply the time-dependent Hartree-Fock machinery to evaluate the response to an external ﬁeld Vext(q, t) which couples to the density. We readily ﬁnd that the equation of motion of the average values

† ∆k,k+q; σ = hckσck+qσi, is, at linear order,

(i~∂t − k+q HF + k HF ) ∆k,k+q; σ =   1 X (n − n ) V (q, t) + ∆ 0 (U(q) − δ 0 U(k − p)) . k k+q  ext V p,p+q; σ σσ  pσ0

Let us solve this equation neglecting the second term, i.e. the Fock contribution, in the right hand side, which corresponds to the so-called Random Phase Approximation (RPA). To be consistent, the RPA approximation also amounts to neglect the interaction correction to the HF eigenvalues, which only derives from the Fock term, so that k HF → k. Within RPA we ﬁnd that   nk − nk+q U(q) X ∆k,k+q; σ = Vext(q, t) + ∆p,p+q; σ0  . (4.172) ω − k+q + k V pσ0 The induced charge is deﬁned for as

X † ρind(q, ω) = hckσck+qσi − V nδq0, kσ and satisfy the RPA equation !   X nk − nk+q U(q) X ρind(q, ω) = Vext(q, t) + ∆p,p+q; σ0  ω − k+q + k V kσ pσ0 U(q) ≡ V χ (q, ω) V (q, t) + ρ (q, ω) , eRPA ext V ind where 1 X nk − nk+q χeRPA(q, ω) = (4.173) V ω − k+q + k + iη kσ corresponds to the proper density-density response function within RPA, the inﬁnitesimal η > 0 playing the role of the adiabatic switching rate. χeRPA(q, ω) is also the density-density response function of free electrons, also called the Lindhart function, which also clariﬁes the meaning of the RPA approximation. As we know the

161 proper response function is the response to the external field plus the electric field generated by the same electrons. In order not to double-count this term, the proper response function has to be evaluated neglecting the contribution of the electron-electron Coulomb repulsion which represents the coupling of the electrons to the average electric field generated by the same electrons, which is roughly as if the eletrons were actually non-interacting. We leave as an exercise the evaluation of the Lindhart function in a three-dimensional. We just mention that the RPA approximation preserves the sum-rule (4.147), hence all the results which derives from that. Concerning the limiting value (4.159), within RPA the compressibility turns out to be that one for free electrons. One can perform the same analysis with an external probe which couples to the transverse current. The result would be a transverse current-current response function which, within the RPA approximation, coincides with that of free electrons. Also in this case the exact limiting values we have previously extracted are reproduced. In other words the Random Phase is a consistent Approximation.

Excercises

2 2 (3) Calculate the Lindhart function for an k = ~ k /2m dispersion in three dimensions. (4) Calculate the lim lim χ⊥(q, ω), q→0 ω→0 e of the transverse current-current response function for free electrons. Show that the limit coincides with (4.163) with A = 4. Prove that the diamagnetic term is actually 1/3 of the paramagnetic one.

162 Chapter 5

Feynman diagram technique

Time-dependent Hartree-Fock is a relatively simple way to access excitation properties thus going beyond Hartree-Fock theory. A more systematic scheme, which includes time-dependent Hartree-Fock as an approximation, is to perform perturbation theory. However the perturbation expansion in a many-body problem is not as straightforward as in a single-body case but can be simpliﬁed substantially by means of the so-called Feynman diagram technique. In this chapter we present this technique at ﬁnite temperature.

5.1 Preliminaries

First of all we need some preliminary deﬁnitions and results.

5.1.1 Imaginary-time ordered products Let us introduce ﬁrst the imaginary time evolution of an operator A(x) through

A(x, τ) = eH τ A(x) e−H τ , (5.1) where H is the fully interacting Hamiltonian and τ ∈ [0, β], where β = 1/T is the inverse temperature (we use ~ = 1 and KB = 1). Given two operators A(x) and B(y) we deﬁne the imaginary time Green’s function through

0 0 GAB(x, y; τ − τ ) = −hTτ A(x, τ) B(y, τ ) i = −θ(τ − τ 0) hA(x, τ) B(y, τ 0)i ∓ θ(τ 0 − τ) hB(y, τ 0) A(x, τ)i. (5.2) where Tτ denotes the time-ordered product, and the minus sign applies when one or both the operators are bosonic-like, namely contain bosons or an even product of fermionic operators, while the plus sign when both operators are fermionic-like, i.e. contain an odd product of

163 fermionic operators. Due to time-translation invariance, the Green’s functions only depend on the time diﬀerence τ − τ 0 ∈ [−β, β]. Analogously, given n operators, Ai(xi, τi), i = 1, . . . , n, we can deﬁne the multi-operator Green’s function

GA1,...,An (x1,..., xn; τ1, . . . , τn) = −hTτ (A1(x1, τ1) ...An(xn, τn))i, (5.3) which is the average of the time ordered product, namely operators with earlier times appear on the right of those at later times, having a minus sign if the number of crossings needed to bring fermionic-like operators in the correct time-ordered sequence is even and the sign plus otherwise.

5.1.2 Matsubara frequencies

Let us consider the two-operator Green’s function GAB(x, y; τ), with τ ∈ [−β, β] being the time diﬀerence. Since the time domain is bounded, we can introduce a discrete Fourier transform, with frequencies ωn = (2π/2β) n = π n T , with n an integer. We deﬁne, dropping for simplicity the spatial dependence, Z β 1 iωnτ GAB(iωn) = dτ e GAB(τ), 2 −β and, consequently, X −iωnτ GAB(τ) = T e GAB(iωn). n From the properties of the trace it derives that 1 h i hB(0) A(τ)i = Tr e−β H B eH τ A eH− τ Z 1 h i = Tr e−β H eβ H eH τ A e−H τ e−β H B = hA(β + τ) B(0)i, Z which implies that Z β 1 iωnτ GAB(iωn) = dτ e GAB(τ) 2 −β 1 Z β 1 Z 0 = − dτ eiωnτ hA(τ) B(0)i ∓ dτ eiωnτ hB(0) A(τ)i 2 0 2 −β 1 Z β 1 Z 0 = − dτ eiωnτ hA(τ) B(0)i ∓ dτ eiωnτ hA(β + τ) B(0)i 2 0 2 −β 1 Z β 1 Z β = − dτ eiωnτ hA(τ) B(0)i ∓ e−iωnβ dτ eiωnτ hA(τ) B(0)i 2 0 2 0 Z β 1 −iωnβ iωnτ = 1 ± e dτ e GAB(τ). 2 0

164 n Since exp (−iωnβ) = (−1) then a non-zero Fourier transform requires for bosonic-like operators even integers n’s and for fermionic-like odd n’s. We deﬁne the bosonic Matsubara frequencies

Ωm = 2m π T (5.4) and the fermionic Matsubara frequencies

ωm = (2m + 1) π T, (5.5) so that the bosonic- and fermionic-like Fourier transform of GAB(τ) become, respectively,

Z β Z β iΩmτ iωmτ GAB(Ωm) = dτ e GAB(τ),GAB(ωm) = dτ e GAB(τ), (5.6) 0 0 hence just require the knowledge of the Green’s functions for positive τ.

Connection between bosonic Green’s functions and linear response functions Let us assume that both A and B are observable quantities, which implies they are hermitean, hence bosonic-like operators. Then, for positive τ, we have

1 X 1 G (iΩ ) = e−βEn − e−βEm hn|A|mihm|B|ni. (5.7) AB l Z iΩ − (E − E ) nm l m n Comparing this expression with the Fourier transform in frequency of the linear response function χAB(ω), see Eq. (3.33), we easily realize that the two coincide if we analytically continue GAB(iΩl) on the real axis, iΩl → ω + iη. In other words the knowledge of GAB(iΩl) means that we know the value of the function GAB(z) of the complex variable z on an inﬁnite set of points on the imaginary axis zl = iΩl. This is suﬃcent to determine the full GAB(z) in the complex plane, with the supplementary condition that, since GAB(z) = χAB(z), it has to be analytic in the upper half-plane.

165 Useful formulas There is a very useful trick to perform summation over Matsubara frequencies. Let us start from the fermionic case. Suppose we have to calculate X T F (in), n where we assume that F (z) vanishes faster than 1/|z| for |z| → ∞ and has poles but on the imaginary axis. We notice that the Fermi distribution function, as function of a complex variable, 1 f(z) = , eβ z + 1 has poles at π z = i (2n + 1) = i , n β n namely right at the Matsubara frequencies, with residue −T . Let us consider the integral along the contour shown in Fig. 5.1 Im(z)

Re(z)

Figure 5.1: Integration countour, see text.

I dz I = f(z) F (z). 2πi This countour integral can be calculated by catching all poles of the integrand in the region enclosed by the countour which includes the imaginary axis, shaded area in Fig. 5.1. Since this area is enclosed clockwise the integral gives X X I = − Res (f(zn)) F (zn) = T F (in), zn n

166 which is just the sum we want to calculate. On the other hand, I can be equally calulated by catching all poles of the integrand in the area which does not include the imaginary axis, the non-shaded region, which, being enclosed anti-clockwise gives X I = f(z∗) Res (F (z∗)) .

z∗= poles of F (z) Therefore X X T F (in) = f(z∗) Res (F (z∗)) . (5.8)

n z∗= poles of F (z) Notice that, if F (z) has branch cuts the calculation is slightly more complicated since we have to deform the contour as to avoid branch cuts. In this case, instead of catching poles we have to integrate along branch cuts. For instance, suppose that F (z) has a branch cut at z = x + iω, with x ∈ [−∞, ∞], as shown in Fig. 5.2. Let us consider the non-shaded area enclosed inside Im(z)

Re(z)

Figure 5.2: Integration countour in the presence of a branch cut, see text. the countour depicted in Fig. 5.2. In this area there are no poles hence the contour integral is zero. On the other hand this contour integral is also equal to the contribution of the poles inside the shaded area, which includes the imaginary axis, plus the integral along the contour which encloses the branch cut. Therefore X Z dx T F (i ) = − f(x + iω) F (x + iω + i0+) − F (x + iω − i0+) . (5.9) n 2πi n

In the case in which the summation is performed over bosonic frequencies, we can use similar tricks once we recognize that the poles of the Bose distribution function 1 b(z) = , eβ z − 1

167 coincide with the bosonic Matsubara frequencies, π z = i 2n , n β with residue T .

5.1.3 Single-particle Green’s functions Among the averages of time-ordered products an important role in the diagrammatic technique is played by the so-called single-particle Green’s functions.

Fermionic case

If Ψσ(x, τ) is the imaginary-time evolution of the Fermi ﬁeld, then the single-particle Green’s functions is deﬁned through

† Gσσ0 (x, y; τ) = −hTτ Ψσ(x, τ)Ψσ0 (y) i † † = −θ(τ)hΨσ(x, τ)Ψσ0 (y) i + θ(−τ)hΨσ0 (y) Ψσ(x, τ) i. (5.10)

If we make a spectral representation as before and calculate the Fourier transform using the fermionic Matsubara frequency, we readily ﬁnd that 1 X −βEn −βEm 1 † G 0 (x, y; iω ) = e + e hn|Ψ (x) |mihm|Ψ 0 (y) |ni. σσ l Z iΩ − (E − E ) σ σ nm l m n

Let us take σ = σ0 and x = y, which correspond to the so-called local Green’s function, and introduce the real and positive spectral function

1 X 2 A (x, ) = e−βEn + e−βEm hn|Ψ (x) |mi δ ( − E + E ) , σ Z σ m n nm through which, after continuation in the complex plane iωl → z Z 1 G (x, x; z) = d A (x, ) . σσ σ z −

As function of the complex frequency, G(z) has generally branch cut singularities along the real axis. Indeed

+ + Gσσ(x, x; z = ω + i0 ) − Gσσ(x, x; z = ω − i0 ) = −2π i Aσ(x, ω).

168 What is the physical meaning of the spectral function? Let us rewrite Aσ(x, ω) in the following equivalent way:

X e−βEn 2 A (x, ) = hm|Ψ (x)†|ni δ ( − E + E ) σ Z σ m n nm X e−βEn 2 + hm|Ψ (x) |ni δ ( + E − E ) , (5.11) Z σ m n nm which shows more explicitly that Aσ(x, ω) is just the local density of states for adding, ﬁrst term in the right hand side, or removing, second term, a particle at position x with spin σ. This quantity can be for instance measured in tunneling experiments. Finally we notice that

In all this chapter we are going to assume a grand-canonical ensamble for the fermions, which is equivalent to add to the fermionic Hamiltonian H a chemical potential term, i.e. H → H−µN, where µ is such that the average number of particles has the desired value N0. At T = 0 µ becomes the Fermi energy and selects as the ground state the lowest energy state with N0 particles. Therefore, at T = 0 the sum over n is reduced just to the ground state, |0,N0i, with N0 particles, hence 2 X † Aσ(x, ) = δ + E0(N0) − Em(N0 + 1) hm, N0 + 1|Ψσ(x) |0,N0i m 2 X + δ − E0(N0) + Em(N0 − 1) hm, N0 − 1|Ψσ(x) |0,N0i , m where we explicitly indicate that the ground state energy refers to N0 particles and the ﬁrst term in the right hand side involves moving to the subspace with N0 + 1 electrons, and the second to the subspace with N0 − 1 electrons. The ﬁrst δ-function implies that

= Em(N0 + 1) − E0(N0) = (Em(N0 + 1) − E0(N0 + 1)) + (E0(N0 + 1) − E0(N0)) ≥ 0,

169 because the energy of the state m with N0 + 1 electrons is greater or equal to the ground state energy E0(N0 + 1), and, by deﬁnition, E0(N0 + 1) − E0(N0) ≥ 0 since the lowest energy state for any N occurs at N0. Analogously, in the second δ-function

= − (E0(N0 − 1) − E0(N0)) − (Em(N0 − 1) − E0(N0 − 1)) ≤ 0. In other words, for positive the spectral function is the density of states upon adding an electron and for negative values that one for removing one particle. Let us suppose now that the system is translationally invariant. If we then expand the Fermi ﬁelds in eigenstates of the crystalline momentum, and introduce the space Fourier transform of the Green’s function, we ﬁnd

† Gσσ0 (k, p; τ) = −hTτ ckσ(τ) cpσ0 i = δkp Gσσ0 (k, τ), where † Gσσ0 (k, τ) = −hTτ ckσ(τ) ckσ0 i. (5.12) In this case and for σ = σ0 the Fourier transform in the complex frequency plane would be Z 1 G (k, z) = d A (k, ) , σσ σ z − where 1 X 2 A (k, ) = e−βEn + e−βEm hn|c |mi δ ( − E + E ) , (5.13) σ Z kσ m n nm is the density of states for adding and removing a particle at momentum k. Since A(k, ) is real and positive, G(k, z) has generally branch cuts on the real axis, since

+ + Gσσ(k, + i0 ) − Gσσ(k, − i0 ) = −2πi Aσ(k, ). (5.14) Notice that, as before, the spectral function is normalized to one. In addition one can readily verify that 1 X A (x, ) = A () = A (k, ). (5.15) σ σ V σ k

Let us consider non-interacting electrons described by the Hamiltonian

X † H0 = k ckσckσ, kσ where, as discussed previously, k is measured with respect to the chemical potential µ, which is the Fermi energy at T = 0. In this case it is easy to show that

(0) (0) Gσσ0 (k, iωn) = δσσ0 G (k, iωn),

170 where (0) 1 G (k, iωn) = . (5.16) iωn − k The spectral function is simply (0) A (k, ) = δ( − k), (5.17) and is ﬁnite for > 0 if k is outside the Fermi surface, and for < 0 otherwise. Notice that a small frequency corresponds to small k, i.e. to momenta close to the Fermi surface.

Bosonic case Let us define as Φ(x) the Bose field for spinless bosons. Analogously to the fermionic case we can define a single-particle Green’s function through

† G(x, y; τ) = −hTτ Φ(x, τ) Φ(y) i = −θ(τ)hΦ(x, τ) Φ(y)†i − θ(−τ)hΦ(y)† Φ(x, τ) i.

Let us again assume that the model is translationally invariant and introduce the bosonic oper- † ators aq and aq, as well as the conjugate variables

r1 x = a + a† , q 2 q −q r1 p = −i a − a† . q 2 q −q In most common situations, the object which appears within perturbation theory is, rather than the single-particle Green’s function, the x − x time-ordered product, namely

D(q, τ) = −hTτ (xq(τ) x−q)i. (5.18)

Let us consider the bosonic non-interacting Hamiltonian

X † H0 = ωq aq aq, q with ωq = ω−q. Then, for positive τ, 1 1 D(q, τ) = − ha (τ) a† i − ha† (τ) a i 2 q q 2 −q −q 1 1 = − e−ωqτ (1 + b(ω )) − eωqτ b(ω ), 2 q 2 q

171 where −1 † ωqβ b(ωq) = haq aqi = e − 1 , is the Bose distribution function, and −1 † −ωqβ haq aqi = 1 + b(ωq) = − e − 1 .

Noticing that Z β 1 dτ eiΩmτ e∓ωqτ = e∓ωqβ − 1 , 0 iΩm ∓ ωq one readily ﬁnds that (0) 1 1 1 ωq D (q, iΩm) = − = − 2 2 . (5.19) 2 iΩm − ωq iΩm + ωq Ωm + ωq

5.2 Perturbation expansion in imaginary time

Let us suppose that the full Hamiltonian

H = H0 + V, where H0 is a single particle Hamiltonian which can be diagonalized exactly while V is a perturbation which makes the model not solvable, e.g. it is the electron-electron interaction. We already showed in Section 4.2 that

e−τ H = e−τ H0 S(τ), where, see Eq. (4.34), Z τ S(τ) ≡ S(τ, 0) = Tτ exp − dτ1 V (τ1) , (5.20) 0 being V (τ) the Heisenberg evolution of V with the non-interacting Hamiltonian. It is easy to show that, for τ ≥ τ 0,

" Z τ−τ 0 !# Z τ 0 0 S(τ − τ , 0) = Tτ exp − dτ1 V (τ1) = Tτ exp − dτ1 V (τ1 − τ ) 0 τ 0 τ 0 Z 0 0 0 −τ H0 τ H0 −τ H0 0 τ H0 = e Tτ exp − dτ1 V (τ1) e ≡ e S(τ, τ ) e , τ 0 namely that 0 0 e−(τ−τ ) H = e−τ H0 S(τ, τ 0) eτ H0 . (5.21)

172 Suppose we have to calculate the multi-operator Green’s function (5.3), and further assume that, e.g., τ1 ≥ τ2 ≥ · · · ≥ τn−1 ≥ τn, then

GA1,...,An (τ1, . . . , τn) = −hA1(τ1) ...An(τn)i 1 h i = Tr e−βH eτ1H A e−τ1H eτ2H A e−τ2H ... e−τn−1H eτnH A e−τnH Z 1 2 n 1 h = Tr e−βH0 S(β, τ ) eτ1H0 A e−τ1H0 S(τ , τ ) eτ2H0 A e−τ2H0 ... Z 1 1 1 2 2 τnH0 −τnH0 ...S(τn−1, τn) e An e S(τn, 0) . We readily see that all times remain ordered, so that for a generic time-ordering we would get

GA1,...,An (τ1, . . . , τn) = −hTτ (A1(τ1) ...An(τn))i 1 h i = Tr e−βH0 T (S(β) A (τ ) ...A (τ )) , Z τ 1 1 n n where the time evolution of the Ai’s operators in the last equation is through the non-interacting Hamiltonian. Recalling that Z = Z0 hS(β)i, we therefore conclude that 1 G (τ , . . . , τ ) = − hT (S(β) A (τ ) ...A (τ ))i, (5.22) A1,...,An 1 n hS(β)i τ 1 1 n n where the thermal averages as well as the imaginary-time evolution are done with the non- interacting Hamiltonian H0. This expression is now suitable for an expansion in V .

5.2.1 Wick’s theorem Upon expanding S(β) in powers of the perturbation V , the calculation of any Green’s function reduces to evaluate the average value of a time-ordered product of Fermi or Bose ﬁelds with a non interacting Hamiltonian. It is therefore essential to know how to perform this calculation. Let us suppose to evaluate with a non-interacting Hamiltonian for free fermions the average value 0 † 0 † −hTτ Ψ(x1, τ1) Ψ(x2, τ2) ... Ψ(xn, τn) Ψ(yn, τn) ... Ψ(y1, τ1) i. For any time-ordering this amounts to average a product of creation and annihilation operators over a non interacting Hamiltonian. We already know that this is the sum of the products of all possible contractions of an annihilation with a creation operator. Suppose that the operator 0 † 0 Ψ(xi, τi) is contracted with Ψ(yj, τj) . If all other times but τi and τj are the same, there are 0 two cases: if τi ≥ τj we will have to evaluate the contraction

0 † hΨ(xi, τi) Ψ(yj, τj) i;

173 0 if τi ≤ τj we need to interchange the two operators, which leads to a minus sign hence to

0 † −hΨ(yj, τj) Ψ(xi, τi)i.

Both cases can be described by the single formula

0 −G(xi, yj; τi − τj).

This argument can be extended to all other contractions, justifying the result that The average value with a non-interacting Hamiltonian of the time-ordered product of Fermi ﬁelds is obtained by contracting in all possible ways an annihilation with a creation operator and regarding any contraction as a non-interacting single-particle Green’s function. The sign of each product of contractions is (−1)n (−1)L, where n is the number of annihilation operators, which has to be equal to the number of creation operators, and L is the number of crossings needed to bring each creation operator to the right of the annihilation operator with which it is contracted. This is the so-called Wick’s theorem. For instance

0 † 0 † −hTτ Ψ(x1, τ1) Ψ(x2, τ2) Ψ(y2, τ2) Ψ(y1, τ1) i (0) 0 (0) 0 (0) 0 (0) 0 = −G (x1, y1; τ1 − τ1) G (x2, y2; τ2 − τ2) + G (x1, y2; τ1 − τ2) G (x2, y1; τ2 − τ1).

In the case of non interacting bosons the Wick’s theorem does not hold rigorously. Yet, if we work with a translationally invariant model and if there is no Bose condensation, the Wick’s theorem still works, this time without any phase-factor (−1)L, apart from corrections which vanish in the thermodynamic limit.

5.3 Perturbation theory for the single-particle Green’s function and Feynman diagrams

Let us consider the Hamiltonian for free electrons

X † H0 = k ckσckσ, (5.23) kσ in the presence of an interaction Z 1 X † † H = dx dy Ψ (x) Ψ 0 (y) U(x − y)Ψ 0 (y)Ψ (x) , int 2 σ σ σ σ σσ0

174 which will be our perturbation V . In the S-matrix it appears Z Z Z 1 X † † dτ H (τ) = dx dy dτ Ψ (x, τ) Ψ 0 (y, τ) U(x − y)Ψ 0 (y, τ)Ψ (x, τ) int 2 σ σ σ σ σσ0 Z Z 1 X 0 † 0 † 0 0 = dx dy dτ dτ Ψ (x, τ) Ψ 0 (y, τ ) U(x − y) δ(τ − τ )Ψ 0 (y, τ )Ψ (x, τ) 2 σ σ σ σ σσ0 Z 1 X † † ≡ dx dy Ψ (x) Ψ 0 (y) U(x − y)Ψ 0 (y)Ψ (x) , 2 σ σ σ σ σσ0 where we have introduced the multicomponent coordinates x = (x, τ), y = (y, τ 0) and by definition U(x − y) = U(x − y) δ(τ − τ 0). Therefore " Z !# 1 X † † S(β) = T exp − dx dx Ψ (x ) Ψ 0 (x ) U(x − x )Ψ 0 (x )Ψ (x ) . τ 2 1 2 σ 1 σ 2 1 2 σ 2 σ 1 σσ0 The single-particle Green’s function is, as demonstrated previously, 1 G (x, y) = − hT S(β)Ψ (x)Ψ (y)† i, (5.24) σ hS(β)i τ σ σ where the averages are done with the non interacting Hamiltonian. Let us calculate it up to first order in perturbation theory, namely with Z (0) (1) 1 X † † S(β) ' S (β) + S (β) = 1 − dx dx Ψ (x ) Ψ 0 (x ) U(x − x )Ψ 0 (x )Ψ (x ) . 2 1 2 σ 1 σ 2 1 2 σ 2 σ 1 σσ0 We need to calculate up to first order the numerator and the denominator of (5.24). About the numerator and using Wick’s theorem we find

† (0) −hTτ S(β)Ψσ(x)Ψσ(y) i = Gσ (x, y) 1 X Z h i + dx dx U(x − x ) hT Ψ (x)Ψ (y)† Ψ (x )†Ψ (x )† Ψ (x )Ψ (x ) i 2 1 2 1 2 τ σ σ α 1 β 2 β 2 α 1 αβ (0) (1) = Gσ (x, y) 1 + hS (β)i Z 1 X h (0) + dx dx U(x − x ) δ G(0)(x, x ) G(0)(x , y) G (x , x ) 2 1 2 1 2 σα σ 1 σ 1 β 2 2 αβ (0) (0) (0) i −δσα δαβ Gσ (x, x1) Gσ (x1, x2) Gσ (x2, y)

175 x x y = G(x,y) = U(x−y) x y (0) = G (x,y) y

Figure 5.3: Graphical representation of the Green’s functions and the interaction.

1 X Z h + dx dx U(x − x ) δ G(0)(x, x ) G(0)(x , y) G(0)(x , x ) 2 1 2 1 2 σβ σ 2 σ 2 α 1 1 αβ (0) (0) (0) i −δσβ δβα Gσ (x, x2) Gσ (x2, x1) Gβ (x1, y)

The last two integrals are actually equal since we can interchange x1 ↔ x2, so that

† (0) (1) −hTτ S(β)Ψσ(x)Ψσ(y) i = Gσ (x, y) 1 + hS (β)i Z X h (0) (0) (0) + dx1 dx2 U(x1 − x2) δσα Gσ (x, x1) Gσ (x1, y) Gβ (x2, x2) αβ (0) (0) (0) i −δσα δαβ Gσ (x, x1) Gσ (x1, x2) Gσ (x2, y)

The denominator gives simply 1 + hS(1)(β)i so that, up to ﬁrst order  Z (0) (0) (0) X (0) Gσ(x, y) = Gσ (x, y) + dx1 dx2 U(x1 − x2) Gσ (x, x1) Gσ (x1, y) Gβ (x2, x2) β (0) (0) (0) i − Gσ (x, x1) Gσ (x1, x2) Gσ (x2, y) . (5.25)

Let us give a graphical representation of this result. We represent the fully interacting Green’s function as a line with an arrow, the non-interacting one as a tiner line and the interaction as a wavy line with four legs, see Fig. 5.3. An incoming vertex represent a creation operator, while an outgoing one an annihilation operator. With these notations the Green’s function up to first order in the interaction can be represented as in Fig 5.4. The conventions are that any internal coordinate is integrated, the spin is conserved along a Green’s function as well as at any interaction-vertex, and a wavy line is the interaction U. There are two first order diagrams. The tadpole one has a plus sign while the other one a minus sign. We notice that both diagrams are fully connected. Actually the disconnected pieces cancel with the denominator hS(β)i. One can go on and calculate the second order corrections to infer the rules for constructing diagrams. Here we just quote the final answer.

176 The n-th order diagrams for the single particle Green’s functions are all fully connected and topologically not equivalent diagrams which can be constructed with n interaction lines and 2n + 1 non-interacting Green’s functions, with external points at x and y. All internal coordinates are integrated. The sign of each diagram is (−1)n (−1)L, where L is the number of internal loops. Spin is conserved at each vertex, hence each loop implies a spin sum. Finally, if a Green’s function is connected by the same interaction line it has to be interpreted as the limit of τ → 0−, since in the interaction creation operators are on the left of the annihilation ones. One easily realizes that the following rules reproduce the two ﬁrst order corrections we have just derived. In Fig. 5.5 we draw all topologically inequivalent diagrams at second order in perturbation theory. Let us follow the above rules to ﬁnd the expression of the last two, (h) and (i). The former has no loops, hence its sign is simply (−1)n = (−1)2 = 1. The spin is the same for all lines hence its expression is

Z 4 Y (0) (0) (0) (0) (0) (h) = dxi U(x1 −x3) U(x2 −x4) Gσ (x, x1) Gσ (x1, x2) Gσ (x2, x3) Gσ (x3, x4) Gσ (x4, y). i=1

Diagram (i) has a loop, hence (−1)n (−1)L = (−1)2 (−1)1 = −1. The internal loop implies a spin summation, hence

Z 4 X Y (0) (0) (0) (0) (0) (i) = − dxi U(x1−x3) U(x2−x4) Gσ (x, x1) Gσ (x1, x2) Gσ (x2, y) Gβ (x3, x4) Gβ (x4, x3). β i=1

5.3.1 Diagram technique in momentum and frequency space Let us assume that, besides time-translational invariance, the system is also space translational invariant. Therefore let us derive the perturbation expansion of the Fourier transform of the

x y x y = +

x y x y + x 1 x1 x2

Figure 5.4: Graphical representation of the Green’s function up to ﬁrst order.

177 (b) (c) (a)

(d) (e) (f)

(g)

x x 4 (h) 3 (i)

x y x y x 1 x 2 x 3 x 4 x 1 x2

Figure 5.5: Graphical representation of the second order corrections to the Green’s function.

Green’s function Gσ(p, in). We further need to introduce the Fourier transform of the interaction line as well as we need to indicate a direction to this line, which is the direction along

p, iε n p+q, iε n+ iω l

U(q, i ω l ) q, iω l

ε k+q, iεm + iω l k, i m

Figure 5.6: Graphical representation of the interaction in Fourier space. which momentum and frequency ﬂow, see Fig. 5.6. Notice that the frequency carried by the interaction is the diﬀerence between two fermionic Matsubara frequencies, hence it is a bosonic one, namely an even multiple of π T . Moreover, since U(x − y) = U(x − y) δ(τ − τ 0), the Fourier transform is U(q, iωl) = U(q), independent of frequency. Moving to the perturbation expansion, since the spatial and time coordinates of each internal vertex are integrated out, momentum and frequency are conserved at each vertex, namely the sum of the incoming values is equal to that of the outgoing ones. Therefore the rules are: The n-th order diagrams for the single particle Green’s functions with momentum p and frequency in are all fully connected and topologically not equivalent diagrams which can be constructed with n interaction lines and 2n+1 non-interacting Green’s functions, with external lines at (p, in). At each vertex, the sum of momenta and

178 that of frequencies of the incoming lines must be equal to the those of outgoing lines. All internal momenta and frequency are summed up, sum over momentum, e.g. k, being 1 X , V k and over frequency, e.g. im, X T . m The sign of each diagram is (−1)n (−1)L, where L is the number of internal loops. Spin is conserved at each vertex, hence each loop implies a spin sum. Finally, if a Green’s function is connected by the same interaction line it has to be interpreted as the inverse Fourier transform at τ = 0−, i.e.

− X −im 0 T Gσ(k, im) e . m

For instance up to ﬁrst order the diagrams are those in Fig. 5.7 and are explicitly

X 1 X X (0) − G (p, i ) = G(0)(p, i ) + G(0)(p, i )2 U(0) T G (k, i ) e−im 0 σ n σ n σ n V β m m k β X 1 X − −G(0)(p, i )2 T U(q) G(0)(p − q, i − iω ) e−i(n−ωl) 0 , σ n V σ n l l q where we use the convention that denote fermionic Matsubara frequencies and ω bosonic ones.

p, i ε n p, i ε n = + k, i ε m

q, i ω 0, 0 l

p, i ε n p, i ε n p, i ε n p, i ε n + p−q, i ε n −i ω l

Figure 5.7: Diagrams for the single-particle Green’s function up to ﬁrst order in Fourier space.

179 Σ = = + +

+ +

Figure 5.8: Diagrams for the single-particle self-energy up to second order.

5.3.2 The Dyson equation Among all possible diagrams for the single particle Green’s function, we can distinguish two classes. The first includes all diagrams which, by cutting an internal Green’s function line, transform in two lower-order diagrams of the same Green’s function. These kind of diagrams are called single-particle reducible. For instance diagrams (a), (b), (c) and (d) in Fig 5.5 are single-particle reducible. The other class contains all diagrams which are not single-particle reducible, also called single-particle irreducible. For instance both first order diagrams as well as the second order ones (e) to (i) are irreducible. Let us for convenience introduce a generalized momentum p = (p, in) and let us define as the single-particle self-energy Σσ(p) the sum of all irreducible diagrams without the external legs. For instance, the self-energy diagrams up to second order are drawn in Fig. 5.8, where the self-energy is represented by a rounded box. It

Σ( p ) p p p p = +

Σ( p ) Σ( p ) p p p + + .....

Σ( p ) p p p = +

Figure 5.9: Dyson equation for the single-particle Green’s function. is not diﬃcult to realize that, in terms of the self-energy, the perturbation expansion can be

180 rewritten as in Fig. 5.9, which has the formal solution

G(0)(p, i ) 1 G (p, i ) = σ n = , (5.26) σ n (0) 1 − Gσ (p, in)Σσ(p, in) in − p − Σ(p, in) the last expression being valid for H0 in Eq. (5.23), which, being spin-independent like the interaction, can not produce two diﬀerent self-energies for spin-up and down electrons. This is the so-called Dyson equation for the single-particle Green’s function. There are therefore two possible ways of doing perturbation theory. The simplest is just to do perturbartion theory, say up to order n, directly for the Green’s function. The second is to obtain up to order n the self energy and insert its expression in the Dyson equation (5.26). This provides in principle an approximate Green’s function which contains all order in perturbation theory, so (5.26) actually represents a way to sum up perturbation theory. This latter is in reality the most physical way to proceed. The reason is that the perturbation is going to change the branch cut singularities of the Green’s function, which can be traced easier using (5.26) than directly from the perturbative expansion of G.1

Physical meaning of the self-energy The single particle Green’s function has a branch cut on the real axis, see (5.14), which implies that the real part is continuous but the imaginary part has a jump. This also means that the real part of the self-energy is continuous but the imaginary part is not, so that we can deﬁne

Σ0(k, ) = Re Σ(k, + iη) = Re Σ(k, − iη),

1A simple way to explain it is the following. Let us take the Green’s function for non-interacting fermions 1 G(in) = , in − E(V ) where E(V ) are the single-particle eigenvalues which depend on some Hamiltonian parameter V , so that

X n E(V = 0) = E0 + En V , n and the unperturbed Green’s function at V = 0 is

(0) 1 G (in) = . in − E0 The calculation of the self-energy in perturbation theory gives simply the perturbation expansion of the eigenvalues hence tells us how these eigenvalues changes with V . On the contrary the perturbation expansion of the Green’s function 2 2 3 (0) (0) 2 (0) (0) 2 G = G + V G E1 V1 + V G E2 + G E1 + ..., is of diﬃcult interpretation since at each order in perturbation theory all lower order corrections to the eigenvalue intervene.

181 Σ00(k, ) = Im Σ(k, + iη) = −Im Σ(k, − iη) ≤ 0, where η → 0+ and the last inequality follows from the fact that the spectral function is positive and given by 1 η − Σ00(k, ) A(k, ) = . 0 2 00 2 π − k − Σ (k, ) + η − Σ (k, )

For non-interacting electrons Σ = 0 and the spectral function tends towards a δ-function δ(−k) for η → 0. For interacting electrons this δ-function broadens on a width which is controlled by 00 Σ (k, ). This broadening reﬂects the fact that, for instance, a particle, k > 0, can decay by the interaction into a particle plus several particle-hole pairs provided that momentum is conserved. The product of this decay will have an energy which is spread around the non-interacting value 00 k. Therefore Σ (k, ) can be regarded as the decay rate of a particle with momentum k into an object with the same momentum but energy . This interpretation can be justiﬁed by analyzing the structure of the self-energy diagrams which contribute to the imaginary part. Let us consider for instance the second-order diagram for the self-energy Σ(k, i) drawn in Fig. 5.10, Neglecting

p ,iω 1 1

ε k 1 ,i 1

k −k + p ,i ε −i ε + iω 1 1 1 1

Figure 5.10: Second-order diagram contributing to the imaginary part of the self-energy. the momentum dependence of the interaction, its expression is 1 X X δΣ(k, i) = −2 U 2 T 2 δk − k − p + k V 2 0 1 1 k0 k1 p1 1 ω1 1 1 1 , i1 − k1 iω1 − p1 i + iω1 − i1 − k0 where the δ-function imposes the momentum conservation and the factor 2 comes from the loop spin summation. Let us ﬁrst sum over i1, which gives X 1 1 1 T = f(k1 ) i1 − k1 i + iω1 − i1 − k0 i + iω1 − k1 − k0 1 1 −f(i + iω1 − k0 ) i + iω1 − k0 − k1

182 1 = f(k1 ) − f(−k0 ) , i + iω1 − k0 − k1 where we used the fact the i + iω1 is a bosonic frequency. Notice that for T = 0 this term is

ﬁnite only if k1 and k0 are both particles or holes, namely if k0 k1 > 0. Now let us sum over iω1, X 1 1 1 T = f(p1 ) iω1 − p1 i + iω1 − k0 − k1 i + p1 − k0 − k1 ω1 1 +f(k0 + k1 − i) k0 + k1 − i − p1 1 = f(p1 ) + b(k0 + k1 ) . i + p1 − k0 − k1 Here we used the fact that, since i is fermionic, then β(k0 +k1 −i) β(k0 +k1 ) f(k0 + k1 − i) = e + 1 = − e + 1 = −b(k0 + k1 ).

Since at T = 0 b(x) = −θ(−x) and f(x) = θ(−x), the sum over iω1 vanishes if p1 < 0 and

k0 + k1 < 0 or if p1 > 0 and k0 + k1 > 0. Since k0 k1 > 0 for the sum over i1 to be ﬁnite, the only possibilities are: (1) k0 > 0, k1 > 0 and p1 < 0 or (2) k0 < 0, k1 < 0 but now

p1 > 0. Now we send i → + i0+ with > 0 and just consider the imaginary part, namely

1 1 → + = −iπδ + p1 − k0 − k1 . i + p1 − k0 − k1 + i0 + p1 − k0 − k1 The δ-function implies that

k0 + k1 − p1 = > 0, which is only compatible with the above possibility (1), i.e. k0 > 0, k1 > 0 and p1 < 0, thus leading to

1 X Im δΣ(k, +i0+) = −2π iU 2 δk −k−p +k δ+ − − f(− ) f(− ) f( ). V 2 0 1 1 p1 k0 k1 k1 k0 p1 k0 k1 p1

This expression coincides with the Fermi golden rule for the probability of a particle at momentum k to decay into two particles and one hole with the same total momentum but energy . More generally, the diagrams which contribute to the imaginary part of the self-energy can be represented as in Fig. 5.11, where a particle with momentum k and frequency i decays by

183 . . . . k k

Figure 5.11: Graphical representation of a self-energy diagram which contribute to the imaginary part. a matrix element drawn as a box into a particle plus a certain number, say m ≥ 1, of particle- hole pairs,2 which recombine through the other box into the original particle. Let us neglect for simplicity the momentum and frequency dependence of the boxes, and deﬁne as ki(i) and pi(ωi), i = 1, m, the momenta(frequencies) of the particles and holes, respectively, of the m particle-hole pairs, and as k0(0) that of the additional particle. By momentum and frequency conservation X X k0 + ki − pi = k, 0 + i − ωi = n. i i As before, if we sum over all the 2m independent internal frequencies and ﬁnally send i → + i0+, with > 0, we end up with expression for the imaginary part proportional to

m X X X X Y δ − kj − pj − k0 δ k0 + (kj − pj) − k f(−k0 ) f(−kj ) f(pj ). k0 ki pi j j j=1

Since ki ≥ 0 and pi ≤ 0, the energy conservation implies that, for small , all particles and holes should lye very close to the Fermi surface, on a strip of width at most δk = /vF , where vF is the Fermi velocity. Therefore the phase space available for the decay process grows at most 2m like , since there are only 2m free summations over momentum, k0 being ﬁxed by momentum conservation. The ﬁnal conclusion is therefore that, within perturbation theory,

lim Σ00(k, ) = lim Im Σ(k, + iη) = 0, (5.27) →0 →0

2Two lines which propagate in opposite directions correspond to two Green’s functions one at positive time and the other at negative time. Positive time means that ﬁrst we create an electron and later we annihilate it, which denote a so-called particle excitation. On the contrary, negative time means that ﬁrst we annihilate an electron and later we create it back, so-called hole excitation.

184 5.4 Other kinds of perturbations

So far we have just constructed and analysed the perturbation expansion in terms of the electron- electron interaction. Let us consider now what changes for diﬀerent types of perturbations. In particular we will just brieﬂy mention two perturbations: a scalar potential and the coupling to bosonic modes.

5.4.1 Scalar potential Let us suppose to have non-interacting electrons identiﬁed by a quantum label a, with non- interacting Green’s functions

(0) (0) Gab (x, y; τ) = δab Ga (x, y; τ), in the presence of the scalar potential X Z V = dx Ψa(x) V (x)ab Ψb(x). (5.28) ab The S-matrix is now !! X Z β Z S(β) = Tτ exp − dτ dx Ψa(x, τ) V (x)ab Ψb(x, τ) ab 0 !! X Z ≡ Tτ exp − dx Ψa(x) V (x)ab Ψb(x) , ab where, as before, we have introduced the four-dimensional coordinate x = (x, τ). Upon expanding S(β) up to ﬁrst order and keeping only connected diagrams, one can readily obtain the Green’s function expansion Z (0) X † † Gab(x, y) = δab Ga (x, y) + dz V (z)cd hTτ Ψa(x)Ψb(y) Ψc(z) Ψd(z) iconn cd Z (0) (0) (0) = δab Ga (x, y) + dz V (z)ab Ga (x, z) Gb (z, y). (5.29)

If we represent graphically the perturbation as in Fig. 5.12(a), then the Green’s function (5.29) up to ﬁrst order can be drawn as the ﬁrst two diagrams in the right hand side of Fig. 5.12(b). Higher order terms can be simply obtained by inserting other potential lines, as the second order term in Fig. 5.12(b). All diagrams have a positive sign. The Dyson equation can be easily read out: Z (0) X (0) Gab(x, y) = δab Ga (x, y) + + dz V (z)ac Ga (x, z) Gcb(z, y), (5.30) c and is drawn in Fig. 5.12(b).

185 Vab

a b a b a a b a c b a a c b = + + + ... = + (a) (b)

Figure 5.12: (a) Graphical representation of the scalar potential perturbation. (b) Perturbation expansion of the Green’s function and Dyson equation.

5.4.2 Coupling to bosonic modes

p+q p p+q p g(q)

x q = |g(q)| 2 D (0) (q)

x −q

k g(−q) k+q k k+q

Figure 5.13: On the left: electron-phonon vertices at q and −q. The external dotted vertex line represents phonon coordinates. On the right: the eﬀective electron-electron interaction after contracting phonons.

Let us imagine now that our system of electrons is coupled to phonons described by the free Hamiltonian X † Hphon = ωq aqaq. q The coupling term is assumed to be X V = g(q)xq ρ(−q), (5.31) q

186 where g(q)∗ = g(−q) is the coupling constant, the phonon coordinate is

r1 x = a + a† , q 2 q −q and X X † ρ(q) = ckσck+qσ, σ k is the density operator. This perturbation is actually more general and describes any kind of electron-boson coupling.

= +

+ +

+ + . . .

Figure 5.14: First orders in the diagrammatic perturbation expansion of D(q). The exact D(q) is represented by a bold dashed line, while D(0)(q) by a thiner dashed line. The solid lines are electron Green’s functions.

We can perform perturbation theory in (5.31) using the Wick theorem both for the electrons and for the bosons. The latter amounts to contract an xq with x−q which leads to the free (0) propagator D (q, iΩn) of Eq. (5.19). Therefore, if we represent the electron-phonon coupling as in Fig. 5.13, in which a dotted vertex line represents the phonon-coordinate, by contracting two vertices one recovers an eﬀective electron-electron interaction, see also Fig. 5.13, which is mediated by the phonons and given by

2 (0) 2 ωq |g(q)| D (q, iωl) = −|g(q)| 2 2 . (5.32) ωl + ωq

187 This eﬀective interaction is retarded, namely depends on the frequency, and attractive. More speciﬁcally, if we move on the real axis, then

2 (0) 2 ωq |g(q)| D (q, iωl → ω) → −|g(q)| 2 2 , ωq − ω which is attractive if |ω| < ωq and repulsive otherwise. The rules for constructing diagrams are therefore the same as for the electron-electron interaction, with the additional complication that interaction is frequency dependent. However, in this particular case one may also investigate the effects of the electron-phonon coupling to the phonon Green’s function D(q, iωl). The perturbation expansion is shown up to second order in Fig. 5.14. Two kinds of diagrams can be identified. The first class includes diagrams which can be divided into two diagrams of the same expansion by cutting non-interacting D(0)(q)-line, like the first second order diagram shown in Fig. 5.14. These diagrams are reducible. The other class includes all other diagrams which are irreducible. If we define a 2 self-energy |g(q)| Π(q, iωn) through these irreducible diagrams, shown up to second order in Fig. 5.15, we can easily derive the Dyson equation, also shown in the same figure, whose solution is D(0)(q, iω ) D(q, iω ) = n . (5.33) n 2 (0) 1 − |g(q)| D (q, iωn) Π(q, iωn)

= + + + + . . .

= +

Figure 5.15: Upper panel: Phonon self-energy, represented by a ﬁlled circle, up to second order. Lower panel: Dyson equation for D(q).

188 5.5 Two-particle Green’s functions and correlation functions

Let us consider back the Hamiltonian of interacting electrons, which we assumed to be

X 1 X X H = c† c + U(q) c† c† c c . (5.34) k kσ kσ 2V pα k+qβ kβ p+qα kσ kpq αβ

The Heisenberg imaginary time evolution of an annihilation operator is

∂c (τ) − kσ = [c (τ),H(τ)] = c (τ) ∂τ kσ k kσ 1 X + U(q) c† (τ)c (τ)c (τ). V p+qα pα k+qσ pqα

Therefore the Green’s function satisﬁes the equation of motion ∂ ∂ h i − G (k, τ) = − −hT c (τ) c† i ∂τ σ ∂τ τ kσ kσ ∂ h i = − −θ(τ) hc (τ) c† i + θ(−τ) hc† c (τ)i ∂τ kσ kσ kσ kσ ∂c (τ) = δ(τ) − hT kσ c† i τ ∂τ kσ 1 X = δ(τ) + G (k, τ) − U(q) hT c† (τ) c (τ) c (τ) c† i. k σ V τ p+qα pα k+qσ kσ pqα

This equation can be written as

∂ 1 X − − G (k, τ 0) = δ(τ 0) − U(q) hT c† (τ 0) c (τ 0) c (τ 0) c† i. (5.35) ∂τ 0 k σ V τ p+qα pα k+qσ kσ pqα

The non-interacting Green’s function satisﬁes on the contrary

∂ − − G(0)(k, τ 0) = δ(τ 0). ∂τ 0 k σ

(0) 0 0 If we multiply both sides of Eq. (5.35) by Gσ (k, τ − τ ) and integrate over τ , we obtain

Z ∂ dτ 0 G(0)(k, τ − τ 0) − − G (k, τ 0) = G(0)(k, τ) σ ∂τ 0 k σ σ 1 X Z − U(q) dτ 0 G(0)(k, τ − τ 0) hT c† (τ 0) c (τ 0) c (τ 0) c† i, V σ τ p+qα pα k+qσ kσ pqα

189 and upon integrating by part the left hand side, one obtains (0) Gσ(k, τ) = Gσ (k, τ) 1 X Z − U(q) dτ 0 G(0)(k, τ − τ 0) hT c† (τ 0) c (τ 0) c (τ 0) c† i(5.36). V σ τ p+qα pα k+qσ kσ pqα Therefore the single-particle Green’s function can be expressed in terms of a two-particle Green’s function. Namely, let us deﬁne the two-particle Green’s function

† † Kσ1σ2;σ3σ4 (p1τ1, p2τ2; p3τ3, p4τ4) = −hTτ cp1σ1 (τ1)cp2σ2 (τ2) cp3σ3 (τ3)cp4σ4 (τ4) i, (5.37) where, by momentum conservation,

p1 + p2 = p3 + p4, in terms of which (0) Gσ(k, τ) = Gσ (k, τ) 1 X Z − U(q) dτ G(0)(k, τ − τ 0) K (k + q τ 0, p τ 0; p + q τ 0, k 0).(5.38) V σ σα;ασ pqα

5.5.1 Diagrammatic representation of the two-particle Green’s function Let us write formally

Kσ1σ2;σ3σ4 (p1τ1, p2τ2; p3τ3, p4τ4) = −δσ1σ4 δσ2σ3 δp1p4 δp2p3 Gσ1 (p1, τ1 − τ4)Gσ2 (p2, τ2 − τ3)

+δσ1σ3 δσ2σ4 δp1p3 δp2p4 Gσ1 (p1, τ1 − τ3)Gσ2 (p2, τ2 − τ4) Z 4 Y 0 0 0 0 0 + dτi Gσ1 (p1, τ1 − τ1)Gσ2 (p1, τ2 − τ2)Gσ3 (p3, τ3 − τ3)Gσ4 (p4, τ4 − τ4) i=1 0 0 0 0 Γσ1σ2;σ3σ4 (p1τ1, p2τ2; p3τ3, p4τ4), (5.39) which is represented graphically to Fig. 5.16. We notice that, in the absence of interaction, only the first two terms survive with the Green’s functions being the non-interacting ones. If we start doing perturbation theory, which has the same rules as before, we will (1) dress the non- interacting Green’s functions, which explains the first two terms; (2) couple them by interaction lines, which justify the last contribution and provides the definition of the interaction vertex Γ, which in lowest order perturbation theory is shown in Fig. 5.17. In conclusion the equation (5.38) of the single particle Green’s function in terms of the two- particle one can be expressed as function of the single particle Green’s function itself and the interaction vertex as shown in Fig. 5.18. Notice that the interaction vertex acts as the bare interaction, so it carries a (-1) sign. Since there is a loop in the third diagram of Fig. 5.18, the overall sign is plus. This result also provides an expression for the single-particle self-energy, as shown in Fig. 5.19.

190 1 4 1 4 K(1,2;3,4) = − + 2 3 2 3 1 4 + 2 3 Γ(1,2;3,4)

Figure 5.16: Graphical representation of the two-particle Green’s function.

p+q p p+q p p+q p

= U(q) − U(p−k) +

k k+q k k+q k k+q p+q p p+q h+q p

h+q h + + + .....

k p+k−h k+q k k+q

Figure 5.17: Lowest order expansion of the interaction vertex. p, k, p + q and k + q label both momenta and frequencies.

5.5.2 Correlation functions We already showed that the average value of the imaginary-time ordered product of two bosonic- like operators can provide, after analytic continuation on the real axis, the linear response functions. Let us then consider two single-particle density operators:

X A † X B † A(q) = λk,k+q;αβ ckαck+qβ,B(q) = λk,k+q;αβ ckαck+qβ. k αβ k αβ

191 = +

+ +

Figure 5.18: Dyson equation in terms of the interaction vertex Γ.

Σ = + + Figure 5.19: Single-particle self-energy in terms of the interaction vertex Γ.

Their Green’s function, usually called correlation function, is (we use the notation GAB = χAB since we know that the two coincide on the frequency real axis):3

1 1 X X χ (q, τ) = − hT (A(q, τ) B(−q))i = − λA λB AB V τ V k,k+q;αβ p+q,p;γδ k αβ p γδ † † hTτ ckα(τ)ck+qβ(τ)cp+qγcpδ i 1 X X = λA λB K (k + q τ, p 0; p + q 0, k τ) V k,k+q;αβ p+q,p;γδ βδ;γα k αβ p γδ ! ! 1 X X = − λA G (k, 0−) λB G (p, 0−) V k,k;αα α p+q;γγ γ k α p γ 1 X + λA λB G (k + q, τ) G (k, −τ) V k,k+q;αγ k+q,k;γα α γ k αγ

3 In the equation the Green’s functions at equal times have to be interpreted as the limit of τ → 0−, since in the deﬁnition of A and B creation operators are on the left of the annihilation ones.

192 4 1 X Z Y + λA λB dτ G (k + q, τ − τ )G (p, 0 − τ ) V k,k+q;αγ k+q,k;γα i β 1 δ 2 k αγ i=1

Gγ(p + q, τ3 − 0)Gα(k, τ4 − τ)Γβδ ; γα(k + q τ − τ1, p − τ2; p + q τ3, k τ4 − τ) 1 = − hA(q)i hB(−q)i V 1 X + λA λB G (k + q, τ) G (k, −τ) V k,k+q;αγ k+q,k;γα α γ k αγ 4 1 X Z Y + λA λB dτ G (k + q, τ − τ )G (p, 0 − τ ) V k,k+q;αγ k+q,k;γα i β 1 δ 2 k αγ i=1

Gγ(p + q, τ3 − 0)Gα(k, τ4 − τ)Γβδ ; γα(k + q τ − τ1, p − τ2; p + q τ3, k τ4 − τ). The disconnected term hA(q)i hB(−q)i vanishes for q 6= 0. In general it can be moved on the left hand side, in which case the correlation function is the average of the time-ordered product minus the product of the averages. With this deﬁnition, the correlation function has the graphical representation shown in Fig. 5.20. In the ﬁgure, the triangular vertices represent

k+q k+q p+q

χAB(q) = λ A λ B + λ A λ B (k+q,p;p+q,k)

k k Γ p

Figure 5.20: Graphical representation of the correlation function χAB(q), where q includes both momentum q and a bosonic frequency iωl.

A(B) the matrix elements λk,k+q;αγ and the loops imply as usual summation over the spin-indices. Perturbation theory has the same rules as before with the only exception that the loop phase- factor is now (−1)L−1 due to the minus sign in the deﬁnition of χ. If we Fourier transform also in imaginary time, the correlation function

χAB(q, iωl), depends on the transferred momentum, q, and frequency, iωl. Since the operators are bosonic- like, iωl is a bosonic Matsubara frequency. By time-translation invariance, the frequency is conserved at any vertex, as usual, and the interaction vertex becomes consequently

Γβδ ; γα(k + q in + iωl, p im; p + q im + iωl, k in), (5.40) as shown in Fig. 5.21.

193 k+q, i + i l i ε n ω β α k, ε n

Γ p, i εm p+q, i εm+ i ω l

δ γ

Figure 5.21: Graphical representation of the interaction vertex of Eq. (5.40).

Non-interacting values In the absence of interaction only the ﬁrst term in Fig. 5.20 survives with the Green’s function lines being the non-interacting ones. Since there is a loop, according to what we said before, the sign is (−1)L−1 = (−1)1−1 = 1 hence

(0) 1 X X X χ (q, iω ) = T λA λB G(0)(k + q, i + iω ) G(0)(k, i ) AB l V k,k+q;αγ k+q,k;γα α n l γ n k n αγ 1 X X X A B 1 1 = T λk,k+q;αγ λk+q,k;γα . (5.41) V in + iωl − k+q in − k k n αγ By means of Eq. (5.8) we can easily perform the sum over frequencies in Eq. (5.41), which gives X 1 1 1 1 T = f( − iω ) + f( ) i + iω − i − k+q l − iω − k iω − + n n l k+q n k k+q l k l k+q k f( ) − f( ) = k k+q , iωl − k+q + k where we used the fact that, if iωl is bosonic, then f(±iωl) = f(). The ﬁnal result is therefore

(0) 1 X X A B f(k) − f(k+q) χAB(q, iωl) = λk,k+q;αγ λk+q,k;γα . (5.42) V iωl − k+q + k k αγ

The linear response function is simply obtained by iωl → ω + iη.

5.6 Coulomb interaction and proper and improper response functions

Let us consider the case in which the electron-electron interaction is just the Coulomb repulsion 4πe2 U(q) = , q2

194 which is singular for q → 0. In order to cure this singularity, it is convenient to recast perturbation theory in a diﬀerent manner, which naturally brings to identify proper and improper response functions. In Fig. 5.22 we draw the diagrammatic expansion of the density-density correlation function χ(q, iωn), which is also the improper response function, up to ﬁrst order. We can distinguish already two kinds of diagrams: those which can be cut into two by cutting

χ(q )= + + + + + . . .

Figure 5.22: Diagrammatic expansion of the density-density correlation function up to first order. an interaction line, as the last diagram, and those which can not, all the others, which we call irreducible with respect to the interaction. Let us define as χe(q, iωn) the sum of all irreducible diagrams, in the above sense, whose first order corrections are drawn in Fig. 5.23. We can formally define χe(q, iωn) as X 1 X χ(q, iωn) = 2T G(k + q, im + iωn) G(k, im) Λ(e k + q im + iωn , k im; q iωn), (5.43) e V m k by introducing the proper density-vertex function Λ(e k, im; q, iωn), also shown in the same figure, so to distinguish between Green’s function corrections and vertex corrections. Notice that without interaction the density-vertex function is the identity matrix in spin space. In terms of χe(q, iωn) the perturbation expansion of χ(q, iωn) can be recast as in the lower panel of Fig. 5.23, which has the formal solution4

χ(q, iω ) χ(q, iω ) = e n , (5.44) n 4πe2 1 − χ(q, iω ) q2 e n proving that χe(q, iωn) is actually the proper response function.

5.6.1 Screened interaction and corresponding Dyson equation The proper density-density correlation function allows us to deﬁne a screened Coulomb interaction and a dielectric constant through

4πe2 4πe2 1 W (q, iω ) = = , (5.45) n q2 (q, iω ) q2 4πe2 n 1 − χ(q, iω ) q2 e n

195 ~ χ(q ) = = + + + + . . .

k+q ~ Λ(k+q,k;q ) = = + + . . . k

χ( q )= = +

Figure 5.23: Upper panel: The proper χe(q) up to ﬁrst order. Middle panel: The proper density- vertex function up to ﬁrst order. Lower panel: Perturbation expansion of the improper χ(q, iωn) in terms of the proper χe(q).

= +

Γ = − +

Figure 5.24: Upper panel: Screened Coulomb interaction W . Lower panel: Interaction vertex in terms of the screened interaction and the proper vertex. which is drawn as a bold wavy line in Fig. 5.24. The interaction vertex Γ which we previously introduced can be also expressed in terms of the screened Coulomb interaction and the proper density-vertex, as also shown in Fig. 5.24. Using this definition for the interaction vertex in the expression of the self-energy, see Fig. 5.19, one finds the alternative definition of the self-energy which is drawn in Fig. 5.25

4 Notice both the interaction and χe(q, iωn), which is a loop, bring a minus sign, hence the sign is plus.

196 Σ = +

Figure 5.25: Self energy in terms of the screened W and the proper density-vertex.

5.7 Irreducible vertices and the Bethe-Salpeter equations

Let us consider again the interaction vertex (5.40). Let us for instance focus on the upper or the lower incoming and outgoing lines which differ by the momentum/frequency transfer q. This identifies the particle-hole channel at momentum/frequency transfer q. By this definition, we can distinguish two classes of diagrams in the perturbation expansion. The first includes those diagrams which can be divided in two by cutting two fermionic lines within the same particle- hole channel at momentum/frequency q. These type of diagrams are called reducible in the particle-hole channel. For instance the third diagram in Fig. 5.17 belongs to this class. The other class includes all other diagrams which are called irreducible, as the fourth diagram in Fig. 5.17. Let us denote the sum of all irreducible diagrams as

0 Γβδ ; γα(k + q in + iωl, p im; p + q im + iωl, k in). (5.46) In terms of Γ0 the perturbation expansion of Γ can be formally written as

Γαβ ; γδ(k + q in + iωl, p im; p + q im + iωl, k in) 0 = Γαβ ; γδ(k + q in + iωl, p im; p + q im + iωl, k in) 1 X X X + T Γ0 (k + q i + iω , h i ; h + q i + iω , k i ) G (h + q, i + iω ) V αµ ; νδ n l h h l n ν h l h h µν Gµ(h, ih)Γνβ ; µδ(h + q ih + iωl, p im; p + q im + iωl, h ih), (5.47) which is graphically shown in Fig. 5.26. This is the so-called Bathe-Salpeter equation which relates the fully reducible interaction vertex Γ with its irreducible part Γ0 in the particle-hole channel. Actually, we can proceed in a diﬀerent way. Namely, instead of selecting the particle- hole channel at momentum transfer q, we can select the two incoming lines, particle-particle channel, which have total momentum/frequency P = k + p + q. P is also conserved in the scattering process represented by the interaction vertex. Then we can introduce the concept of reducibility/irreducibility in the particle-particle channel at total momentum/frequency P and identify the irreducible vertex in that channel. Eventually we end up to another Bethe-Salpeter equation, which is shown in Fig. 5.27. In this ﬁgure we have denoted the irreducible vertex in

197 k+q p+q k+qp+q k+q h+q p+q

Γ = Γ 0 + Γ 0 Γ

h k pp k p k p

Figure 5.26: Bethe-Salpeter equation for the interaction vertex in the particle-hole channel. P−k P−p P−k P−p P−k P−h P−p

Γ = Γ 0 + Γ 0 pp pp Γ

h k pp k p k p

Figure 5.27: Bethe-Salpeter equation for the interaction vertex in the particle-particle channel.

0 0 the particle-particle channel as Γpp to distinguish it from Γ .

5.7.1 Bethe-Salpeter equation for the vertex functions Let us now introduce the vertex function for a density operator A(q), which is a particle-hole operator, through

Z 2 † Y 0 0 0 hTτ ckα(τ1)ck+qβ(τ2) A(q, τ) i ≡ dτi Gα(k, τ1 − τ1) Gβ(k + q, τ2 − τ2) i=1 A 0 0 Λαβ(k τ1 , k + q τ2 ; q , τ). (5.48)

In the absence of interaction, one readily realizes that

A 0 0 0 0 A Λαβ(k τ1 , k + q τ2 ; q , τ) = δ(τ1 − τ) δ(τ2 − τ) λk,k+q;αβ.

In the presence of interaction the formal expression of Λ (in what follows we drop the label A) in terms of λ and the interaction vertex can be readily inferred from the expression of the correlation function, and it is graphically shown in Fig. 5.28.

198 Λ = + Γ λ λ

= 0 + 0 λ λ+ Γ λ Γ Γ

=λ + Γ 0 Λ

Figure 5.28: Bethe-Salpeter equation for the particle-hole vertex function Λ.

In Fourier space it reads:

Λαβ(k in , k + q in + iωl ; q , iωl) = λk,k+q;αβ 1 X X X + T Γ0 (k i , p + q i + iω ; p i , k + q i + iω ) V αγ;δβ n m l m n l p m γδ

Gγ(p + q, im + iωl) Gδ(p, im)Λδγ(p im , p + q im + iωl ; q , iωl) (5.49) 1 X X X = λ + T Γ (k i , p + q i + iω ; p i , k + q i + iω ) k,k+q;αβ V αγ;δβ n m l m n l p m γδ

Gγ(p + q, im + iωl) Gδ(p, im) λp,p+q;δγ. (5.50)

Notice that, in the case of a Coulomb repulsion, the above vertex function for the density A operator, i.e. λk,k+q;αβ = δαβ, has not to be confused with the proper density-vertex shown in Fig. 5.23. The latter is in fact irreducible with respect to cutting an interaction line, while the former is not. One can easily show that Λe satsiﬁes the Bethe-Salpeter equation graphically shown in Fig. 5.29.

~ Λ(k+q,k;q ) = = + Γ0 −

Figure 5.29: Bethe-Salpeter equation for the proper density-vertex function Λ.e

199 5.8 The Ward identities

Let us suppose that our model has a set of conserved quantities, and let us select one of them, whose density operator we deﬁne as

X X † 0 J0(q) = cka λk,k+q;ab ck+qb, (5.51) k ab where to be as general as possible we assume that the roman indices which identify the fermionic operators include both spin as well as other internal labels, like e.g. the band index. We will further assume that the single-particle Green’s function are generally non-diagonal in these indices. We can associate to the density operator (5.51) a current density operator J = (J1,J2,J3) given by X X † J(q) = cka λk,k+q;ab ck+qb, (5.52) k ab such that the continuity equation in imaginary time is satisﬁed: ∂ J (q, τ) + q · J(q, τ) = 0. ∂τ 0

If we consider J0 and J as the time and space components of a four dimensional current, (J0,J1,J2,J3), we can introduce for each component a vertex function through

† hTτ cka(τ1) ck+qb(τ2) Ji(q, τ) i.

Let us take the time derivative with respect to τ of the zeroth component. The derivative acts either directly on the charge density operator or on the θ-functions which deﬁne the time-ordered product. The latter is, dropping for simplicity all indices,

† [∂τ θ(τ1 − τ2)θ(τ2 − τ)] hc (τ1) c (τ2) J0(τ)i † + [∂τ θ(τ1 − τ)θ(τ − τ2)] hc (τ1) J0(τ) c (τ2)i † + [∂τ θ(τ − τ1)θ(τ1 − τ2)] hJ0(τ) c (τ1) c (τ2)i † − [∂τ θ(τ2 − τ1)θ(τ1 − τ)] hc (τ2) c (τ1) J0(τ)i † − [∂τ θ(τ2 − τ)θ(τ − τ1)] hc (τ2) J0(τ) c (τ1)i † − [∂τ θ(τ − τ2)θ(τ2 − τ1)] hJ0(τ) c (τ2) c (τ1)i † = −θ(τ1 − τ2)δ(τ2 − τ) hc (τ1) c (τ2) J0(τ)i † + [−δ(τ1 − τ)θ(τ − τ2) + θ(τ1 − τ)δ(τ − τ2)] hc (τ1) J0(τ) c (τ2)i † +δ(τ − τ1)θ(τ1 − τ2) hJ0(τ) c (τ1) c (τ2)i † +θ(τ2 − τ1)δ(τ1 − τ) hc (τ2) c (τ1) J0(τ)i † − [−δ(τ2 − τ)θ(τ − τ1) + θ(τ2 − τ)δ(τ − τ1)] hc (τ2) J0(τ) c (τ1)i

200 † −δ(τ − τ2)θ(τ2 − τ1) hJ0(τ) c (τ2) c (τ1)i h † i = δ(τ − τ2) hTτ c (τ1) J0(τ), c (τ) i

h i † +δ(τ − τ1) hTτ J0(τ), c (τ) c (τ2) i. It can be readily shown that

X X h † 0 † i X † 0 cpc λp,p+q;cd cp+qd , ck+qb = ckc λk,k+q;cb p cd c X X h † 0 i X 0 cpc λp,p+q;cd cp+qd , cka = − λk,k+q;ad ck+qd, p cd d so that the above time-derivative becomes X † 0 δ(τ − τ2) hTτ cka(τ1) ckc(τ) λk,k+q;cb i c X 0 † − δ(τ − τ1) hTτ λk,k+q;ad ck+qd(τ) ck+qb(τ2) i d X 0 = − δ(τ − τ2) Gac(k, τ1 − τ) λk,k+q;cb c X 0 + δ(τ − τ1) λk,k+q;ad Gdb(k + q, τ − τ2). d In conclusion we ﬁnd, making use of the continuity equation, ∂ ∂J (q, τ) hT c (τ ) c† (τ ) J (q, τ) i = hT c (τ ) c† (τ ) 0 i ∂τ τ ka 1 k+qb 2 0 τ ka 1 k+qb 2 ∂τ X 0 − δ(τ − τ2) Gac(k, τ1 − τ) λk,k+q;cb c X 0 + δ(τ − τ1) λk,k+q;ad Gdb(k + q, τ − τ2) d † = −hTτ cka(τ1) ck+qb(τ2) q · J(q, τ) i X 0 − δ(τ − τ2) Gac(k, τ1 − τ) λk,k+q;cb c X 0 + δ(τ − τ1) λk,k+q;ad Gdb(k + q, τ − τ2). d Upon introducing the vertex functions for the charge and current densities, Λ0 and Λ, respectively, and Fourier transforming in frequency we ﬁnd X Gac(k, in) Gdb(k + q, in + iωl) cd

201 0 − iωl Λcd(k in , k + q in + iωl ; q iωl) + q · Λcd(k in , k + q in + iωl ; q iωl) X 0 X 0 = − Gac(k, in) λk,k+q;cb + λk,k+q;ad Gdb(k + q, in + iωl). (5.53) c d By means of the Bethe-Salpeter equation for the vertex functions (5.49) we can rewrite this equation as X Gac(k, in) Gdb(k + q, in + iωl) cd 0 − iωl Λcd(k in , k + q in + iωl ; q iωl) + q · Λcd(k in , k + q in + iωl ; q iωl)

X 0 = Gac(k, in) Gdb(k + q, in + iωl) − iωl λk,k+q;c d + q · λk,k+q;c d cd 1 X X X + T G (k, i ) G (k + q, i + iω ) V ac n db n l p m cdefgh 0 Γce;fd(k in, p + q im + iωl; p im, k + q in + iωl) Gfg(p, im) Ghe(p + q, im + iωl) 0 − iωl Λgh(p im , p + q im + iωl ; q iωl) + q · Λgh(p im , p + q im + iωl ; q iωl) X 0 X 0 = − Gac(k, in) λk,k+q;cb + λk,k+q;ad Gdb(k + q, in + iωl) c d X 0 = Gac(k, in) Gdb(k + q, in + iωl) − iωl λk,k+q;c d + q · λk,k+q;c d cd 1 X X X + T G (k, i ) G (k + q, i + iω ) V ac n db n l p m cdef 0 Γce;fd(k in, p + q im + iωl; p im, k + q in + iωl) X 0 X 0 − Gfg(p, im) λp,p+q;ge + λp,p+q;fh Ghe(p + q, im + iωl) . g h

−1 −1 5 If we multiply both sides by the matrix product Gˆ (k, in) Gˆ (k + q, in + iωl) , we get

0 − iωl Λab(k in , k + q in + iωl ; q iωl) + q · Λab(k in , k + q in + iωl ; q iωl)

0 = − iωl λk,k+q;a b + q · λk,k+q;a b 1 X X X + T Γ0 (k i , p + q i + iω ; p i , k + q i + iω ) V ae;fb n m l m n l p m ef

5 The Green’s function Gab can be interpreted as the ab component of a matrix Gˆ

202 X 0 X 0 − Gfg(p, im) λp,p+q;ge + λp,p+q;fh Ghe(p + q, im + iωl) g h X 0 −1 X −1 0 = − λk,k+q;adGdb (k + q, in + iωl) + Gac (k, in) λk,k+q;cb. d c

We notice that the non-interacting Green’s functions satisfy the same equation with Γ0 = 0, namely

−1 0 X 0 (0) − iωl λk,k+q;a b + q · λk,k+q;a b = − λk,k+q;ad G (k + q, in + iωl) db d −1 X (0) 0 + G (k, in) λk,k+q;cb, ac c so that we can write 1 X X X T Γ0 (k i , p + q i + iω ; p i , k + q i + iω ) V ae;fb n m l m n l p m ef X 0 X 0 − Gfg(p, im) λp,p+q;ge + λp,p+q;fh Ghe(p + q, im + iωl) g h −1 X 0 −1 (0) = − λk,k+q;ad Gdb (k + q, in + iωl) − G (k + q, in + iωl) db d −1 X −1 (0) 0 + Gac (k, in) − G (k, in) λk,k+q;cb. ac c

By deﬁnition −1 Gˆ−1 − Gˆ(0) = −Σˆ, where Σˆ is the self-energy matrix, so that

X 0 0 λk,k+q;ac Σcb(k + q, in + iωl) − Σac(k, in) λk,k+q;cb c 1 X X X = T Γ0 (k i , p + q i + iω ; p i , k + q i + iω ) V ac;db n m l m n l p m cd X 0 0 − Gde(p, im) λp,p+q;ec + λp,p+q;de Gec(p + q, im + iωl) . (5.54) e

203 This is the so-called Ward identity which is a consequence of conservation laws. There are other two equivalent ways to rewrite this identity. Indeed, through (5.53), we can also write

X 0 0 λk,k+q;ac Σcb(k + q, in + iωl) − Σac(k, in) λk,k+q;cb c 1 X X X = T Γ0 (k i , p + q i + iω ; p i , k + q i + iω ) V ac;db n m l m n l p m cd Gde(p, im) Gfc(p + q, im + iωl) (5.55) 0 − iωl Λef (p im, p + q im + iωl; q iωl) + q · Λef (p im, p + q im + iωl; q iωl) .

Futhermore, using the fully reducible vertex Γ instead of Γ0, see Eq. (5.50), we also ﬁnd

X 0 0 λk,k+q;ac Σcb(k + q, in + iωl) − Σac(k, in) λk,k+q;cb c 1 X X X = T Γ (k i , p + q i + iω ; p i , k + q i + iω ) V ac;db n m l m n l p m cd 0 Gde(p, im) Gfc(p + q, im + iωl) − iωl λp,p+q;ef + q · λp,p+q;ef . (5.56)

5.9 Consistent approximation schemes

Since it is usually impossible to sum up all orders in perturbation theory, one is forced to make some approximation on the self-energy. In doing that, one would like not to spoil any conservation law. In this section we show what is the proper way to deﬁne an approximation scheme which is consistent with the conservation laws. In general we can recast the perturbation expansion for the self-energy in terms of the fully interacting Green’s functions instead of the non-interacting ones. This kind of expansion is called skeleton expansion, the diagrams up to second order being drawn in Fig. 5.30. This allows us

Σ[G] = + + + + ....

Figure 5.30: Skeleton expansion for the self-energy up to second order in the interaction. to assume formally that the self-energy is a functional of the Green’s function Σ[G]. Let us

204 calculate the functional derivative of Σ with respect to G. Let us imagine to do it graphically, as shown in Fig. 5.31. Namely, we take out of the rounded box, which represents the whole

δG k+q k c d c d a b = a b p p+q

Figure 5.31: Graphical representation of δΣ/δG. skeleton expansion of Σ, a Green’s function G and we vary it, δG. Therefore δΣ/δG is a four leg vertex. What can it be? One easily realizes that, since the functional derivative is done with respect to the fully interacting Green’s function, this four leg vertex is just the irreducible Γ0 in the particle-hole channel. Therefore

δΣab(p, p + q) 0 = Γac;db(p, k + q; k, p + q). (5.57) δGdc(k, k + q) We further notice that

X h (0) −1 i δGdc(k, k + q) = − Gde(k) δ Gef (k, k + q) − δΣef (k, k + q) Gfc(k + q), ef which inserted into (5.57) and solving for δΣ leads to Z 0 δΣab(p, p + q) = dk Γac;db(p, k + q; k, p + q) δGdc(k, k + q) (5.58) Z X (0) −1 = − dk Γac;db(p, k + q; k, p + q) Gde(k) δGef (k, k + q) Gfc(k + q), ef where we used the Bethe-Salpeter equation for Γ versus Γ0.

We notice that (5.57) is perfectly consistent with the Ward identity (5.54). Since the Ward identities are consequence of conservation laws, we get to the following conclusion:

If the self-energy is approximated with a functional Σappx[G], this approximation is consistent with the conservation laws if the irreducible vertex is also approximated by δΣ [G] Γ0 = appx . appx δG

205 Finally we notice that, given an approximate Σappx[G], then the actual Green’s function is obtained by solving the self-consistency equation:

−1 −1 (0) G = G − Σappx[G]. (5.59)

5.9.1 Example: the Hartree-Fock approximation Let us consider the interacting Hamiltonian X 1 X H = t c† c + U c† c† c c . ab a b 2 acdb a c d b ab abcd Let us approximate the skeleton expansion of the self-energy with the two ﬁrst-order diagrams, as shown in Fig. 5.32. This amounts to the following expression of the self-energy

c d

Σ HF [G ] = a b + a b d c

Figure 5.32: Skeleton expansion in the Hartree-Fock approximation.

− − X X −im0 X X −im0 Σab(in) = T Uacdbe Gdc(im) − T Uacbd e Gdc(im). (5.60) m cd m cd We notice that − X −im0 − † T e Gdc(im) = Gdc(τ = 0 ) = hcc cdi ≡ ∆cd. m The Green’s function satisﬁes the self-consistency equation (5.59), which in this case reads

−1 X G ab = in − tab − Σab = in − tab − ∆cd (Uacdb − Uacbd) , cd which is diagonalized by diagonalizing X tab + ∆cd (Uacdb − Uacbd) = (HHF )ab , cd which is nothing but the Hartree-Fock Hamiltonian. Namely, the self-consistency requirement (5.59) is just the Hartree-Fock self-consistency equation.

206 In order to have a consistent scheme, we need to approximate Γ0 as

0 δΣab Γacdb = = Uacdb − Uacbd. δGdc One can readily show that the correlation functions calculated with the above Γ0 coincide with the time-dependent Hartree-Fock, which is therefore a consistent approximation.

207 5.10 Some additional properties and useful results

In this ﬁnal section we derive some properties that may be useful in several contexts.

5.10.1 The occupation number and the Luttinger-Ward functional By the deﬁnition of the single-particle Green’s function in momentum space, one can derive the average values † ∆k ab = hckb ckai, where the labels a and b include spin and additional internal indices, through

− − X −in0 ∆ˆ k = Gˆ(k, τ = 0 ) = T Gˆ(k, in) e n X 1 − = T e−in0 , ˆ n in − ˆk − Σ(k, in) where ∆ˆ k, ˆk and Σ(ˆ k, in) are matrices in the a-space. We define 0 = πT such that, for T → 0, 0 → 0, which allows to formally introduce the derivative with respect to the Matsubara frequencies. One finds that 1 Gˆ(k, in) = in − ˆk − Σ(ˆ k, in) −1 −1 ln Gˆ (k, in + i0) − ln Gˆ (k, in) Σ(ˆ k, in + i0) − Σ(ˆ k, in) = + Gˆ(k, in) i0 i0 −1 ∂ ln Gˆ (k, in) ∂Σ(ˆ k, in) = + Gˆ(k, in) , ∂in ∂in hence ˆ−1 ˆ X ∂ ln G (k, in) ∂Σ(k, in) − ∆ˆ = T σ + Gˆ(k, i ) e−in0 . (5.61) k ∂i n ∂i n n n Let us concentrate on the second term in (5.61). By the definition of the derivatives it follows that ˆ ˆ X ∂Σ(k, in) X ∂G(k, in) T Gˆ(k, i ) = −T Σ(ˆ k, i ). n ∂i ∂i n n n n n The interacting Hamiltonian allows for several conserved quantities. Let us write a generic one of them like X X † M = cka Mab(k) ckb. k ab

208 It turns out that the average value of M is given by X hMi = mk, k where

" ˆ−1 ! X ∂ ln G (k, in) m = Tr M(k) ∆ = T Tr M(k) k k ∂i n n ! # ˆ − ∂G(k, in) −in0 −Tr M(k) Σ(ˆ k, in) e . ∂in

We draw in Fig. 5.33 the ﬁrst orders in the skeleton expansion of the functional of the fully- interacting Green’s function X[G], introduced originally by Luttinger and Ward. It is obtained from the skeleton expansion of the self-energy by connecting the external vertices with a fully- interacting Green’s function and dividing each term by 1/2n, where 2n is the number of Green’s functions, n being the number of interaction lines. 6 By its deﬁnition, one can readily verify

X[G] = 1/2 + 1/2 + 1/4 + 1/4 + ....

Figure 5.33: Graphical representation of the functional X[G]. that X[G] satisﬁes X δX[G(k)] = T Tr Σ(ˆ k, in) δGˆ(k, in) , (5.62) n namely the self-energy is the functional derivative of X. One can readily show that, for any

6The self-energy itself is a functional of G, which guarantees that X is a well deﬁned functional.

209 conserved quantity, the following result holds:7 " ˆ # X ∂G(k, in) δX[G(k)] = T Tr Σ(ˆ k, i ) M(k) δi = 0. (5.63) n ∂i 0 n 0

7To prove Eq. (5.63), let us assume to rotate the basis into the one in which the hermitean matrix Mˆ (k) is † diagonal, with eigenvalues mα(k). The fermion operators in the new basis are accordingly ckα and ckα, hence the conserved operator becomes X † M = mα(k) ckα ckα. kα The fully interacting Green’s function must be diagonal in α for M to commute with the Hamiltonian h i h i h i 0 = H, M = H0, M + Hint, M .

Note that, because the ﬁrst commutator is a one-body operator while the second a two-body one, it follows that h i h i H0, M = 0, Hint, M = 0.

Now suppose that, within the perturbative calculation of the Luttinger-Ward functional X, which just amounts to calculate the average value of products of Hint(τ) in the interaction representation at diﬀerent times, each interaction vertex H0τ −H0τ Hint(τ) = e Hint e , is transformed into 0 i0 Mτ H0τ −H0τ −i0 Mτ Hint(τ) = e e Hint e e , 0 where 0 = π T . Since Hint and H0 commute separately with M, it follows that Hint(τ) = Hint(τ); indeed

0 i0 Mτ H0τ −H0τ −i0 Mτ Hint(τ) = e e Hint e e H0τ i0 Mτ −i0 Mτ −H0τ = e e Hint e e H0τ −H0τ = e Hint e = Hint(τ). Therefore the Luttinger-Ward functional X is invariant to this transformation. On the other hand, instead of transforming the interaction as a whole, I can also transform each fermionic operator inside the interaction separately, that implies, in the Heisenberg representation,

Hτ −Hτ i0 Mτ −i0 Mτ ckα(τ) = e ckα e → e ckα(τ) e −i0mα(k)τ = e ckα(τ), and accordingly that the Green’s function changes into

0 0 −i0mα(k)(τ−τ ) 0 Gα(k, τ − τ ) → e Gα(k, τ − τ ), namely Gα(k, in) → Gα (k, in + i0 mα(k)) . Since the ﬁnal result must not change, we get to the conclusion that

δX[G] = X [Gα (k, in + i0 mα(k))] − X [Gα(k, in)]

X X ∂Gα(k, in) = T Σα(k, in) mα(k) δi0 = 0. ∂i0 n α After back-rotation in the original basis, we recover Eq. (5.63).

210 As a result we obtain that ˆ−1 ! X ∂ ln G (k, in) − m = T Tr M(k) e−in0 . (5.64) k ∂i n n We already showed that the Green’s function, hence also its logarithm, has generally branch cuts on the real axis. By means of (5.9) we then ﬁnd8

ˆ−1 ! X ∂ ln G (k, in) − m = T Tr M(k) e−in0 k ∂i n n ! Z d ∂ = − f() Arg Tr M(k) ln Gˆ−1(k, + i0+) (5.65) π ∂ Z d ∂f() = − Im Tr M(k) ln Gˆ(k, + i0+) . (5.66) π ∂ We emphasize that this result holds only for conserved quantities.

5.10.2 The thermodynamic potential A simple way to determine the thermodynamic potential is by means of the Hellmann-Feynman theorem, according to which, if the Hamiltonian depends on some parameter λ, i.e. H = H(λ), then the derivative of the thermodynamic potential Ω(λ) with respect to λ is ∂Ω(λ) ∂H(λ) = h i , (5.67) ∂λ ∂λ λ where h... iλ means quantum and thermal average with the Hamiltonian at ﬁnite λ. Let us consider for simplicity the case of interacting electrons, the cases of bosons or electrons plus bosons being a straightforward generalization. The Hamiltonian is the free electron Hamiltonian

X † H0 = k ckσckσ, kσ plus the electron-electron interaction 1 X X H = U(q) c† c† c c . int 2V k+qα pβ p+qβ kα kpq αβ Let us consider now a λ-dependent Hamiltonian

H(λ) = H0 + λ Hint,

8The exponential factor in (5.64) guarantees that the contour integral is well behaved at Re z → −∞, while when Re z → +∞ is the Fermi distribution function which does the job.

211 then, through the Hellmann-Feynman theorem, it holds that ∂Ω(λ) ∂H = h int i ∂λ ∂λ λ 1 1 X X = λU(q) h c† c† c c i . λ 2V k+qα pβ p+qβ kα λ kpq αβ One might in principle calculate directly this average value by using the ﬂuctuation-dissipation theorem that allows to relate the average value of a four fermion operator with the sum over frequency of the imaginary part of appropriate correlation functions, that can be accessed perturbatively. Alternatively, by means of the equation of motion of the Green’s function, Eq. (5.35), we can write

1 X X 1 X X h i + λU(q) h c† c† c c i = T (i − ) G (k, i ) − 1 ein 0 2V k+qα pβ p+qβ kα λ 2 n k α n kpq αβ n kα

−1 h i + X X (0) (0) in 0 = T Gα (k, in) Gα(k, in) − Gα (k, in) e , n k having made use of the fact that X X lim δ(τ) = lim T einτ = T . τ→0 τ→0 n n Therefore we ﬁnally obtain that

∂Ω(λ) T X X −1 h i + = G(0)(k, i ) G (k, i ) − G(0)(k, i ) ein 0 ∂λ 2λ α n α n α n n kα T X X + = Σ (k, i ) G (k, i ) ein 0 , (5.68) 2λ α n α n n kα where the interacting Green’s function and self-energy are calculated at ﬁnite λ, and we used the Dyson equation. Since for λ = 0, the thermodynamic potential is that one of free electrons, Ω0, that is known, we can evaluate the interacting potential Ω = Ω(λ = 1) by Z 1 T X X −1 h i + Ω = Ω + dλ G(0)(k, i ) G(k, i ) − G(0)(k, i ) ein 0 . 0 λ n n n 0 n k

Now, let us deﬁne a functional Xe[Σ] through

+ X X (0) in 0 Xe [Σ] = −T ln 1 − Gσ (k, in)Σσ(k, in) + Σσ(k, in) Gσ(k, in) e + X G[Σ] , n kσ (5.69)

212 where X is the Luttinger-Ward function, see Fig. 5.33, assuming now that G is functional of Σ (we have explicitly indicated the spin label, eventhough each term is spin-independent). By means of (5.62) we ﬁnd that

δXe G(0) δG δX = σ − G − Σ + (0) δΣσ 1 − Gσ Σ δΣσ δΣσ G(0) δG δX δG = σ − G − Σ + σ (0) 1 − Gσ Σ δΣσ δGσ δΣσ G(0) δG δG = σ − G − Σ + Σ σ (0) 1 − Gσ Σ δΣσ δΣσ 1 = −1 − G. (0) Gσ − Σ

The last expression is nothing but the Dyson equation. This means that Xe is stationary

δXe = 0, δΣσ provided G satisﬁes the Dyson equation. Now, once again, we multiply the interaction by a parameter λ and assume we have calculated all quantities at ﬁnite λ, so that the self-energy becomes function of λ, i.e. Σ(λ). We note that

∂Xe δXe ∂Σ ∂X ∂X = + = , ∂λ δΣ ∂λ ∂λ Σ ∂λ Σ where we have assumed that Σ is the stationary point, and the last term implies that the derivative is at ﬁxed Σ, namely one does not need to derive anymore Σ(λ) with respect to λ. The only other place in which λ appears in X is in the interaction lines that are now λ U(q). Since for the n-th order diagram X(n) there are n interaction lines, the derivative is ! ∂X(n) n = X(n)(λ). ∂λ λ Σ Therefore the perturbation expansion of the derivative of X is just the same as of X itself without the 1/n pre-factor of the n-th order term but with a 1/λ pre-factor. Hence, it is easy to realize, see Fig. 5.33, that

∂Xe ∂X 1 X X = = T Σ (k, i ) G (k, i ). (5.70) ∂λ ∂λ 2λ σ n σ n Σ n kσ

213 Comparing with (5.68) we ﬁnd that ∂Ω(λ) ∂Xe = . (5.71) ∂λ ∂λ Since, for λ = 0, Xe(0) = 0, we immediately ﬁnd that, at λ = 1,

Ω = Ω0 + Xe [Σ]

+ X X (0) in 0 = Ω0 − T ln 1 − Gσ (k, in)Σσ(k, in) + Σσ(k, in) Gσ(k, in) e + X [G] n kσ

+ X X in 0 = T ln Gσ(k, in) − Σσ(k, in) Gσ(k, in) e + X [G] , (5.72) n kσ where we use the expression

+ X X (0) in 0 Ω0 = T ln Gσ (k, in) e . n kσ

5.10.3 The Luttinger theorem As usually let us assume that the Hamiltonian as well as the ground state are spin-rotationally invariant, which also implies that the Green’s function is spin-independent. In this case the total number of electrons per unit volume at zero temperature is

N 1 X X 2 X Z d ∂ = = n = − f() Arg ln G−1(k, + i0+) V V kσ V π ∂ k σ k 2 X Z 0 d ∂ = − Arg ln G−1(k, + i0+) V π ∂ k −∞ 2 X = − Arg ln G−1(k, 0 + i0+) − Arg ln G−1(k, −∞ + i0+) πV k 2 X = − φ(k, 0) − φ(k, −∞), (5.73) πV k where Im G(k, + i0+) −Im G−1(k, + i0+) φ(k, ) = tan−1 = tan−1 , (5.74) Re G(k, + i0+) Re G−1(k, + i0+) is the phase of the Green’s function approaching the real axis from above. From Eq. (5.14) it derives that

Im G(k, + i0+) = −Im G(k, − i0+) = |G(k, )| sin φ() = −π A(k, ) ≤ 0,

214 which implies that φ is defined within [−π, 0]. For infinite frequency a constant interaction can not be effective, hence 1 lim G(k, + i0+) = lim G(0)(k, + i0+) → , ||→∞ ||→∞ + i0+ which implies that 0− φ(k, −∞) = tan−1 = −π. −∞ We will show later that, in the most common situations, Im G−1(k, +i0+ → 0+i0+) → 0+ so that G−1(k, 0) is purely real hence

0− φ(k, 0) = tan−1 G−1(k, 0) is equal to π if G−1(k, 0) < 0 and 0 otherwise. Using these results to evaluate (5.73) we ﬁnally obtain N 2 X = θ G−1(k, 0) . (5.75) V V k This expression, also known as Luttinger’s theorem, states that the number of electrons is equal to the volume in the Brillouin zone which includes all momenta such that the Green’s function at zero real-frequency is positive.

Remarks Let us discuss in what cases the above result applies. The simplest example is when the single-particle local spectral function A() has a gap or a pseudo-gap, i.e. vanishes as a power-law, at the chemical potential, namely at = 0. Since

1 X A() = A(k, ), V k and both A() and A(k, ) are by deﬁnition real and positive, this implies that for any k A(k, → 0) → 0, hence also Im G(k, + i0+ → 0 + i0+) = 0. Then, through (5.74), we ﬁnd that

0− φ(k, 0) = tan−1 . Re G(k, 0)

Since the sign of the real part of G is the same as that of G−1, the Luttinger sum rule (5.75) is recovered. This would be for instance the case of a superconductor.

215 On the contrary, in the case of a metal we do expect a ﬁnite spectral function at the chemical potential A(0) 6= 0, so the above demonstration does not work. We notice that

−1 + + Re G (k, + i0 ) = − k − Re Σ(k, + i0 ), Im G−1(k, + i0+) = 0+ − Im Σ(k, + i0+), so that the question reduces to know how it behaves the imaginary part of the self-energy upon approaching the real axis. We have already shown, see Eq. (5.27), that within perturbation theory lim Im Σ(k, + i0+) = 0−, →0 which justiﬁes the Luttinger sum rule also in this case.

216 Chapter 6

Landau-Fermi liquid theory: a microscopic justiﬁcation

The Landau-Fermi liquid theory, that we have introduced at the beginning as a phenomeno- logical description of the low energy excitations of an interacting electron gas, can be justiﬁed microscopically using the Feynman diagram technique that we have just discussed.

6.1 Preliminaries

We are going to discuss the Landau-Fermi liquid theory in a model of N degenerate species of fermions, a = 1,...,N. At the moment we shall consider only short-range interactions; later we will extend the analysis to the Coulomb case. The non-interacting Green’s function is deﬁned by (0) (0) 1 Ga (k, in) = G (k, in) = 0 , in − k and the interacting one by 1 Ga (k, in) = G (k, in) = 0 , in − k − Σ(k, in) and both are a-independent. All energies are measured with respect to the chemical potential. We know that, provided perturbation theory is valid, the self-energy analytically continued on the real axis, Σ(k, ), has an imaginary part that vanishes at least like 2 for → 0. This implies that, for very small , 1 G (k, ) ' 0 , − k −

217 which we are going to assume has a simple pole on the real axis at = k such that

0 k − k −

Zk G (k, z) ' + Ginc (k, z) , (6.1) z − k where Ginc (k, z) is an incoherent component that has no singularity for small |z|. Within this assumption, the surface identiﬁed by k = 0 corresponds by the Luttinger theorem to the Fermi surface.

Let us now consider the Bethe-Salpeter equation (5.47), which can be written formally as

Γ = Γ0 + Γ0 R Γ, (6.2) where means the sums over internal indices, including the sum over momenta and Matsubara frequencies, and R (k, in; q, iωl) = G (k + q, in + iωl) G (k, in) . (6.3)

We note that, for free electrons and for small q and ωl

X (0) (0) X 1 1 T G (k + q, in + iωl) G (k, in) F (in) = T 0 0 F (in) n n in + iωl − k+q in − k I dz 1 1 = f(z) 0 0 F (z) 2πi z + iωl − k+q z − k 0 0 I 2 f k − f k+q 0 dz 1 ' 0 0 F k + f(z) 0 F (z) iωl − k+q + k 2πi z − k sing 0 0 0 I 2 ∂f k k+q − k 0 dz 1 ' − 0 0 0 F k + f(z) 0 F (z) ∂k iωl − k+q + k 2πi z − k sing 0 0 I 2 0 k+q + k dz 1 ' δ k 0 0 F (0) + f(z) 0 F (z) , iωl − k+q + k 2πi z − k sing

218 where F (z) means that one has to catch in the contour integral only the singularities of sing F . In the above equation, the term

0 0 vk · q δ k 0 iωl − vk · q has a singular behavior for small q and ωl. If q = 0 and ωl ﬁnite, it vanishes, while if ωl = 0 at 0 ﬁnite q, it is −δ k . In other words, the function

(0) (0) (0) R (k, in; q, iωl) = G (k + q, in + iωl) G (k, in) regarded as a distribution in in, has a singular part at small q and ωl given by

0 0 ∂f k vk · q − 0 δ (in) 0 . ∂k iωl − vk · q

The formal meaning of δ (in) for discrete Matsubara frequencies is 1 δ (i ) = δ . n T n0

Seemingly, the product of two interacting Green’s functions of the form (6.1), regarded as a distribution, can be represented at small q and ωl as

R (k, in; q, iωl) = G (k + q, in + iωl) G (k, in)

∂f (k) 2 k+q − k ' − δ (in) Zk + Rinc (k, in; q, iωl) ∂k iωl − k+q + k ≡ ∆ (k, in; q, iωl) + Rinc (k, in; q, iωl) , (6.4) where Rinc (k, in; q, iωl) is non-singular for q → 0 and ωl → 0, and

∂f (k) 2 k+q − k ∆ (k, in; q, iωl) = − δ (in) Zk . (6.5) ∂k iωl − k+q + k

We note that, if we send ﬁrst q → 0 and then ωl → 0, so-called ω-limit, then

ω lim lim R = R = Rinc, (6.6) ωl→0 q→0 while, in the opposite case, so called q-limit,

q ∂f (k) 2 q ω lim lim R = R = δ (in) Zk + Rinc = ∆ + R . (6.7) q→0 ωl→0 ∂k

219 Going back to the Bethe-Salpeter equation (6.2), we note that the irreducible vertex Γ0 is by deﬁnition non-singular at small q and ωl, in other words its ω-limit coincides with the q-limit. Therefore, in both limits,

Γq = Γ0 + Γ0 Rq Γq, (6.8) Γω = Γ0 + Γ0 Rω Γω, (6.9) with the same Γ0. Solving for Γ0 we ﬁnd that

Γ = Γq + Γq (R − Rq) Γ ≡ Γq + Γq ∆e Γ, (6.10) where

∂f (k) 2 ∆e (k, in; q, iωl) = ∆ (k, in; q, iωl) − δ (in) Zk ∂k ∂f (k) 2 iωl = − δ (in) Zk , (6.11) ∂k iωl − k+q + k or, alternatively, Γ = Γω + Γω (R − Rω) Γ = Γω + Γω ∆ Γ. (6.12) 0 In this way we have been able to absorb the unknown Rinc and Γ into two scattering vertices, Γq and Γω.

6.1.1 Vertex and Ward identities

The fermionic operators cka, labelled by generic quantum numbers a = 1,N, have, as we assumed, interacting Green’s functions diagonal in these numbers and independent of them. Let us now add a perturbation of the form

X X † δH = h(−q) · mab(k, q) ckack+qb ≡ h ·M, k ab in the limit of q → 0. In this limit it is convenient for what follows to rotate the basis cka into the one ckα that diagonalizes the matrixm ˆ (k) with elements mab(k), i.e. mab(k) → mα(k) δαβ. However, even though we are going to use this diagonal basis to derive explicitly several identities, we will always provide for each of these identities also a matrix representation, which is obviously independent of the basis set. In the diagonal basis the non interacting Green’s function is also diagonal,

−1 (0) Gα (k, in) = in − k − h mα(k),

220 while the interacting one might not be so

−1 (Gα(k, in))αβ = in − k − h mα(k) δαβ − Σαβ(k, in). (6.13) If the operator M refers to a conserved quantity of the fully interacting Hamiltonian, then the interacting self-energy is diagonal, hence also the Green’s function. In the presence of the external ﬁeld, the average value of the operator M, namely X 1 X X X hMi = m (k) n = T m (k) G (k, i ), α kα V α αα n α k n α becomes h dependent, which allows to deﬁne the thermodynamic susceptibility

∂hMi 1 X X X ∂Gαα(k, in) κ = = T mα(k) . (6.14) ∂h | h=0 V ∂h | h=0 k n α

We note that, considering Gαβ as elements of a matrix Gˆ, as well as mα of a diagonal matrix mˆ , and so on, the following relation holds: ! ∂Gˆ(k, i ) ∂Σ(ˆ k, i ) n = Gˆ(k, i ) mˆ (k) + n Gˆ(k, i ). (6.15) ∂h n ∂h n

This equation is actually independent of the basis set, namely in holds also in the original representation wherem ˆ (k) has elements mab(k). On the other hand, through Eq. (5.58) we know that

∂Σαβ(k, in) X X 1 X = T Γ (k i , p i ; p i , k i ) G (p, i ) m (p) G (p, i ), ∂h V αγ;δβ n m m n δν m ν νγ m m γ p where the product of the two Green’s functions must be interpreted as

lim Gδν(p + q, im) Gνγ(p, im), q→0 namely as the q-limit. In the limit h → 0, that we will consider hereafter, all Green’s functions become diagonal and independent of the index α, i.e. Gαβ = δαβ G, hence

∂Σαβ(k, in) mα(k) δαβ + = mα(k) δαβ (6.16) ∂h | h=0 X X 1 X +T Γ (k i , p i ; p i , k i ) Rq(p i ) m (p), V αγ;γβ n m m n m γ m γ p where we recall that q R (p im) = lim G(p + q, im) G(p, im). q→0

221 We can also introduce the triangular vertex Λαβ(k + q in + iωl, k in; q iωl) corresponding to the perturbation, with non-interacting value at small q

(0) Λαβ (k + q in + iωl, k in; q iωl) = mα(k) δαβ. As we know Λ satisﬁes the Bethe-salpeter equation

Λ = Λ(0) + Γ R Λ(0).

Similarly to the above discussion of the interaction vertices, we can deﬁne the q and ω limits of the triangular vertex

Λq = Λ(0) + Γq Rq Λ(0), Λω = Λ(0) + Γω Rω Λ(0).

Solving for Λ(0) one ﬁnds that the vertex is related to its q or ω limits by

Λ = Λω + Γω ∆ Λ = Λω + Γ ∆ Λω, (6.17) Λ = Λq + Γq ∆e Λ = Λq + Γ ∆e Λq, (6.18) where ∆ and ∆e have been deﬁned in Eqs. (6.5) and (6.11), respectively. Through Eq. (6.16) and by the deﬁnition of the triangular vertex, we obtain the following identity: ∂Σ (k, i ) m (k) δ + αβ n = Λq (k, i ) , (6.19) α αβ ∂h αβ n or, in a representation independent of the basis choosen,

∂Σ(ˆ k, i ) mˆ (k) + n = Λˆ q(k, i ) . (6.20) ∂h n

Let us assume now that the operator M is a conserved quantity, which implies that, even if h 6= 0, the Green’s functions remain diagonal in α. In this case the Ward identity (5.56) holds, which reads in this case and for q = 0

Σα(k, in + iωl) − Σα(k, in) 1 X X X mα(k) = − T iωl V p m β

Γαβ;βα(k in, p im + iωl; p im, k in + iωl) Gβ(p, im) Gβ(p, im + iωl) mβ(p).

In the limit of ωl → ω0 1 we can write

∂Σα(k, in) 1 X X X mα(k) = − T Γαβ;βα(k in, p im + iω0; p im, k in + iω0) ∂in V p m β

222 Gβ(p, im) Gβ(p, im + iω0) mβ(p) ω = −Λαα(k, in) + mα(k). (6.21)

In other words, another identity holds, namely:

∂Σα(k, in) ω mα(k) − mα(k) = Λα(k, in) , (6.22) ∂in or, in matrix form, ˆ ∂Σ(k, in) ω mˆ (k) − mˆ (k) = Λˆ (k, in) . (6.23) ∂in

In the limit of vanishing h and for small n, this identity becomes

ˆ (0) −1 ˆ ω Λ (k) Zk = Λ (k, in) . (6.24)

We emphasize that this result holds only for a conserved operator.

6.2 Correlation functions

Two single particle operators

X X † (0) X X † (0) A(q) = cka ΛA ab(k, q) ck+qb, B(q) = cka ΛB ab(k, q) ck+qb, k ab k ab can be associated to a correlation function χAB(q, iωl) that is deﬁned by

(0) (0) (0) (0) (0) χAB = Tr ΛA R ΛB + Tr ΛA R Γ R ΛB = Tr ΛA R ΛB , (6.25) where all quantities are calculated at zero external ﬁelds.1 Once more we can deﬁne q and ω limits by (we drop for simplicity the subscripts A and B, keeping in mind that the triangular vertex on the left refers to A, and that one on the right to B): χω = Tr Λ(0) Rω Λω ,

1The actual meaning of the trace over the internal indices α’s is

(0) X (0) Tr ΛA R ΛB = ΛA αβ Gβγ ΛB γδ Gδα

X (0) 2 = ΛA αβ G ΛB βα since at zero ﬁeld the Green’s functions are diagonal and independent of α.

223 χq = Tr Λ(0) Rq Λq .

Through Eqs. (6.17) and (6.18), we can rewrite (6.25) in the following way: h i χ = Tr Λ(0) R (Λω + Γ ∆ Λω) = Tr Λ(0) R Λω + Tr Λ(0) R Γ ∆ Λω h i = Tr Λ(0) Rω Λω + Tr Λ(0) (R − Rω)Λω + Tr Λ(0) R Γ ∆ Λω = χω + Tr Λ(0) ∆ Λω + Tr Λ(0) R Γ ∆ Λω = χω + T r Λ ∆ Λω h i = χω + Tr (Λω + Λω ∆ Γ) ∆ Λω = χω + Tr Λω ∆ Λω + Tr Λω ∆ Γ ∆ Λω . (6.26)

Similarly, it also holds that χ = χq + Tr Λq ∆e Λq + Tr Λq ∆e Γ ∆e Λq . (6.27)

We note that both ∆(k, in; q, iωl) and ∆(e k, in; q, iωl) contain a delta function in frequency, δ(n) and a derivative of the Fermi function with respect to its argument. This implies first that all sums over the internal Matsubara frequencies drop out, all frequencies being fixed to the lowest one 0 = π T , and that all momenta are very close to the Fermi surface. We define, for small q and ωl ω ω λk = Zk Λ (k, 0; q = 0, iωl → 0) , (6.28) q q λk = Zk Λ (k, 0; q → 0, iωl = 0) , (6.29) Akp (q, iωl) = Zk Zp Γ(k + q 0 + iωl, p 0; p + q 0 + iωl, k 0) ,, (6.30) ω fkp = Zk Zp Γ (k 0, p 0; p 0, k 0), (6.31) as well as

∂f (k) k+q − k δk (q, iωl) = − , (6.32) ∂k iωl − k+q + k ∂f (k) iωl δek (q, iωl) = − . (6.33) ∂k iωl − k+q + k With these deﬁnitions, A = f + f δ A, (6.34) and Eqs. (6.26) and (6.27) can be written as

ω ω ω ω ω χ(q, iωl) = χ + Tr λ δ λ + Tr λ δ A δ λ

224 = χq + Tr λq δe λq + Tr λq δe A δe λq , (6.35) where the trace does not include anymore the sum over Matsubara frequencies. The explicit expressions read

ω 1 X X ω ω ∂f (k) k+q − k χ(q, iωl) = χ − λk αβ λk βα V ∂k iωl − k+q + k k αβ

1 X X ∂f (k) ∂f (p) ω ω + 2 λk αβ Ak β,p γ;p δ,k α(q, iωl) λp δγ V ∂k ∂p kp αβγδ − − k+q k p+q p (6.36) iωl − k+q + k iωl − p+q + p

q 1 X X q q ∂f (k) iωl = χ − λk αβ λk βα V ∂k iωl − k+q + k k αβ

1 X X ∂f (k) ∂f (p) q q + 2 λk αβ Ak β,p γ;p δ,k β(q, iωl) λp δγ V ∂k ∂p kp αβγδ iω iω l l . (6.37) iωl − k+q + k iωl − p+q + p

We just note that, through Eqs. (6.20), (6.13) and (6.15), we can relate χq

" # q 1 X X (0) (0) ∂Σ(k, in) χ = T Tr ΛA (k, 0) G(k, in) ΛB (k, 0) + G(k, in) V ∂hB k n | h=0 " # 1 X X (0) ∂G(k, in) = T Tr ΛA (k, 0) = κAB, V ∂hB k n | h=0 to the thermodynamic susceptibility, κAB, of the operator A to an external ﬁeld hB that couples to B. So far we have been able to express any correlation function in a form similar to that of free particles (quasi-particles), with R replaced by the “free-particle” quantities δ and δe, all interaction eﬀects being absorbed into the unknown three- and four-leg vertices. However, when both A and B refer to conserved quantities, a lot of simplications arise.

6.2.1 Conserved quantities Suppose that A is conserved (the same result would hold if B, or both were conserved). Then

ω ω ω (0) χAB = Tr ΛA R ΛB .

225 Through Eq. (6.23), we ﬁnd that2

" # ω 1 X X (0) ∂Σ(k, in) (0) χAB = T Tr ΛA (k, q = 0) 1 − G(k, in + iωl) G(k, in)ΛB (k, q = 0) V ∂in k n 1 X h (0) (0) i X ∂G(k, in) = Tr ΛA (k, q = 0)ΛB (k, q = 0) T = 0, (6.38) V ∂in k n

ω namely χAB vanishes. Therefore, by means of Eq. (6.24), we ﬁnally obtain, when both A and B are conserved, and for small q and ωl,

1 X X (0) (0) ∂f (k) k+q − k χAB(q, iωl) = − ΛA ab(k, 0)ΛB ba(k, 0) V ∂k iωl − k+q + k k ab 1 X X (0) (0) + Λ (k, 0) A Λ (k, 0) V 2 A ab k b,p c;p d,k a B dc kp αβ ∂f ( ) ∂f ( ) − − k p k+q k p+q p . (6.39) ∂k ∂p iωl − k+q + k iωl − p+q + p We immediately recognize that the above expression coincides with that one obtained within the Landau-Fermi-liquid theory, in the general case of quantum numbers a = 1,...,N. There- fore, when considering conserved quantities, all interaction eﬀects can be absorbed into the quasiparticle energy k and into the scattering vertices A.

6.3 Coulomb interaction

So far we have only taken into account short-range electron-electron interaction. What does it change when the interaction is instead long-ranged? Let us assume therefore that the electrons interact mutually by a long-range Coulomb repulsion 1 X X H = U(q) c† c† c† c† , int V k+qa pb p+qb ka kpq ab with 4πe2 U(q) = . q2 Furthermore we shall assume that there is a background positive charge that compensate exactly the electron one. This implies that the self-energy, see Fig. 5.19, does not include the Fock term, which otherwise would give an inﬁnite contribution since U(q → 0) → ∞. The problem with

2 Recall that at zero ﬁeld the Green’s functions Gab = δab G.

226 a long-range interaction is that, besides the singular behavior for small q and ωl due to R, additional singularities arise at q → 0 by the interaction. In order to disentagle both sources of singularities, it is convenient to adopt the same approach as in section 5.6. Therefore, in the perturbative expansion of the four-leg vertex Γ(k + q, p; p + q, k), we identify a subset of proper diagrams that do not include any interaction line with momentum q. This subset defines the proper vertex Γ,e which by definition does not include any singularity for q → 0 arising from the singular behavior of U(q). The proper vertex allows to define a proper density-density correlation function3 χe = Tr (R) + Tr R Γe R , as well as a proper three-leg vertex

Λe = Λ(0) + Γe R Λ(0).

By exploiting the analytical properties of R, we can follow precisely what we did before, this time for proper vertices. Notice that, since the self-energy does not include the Fock-term, all vertex and Ward identities still hold in terms of proper vertices. In particular, the expression (6.39) remains valid for the proper correlation function with

ω Akp = Zk Zp Γe (k, p; p, k).

Once we have been able to express proper correlation functions in terms of few parameters, the improper ones can be derived straightforwardly just like in section 5.6. The results obviously coincide with the Landau-Silin theory.

3 (0) Note that in the charge-density channel the non-interacting vertex Λab = δab.

227 Chapter 7

Kondo eﬀect and the physics of the Anderson impurity model

The behavior of magnetic impurities (Fe, Mn, Cr) diluted into non-magnetic metals (e.g. Cu, Ag, Au, Al) is likely the simplest manifestation of a breakdown of the conventional band-structure theory due to strong electron-electron correlations. 4 We know that the behavior of a good metal with a large Fermi temperature, TF ∼ 10 K, is dominated at low temperature T TF by the Pauli principle. For instance its magnetic susceptibility is roughly constant, χ ∼ 1/TF , and one should in principle heat the sample to very high temperatures T TF to release the spin entropy and recover a Curie-Weiss behavior χ ∼ 1/T . Moreover the resistivity is an increasing function of temperature, since the channels which may dissipate current, the coupling to the lattice as well as to multi particle-hole excitations, become available only upon heating At odds with this expectation, if one introduces a very diluted (few part per million) concentration of magnetic impurities the above behavior changes drastically. We just mention three distinct features.

(1) The magnetic susceptibility shows a Curie-Weiss behavior well below TF , proportional to the impurity concentration ni and roughly with the same g-factor of the isolated magnetic impurity, apart from corrections due to the crystal ﬁeld. Around a very low temperature, called Kondo-temperature TK , the Curie-Weiss behavior turns into a logarithmic behavior and ﬁnally the susceptibility saturates at low temperature to a value χ(0) ∼ ni/TK , with χ(T ) − χ(0) ∼ −T 2.

(2) The resistivity R(T ) displays a minimum around TK , followed at T < TK by a logarithmic increase. At very low temperatures R(T ) approaches a constant value R(0) ∝ ni with R(T ) − R(0) ∼ −T 2. The value of the residual resistivity R(0) suggests very strong scattering potential, near the so-called unitary limit.

228 (3) The entropy which is released above the Kondo temperature, which can be obtained by measuring the specific heat, includes only spin and eventually orbital degrees of freedom of the isolated impurity but not the charge degrees of freedom. This indicates that the magnetic impurities behave as local moments above TK . Explaining this behavior amounts actually to understand three different problems. The first concerns the region TK T TF and can be formulated as follows: how is it possible to sustain a local moment inside a metal ? We will see that to answer this question one is obliged to abandon the conventional band-structure theory. The second problem regards what happens around the Kondo temperature, namely to understand the logarithmic crossover. Finally, the last question concerns the low temperature behavior and the way the local moments get screened and why the resistivity is a decreasing function of temperature. Last two questions turn out to represent a complicated many-body problem which has re- quired the full machinary of renormalization group to be solved for the first time by Wilson. Nowadays one has at his disposal other sophisticated methods like the Bethe-Ansatz, which provides an exact solution, the abelian bosonization and the Conformal Field Theory.

7.1 Brief introduction to scattering theory

We start by showing why the existence of local moments in a metal for T TF is so puzzling. Let us consider an impurity imbedded in a normal metal. We only consider the valence band, described by the Hamiltonian ˆ X † H0 = k ckσckσ. (7.1) k,σ The scattering potential provided by the impurity has the general form

ˆ X X † V = Vkp ckσcpσ, (7.2) σ k,p where Vkp are the matrix elements of the impurtity potential onto the valence band Block waves. The single-particle Green’s function in complex frequency for the full Hamiltonian Hˆ = Hˆ0 + Vˆ can be formally written as the matrix 1 Gˆ(z) = . (7.3) z − Hˆ Analogously the unperturbed one is 1 Gˆ0(z) = . (7.4) z − Hˆ0

229 The above operators have the following interpretation. If we consider for instance the inverse of the Green’s function Gˆ, namely Gˆ(z)−1 = z − H,ˆ in the basis set of the valence band Block waves, this is the matrix

ˆ −1 G(z) = δσσ0 [δkp z − δkp k − Vkp] , (7.5) kσ,pσ0 hence the Green’s function is the inverse of the above matrix. Suppose that we have diagonalized the full Hamiltonian and got the eigenvalues a’s. In the diagonal basis also the Green’s function is diagonal with matrix elements 1 G(z)ab = δab . z − a Therefore we readily ﬁnd that 1 1 X 1 − lim Im Tr Gˆ(z = ω + iδ) = − lim Im π δ→+0 π δ→+0 ω − + iδ aσ a X = δ(ω − a) ≡ ρ(ω), aσ where ω is a real frequency and ρ(ω) is the so-called density of states (DOS). Since the trace is invariant under unitary transformations, hence also under the transformation which diagonalizes the Hamiltonian, it is generally true that 1 h i − lim Im Tr Gˆ(z = ω + iδ) = ρ(ω). (7.6) π δ→+0 Analogously 1 h i X − lim Im Tr Gˆ0(z = ω + iδ) = ρ0(ω) = δ(k − ω),. (7.7) π δ→+0 kσ gives the DOS of the host metal. Let us formally write the full Green’s function as 1 1 Gˆ = = ˆ ˆ ˆ−1 ˆ z − H0 − V G0 − V 1 h i−1 = Gˆ0 = Gˆ0 + Gˆ0 Vˆ 1ˆ − Gˆ0 Vˆ Gˆ0 1ˆ − Gˆ0 Vˆ

≡ Gˆ0 + Gˆ0 Tˆ Gˆ0, (7.8) which provides the deﬁnition of the so-called T -matrix, namely h i−1 Tˆ(z) = Vˆ 1ˆ − Gˆ0 Vˆ . (7.9)

230 From now on we will assume z = ω + iδ with δ a positive inﬁnitesimal number, so we will not explicitly indicate the limδ→+0. We notice that ∂ ln Gˆ(z) = −Gˆ(z), ∂z so that 1 ∂ n o ρ(ω) = ImTr ln Gˆ(z) . (7.10) π ∂z We are interested in the variation of the DOS induced by the impurity, which, by making use of (7.8) and (7.9), is given by

∆ρ(ω) = ρ(ω) − ρ0(ω) 1 ∂ n h io = ImTr ln Gˆ(z) Gˆ (z)−1 π ∂z 0 ( " #) 1 ∂ 1 = ImTr ln π ∂z 1ˆ − Gˆ0(z) Vˆ 1 ∂ n o = ImTr ln Vˆ −1 Tˆ(z) . (7.11) π ∂z We deﬁne the matrix of the scattering phase shifts δˆ(z) = Im ln Vˆ −1 Tˆ(z) = Arg Vˆ −1 Tˆ(z) , (7.12) through which 1 ∂ ∆ρ(ω) = Trδˆ(ω). (7.13) π ∂ω

We notice that for |z| → ∞, Gˆ0(z) → 1/z → 0, hence Tˆ(z) → Vˆ and δˆ(z) → 0. The variation of the total number of electrons, ∆Nels, induced by the impurity at fixed chemical potential µ is therefore Z µ 1 ˆ ∆Nels = dω ∆ρ(ω) = Trδ(µ), (7.14) −∞ π which is the so-called Friedel sum rule. Let us go back to the definition of the T -matrix (7.9). One readily finds that its inverse is given by −1 h −1 i Tˆ(z) = Vˆ − Gˆ0(z) . Since the Hamiltonian is hermitean, then

† h −1i h −1 ∗ i Tˆ(z) = Vˆ − Gˆ0(z ) ,

231 so that † h −1i −1 h ∗ i Tˆ(z) − Tˆ(z) = Gˆ0(z) − Gˆ0(z ) . Therefore h i† Tˆ(z)† Tˆ(z)−1 − Tˆ(z)−1 Tˆ(z) = Tˆ(z) − Tˆ(z)†

† h ∗ i = Tˆ(z) Gˆ0(z) − Gˆ0(z ) Tˆ(z). (7.15)

Since ∗ Gˆ0(z) − Gˆ0(z ) = −2πiδ ω − Hˆ0 , the Eq. (7.15) implies the following identity

† X † Tkp(ω) − Tkp(ω) = −2πi Tkq(ω) δ(ω − q) Tqp(ω), (7.16) q which is the so-called optical theorem. It also shows that the imaginary part of the T -matrix is ﬁnite only within the conduction band. Let us now analyse the on-shell T -matrix

Tkp(), where k = p = . We can rewrite the on-shell optical theorem as follows

h † i −2πi δ (k − p) Tkp() − Tkp() 2 X † = (−2πi) δ (k − p) Tkq() δ(k − q) Tqp(). q

The above equation implies that, if we introduce the so-called on-shell S-matrix through

Skp() = δkp − 2πi δ (k − p) Tkp(), (7.17) where, as before, k = p = , then it follows that the S-matrix is unitary, i.e

Sˆ Sˆ† = I.ˆ (7.18)

Skp() is the transition probability that an electron in state k scatters elastically into state p. Since it is unitary, it follows that only elastic scattering survives.

Let us consider the simpler case of a spherical Fermi surface, i.e. k = k depending only on the modulous of the wavevector. In addition we assume that the matrix elements Vkp only

232 depend on the angle between the two wavevectors, θkp, hence can be expanded in Lagrange polynomials X Vkp = Vl (2l + 1) Pl (cos θkp) . (7.19) l Then also the T -matrix has a similar expansion X Tkp(z) = tl(z) (2l + 1) Pl (cos θkp) . (7.20) l Since Z dΩq 1 P (cos θ ) P 0 (cos θ ) = δ 0 P (cos θ ) , 4π l kq l qp 2l + 1 ll l kp the optical theorem transforms into ∗ 2 tl (ω) − tl (ω) = 2iIm tl (ω) = −2πi ρ(ω) |tl (ω)| , (7.21) where ρ(ω) is the bare conduction electron density of states per spin. Since

iδl(ω) tl(ω) = |tl(ω)| e , it derives that 1 t (ω) = − sin δ (ω) eiδl(ω). (7.22) l πρ(ω) l If the concentration of impurity is very low, then the scattering rate suﬀered by the electrons is given by the Fermi-golden rule with the T -matrix, namely 1 X = 2π n T † (ω) δ(ω − ) T (ω) (1 − cos θ ) , τ i kq q qk kq k q where ni is the impurity concentration and the last term is a geometric factor which guarantees that forward scattering, θ = 0, does not contribute to the current dissipation. Since

(2l + 1) z Pl(z) = (l + 1) Pl+1(z) + l Pl−1(z), we obtain Z 1 X ∗ dΩq = 2π ρ(ω) tl (ω) tl0 (ω) (2l + 1) Pl(cos θkq) ni τk(ω) 4π ll0 0 0 0 (2l + 1) Pl0 (cos θkq) + (l + 1) Pl0+1(cos θkq) + l Pl0−1(cos θkq) X 2 ∗ ∗ = 2π ρ(ω) (2l + 1) |tl(ω)| + l tl (ω) tl−1(ω) + (l + 1) tl (ω) tl+1(ω), (7.23) l which as expected gives a scattering rate independent upon momentum. The zero-temperature resistivity turns out to be proportional to 1 R(0) ∝ . (7.24) τ(0)

233 7.1.1 General analysis of the phase-shifts Let us keep assuming a spherically symmetric case. In addition we assume that the scattering components Vl are non-zero only for a well deﬁned l = L. For instance, in the case of magnetic impurities with partially ﬁlled d-shell, L = 2. Then the only non-zero phase shift is

−1 Im tL(ω) δL(ω) = tan . Re tL(ω) An important role is played by the frequencies at which the real part vanishes. The ﬁrst possibility is that this occurs outside the conduction band, either below or above. In this case we know that the imaginary part is zero. This implies that the phase-shift jumps by π at these energies so that the variation of the DOS is δ-like. In this case one speaks about bound states which appear outside the conduction band. Clearly this possibility can not explain a Curie-Weiss behavior for T TF . Indeed at very low temperature the bound state is either doubly occupied, if below the conduction band, or empty, if above, and one needs a temperature larger than the conduction bandwidth, hence larger that TF , to release its spin entropy. The other possibility is that the real part vanishes at a frequency ω∗ within the conduction band where the imaginary part is non zero. Around ω∗ and assuming a real part linearly vanishing and an imaginary part roughly constant we get

−1 Γ δL(ω) ' tan , ω − ω∗ leading to a Lorentian DOS variation due to the impurity 1 Γ ∆ρ(ω) ∼ 2 2 , (7.25) π (ω − ω∗) + Γ which is called a resonance. Clearly, in other for this resonance to contribute at T TF , ω∗ should be very close to the chemical potential, in particular |ω∗ − µ| ≤ TK , and, in addition, Γ ' TK .

Let us therefore assume Γ = TK and for further simplicity that ω∗ = 0. The contribution of the resonant state to the magnetic susceptibility is then given by Z ∂f() 1 TK ∆χ(T ) = −µB g (2L + 1) d 2 2 , (7.26) ∂ π + TK where −1 f() = 1 + eβ , is the Fermi distribution. One readily ﬁnds that for T TK , 2L + 1 ∆χ ' µ g , B T

234 while for T TK 2L + 1 ∆χ ' µB g , πTK similar to the observed behavior. In addition the resistivity would be given through (7.23) and (7.24) by (ρ(ω) ∼ ρ(0) ≡ ρ0)

1 2(2L + 1) 2 R(0) ∝ = ni sin δL(0) τ(0) πρ0 2(2L + 1) = ni , (7.27) πρ0 again compatible with the almost unitary limit, δ(0) = π/2, observed experimentally. Therefore the existence of a narrow resonance near the chemical potential with width exactly given by TK would seem the natural explanation to what experiments find out. However this is evidently very strange, since the Kondo behavior is observed in many different host metals as well as for different magnetic impurities, hence it would be really surprising that in all these cases a resonance appear always pinned near the chemical potential. Moreover this simple single-particle scenario fails to explain the entropy released above TK . Indeed, for T TK the resonance is effectively like an isolated level which can be empty, singly occupied with a spin up or down, or doubly occupied. Therefore its entropy should by S = 2(2L+1) ln 2. On the other hand the experiments tells us that only spin and orbital degrees of freedom are released above TK , which amounts in the above simplified model to an entropy S = 2(2L + 1) ln 2. In other words a resonance at the chemical potential at temperatures larger than its broadening is not at all the same as a local moment, since the former does have valence fluctuations which are absent in the latter. Therefore, although the resonance scenario is suggestive, it is not at all the solution to the puzzle.

7.2 The Anderson Impurity Model

Something which is obviously missing in the above analysis are local electron-electron correlations around the magnetic ions. A narrow resonance induced by an impurity is quite localized around the same impurity and is essentially an atomic level broadened by the hybridization with the conduction electrons of the host metal. Therefore, like an atomic level, such a resonance is likely to accomodate only a fixed number of electrons, say N, paying a finite amount of energy in adding or removing electrons with respect to the reference valency N. As usual it is convenient to define a so-called Hubbard repulsion U through

U = E(N + 1) + E(N − 1) − 2E(N), (7.28)

235 where E(M) is the total energy when the resonance is forced to have M electrons, N being the equilibrium value. In principle one might include this additional ingredient by adding to the Hamiltonian a term U (ˆn − N)2 , (7.29) 2 withn ˆ the occupation number operator of the resonance. If it were possible to define such an operator, then (7.29) would actually solve one of the puzzles. Indeed, if U TK , for temperatures TK T U valence fluctuations with respect to the equilibrium value N would be suppressed and the only degrees of freedom contributing to the entropy would be those related to the degeneracy of the N-electron configurations at the resonance. Namely, if the resonance is 2(2L+1) degenerate, 2 from spin and (2L+1) from the orbital, the degeneracy of the N-electron state is the binomial 2(2L + 1) (4L + 2)! C2(2L+1) = = , N N N! (4L + 2 − N)!

2(2L+1) with entropy S = ln CN . In other words the resonance would actually behave like a local moment. Unfortunately the resonance occupation number operator is an ill defined object. To overcome this difficulty, Anderson had the idea to represent the resonance as an additional electronic-level inside the conduction band, even if this would be, rigorously speaking, an over- complete basis. This leads to the so-called Anderson Impurity Model (AIM) defined in the simplest case of a single-orbital impurity L = 0 as

X † X † H = k ckσckσ + Vk ckσdσ + H.c. kσ kσ U + (ˆn − 1)2 + nˆ . (7.30) 2 d d d

This model describes a band of conduction electrons with energy dispersion k, hybridized with a single level, dσ, with energy d, both k and d being measured with respect to the chemical P † potential, andn ˆd = σ dσdσ. This level suﬀers from an interaction U which tends to lock no more no less than a single electron.

Let us start from the non-interacting model, U = 0, to show that it indeed describes a resonance. Before switching the hybridization Vk the conduction electron Green’s function in complex frequency z is (0) 1 Gk (z) = = (7.31) z − k and is represented as a dashed tiny line and the impurity one is 1 G(0)(z) = = (7.32) z − d

236 * V k Vk = + + ..... = k * V k Vk + k

k p Vk V * k k p p = + k V k k k =

V* k k k =

Figure 7.1: Dyson equations for the Green’s functions. Wavy (red) lines are the Green’s functions in the presence of hybridization, represented as a circle, while tiny (black) lines are unperturbed Green’s functions.

and represented as a solid tiny line. The corresponding Green’s functions in the presence of Vk,

Gkp(z) = G(z) = , are represented as wavy lines. Notice that the conduction electron Green’s function is nomore diagonal in momentum. In addition one has to introduce the mixed Green’s functions

† Gkd(τ) = −hT ckσ(τ) dσ i = † Gdk(τ) = −hT dσ(τ) ckσ(τ) i =

Through the Dyson equation shown diagramatically in Fig. 7.1 we ﬁnd that 1 G(z) = , (7.33) z − d − ∆(z) where the so-called hybridization function is deﬁned through

2 X |Vk| ∆(z) = . (7.34) z − k k

237 The conduction electron Green’s function is instead

(0) (0) (0) Gkp(z) = δkp Gk (z) + Gk (z) Tkp(z) Gp (z), (7.35) where the T -matrix is 1 ∗ Tkp = Vk Vp . (7.36) z − d − ∆(z) In addition

Vk 1 Gkd(z) = , (7.37) z − k z − d − ∆(z) ∗ Vk 1 Gdk(z) = . (7.38) z − k z − d − ∆(z)

We notice that the hybridization function has a branch cut on the real axis since

+ X 2 ∆(z → ± i0 ) = ∓ iπ |Vk| δ ( − k) ≡ ∓ iΓ(). (7.39) k In terms of Γ() > 0 we can write

Z d Γ() ∆(z) = . π z −

Since we are interested in temperatures/frequencies much smaller than the conduction electron Fermi energy, we will assume for simplicity that Γ() = Γ for || < W and Γ() = 0 otherwise, with W of order TF . Then, if |z| W one readily ﬁnds that

∆(z) = −iΓ sgn Im z. (7.40)

In this approximation the impurity DOS is 1 1 Γ ρd() = − Im G( + i0+) = 2 2 , (7.41) π π ( − d) + Γ which is just a resonant state trapping a number of electrons of given spin given by

Z +∞ Nd = d f() ρd(). (7.42) −∞

238 7.2.1 Variation of the electron number

Quite generally the Green’s function, e.g. of the dσ-electron, in Matsubara frequency is given by −βEM −βEN 1 X † e + e G(iωn) = hN|dσ|MihM|dσ|Ni . (7.43) Z iωn − (EM − EN ) NM P We want to calculate T n G(iωn). We notice that the Fermi function in the complex plane −1 f(z) = eβz + 1 , has poles on the imaginary axis just at the Matsubara frequencies iωn with residue −T . Therefore if we consider a contour which encircles anticlockwise the imaginary axis then I dz X f(z) G(z) = T G(iω ). 2πi n n On the other hand the contour integral is also equivalent to the clockwise contour which runs avoiding the imaginary axis, which amounts to sum over the poles of the denominator in (7.43). As a result one gets X 1 X T G(iω ) = hN|d |MihM|d† |Ni f(E − E ) e−βEM + e−βEN n Z σ σ M N n NM 1 X = hN|d |MihM|d† |Ni e−βEM Z σ σ NM 1 X = e−βEM hM|d† d |Mi = hd† d i. Z σ σ σ σ M Therefore the sum over Matsubara frequencies gives the occupation number, hence the variation of the electron number per spin due to the impurity is simply given by " # X X (0) ∆Nels = T G(iωn) + Gkk(iωn) − Gkk(iωn) n k " 2# X 1 X 2 1 = T 1 + |Vk| iωn − d − ∆(iωn) iωn − k n k X ∂ = T ln (iω − − ∆(iω )) , (7.44) ∂iω n d n n n

Let us consider the contour drawn in Fig. 7.2. The function S(z) = ln(z − d − ∆(z)) has a branch cut on the real axis and it is analytic everywhere else. Therefore I dz ∂ f(z) S(z) = 0, 2πi ∂z

239 Im z

Re z

Figure 7.2: Integration contour. because in the region which excludes both axes the function is analytic. On the other hand I dz ∂S(z) X ∂ 0 = f(z) = T ln (iω − − ∆(iω )) 2πi ∂z ∂iω n d n n n Z +∞ d ∂ + f() S( + i0+) − S( + i0+) −∞ 2πi ∂ Therefore we ﬁnally get Z +∞ d ∂ + + ∆Nels = − f() S( + i0 ) − S( − i0 ) . (7.45) −∞ 2πi ∂ Within the previous approximation

+ + −1 Γ S( + i0 ) − S( − i0 ) = ln ( − d + iΓ) − ln ( − d − iΓ) = 2i tan , − d hence, through Eq. (7.42), Z +∞ 1 Γ ∆Nels = d f() 2 2 = Nd. (7.46) −∞ π ( − d) + Γ In other words the change in the number of electrons only comes from those electrons occupying the impurity level. If for instance d↑ is diﬀerent from d↓, at zero temperature the number of electrons at the impurity with spin σ is 1 1 N = − tan−1 dσ . (7.47) dσ 2 π Γ

7.2.2 Energy variation For later convenience, we calculate here the impurity contribution to the total energy. Suppose that we change Vk → λ Vk. The total energy becomes a function of λ, E(λ), which, by the Helmann-Feynmann theorem, satisﬁes ∂E(λ) ∂H = hΨ | |Ψ i ∂λ λ ∂λ λ

240 X † X X ∗ = hΨλ|Vk ckσdσ + H.c. |Ψλi = 2T Vk Gdk(iωn; λ) + Vk Gkd(iωn; λ). kσ n k

Here |Ψλi is the ground state wavefunction at a ﬁxed λ, and G(iωn; λ) are the corresponding Green’s functions. Speciﬁcally

Vk 1 Gkd(z; λ) = λ 2 , (7.48) z − k z − d − λ ∆(z) ∗ Vk 1 Gdk(z; λ) = λ 2 , (7.49) z − k z − d − λ ∆(z) with ∆(z) deﬁned in Eq. (7.34). Therefore

2 ∂E(λ) X X |Vk| 1 = 4T λ 2 ∂λ iωn − k iωn − d − λ ∆(iωn) n k X 1 X ∂ = 4T λ ∆(iω ) = −2T ln iω − − λ2 ∆(iω ) , n iω − − λ2 ∆(iω ) ∂λ n d n n n d n n so that the energy variation due to the impurity is 2 X ∂ iωn − d − λ ∆(iωn) E(λ) − E(0) = −2T ln . (7.50) ∂λ iω − n n d In addition the average value of the hybridization at ﬁxed λ is X ∂E(λ) E (λ) = hΨ |λ V c† d + H.c. |Ψ i = λ , (7.51) hyb λ k kσ σ λ ∂λ kσ so that the change in the bath energy is given by ∂E(λ) E (λ) − E (0) = E(λ) − E(0) − E (λ) = E(λ) − E(0) − λ . (7.52) bath bath hyb ∂λ

To calculate (7.50) we use the same integration contour as in Fig. 7.2. By noticing that

+ + + −1 0 ln( − d + i0 ) − ln( − d − i0 ) = 2i tan = 2π i θ(d − ), − d we ﬁnd Z 2 d −1 λ Γ E(λ) − E(0) = 2 f() tan − π θ(d − ) . (7.53) π − d

Let us evaluate this integral in the simple case d = 0 at zero temperature, where Z 0 2 d −1 λ Γ 2 2 W e E(λ) − E(0) = 2 tan − π = − λ Γ ln 2 . (7.54) −∞ π π λ Γ

241 − σ

Σ [G] = = U N σ d − σ

Figure 7.3: Self-energy within the Hartree-Fock approximation.

Therefore the full energy gain is 2 W e ∆E = − Γ ln , (7.55) π Γ which derives from an hybridization gain 4 W E = − Γ ln , (7.56) hyb π Γ and a bath energy cost 2 W ∆E = Γ ln . (7.57) bath π e Γ

7.2.3 Mean-ﬁeld analysis of the interaction

Let us consider the case d = 0 and switch on the interaction U U (ˆn − 1)2 = U nˆ nˆ − nˆ . 2 d d↑ d↓ 2 d

The second term gives an d = −U/2. Let us treat the ﬁrst term within the Hartree-Fock approximation. The self-energy for the spin σ impurity electrons is given by, see Fig. 7.3,

Σσ = UNd −σ, (7.58) where Nd −σ = hnˆd −σi, is the average value, which has to be evaluated self-consistently. Therefore the impurity Green’s function is 1 1 G (z) = = , (7.59) σ z − − ∆(z) − Σ (z) U d σ z + 2 − ∆(z) − UNd −σ as if dσ = −U/2 + UNd −σ. By writing Nd σ = 1/2 + m σ we ﬁnd through (7.47) the self- consistency condition 1 U m m = tan−1 . (7.60) π Γ There are two possibilities. If U ≤ πΓ the only solution is m = 0. It describes a paramagnetic impurity and corresponds to the previous analysis at U = 0. If U > πΓ, there are two stable

242 DOS

−U/2 0 U/2

Figure 7.4: DOS of the AIM within the Hartree-Fock approximation.

solutions at finite m = ±m0, which describe a partially polarized impurity. Since these two solutions are degenerate, the entropy S = ln 2 just reflects the spin degrees of freedom. In other words, the magnetic state m 6= 0 describes indeed a local moment. If we assume a density matrix which is the average of the two degenerate ground states, then the impurity Green’s function reads: 1 1 1 G(z) = + , (7.61) 2 z + U m + iΓsign Im z z − U m + iΓsign Im z and the DOS is composed by two lorentians, one below and one above the chemical potential, which are called the Mott-Hubbard bands. In particular, if U Γ, m → 1/2 and the two lorentians are separated by an energy roughly of order U, see Fig. 7.4. The mean-field solution for large U of the Anderson impurity model seems to account for the behavior above the Kondo temperature and explains how local moments appear. The key ingredient is a large Coulomb repulsion around the magnetic impurity in comparison with the resonance broadening Γ. Yet we have to understand what happens as the temperature is lowered.

243 7.3 From the Anderson to the Kondo model

The Hartree-Fock solution of the Anderson impurity model does explain in the large U/Γ-limit the existence of local free moments in a metal well below its Fermi temperature. Nevertheless the degeneracy of the mean-field solution is an artifact of mean-field theory, since the two solutions, in the above example, are not orthogonal and can be coupled through the conduction electrons. In order to understand the role of this coupling, it is more convenient to introduce the conduction wave-function at the impurity site through r 1 X c = V ∗ c , (7.62) 0σ V k kσ k where 2 X 2 V = |Vk| , k so that the impurity Hamiltonian plus the hybridization becomes X U V c† d + H.c. + (ˆn − 1)2 . (7.63) 0σ σ 2 d σ In the large U limit the impurity traps a single electron, either with spin up or down, and we can treat the hybridization as a small perturbation lifting the degeneracy between the two spin direction. By second order perturbation theory through intermediate states in which the impurity is empty or doubly occupied, with energy difference equal to U/2 for large U, one get an effective operator lifting the impurity degeneracy given by, apart from constant terms,

2V 2 X − c† d d† c + d† c c† d U 0σ σ σ0 0σ0 σ0 0σ0 0σ σ σσ0 = JK S0 · Sd where 8V 2 J = , K U and 1 X S = c† σ c 0 2 0α αβ 0β αβ 1 X = V V ∗ c† σ c , (7.64) 2V 2 k p kα αβ pβ kpαβ 1 X S = d† σ d . (7.65) d 2 α αβ β αβ

244 In other words in the large U limit the AIM reduces to the so-called Kondo model describing conduction electrons antiferromagnetically coupled to a local moment, a spin-1/2 in our single- orbital AIM example. The Kondo Hamiltonian is therefore

X † HK = k ckσckσ + JK S0 · Sd. (7.66) kσ First of all we notice that this model has built-in a local moment, hence by construction correctly describes the local moment regime for TK T TF . This local moment provide a scattering potential for the conduction electrons which the major diﬀerence with respect to a conventional scalar potential that the impurity has internal degrees of freedom. We will show now the consequences of this diﬀerence.

7.3.1 The emergence of logarithmic singularities and the Kondo temperature Let us analyse the role of the Kondo exchange

JK S0 · Sd, (7.67) in perturbation theory. Since (7.67) conserves independently the number of conduction and d-electrons, we use a trick due to Abrikosov and treat the impurity spin in terms of d-electrons with an unperturbed Green’s function 1 G(0)(z) = , z namely as a resonant state right at the chemical potential with a vanishing broadening, which we already know contains in the “unperturbed state” just one electron, number which is not going to be changed by (7.67). Since a single electron has a well deﬁned spin, it acts indeed like a local spin-1/2 moment. Moreover the Green’s function for the conduction electron c0 is 2 (0) 1 X |Vk| 1 G0 (z) = 2 = 2 ∆(z). V z − k V k For convenience let us rewrite (7.67) in a spin-asymmetric form as

X i i Ji S0 Sd, (7.68) i=x,y,z and calculate the ﬁrst order corrections to the exchange as given by the diagrams in Fig. 7.5. For simplicity we assume that all external lines are at zero frequencies. Each diagram is multiplied by (-1) because of ﬁrst order perturbation theory. The explicit expression of the diagram (a) is

X X i i j j X (0) (0) (a) = − Ji Jj Sαγ Sac Sγβ ScbT G0 (in) G (−in) i,j=x,y,z γc n

245 γ α β −ε J (a) J i j ε b a c

γ β α ε J J (b) i j ε a c b

Figure 7.5: First order corrections to the Kondo exchange. Solid and dashed lines indicate impurity and conduction electron Green’s functions. Also indicated are the spin labels. All external lines are assumed to be at zero frequency = 0. The internal frequency, which is going to be summed over, is also indicated.

X JiJj X X ∆(in) = − Si Si Sj Sj T V 2 αγ ac γβ cb −i i,j=x,y,z γc n n X JiJj X = −I(T ) Si Sj Si Sj , V 2 αγ γβ ac cb i,j=x,y,z γc where the function of temperature I(T ) is deﬁned through

X ∆(in) I(T ) = −T . i n n The diagram (b) is analogously

X X i i j j X (0) (0) (b) = − Ji Jj Sαγ Scb Sγβ SacT G0 (in) G (in) i,j=x,y,z γc n X JiJj X = I(T ) Si Sj Sj Si , V 2 αγ γβ ac cb i,j=x,y,z γc hence the sum is

X JiJj X X (a) + (b) = I(T ) Si Sj Sj Si − Si Sj V 2 αγ γβ ac cb ac cb i,j=x,y,z γ c X JiJj X = I(T ) Si Sj i Sk , V 2 αγ γβ jik ab i,j,k=x,y,z γ

246 where we used the commutation relations between the spin operators i j k S ,S = i ijk S , where ijk is the Levi-Civita antisymmetric tensor. Since we sum over i and j, we can also write

1 X JiJj X (a) + (b) = I(T ) Si Sj i Sk + (i ↔ j) 2 V 2 αγ γβ jik ab i,j,k=x,y,z γ 1 X JiJj X = I(T ) i Sk Si Sj − Sj Si 2 V 2 jik ab αγ γβ αγ γβ i,j,k=x,y,z γ   X 1 X JiJj = Sk Sk I(T ) | |2 , ab αβ 2 V 2 ijk  k=x,y,z i,j which provides the ﬁrst order correction to the exchange constants according to

1 X JiJj J + δJ = J + I(T ) | |2. (7.69) k k k 2 V 2 ijk i,j In order to evaluate I(T ) we use the same integration contour as in Fig. 7.2. We recall that the hybridization function has a branch cut approximately given by ∆( + i0+) − ∆( − i0+) = −2 i Γ θ(W − ||), where θ(W − ||) is the Heavyside function and W a cut-off of the order of the conduction bandwidth. We then find that Z W d Γ Γ W I(T ) = − f() ' ln , −W π π T so that e.g. in the isotropic case Jx = Jy = Jz = J we finally get Γ W J + δJ = J + J 2 ln . π V 2 T In other words the perturbation theory in J generates logarithmic singularities which become visible roughly around a temperature, which is therefore to be identified with the Kondo temperature, when the correction becomes of the same order as J, namely

Γ 2 W J = 2 J ln , π V TK leading to πV 2 πU T = W exp − = W exp − W ∼ T , (7.70) K ΓJ 8Γ F since U Γ. In agreement with experiments we do ﬁnd that the Kondo temperature is much smaller than the host-metal Fermi temperature. Since perturbation theory becomes meaningless below TK , the next obvious question is how to proceed further.

247 7.3.2 Anderson’s scaling theory What we have calculated before is the ﬁrst order correction to the Kondo exchange for conduction and impurity electrons at zero frequency, which we found to be Γ W J [W ] + δJ [W ] = J [W ] + J [W ]2 ln , (7.71) z z z π V 2 ⊥ T Γ W J [W ] + δJ [W ] = J [W ] + J [W ] J [W ] ln . (7.72) ⊥ ⊥ ⊥ π V 2 ⊥ z T

In (7.71) and (7.72) we have assumed an easy-axis asymmetry Jz 6= Jx = Jy ≡ J⊥, and we have explicitly indicated the dependence upon the bandwidth cut-off W . Clearly the result does not change for finite external frequencies provided they are much smaller than the temperature, otherwise they would cut-off the log-singularity instead of T . Now suppose we have another model with a smaller conduction electron bandwidth cut-off W (s) = W/s with s > 1, different J’s but equal Γ/V 2. In this case, for electrons close to the chemical potential, specifically much closer than the temperature, we would get the first order correction Γ W J [W (s)] + δJ [W (s)] = J [W (s)] + J [W (s)]2 ln , z z z π V 2 ⊥ sT Γ W J [W (s)] + δJ [W (s)] = J [W (s)] + J [W (s)] J [W (s)] ln . ⊥ ⊥ ⊥ π V 2 ⊥ z sT On the other hand, it makes not really a big difference for the low energy behavior whether these electrons which lie so close to the chemical potential derive from a band with width W or W (s) provided they suffer the same scattering off the impurity. Therefore we can ask the following question: what J [W (s)] should be in order for the effective exchange up to first order to be the same as that one with bandwidth W ? The answer is quite simple, since we can re-write e.g. (7.71) as Γ Γ W J [W ] + δJ [W ] = J [W ] + J [W ]2 ln(s) + J [W ]2 ln z z z π V 2 ⊥ π V 2 ⊥ sT Γ ' J [W ] + J [W ]2 ln(s) z π V 2 ⊥ Γ Γ 2 W + J [W ] + J [W ]2 ln(s) ln + O J 3 , π V 2 z π V 2 ⊥ sT the equality being valid up to the order at which we have stopped the expansion. Therefore the two models with W and W/s have the same spin exchange up to first order in perturbation theory provided the bare exchange constants satisfy Γ J [W (s)] = J [W ] + J [W ]2 ln(s), z z π V 2 ⊥

248 Γ J [W (s)] = J [W ] + J [W ] J [W ] ln(s), ⊥ ⊥ π V 2 ⊥ z which can be cast in a differential form dj (s) z = j (s)2, (7.73) d ln s ⊥ dj (s) ⊥ = j (s) j (s), (7.74) d ln s ⊥ z having defined Γ J [W (s)] ≡ j (s). i π V 2 i These equations describe how the bare exchange constants have to be modified in order for the model with a reduced bandwidth W/s to have the low-energy scattering amplitudes equal to those of the original model. The idea behind is that, if we are able to follow the evolution of ji(s) until W/s ∼ T , at this point we can rely on perturbation theory since ln(W/sT ) ' 0. This is more or less how Anderson formulated his poor man’s scaling theory for the Kondo problem in 1970 which is in essence the first implementation of a Renormalization Group transformation. Let us study the scaling equations. We readily find that, ln s = x,

d2j dj dj dj2 z = 2j ⊥ = 2 j z = z . dx2 ⊥ dx z dx dx By integrating from x = ln s = ln 1 = 0, the initial condition W (s) = W , to same x > 0 we get

dj (x) dj (0) dj (x) z − z = z − j (0)2 = j (x)2 − j (0)2, dx dx dx ⊥ z z namely dj (x) z = j (x)2 − j (0)2 + j (0)2, dx z z ⊥ d ln j (x) ⊥ = j (x). dx z

In Fig. 7.6 it is shown the scaling flow of the exchange constants jz(x) and j⊥(x) as x → ∞. Since the sign of J⊥ can be always changed by rotating of an angle π the impurity spin around the z-axis, we have just drawn the flow with J⊥ > 0, Jz > 0 and Jz < 0 implying antiferro- and ferro-magnetic Kondo exchange, respectively. We notice that for ferromagnetic exchange Jz < 0 with |Jz| ≥ J⊥, the couplings flow to a line of fixed points, where perturbation theory become valid. In particular for isotropic couplings Jz = J⊥ = J < 0, the value at x = ln W/T 1 is j 1 j(T ) = ' − → 0. (7.75) 1 − j ln W/T ln W/T

249 J

J z

Figure 7.6: Scaling ﬂow of the Kondo exchange constants as the cut-oﬀ is sent to zero.

This suggests that the impurity asymptotically decouples from the conduction electrons. For instance the impurity magnetic susceptibility, which can be calculated perturbatively in this case, behaves at low temperatures as 1 χ(T ) ∼ , T ln W/T almost the Curie-like behavior of a free spin.

For antiferromagnetic Jz > 0, as well as for Jz < 0 but |Jz| < J⊥, the exchange constants flow to strong-coupling jz ' j⊥ → +∞. This implies that the electrons very close to the chemical potential feel a very strong antiferromagnetic exchange with the impurity spin. It is therefore reasonable to assume that one has to diagonalize the Kondo spin-exchange first and only then treats what is left within perturbation theory. It is convenient to rewrite the Kondo Hamiltonian (7.66) in a form more suitable for perturbation theory in 1/JK . We start from the definition of c0σ given in Eq. (7.62), and introduce the single-particle wave-function (we drop for the meanwhile the spin index) † |φ0i = c0 |0i. We then define a new single-particle wave-function through

t0 |φ1i = (Hbath − 0) |φ0i ≡ Hbath |φ0i − |φ0i hφ0| Hbath |φ0i

Next we introduce another state orthogonal to the previous ones through

Hbath |φ1i = 1 |φ1i + t0 |φ0i + t1 |φ2i, where t1 |φ2i = Hbath |φ1i − |φ1i hφ1| Hbath |φ1i − |φ0i hφ0| Hbath |φ1i.

250 We notice that applying Hbath to |φ2i we do not get any component on |φ0i. In other words

X † † HAIM = n cnσcnσ + tn cnσcn+1σ + H.c. σ,n≥0

+JK S · S0, (7.76) which describes a one-dimensional chain with nearest neighbor hopping where the ﬁrst site, site “0”, is spin-coupled to the impurity spin S. For simplicity let us assume that the model is particle-hole invariant, which means that the Hamiltonian is invariant under the transformation

n † n † cn↑ → (−1) cn↓, cn↓ → −(−1) cn↑. One readily finds that the hopping term as well as the Kondo exchange are invariant under this transformation. On the contrary the site-energy term is not, hence we have to assume n = 0 for all n’s. In the spirit of the scaling approach, we assume that JK is much larger than any tn’s hence we diagonalize the two-site problem which includes the impurity spin and site “0”, and treats the hopping t0 within perturbation theory. By diagonalizing the Kondo exchange we find that the ground state is obtained by coupling the impurity spin-1/2 with a single electron at site “0” into a singlet state. All other states are above at least of energy JK . The hopping term t0 couples the singlet state with high energy states, hence, since JK t0, we have to conclude that t0 is an irrelevant perturbation which as first approximation can be simply dropped out. In other words the impurity spin binds one electron in a singlet state making the site “0” no more available to other electrons. This singlet state is inert hence the net result is that the asymptotic Hamiltonian flows to Eq. (7.76) with t0 = 0. The electron binding implies from the Friedel’s sum rule that

251 Chapter 8

Introduction to abelian Bosonization

In one dimension, many-body models can be handled with by a very powerful tool known as bosonization. In what follows we shall give a brief introduction to its abelian version. 1

8.1 Interacting spinless fermions

Let us consider a one dimensional model of spinless fermions described by a non interacting Hamiltonian of the form X † H0 = kckck (8.1) k with a band dispersion of the form in ﬁgure 8.1. If the chain has L sites, then

2πn L L k = , n ∈ − , for periodic boundary conditions La 2 2 2π L L k = (n + 1), n ∈ − , for antiperiodic boundary conditions. La 2 2

The scope is to study the model Eq. (8.1) in the presence of a generic interaction

1 X H = V (q)ρ(q)ρ(−q) (8.2) int 2L q

P † where ρ(q) = k ckck+q is the Fourier transform of the density. We will adopt an approach which is substantially similar to what is usually done to study quantum ﬂuctuations in models which develop, at the mean ﬁeld level, an order parameter.

1I am grateful to Angelo Russomanno who wrote in LaTex my lecture notes on this topic

252 Figure 8.1:

Let us start by the non interacting Hamiltonian (8.1). If N fermions are present, the ground state is obtained by occupying the momentum eigenstates up to a Fermi momentum such that

k XF L = N ' k if N and L are large. π F k=−kF This ground state (Fermi sea, see ﬁgure 8.2) has a kind of order parameter, namely the average of the occupation number in momentum space

D † E ckck ≡ hnˆki = θ(kF − |k|) . We assume that the average value of the band energy 1 hH i ' N 0 F is much bigger than the interaction energy between two electrons at their typical distance, i.e. V (q) at q ∼ kF ; then the only degrees of freedom affected by the interaction are those around each Fermi point ±kF . We define as right moving fermions those with positive momentum (cR k) and left moving fermions those with k < 0, (cL k). Analogously, the Right and Left moving densities are defined through

X † X † ρR(q) = cR kcR k+q ρL(q) = cL kcL k+q . k k

253 Figure 8.2:

Since the interaction is weak compared to F , the Fourier components of the densities which are aﬀected by the interaction are those with

|q| kF .

In this limit, they are also well defined the Right and Left mover densities in real space ρR(x) and ρL(x), they describe the distribution functions R and L of electrons with momentum k ∼ ±kF which vary in space over distances ∆x ∼ |1/q| such that, from ∆x∆k ∼ 1 1 ∆k ∼ ∼ |q| k . ∆x F The perturbed ground state is identified by the “order parameters” D † E hnR ki = cR kcR k k > 0 D † E hnL ki = cL kcL k k < 0 . Let us study the role of interaction in the same spirit as one studies the role of quantum fluctuations on models which develop an order parameter at the mean field level. The idea is to derive the equations of motion by making the approximation that, whenever a commutator of two operators has a component proportional to the order parameter, we substitute it with its average value on the ground state. Let us show how this procedure works in the well known spin wave theory.

254 8.1.1 Spin wave theory

On each site R there is a spin with components Sx(R), Sy(R), Sz(R). The classical ground state has an order parameter hSz(R)i = M(R) , then, within spin wave approximation

0 Sx(R),Sy(R ) = iδR,R0 Sz(R) ' iδR,R0 M(R) 0 Sx,y(R),Sz(R ) = ±iδR,R0 Sy,x(R) . (8.3) Therefore, the third component of the spin is equivalent to 1 1 S (R) ≡ − S (R)2 + S (R)2 + + M(R) . z 2M(R) x y 2 Let us apply this idea to our case. We start by calculating

0 X h † † i ρR(q), ρR(q ) = cR kcR k+q, cR pcR p+q0 k, p X † † = δp, k+q cR kcR p+q0 − δk, p+q0 cR pcR k+q . k, p By a change of variables, it is equal to

X † † δp, k+qcR kcR k+q+q0 − δp, k+q0 cR kcR k+q+q0 . k, p If the sums run for all k’s, the above term would be zero. However, since all the fermions which are involved should be right movers, with positive momentum, hence the ﬁrst δ-function implies k + q > 0 while the second k + q0 > 0, then one ﬁnds

max(0,−q0,−q−q0) min(π/a,π/a−q,π/a−q−q0) 0 X X † ρR(q), ρR(q ) = + cR kcR k+q+q0 . k=max(0,−q,−q−q0) k=min(π/a,π/a−q0,π/a−q−q0)

0 Since all q’s are small, for q 6= −q , this operator involves excitations very far from kF , either close to k ∼ 0 or k ∼ π/a. Since we expect the ground state occupation number not to be modiﬁed for k ∼ 0 and k ∼ π/a, we can safely assume the commutator to be zero. On the contrary, for q = −q0, the commutator becomes related to the order parameter, namely

max(0,q) min(π/a,π/a−q) X X [ρR(q), ρR(−q)] = + nR k . max(0,−q) min(π/a,π/a+q)

255 In the spirit of what we previously said, we approximate the operator with its average value on the Fermi sea; namely

max(0,q) X qL [ρ (q), ρ (−q)] ≈ h[ρ (q), ρ (−q)]i = θ(k − k) = , (8.4) R R R R F 2π max(0,−q) therefore 0 qL ρ (q), ρ (q ) = δ 0 . (8.5) R R q, −q 2π Analogously for the left movers one ﬁnds

0 qL ρ (q), ρ (q ) = −δ 0 . (8.6) L L q, −q 2π In real space

1 X 0 [ρ (x), ρ (y)] = eiqx eiq y ρ (q), ρ (q0) e−α|q| R R L2 R R q, q0 1 X i X = q eiq(x−y)−α|q| = − ∂ eiq(x−y)−α|q| 2πL 2π x q q i = − ∂ δ (x − y) , 2π x α where α is a cut oﬀ (recall that |q| kF was a starting condition) as well as i [ρ (x), ρ (y)] = ∂ δ (x − y) . (8.7) L L 2π x α

8.1.2 Construction of the eﬀective Hamiltonian We start with the commutator of the density in momentum space with the Hamiltonian

X h † † i [ρR(q),H0] = p cR kcR k+q, cR pcR p k, p>0 X † † = p δp, k+q cR kcR p − δk, p cR pcR k+q k, p>0 X † = (k+q − k) cR kcR k+q ' vF q ρR(q) , (8.8) k where we have deﬁned the Fermi velocity as v ≡ ∂ | and we exploited that only the F k k k=kF electrons near the Fermi momentum k are involved and that |q| k . Since ∂ | = −v F F k k k=−kF F

[ρL(q),H0] = −vF q ρL(q) . (8.9)

256 Thanks to the above commutation relations we can write the eﬀective Hamiltonian π X H = v (ρ (q)ρ (−q) + ρ (q)ρ (−q)) 0 L F R R L L q Z = πvF dx (ρR(x)ρR(x) + ρL(x)ρL(x)) . (8.10)

Moreover, the general continuity equation for Right and Left movers implies

i~ρ˙R(q) = [ρR(q),H0] = qJR(q) i~ρ˙L(q) = [ρL(q),H0] = qJL(q) (8.11) which, compared with Eqs. (8.8) and (8.9), gives the important relations for the current operators

JR(q) = vF ρR(q) (8.12)

JL(q) = −vF ρL(q) . (8.13)

The commutation relations Eqs. (8.5) and (8.6) provide a bosonic representation of the long wavelength density ﬂuctuations. We deﬁne, for q > 0,

r 2π r 2π b = −i ρ (q) b† = i ρ (−q) (8.14) R q qL R R q qL R as well as r 2π r 2π b = i ρ (−q) b† = −i ρ (q) , (8.15) L q qL L L q qL L which satisfy, for {a, b} = {R,L}

h † i ba q, bb q0 = δa, b δq, q0 ba q, bb q0 = 0 .

Notice that, for q = 0, ρR(q = 0) and ρL(q = 0) correspond to the densities of right R and left L moving fermions, NR/L and NL/L respectively, where NR,L are number operators. Therfore, we can write r i X q L NR ρ (x) = eiqxb − e−iqxb† + (8.16) R L 2π R q R q L q>0 r i X q L NL ρ (x) = − eiqxb − e−iqxb† + . (8.17) L L 2π L q L q L q>0

257 Figure 8.3:

We can also introduce fields which are conjugate of ρR,L; indeed the fields s 1 X 2π L π φ (x) = eiqxb + e−iqxb† + k x − θ + N R L q R q R q F,R R 2 L q>0 (8.18) s 1 X 2π L π φ (x) = e−iqxb + eiqxb† + k x + θ + N L L q L q L q F,L L 2 R q>0 (8.19) where kFR = 2πNR/L, kFL = 2πNL/L are the right and left Fermi momenta, corresponding to the deformation shown in figure 8.3. They satisfy 1 [φ (x), ρ (y)] = −iδ(x − y) , ∂ φ (x) = ρ (x) , (8.20) R R 2π x R R 1 [φ (x), ρ (y)] = iδ(x − y) , ∂ φ (x) = ρ (x) . (8.21) L L 2π x L L

Moreover [θR,NR] = [θL,NL] = i, therefore the last term in Eqs. (8.18) and (8.19) implies that

[φR(x), φL(y)] = −iπ (8.22) which will be useful later. We emphasize that the θ operators which are conjugate of NR,L are needed because the sum over q does not include q = 0. Since φR and φL are conjugate of the densities, one can easily construct operators which create or destroy particles. Indeed

e−iφR(x) creates a Right fermion in x since

258 iφR(x) −iφR(x) e ρR(y)e = ρR(y) + i [φR(x), ρR(y)] = ρR(y) + δ(x − y) and analogously eiφL(x). In turns, these operators can be used to construct Fermi fields. It is customary to define shifted φ’s. Given a reference state with N N hN i = hN i = k0 = k0 = π ≡ k , R L 2 FR FL L F we define new φR and φL like in Eqs. (8.18) and (8.19), by substituting the number operators by N N → ∆N = N − . R,L R,L R,L 2 Accordingly, we have also to perform the substitution N k → ∆k = k − k = k − π F, r F, r F, r F F, r L

1 N implying that ρR,L = 2π ∂xφR,L + 2L . Then the operators 1 ikF x+iφR(x) ΨR(x) ≡ √ e (8.23) 2πα 1 −ikF x−iφL(x) ΨL(x) ≡ √ e (8.24) 2πα behave like Fermi ﬁelds which annihilate a right or left fermion at position x. To show it, we recall that, given two operators A and B both commuting with [A, B], then the BCH formula holds

eAeB = eBeAe−[B,A] A+B A B − 1 [A,B] B A − 1 [B,A] e = e e e 2 = e e e 2 . (8.25)

Therefore (using Eq. (8.22)) 1 Ψ (x)Ψ (y) = eikF (x−y)eiφR(x)e−iφL(y) R L 2πα 1 = eikF (x−y)e−iφL(y)eiφR(x)e[φL,φR] = −Ψ (y)Ψ (x) . 2πα L R Moreover 1 Ψ (x)Ψ† (y) = eikF (x−y)eiφR(x)e−iφR(y) R R 2πα 1 ik (x−y) i(φ (x)−φ (y)) 1 [φ (x),φ (y)] = e F e R R e 2 R R . (8.26) 2πα

259 The above commutator is not impossible to evaluate

2π X 1 2π [φ (x), φ (y)] = eiq(x−y) − e−iq(x−y) + i(x − y) R R L q L q>0 2π X 1 x − y = eiq(x−y)e−α|q| = 2i atan ' iπ sign(x − y) . L q α q (8.27)

Analogously x − y [φ (x), φ (y)] = −2i atan ' −iπ sign(x − y) . (8.28) L L α Hence

† 1 ik (x−y) i(φ (x)−φ (y)) i π (x−y) Ψ (x)Ψ (y) = e F e R R e 2 while (8.29) R R 2πα † 1 ik (x−y) i(φ (x)−φ (y)) −i π (x−y) Ψ (y)Ψ (x) = e F e R R e 2 . (8.30) R R 2πα Hence, for |x − y| α {ΨR(x), ΨR(y)} = 0 and the same holds for ΨL {ΨL(x), ΨL(y)} = 0 .

What does it happen when we send x → y? We cannot simply take φR(x) − φR(y) → 0 in Eqs. (8.29) and (8.30), since the exponential operator is singular. We ﬁrst have to normal order it and then take the limit x → y s 1 X 2πL φ (x) − φ (y) ' e−iqx − e−iqy b† R R L q R q q>0 s 1 X 2πL (+) (−) + eiqx − eiqy b = φ + φ (8.31) L q R q R R q>0

With this decomposition, using relations (8.25), we can write

i(φ(+)+φ(−)) iφ(+) iφ(−) 1 [φ(+), φ(−)] e R = e R e R e 2 R R where

(+) (−) π X 1 [φ , φ ] = − e−iqx − e−iqy eiqx − eiqy e−αq = R R L q q>0

260 π X 1 1 α2 = − e−αq [2 − 2 cos (q(x − y))] = log . L q 2 α2 + (x − y)2 q>0 (8.32)

The same result is valid for φL. By adding the phase factor 1 x − y [φ (x), φ (y)] = ±i atan 2 R/L R/L α

† evaluated in Eqs. (8.27) and (8.28), we have in ΨR/L(x)ΨR/L(y)

1 α2 x − y α log ± i atan = log . 2 α2 + (x − y)2 α α ∓ i(x − y)

Therefore, introducing the normal ordering symbol as

(+) (−) : e±i(φR/L(x)−φR/L(y)) : ≡ eiφ eiφ , we can deﬁnitely write the anticommutator as

† {ΨR/L(x), ΨR/L(y)} 1 α α ' e±ikF (x−y) : e±i(φR/L(x)−φR/L(y)) : + 2πα α ∓ i(x − y) α ± i(x − y) (8.33)

1 α α→0 = e±ikF (x−y) : e±i(φR/L(x)−φR/L(y)) : −→ δ(x − y) . π α2 + (x − y)2 (8.34)

1 α (We have here the approximation δα(x − y) = π α2+(x−y)2 to the δ.) So we can see that the † anticommutator gives a δ-function regularized by α. This was predictable since ΨR/L(x) does not really create a fermion in x, but a wave packet around x of width α. The previous calculation also shows that a particular care has to be taken when we consider the product of two bosonized operators at the same point. Given two operators A(x) and B(x), the correct way to do the product is

A(x)B(x) = lim [A(x)B(y)]| |x−y| . x→y α 1

† Example : How to recover ΨR(x)ΨR(x) = ρR(x).

261 † In the spirit of what we have stated few lines above, let us calculate ΨR(x)ΨR(y) and then take the limit. As we can see in Eq. (8.33), 1 α Ψ† (x)Ψ (y) ' e−ikF (x−y) : e−i(φR(x)−φR(y)) : R R 2πα α − i(x − y) 1 1 = : 1 − ik (x − y) − i∂ φ (x)(x − y): . 2π α − i(x − y) F x R (8.35) Hence † h † i 1 kF Ψ (x)ΨR(x) = lim Ψ (x)ΨR(y) = ∂xφR(x) + ≡ ρR(x) . R y→x R |x−y| α 1 2π 2π The equations of motion Eqs. (8.11) for ρR and ρL imply that

−ivF qt −ivF qt bR q(t) = bR qe bL q(t) = bL qe , hence, thanks to Eqs. (8.16) and (8.17), we have

φR(x, t) = φR(x − vF t)

φL(x, t) = φL(x + vF t) . Therefore, analogously to Eq. (8.33), we ﬁnd the following correlation functions D E 1 1 † ±ikF x ΨR/L(x, t)ΨR/L(0, 0) = e 2π α ∓ i(x ∓ vF t) D E 1 1 † ±ikF x ΨR/L(0, 0)ΨR/L(x, t) = e , 2π α ± i(x ∓ vF t) so that the Green function is easily written D←− E 1 1 † ±ikF x GL/R(x, t) = −i > ΨR/L(x, t)ΨR/L(0, 0) = e . 2π ±x − vF t + iα sign(t) We can even write the density-density correlation function (by using directly formulae Eqs. (8.16) and (8.17)) 1 1 1 ρR/L(x, t)ρR/L(0, 0) = 2 ∂xφR/L(x, t) ∂xφR/L(0) = 2 2 . 4π 4π [α ∓ i(x ∓ vF t)] For future utility, let us calculate D E iβRφR(x,t)+iβLφL(x,t) −iβRφR(0)−iβLφL(0) GβR βL = e e .

Since [φR(x), φL(y) − φL(z)] = 0 then D ED E iβRφR(x,t) −iβRφR(0) iβLφL(x,t) −iβLφL(0) GβR βL = e e e e β2 β2 α R α L = (8.36) α − i(x − vF t) α + i(x + vF t)

262 8.2 Interactions

Let us now switch on interaction 1 X H = V (q)ρ(q)ρ(−q) . int 2L q Since we are assuming a weak coupling approach; of all the scattering amplitudes generated by V (q), we keep only those which act among Fermions close to the Fermi momenta (see ﬁgure 8.4). For generic ﬁlling N/L, there are two such scattering processes. In the former we have

V (q) ∼ V (q = 0) (direct interaction) and in the latter

V (q) ∼ V (q = 2kF ) (exchange interaction) . Shortly, we have

g2 = V (0) − V (2kF ) and g2 X H = ρ (q)ρ (−q) . int L R L q Therefore, the low energy Hamiltonian is

πvF X g2 X H = H + H = [ρ (q)ρ (−q) + ρ (q)ρ (−q)] + ρ (q)ρ (−q) , (8.37) 0 int L R R L L L R L q q

263 Figure 8.4:

which, written in terms of the bR q and bL q (dropping an infinite though constant therm) has the form X h i g2 X h i H = v q b† b + b† b + b b + b† b† . F R q R q L q L q 2π R q L q L q R q q>0 q>0 This is a bilinear form which can be easily diagonalized. However, we will adopt a different approach, which is more useful for the following. Let us introduce two new fields 1 Φ(x) ≡ √ (φR(x) + φL(x)) 4π 1 Θ(x) ≡ √ (φL(x) − φR(x)) (8.38) 4π as well as √ Π(x) ≡ ∂xΘ(x) = 4π (ρL(x) − ρR(x)) . Using the commutators Eqs. (8.22), (8.27), (8.28), we can see that 1 [Φ(x), Θ(y)] = − [φ (x), φ (y)] + [φ (x), φ (y)] 4π R R L L 1 x − y + [φ (x), φ (y)] − [φ (x), φ (y)] = −4i atan − 2πi R L L R 4π α 1 ' {−2πi sign(x − y) − 2πi} = −iθ(x − y) , (8.39) 4π where θ is the Heaviside step function. Therefore we have the important relation

[Φ(x), Π(y)] = iδ(x − y) (8.40)

264 stating that that Π and Φ are conjugate variables. From Eq. (8.38) we have √ φL(x) = π (Φ(x) + Θ(x)) √ φR(x) = π (Φ(x) − Θ(x)) . (8.41)

1 kF Since ρR/L(x) = 2π ∂xφR/L(x) + 2π we have v Z g Z H = F dx [∂ φ (x)∂ φ (x) + ∂ φ (x)∂ φ (x)] + 2 dx ∂ φ (x)∂ φ (x) 4π x R x R x L x L 4π2 x R x L Z Z vF g2 g2 = dx 1 + ∂xΦ(x) ∂xΦ(x) + dx 1 − Π(x)Π(x) 2 2πvF 2πvF s 2 Z 1 g2 1 = vF 1 − dx ∂xΦ(x) ∂xΦ(x) + KΠ(x)Π(x) (8.42) 2 2πvF K where we have deﬁned s 1 − g2 K ≡ 2πvF . 1 + g2 2πvF Notice that the density and the current are given by 1 ρ(x) = ρ (x) + ρ (x) = √ ∂ Φ(x) L R π x 1 J(x) = v (ρ (x) − ρ (x)) = −√ ∂ Θ(x) . (8.43) F R L π x The canonical transformation √ Φ(x) = KΦ(e x) 1 Π(x) = √ Π(e x) (8.44) K makes Eq. (8.42) diagonal s Z 2 1 h i g2 H = v dx ∂xΦ(e x) ∂xΦ(e x) + Π(e x)Π(e x) where v ≡ vF 1 − . (8.45) 2 2πvF

This Hamiltonian describes the normal modes of the low energy particle-hole excitations (in Landau theory the zero sound, which exhausts in one dimension the particle-hole spectrum). The big advantage of Bosonization is that we can also calculate single particle correlations

√ √ 1 φR/L(x) = π KΦ(e x) ∓ √ Θ(e x) K

265 1√ 1 = K φeR(x) + φeL(x) ∓ √ φeR(x) − φeL(x) 2 2 K 1 √ 1 1 √ 1 = K ± √ φeR(x) + K ∓ √ φeL(x) . (8.46) 2 K 2 K Using Eqs. (8.36) and (8.45), we can see that

√ 2 √ 2 1 K± √1 1 K∓ √1 D E 1 α 4 K α 4 K Ψ (x, t)Ψ† (0) = R/L R/L 2πα α − i(x − vt) α + i(x + vt) 1 (K+ 1 ±2) 1 (K+ 1 ∓2) 1 α α ∓ i(x ∓ vt) α 4 K α 4 K = 2πα α ∓ i(x ∓ vt) α α − i(x − vt) α + i(x + vt) 1 (K−1)2 1 α α2 4K = . (8.47) 2πα α ∓ i(x ∓ vt) (α − i(x − vt))(α + i(x + vt))

1 2 Therefore the single particle Green’s function acquires an anomalous exponent 4K (K − 1) .

8.2.1 Umklapp processes Let us now consider the half ﬁlling case N/L = 1/2. In this very special case (assuming the lattice spacing a = 1) kF = π/2, and an additional scattering amplitude is allowed close to the Fermi points, namely

∼ V (2kF ) = V (π) (Umklapp scattering) . In real space it corresponds to an interaction term V (x − y) = V (π)(−1)x−yf(|x − y|), where f(r) is some short range function such that R drf(r) = 1 . Hence the Umklapp term gives a contribution to the Hamiltonian Z x−y h † † i Umklapp = V (π) dxdy (−1) f(|x − y|) ΨR(x)ΨR(y)ΨL(y)ΨL(x) + H.c. .

Applying Eqs. (8.23) and (8.25) we can recast this formula as2 Z V (π) h − 1 [φ (x), φ (y)] Umklapp = dxdy e 2 R R (2πα)2 √ 2To be consistent with eqs. (8.23) the argument of the cosine should be 16πΦ(x) + 2πx; nevertheless the additional term is not important because x is an integer number of lattice spacings a = 1.

266 − 1 [φ (y), φ (x)] −i(φ (x)+φ (y)+φ (x)+φ (y)) i × e 2 L L e R R L L + H. c. 2 Z √ ' V (π) dx cos( 16πΦ(x)) , (8.48) (2πα)2 with g3 = V (π). After applying the canonical transformation Eq. (8.44) we have Z 2g3 √ HUmklapp = dx cos 16πKΦ(e x) (2πα)2 and the Hamiltonian reads Z 1 1 4g3 √ H = v dx ∂xΦ(e x) ∂xΦ(e x) + Π(e x)Π(e x) + cos 16πKΦ(e x) (8.49) 2 v (2πα)2 which is a sine-Gordon model. The dimension of the Umklapp is 4K, to be compared with the dimension of ∂xΦ(x) ∂xΦ(x) + Π(x)Π(x) which is 2. Therefore the Umklapp gets marginal at K = 1/2 and relevant for K√ < 1/2. When it is relevant, the Umklapp locks the ﬁeld Φ to a value which minimizes g3 cos 16πΦ(x), and a gap opens in the spectrum.

How to calculate dimensions √ Let us normal order cos 16πKΦ(e x). We get

√ √ + − cos 16πKΦ(e x) = : cos 16πKΦ(e x) : e8πK[Φe (x), Φe (x)]   √ 2π X 1 = : cos 16πKΦ(e x) : exp −4K e−αq .  L q  q>0

Performing the integration over the second Brillouin zone gives an approximation of the argument of the exponential Z 1/α dq L ' −4K = −4K log 2π/L q 2πα and therefore

1 √ 1 4K √ cos 16πKΦ(e x) = (2πα)4K−2 : cos 16πKΦ(e x): . (2πα)2 L

The relevance/irrelevance is controlled by the behaviour under the limit α → 0.

267 8.3 Bosonization of the Heisenberg model

Let us consider a 1d Heisenberg model

L X z y y y z z H = J Si Si + Si Si+1 + ∆Si Si+1 . (8.50) i=1 We can map this model onto a spinless fermion model via the so called Jordan-Wigner transformation. We write 1 1 Sz = − c†c = − n . (8.51) i 2 i i 2 i z 1 Here ni = 0, 1 is the spinless Fermion number at site i, therefore Si = 2 corresponds to an z 1 empty site and Si = − 2 to an occupied site. Let us show that

Pi−1 Pi−1 + iπ j=1 nj − † −iπ j=1 nj Si = e ci ,Si = ci e (8.52) is a good representation for spin operators. Let us deﬁne

i−1 Pi−1 iπ nj X ξi ≡ e j=1 = (1 − 2nj) (8.53) j=1

2 the string operator. Since ξi = 1, we easily check that

+ − h †i z Si ,Si = ci, ci = 1 − 2ni = 2Si + z † Si ,Si = −ξi [ci, ni] = −Si . Moreover, since for i > j ξicj = −cjξi while for i ≤ j ξicj = cjξi , then h + +(−)i (†) (†) Si ,Sj = ciξiξjcj − cj ξjξici = 0 for i 6= j, as it should be. Let us rewrite (8.50) using (8.51), (8.52) and (8.53). We have 1 SxSx + SySy = S+S− + S−S+ i i+1 i i+1 2 i i+1 i i+1 1 = c ξ ξ† c† + c†ξ†ξ c 2 i i i+1 i+1 i i i+1 i+1 1 = c e−iπni c† + c†eiπni c 2 i i+1 i i+1

268 1 = c†c + c† c (8.54) 2 i i+1 i+1 i and 1 1 1 1 1 Sz = − n − n = + n n − n − n . (8.55) i+1 2 i 2 i+1 4 i i+1 2 i 2 i+1 Hence J X X 1 1 1 H = c†c + H. c. + J∆ + n n − n − n . (8.56) 2 i i+1 4 i i+1 2 i 2 i+1 i i In case of an open chain, i.e. site 1 not coupled to L, the last two terms are J∆ X J∆ X − (n + n ) = (n + n ) − J∆ n . 2 i i+1 2 1 L i i i P Since i ni = N, the total number of fermions, is a conserved quantity, this term introduces a scattering potential at edge sites 1 and L. It is therefore better to deal with a closed chain, by adding the bond J S+S− + S−S+ + J∆SzSz . (8.57) 2 1 L 1 L 1 L We notice that PL−1 PL−1 + − −iπ j=1 nj † † −iπ j=1 nj S1 SL = c1e cL = c1cLe PL † −iπ j=1 nj iπnL N † iπnL N † = c1cLe e = (−1) c1cLe = (−1) c1cL . (8.58) Hence, we can recast Eq. (8.57) as J 1 1 1 (−1)N c c† + c† c + J∆ + n n − n − n . (8.59) 2 1 L L 1 4 1 L 2 1 2 L We further apply the transformation i ci → (−1) ci (8.60) such that the Hamiltonian Eq. (8.50) (using also Eq. (8.59)) reads (apart from constant terms) J X X H = − c†c + H. c. + J∆ n n 2 i i+1 i i+1 i i J + (−1)N+L c c† + c† c + J∆n n . (8.61) 2 1 L L 1 1 L The factor (−1)N+L makes the ground state non degenerate, as it should be for bosonic models. For simplicity, let us take N + L odd. In such case, Eq. (8.61) is equivalent to the spinless fermion Hamiltonian X † X H = −t ci ci+1 + H. c. + V nini+1 i i

269 X V X = c† c + cos (qa) ρ ρ , (8.62) k k k L q −q k q with t = J/2, V = J∆ and k = −2t cos(ka), a being the lattice spacing. The Hamiltonian Eq. (8.62) can be analysed with bosonization. In this speciﬁc case

π N π π k = = − Sz , F a L 2a La total F = 2ta sin kF a ,

g2 = 2V (1 − cos 2kF a) .

At half filling, g3 = −2V cos 2kF a = +2V is the Umklapp. As we showed, the Umklapp is irrelevant/marginal/relevant if 1 K > / < / = ; 2 what does K = 1/2 correspond to? Let us consider the bosonized expression of the spin operators. As before, the spin operators will be related to phase field operators which are assumed to be slowly varying, namely varying in distances ∆x 1 ∼ a. In this approximation a kF continuum limit x ≡ j · a = continuous variable is well defined, and

+ + x/a iπ R x−a dyρ(y) Sj ' Sx ' (−1) e [ΨR(x) + ΨL(x)] , (8.63) where (−1)x/a comes from Eq. (8.60), and

1 1 N ρ(y) = ρ (y) + ρ (y) = ∂ φ (y) + ∂ φ (y) + . R L 2π y R 2π y L L More explicitly

(−1)x/a h i S+ ' √ ei(φR(x−a)+φL(x−a))/2eikF (x−a) eikF x+iφR(x) + e−ikF x−iφL(x) x 2πα (−1)x/ae−ikF a h i ' √ e2ikF xei(3φR(x)+φL(x))/2 + ei(φR(x)−φL(x))/2 2πα (−1)x/ae−ikF a h √ √ i = √ e2ikF xei π(2Φ(x)−Θ(x)) + e−i πΘ(x) 2πα (−1)x/ae−ikF a h √ √ √ √ √ i = √ e2ikF xei π(2 KΦ(e x)−Θ(e x)/ K) + e−i πΘ(e x)/ K ; (8.64) 2πα

270 therefore (applying Eqs. (8.33) and (8.36)) we obtain (for |x − y| α)

1 1/(2K) S+S− ∼ (−1)(x−y)/aα1/(2K)−1 . (8.65) x y |x − y|

On the other hand (using Eqs. (8.25), (8.22) and (8.44))

z z † † † † † Si ∼ Sx ' Ψ (x)Ψ(x) = ΨR(x)ΨR(x) + ΨL(x)ΨL(x) + ΨR(x)ΨL(x) + ΨL(x)ΨR(x) 1 N 1 h i = √ ∂ Φ(x) + + e−2ikF xe−iφR(x)e−iφL(x) + H.c. π x Lπ 2πα 1 N i h √ i = √ ∂ Φ(x) + + e−2ikF xe−i 4πKΦ(e x) + H.c. . (8.66) π x Lπ 2πα

z Stot = 0, which includes the ground state for isotropic Hamiltonians, then N = L/2, i.e. z kF = π/2, hence the leading contribution to Sx is i h √ i Sz ∼ (−1)x/a e−i 4πKΦ(e x) + H.c. . x 2πα Therefore, the correlation is

(−1)x/a 1 2K SzSz ∼ α2(K−1) . (8.67) x y 2π2 |x − y|

z If ∆ = 1, spin isotropy is a symmetry of Eq. (8.50); hence the ground state should have Stot = 0 and Eq. (8.67) should equal Eq. (8.65) up to a coeﬃcient; this implies 1 1 2K = =⇒ K = ; 2K 2 therefore for ∆ < 1, K > 1/2 (K = 1 for ∆ = 0) and Umklapp processes are irrelevant. The spin excitations are massless and the correlation functions√ decay as power laws. If ∆ > 1 then K < 1/2 and Umklapp is relevant. A gap opens if 16πΦ is locked to values π mod (2π) (see Eq. (8.49) for g3 > 0). In this case

z z (x−y)/a SxSy ∼ (−1) · constant which corresponds to a N´eelstate. What does it happen if K = 1/2, namely at the isotropic point? In this case Umklapp is marginal and V = 2t (see Eq. (8.62)). This implies a strong coupling limit which is outside the limit of applicability of bosonization model (Eq. (8.62)). We must resort to better controlled models which represent Eq. (8.50).

271 8.4 The Hubbard model

This model is described by the Hamiltonian

X † X H = −t ci σci σ + U ni ↑ni ↓ . (8.68) i, σ i The interaction term represents an on-site Coulomb repulsion which prevents two electrons, with opposite spin, to be on the same site. If the number of electrons N is equal to the number of sites L and U t, the Hubbard model is insulating, each site being occupied by an electron which does not move, since it would cost energy U t. Therefore, at energy U, charge degrees of freedom are frozen and only the spin degrees of freedom matter. Indeed, it can be mapped onto the Heisenberg antiferromagnet, namely spin isotropic, with J = 4t2/U. However, at half filling, the model Eq. (8.68) is insulating, not only for U t, but for any value of U > 0. This happens because of the nesting property of the Fermi surface together with the filling being commensurate. Therefore, charge degrees of freedom are frozen for any U and since there is no phase transition as a function of U, we expect the spin degrees of freedom to still be described by an Heisenberg model. Therefore, we have the possibility to access the spin isotropic point of the Heisenberg model in the weak coupling, U t, regime of the Hubbard model. Let us bosonize Eq. (8.68). We have to introduce four fields φR ↑, φL ↑, φR ↓ and φL ↓. Moreover, to preserve the proper anticommutation relations among the Fermi fields we must impose that [φp(x), φq(y)] = ±iπ for p 6= q p, q = R ↑,L ↑,R ↓,L ↓ . We write (σ =↑, ↓) s 1 X 2π L φ (x) = eiqxb + e−iqxb† + k x − θ + A R σ L q R σ q R σ q F, R σ R σ R σ q>0 (8.69) s 1 X 2π L φ (x) = e−iqxb + eiqxb† + k x + θ + A L σ L q L σ q L σ q F, L σ L σ L σ q>0 (8.70) where π A ≡ (N + N + N ) R ↑ 2 L ↑ R ↓ L ↓ π A ≡ (N − N − N ) L ↑ 2 R ↑ R ↓ L ↓ π A ≡ (−N − N + N ) R ↓ 2 R ↑ L ↑ L ↓

272 π A ≡ (N + N + N ) . L ↓ 2 R ↑ L ↓ R ↓ This choice implies that (remembering the relation θL/R, σ,NL/R, σ = −iπ)

[φR ↑, φL ↑] = [φR ↓, φL ↓] = −iπ

[φR ↑, φR ↓] = [φL ↑, φL ↓] = iπ

[φR ↑, φL ↓] = [φL ↑, φR ↓] = −iπ .

The advantage of this choice is that, if we introduce 1 1 Φσ ≡ √ (φR σ + φL σ);Θσ ≡ √ (φL σ − φR σ) 4π 4π we have [Φ↑, Φ↓] = [Φ↑, Θ↓] = [Θ↑, Φ↓] = 0 while [Θ↑, Θ↓] = i . We introduce the charge and spin combinations 1 1 Φc/s = √ (Φ↑ ± Φ↓) , Θc/s = √ (Θ↑ ± Θ↓) (8.71) 2 2 as well as Πc/s = ∂xΘc/s which satisfy, thanks to our choice,

[Φc, Φs] = [Πc, Πs] = [Φc, Πs] = [Πc, Φs] = 0 Φc/s(x), Πc/s(y) = iδ(x − y) .

In terms of these ﬁelds, the non interacting part of the Hamiltonian can be written as Z vF X H = dx (∂ Φ (x)∂ Φ (x) + Π (x)Π (x)) , (8.72) 0 2 x p x p p p p=c,s with vF = 2ta sin kF a. Let us bosonize the interaction X U X H = U n n = ρ ρ . int i ↑ i ↓ L q ↑ −q ↓ i q

The term with q ∼ 0 is

Z 1 2 Ua dx (∂ φ (x) + ∂ φ (x))(∂ φ (x) + ∂ φ (x)) 2π x R ↑ x L ↑ x R ↓ x L ↓ Ua Z = dx (∂ Φ (x)∂ Φ (x) − ∂ Φ (x)∂ Φ (x)) . (8.73) 2π x c x c x s x s

273 There are two processes at q ∼ 2kF . The ﬁrst is the so called backscattering

. By bosonization this term reads 2Ua Z √ dx cos( 8πΦ (x)) . (8.74) (2πα)2 s

The other process is the Umklapp

which reads 2Ua Z √ − dx cos 8πΦ (x) + 4k x . (8.75) (2πα)2 c F

At half filling, kF = π/2a, and 4kF x = 2πx/a can be dropped out of Eq. (8.75) since x/a = integer. The full bosonized Hamiltonian at half filling is therefore Z vF X H = dx (∂ Φ (x)∂ Φ (x) + Π (x)Π (x)) 2 x p x p p p p=c,s g Z g Z + s dx∂ Φ (x)∂ Φ (x) + s dx∂ Φ (x)∂ Φ (x) 2π x c x c 2π x s x s 2g Z √ 2g Z √ + 1 dx cos( 8πΦ (x)) − 3 dx cos( 8πΦ (x)) , (2πα)2 s (2πα)2 c (8.76) where gc ≡ Ua, gs = −Ua, g1 = g3 = Ua√. Both charge and spin fields are described by a sine-Gordon Hamiltonian with marginal cos( 8πΦ). Since spin isotropy holds, the sine-Gordon model in the spin sector has to be equivalent to the Heisenberg model with ∆ = 1, which

274 corresponds in the spinless fermion model to a strong coupling point while in this model to a weak coupling one. The bilinear terms can be diagonalised by a canonical transformation like Eq. (8.44) where

1 gc Ua Kc = q ' 1 − = 1 − < 1 1 + gc 2πvF 2πvF πvF

1 gs Ua Ks = q ' 1 − = 1 + > 1 , 1 + gs 2πvF 2πvF πvF and the velocity of charge/spin collective excitations is

r Ua vc/s = vF 1 ± . πvF Since all interaction terms in Eq. (8.76) are marginal, one can do Renormalization group analysis, which leads to the following equations

1 2 1 g˙s = g1, g˙1 = g1gs πvs πvs 1 2 1 g˙c = g3, g˙3 = g3gc πvc πvc whose flux is represented in figure 8.5. From the starting values, one finds that the Umklapp is marginally√ relevant coupling; therefore a charge gap opens and the field Φc is locked to 0 mod (2π/ 8π). As expected, the model is insulating for any U > 0. On the contrary, the backscattering is marginally irrelevant. The fixed point corresponds to ∗ ∗ ∗ gs = g1 = 0, hence Ks = 1. Namely, the spin fixed point Hamiltonian reduces simply to v Z H∗ = s dx(∂ Φ (x)∂ Φ (x) + Π (x)Π (x)) s 2 x s x s s s which describes gapless spin excitations. This analysis gives us the opportunity to derive the expression of the spin operators directly at the isotropic point.

+ † † † † † S (x) = Ψ↑(x)Ψ↓(x) = ΨR ↑(x)ΨR ↓(x) + ΨL ↑(x)ΨL ↓(x) + ΨR ↑(x)ΨL ↓(x) + ΨL ↑(x)ΨR ↓(x) .

1 i √ Ψ† (x)Ψ (x) = e−iφR ↑ eiφR ↓ = e−i 2π(Φs−Θs) R ↑ R ↓ 2πα 2πα (8.77) i √ Ψ† (x)Ψ (x) = ei 2π(Φs+Θs) R ↑ R ↓ 2πα

275 Figure 8.5:

(8.78) i √ √ Ψ† (x)Ψ (x) = (−1)xe−i 2πΦc ei 2πΘs . R ↑ L ↓ 2πα Since the charge ﬁeld is locked to Φ = n √π , then c 2π i √ D √ E Ψ† (x)Ψ (x) = (−1)xCei 2πΘs ,C ≡ e−i 2πΦc . R ↑ L ↓ 2πα

† An analogous result holds for ΨL ↑ΨR ↓, so that i √ i √ i √ S†(x) = e−i 2π(Φs−Θs) + ei 2π(Φs+Θs) + (−1)xCei 2πΘs . 2πα 2πα πα Finally 1 Sz(x) = (Ψ†Ψ − Ψ†Ψ ) 2 ↑ ↑ ↓ ↓ 1 = (Ψ† Ψ + Ψ† Ψ − Ψ† Ψ − Ψ† Ψ ) 2 R ↑ R ↑ L ↑ L ↑ R ↓ R ↓ L ↓ L ↓ 1 + (Ψ† Ψ − Ψ† Ψ + H.c.) 2 R ↑ L ↑ R ↓ L ↓

276 1 (−1)x √ √ = √ ∂xφs + cos( 2πΦc) sin( 2πΦs) 2π πα 1 (−1)xC √ ' √ ∂xφs + sin( 2πΦs) (8.79) 2π πα

277