3. Greens functions. H = H 0 + V t. V t = ˆBF t. e βh. ρ 0 ρ t. 3.1 Introduction

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1 3. Greens functions 3.1 Introduction Greens functions appear naturally as response functions, i.e. as answers to the function how a quantum mechanical system responds to an external perturbation, like for example electrical or magnetic fields; the corresponding response functions would then describe the electrical conductivity or the magnetic susceptibility of a system. Here we will be concerned with small perturbations and thus only the linear response of the system. We describe the system by a Hamiltonian H = H 0 + V t (3.1) where V t represents the interaction with an external field. H 0 describes the system with the external field switched off; due to interactions H 0 is not necessarily exactly solvable. The external field F t couples to the observable ˆB of the system: V t = ˆBF t (3.2) Here, ˆB is an operator and F t is a complex number. We now consider an observable  of the system that is not explicitly time dependent and ask how the dynamic expectation value  reacts to the perturbation V t. Without field we have  0 = Tr(ρ 0Â) (3.3) where ρ 0 is the density matrix of the system without external fields: ρ 0 = e βh Tre βh (3.4) in the grand canonical ensemble H = H µ ˆN (with chemical potential µ, particle number operator ˆN). The density matrix will change if we switch on the field: ρ 0 ρ t (3.5) 29

2 This means for the expectation value of   t = Tr(ρ tâ) (3.6) In the Schrödinger picture, the equation of motion of the density matrix (the statistical operator) is determined by the von Neumann equation i h ρ t t = [ H + V t, ρ t ] (3.7) We consider a perturbation that is switched on at some time so that the boundary condition for our first order differential equation is an unperturbed system for t lim ρ t = ρ 0. (3.8) t We now switch to the Dirac picture where we have ρ D t (t) = e i h H 0t ρ t e i h H 0t (3.9) with the equation of motion ρ D t (t) = i h[ ρ D t, V D t Integrating with the boundary condition leads to ] (t). (3.10) lim t ρd t (t) = ρ 0 (3.11) ρ D t (t) = ρ 0 i h t dt [ V D t (t ), ρ D t (t ) ] (3.12) This equation can be solved by iteration (by substituting ρ D t (t) repeatedly on the right hand side): ρ D t (t) = ρ 0 + ρ D(n) t (t) = n=1 ( i h ) n t ρ D(n) t (t) with dt 1 t1 tn 1 dt 2 dt n [ V D t 1 (t 1 ), [ V D t 2 (t 2 ), [ [V D t n (t n ), ρ 0 ] ] 30 ] ] (3.13)

3 While this formula is exact, it is not practical. For sufficiently small external perturbations, we can restrict to linear terms in the perturbation V t which is called linear response: ρ t ρ 0 i h t dt e i h H 0t [ V D t (t ), ρ 0 ] e i h H 0t (3.14) Here, we have returned to the Schrödinger representation for the density matrix. We can use this result to determine the perturbed expectation value of (3.6):  t = Tr(ρ tâ) =  0 i h =  0 i h =  0 i h t t t dt Tr {e [ ] i h H 0t Vt D (t ), ρ 0 { ] dt F t Tr [ˆB D (t ), ρ 0 } e i h H 0t ÂD (t) =ˆBρ0  ρ 0ˆBÂ=ρ 0 ˆB ρ0ˆbâ { dt F t Tr ρ [ÂD 0 (t), ˆB D (t ) ] } } (3.15) Here, cyclic invariance of the trace was exploited. This shows how the system reacts to the external perturbation, as measured from the observable Â: A t =  t  0 = i h t dt F t [ÂD (t), ˆB D (t )] 0 (3.16) This response is determined by an expectation value of the unperturbed system. The Dirac representation of the operators ÂD (t), ˆB D (t ) corresponds to the Heisenberg representation when the field is switched off. Now we define the retarded two-time greens function G r AB(t, t ) = A(t); B(t ) = iθ(t, t ) [ A(t), B(t ) ] (3.17) 0 The operators are to be taken in Heisenberg representation of the field free system. The retarded Greens function describes the response of a system as manifested in observable  when the perturbation couples to observable ˆB: A t = 1 h dt F t G r AB(t, t ) (3.18) 31

4 It is called retarded because due to the Heaviside function, only perturbations for t < t contribute. With the Fourier transform F(ω) of the perturbation F t = 1 2π dω F(ω)e i(ω+iδ)t (3.19) where δ > 0 is infinitesimally small and using the later result that with a Hamiltonian that is not explicitly time dependent the Greens function depends only on time differences t t, we can rewrite (3.18) in the form of the Kubo formula A t = 1 dω F(ω)G r 2π h AB(ω + iδ)e i(ω+iδ)t (3.20) The δ > 0 in the exponent enforces the boundary condition (3.8). We will now look into two applications of response functions. Magnetic Susceptibility The perturbation is a spatially homogeneous magnetic field that oscillates in time: B t = 1 dω B(ω)e i(ω+iδ)t, (3.21) 2π which couples to the magnetic moment m = m i = gµ B S i. (3.22) h i i Thus, the perturbing potential term in the Hamiltonian becomes V t = m B t = 1 dω m α B α (ω)e i(ω+iδ)t (3.23) 2π α where α = x, y, z are Cartesian directions. An interesting quantity is now the magnetization in response to the applied field. As it is M = 1 V m = gµ B S i, (3.24) hv i we have to choose the magnetic momentum operator for both  and ˆB operators in the Kubo formula: M β t Mβ 0 = 1 dt B α t m β (t); m α (t ). (3.25) V α 32

5 Only in a ferromagnet there is a finite magnetization M β 0 Eq. (3.25) defines the magnetic susceptibility tensor without a field. χ βα ij (t, t ) = µ 0 gµ 2 B S β V h 2 i (t); Sα j (t ) (3.26) as a retarded Greens function. Thus M β t = 1 dt B α t µ χβα ij (t, t ) (3.27) 0 ij or in terms of frequency M β t = 1 2πµ 0 ij α dω e i(ω+iδ)t χ βα ij (ω)b α (ω) (3.28) We have implicitly assumed that the system we consider has permanent localized moments. Two types of susceptibilities are interesting: The longitudinal susceptibility χ zz ij (ω) = µ 0 gµ 2 B S z V h 2 i ; S z j (3.29) ω where the index indicates the Fourier transform of the retarded Greens function. This can be used to obtain information about the stability of magnetic orderings. For the paramagnetic phase, one calculates the spatial Fourier transform χ zz q (ω) = 1 N ij ( Ri ) χ zz ij (ω)e i q R j (3.30) At the singularities of this response function, an infinitesimally small field is sufficient to create a finite magnetization, i.e. a spontaneous ordering of the moments. For that purpose, the conditions under which { lim ( q,ω) χ zz q (ω)} 1 = 0 (3.31) are studied; they indicate the paramagnetic ferromagnetic transition. The other interesting case is the transversal susceptibility χ + ij (ω) = µ 0 gµ 2 B S + V h 2 i ; S j ω where S± i = S x i ± is y i (3.32) 33

6 Poles of this susceptibility correspond to spin wave (magnon) energies: { } 1 (ω) = 0 ω = ω( q). (3.33) χ + q The examples show that linear response theory not only treats weak external perturbations but also yields information about the unperturbed system. Electrical conductivity Now we consider a spatially homogeneous electrical field that oscillates in time: E t = 1 2π dω E(ω)e i(ω+iδ)t. (3.34) The electrical field couples to the electrical dipole moment P P = d 3 r rn( r). (3.35) We consider N point changes q i at positions r i (t); the charge density is n( r) = N q i δ( r r i ). (3.36) i=1 This gives a dipole moment operator P = N i=1 q i r i. (3.37) The electrical field causes the additional external potential term in the Hamiltonian V t = P E t = 1 dω P α E α (ω)e i(ω+iδ)t. (3.38) 2π α An interesting quantity is the response of the current density to the external field: j = 1 V N i=1 q i r i = 1 P. (3.39) V 34

7 Its expectation value without field disappears: j 0 = 0. (3.40) After switching the field on, we have j β t = 1 h dt E α t j β (t); P α (t ). (3.41) α In terms of the Fourier transforms this becomes j β t = 1 dω e i(ω+iδ)t σ βα (ω)e α (ω) (3.42) 2π α This is Ohms law, defining the electrical conductivity tensor σ βα (ω) j β ; P α ω (3.43) that has retarded Greens functions as components. This can be rewritten as σ βα (ω) = i N q 2 V m(ω + iδ) δ αβ + i h j β ; j α (3.44) ω + iδ The first term represents the conductivity of a noninteracting electron system as given by classical Drude theory, and the second one involving a retarded current-current Greens function represents the interaction between the particles. 3.2 Matsubara method In the solid state theory class, Greens functions were introduced as response functions; they can be used to determine the quasiparticle density of states. They also appear as correlation functions, and they give us access to excitation energies. But so far, everything was done at zero temperature. The ojective is now to extend the methods for T = 0 to finite temperatures. 1 As a reminder, the two-time Greens functions were defined as (with ε = 1 for fermionic operators, ε = +1 for bosonic operators) retarded: G r AB(t, t ) A(t); B(t ) r = iθ(t t ) [A(t), B(t )] ε advanced: G a AB(t, t ) A(t); B(t ) a = iθ(t t ) [A(t), B(t )] ε causal: G c AB(t, t ) A(t); B(t ) c = i T ε (A(t)B(t )) 1 This chapter is based on K. Elk, W. Gasser, Die Methode der Greenschen Funktionen in der Festkörperphysik, Akademie-Verlag Berlin 1979, and W. Nolting, Grundkurs Theoretische Physik 7, Viel-Teilchen-Theorie, Springer

8 with the Wick time ordering operator T ε : (3.45) T ε (A(t)B(t )) = θ(t t )A(t)B(t ) + εθ(t t)b(t )A(t) (3.46)... indicates an average over the grand canonical ensemble X = Tr(e βh X) Tr(e βh ) where β = 1, H = H µn (3.47) k B T If the Hamiltonian has no explicit time dependence, the Greens functions are homogeneous in time: H t = 0 Gα AB(t, t ) = G α AB(t t ), α = r, a, c (3.48) This can be shown using the cyclic invariance of the trace. For the correlation functions that are needed for G r and G a, we have (A(t)B(t ) = A(t t )B(0) = A(0)B(t t) (B(t )A(t) = B(t t)a(0) = B(0)A(t t ) (3.49) We now allow the time variables to formally take complex values: Tre βh A(t i hβ)b(t ) = Tr { e βh e i h H(t i hβ) A(0)e i h H(t i hβ) B(t ) } = Tr { B(t )e βh e +βh e i h Ht A(0)e i h Ht e βh} = Tr { e βh B(t )A(t) } A(t i hβ)b(t ) = B(t )A(t) (3.50) As two different correlation functions become related in this way, the extension of the Greens function to complex time seems to be advantageous. In particular, in perturbation theory in V where H = H 0 + V, V would, for finite temperatures, appear in two places, in the Heisenberg representation of time dependent operators e ± i h Ht and in the density operator of the grand canonical averaging e βh ; two perturbation expansions would be necessary. Therefore, we join the exponential functions by introducing a complex time. The Matsubara method introduces purely imaginary times so that the quantity τ = it is real. This leads to a modified Heisenberg representation of operators: A(τ) = e 1 h Hτ A(0)e 1 h Hτ (3.51) 36

9 Note that the operator e i h Hτ creating imaginary time shifts is not unitary. The equation of motion (EOM) for an operator A(τ) h τ A(τ) = h τ [e 1 h Hτ A(0)e 1 h Hτ ] = HA(τ) + A(τ)H (3.52) thus becomes: h τ A(τ) = [ A(τ), H ] (3.53) Here, we use the (conventional) step function θ(τ) = { 1 for τ > 0 (t = τi negative imaginary) 0 for τ < 0 (t = τ i positive imaginary) (3.54) It can be used to introduce the time ordering operator. T τ { A(τ)B(τ ) } = θ(τ τ )A(τ)B(τ ) + ε p θ(τ τ)b(τ )A(τ) (3.55) where p is the number of permutations of creation operators. We assume pure Fermi/Bose operators so that p = 1. The definition of the Matsubara function is: G M AB(τ, τ ) = A(τ); B(τ ) M = T τ (A(τ)B(τ )) (3.56) Using (3.53) and (3.55) we get the EOM h τ GM AB(τ, τ ) = hδ(τ τ ) [A, B] ε + [A(τ), H] ; B(τ ) (3.57) Properties of the Matsubara Greens function introduced in this way are: 1) it depends only on time differences G M AB(τ, τ ) = G M AB(τ τ, 0) = G M AB(0, τ τ) (3.58) 37

10 2) it is periodic in time: Take hβ > τ τ + n hβ > 0, n Z; then Tre βh G M AB(τ τ + n hβ) >0 = Tr { e βh T τ ( A(τ τ + n hβ)b(0) )} = Tr { e βh A(τ τ + n hβ)b(0) } = Tr { e βh e H h (τ τ +n hβ) A(0)e H h (τ τ +n hβ) B(0) } = Tr { e H h (τ τ +(n 1) hβ) A(0)e H h (τ τ +(n 1) hβ) e βh B(0) } = Tr { } e βh B(0)A(τ τ + (n 1) hβ) = Tr { e βh T τ ( B(0)A(τ τ + (n 1) hβ) )} <0 = εtr { e βh T τ ( A(τ τ + (n 1) hβ)b(0) )} (3.59) This gives us the important result: ( hβ > τ τ +n hβ > 0 : τ τ +n hβ ) = εgab( M τ τ +(n 1) hβ ) In particular, for n = 1: G M AB (3.60) G M AB( τ τ + hβ ) = εg M AB( τ τ ) for hβ < τ τ < 0 (3.61) Thus the Matsubara Greens function is periodic in an interval 2 hβ; it is enough to consider the time interval hβ < τ τ < 0. This periodicity allows the Fourier expansion: G M (τ) = 1 [ 2 a 0 + a n cos nπ hβ τ + b n sin nπ ] hβ τ a n = 1 hβ b n = 1 hβ n=1 hβ hβ Now we define discrete energies E n = nπ β ( nπ ) dτ G M (τ) cos hβ τ dτ G M (τ) sin ( nπ hβ τ ) (3.62) (3.63) and the Matsubara Greens function on the imaginary energy (frequency) axis G(E n ) = 1 2 hβ(a n + ib n ) (3.64) 38

11 Then G M (τ) = 1 hβ G M (E n ) = 1 2 e i h Enτ G M (E n ) n= hβ dτ G M (τ)e i h E nτ (3.65) Still, we can simplify a bit more: G M (E n ) = 1 2 = dτ G M (τ)e i h E nτ dτ G M (τ)e i h E nτ = [ 1 + εe ie nβ ] hβ 0 dτ G M (τ)e i h E nτ dτ G M (τ)e i h E nτ dτ G M (τ hβ)e i h E nτ e ie nβ (3.66) where τ = τ + hβ was introduced. The bracket [... ] disappears for fermions (ε = 1) if n is even, for bosons (ε = 1) if n is odd. Thus with G M (τ) = 1 hβ G M (E n ) = E n = n= 0 e i h E nτ G M (E n ) dτ G M (τ)e i h E nτ { 2n π β for bosons (2n + 1) π β for fermions (3.67) (3.68) Now we need to work out a spectral representation for the Matsubara Greens function in order to relate it to the retarded Greens function. We first consider the correlation function 1 A(τ)B(0) = Tre βh E n A(τ)B(0) E n e βe n n 1 = Tre βh E n A E m E m B E n e βe n e 1 h (E n E m )τ (3.69) n 39

12 where we introduced the eigenstates of the Hamiltonian E n : H E n = E n E n. We recall the spectral density on the real energy axis: S AB (E) = h Tre βh E n A E m E m B E n e βe n (1 εe βe )δ ( E (E m E n ) ) n,m A(τ)B(0) = 1 h de S AB(E) 1 εe βee 1 h Eτ (3.70) In the integration interval in (3.67), τ is positive; we need to evaluate G M AB(E n ) = Now we use the integral 0 0 dτ e 1 h (ie n E)τ = dτ e i h E nτ A(τ)B(0) (3.71) h ( e iβe n e βe 1 ) = ie n E and put Eq. (3.70) into Eq. (3.71) to obtain G M AB(E n ) = h ( εe βe 1 ) ie n E (3.72) de S AB(E ) ie n E (3.73) If we compare this to the spectral representation of the retarded Greens function, we only need to replace ie n E+i0 + ; thus, the retarded Greens function can be obtained from the Matsubara Greens function by analytic continuation! 3.3 Some methods for Matsubara axis functions The Green s function on the Matsubara axis is fairly smooth and featureless so that the determination of the sums poses no problems. But there are two points to be considered: (1) Terms falling off as 1/(iω n ) will be badly represented by any number of frequency points; therefore, high frequency corrections are needed. (2) In order to obtain a spectral function, we need to analytically continue the Greens function to the real axis. This is done by the Padé method. 40

13 High frequency correction The idea is to subtract the leading terms of the high frequency expansion and calculate them analytically, e.g. F m (iω n ) = 1 n f + n f ( N 1 ) + O iω n (iω n ) 2 (3.74) n f = T (F m (iω n) 1 n f + n f N )e iω n0 + iω + 1 n f + n f N iω n n 2 (3.75) The last term of Eq. (3.75) is obtained by explicit calculation of the sum: 1 T e iωnτ 2i sin(ω n τ) = T iω n= n iω n=0 n = 2 sin [ (2n + 1)πTτ ] = 1 (3.76) π 2n n=0 because τ = 0 + ]0; 1/T[ and 4 sin [ (2n + 1)x ] 1 for π < x < 0 = 0 for x = π, 0, π π 2n + 1 n=0 1 for 0 < x < π In practice, the 1/(iω n ) 2 is taken into account in the same way. (2) Pade approximation This method for analytic continuation can be used (3.77) 1. when the function to be continued is not given analytically (in that case, use iω n ω + iδ) 2. when the function is given without statistical errors (for Quantum Monte Carlo Green s functions, use Maximum Entropy Method) One has to keep in mind one drawback of the Padé approximation: it is a polynomial representation that has limited precision for functions which are hard to approximate by a polynomial. It can show the wiggles that are typical for polynomial interpolations. The algorithm for calculating the Padé approximant was written down nicely by Vidberg and Serene (H. J. Vidberg and J. W. Serene, Solving the 41

14 Eliashberg Equations by Means of N-Point Padé Approximants, J. Low Temp. Phys. 29, 179 (1977)). Given the values u i of a function at N complex points z i (i = 1,..., N), we define the continued fraction C N (z) = 1 + a 1 a 2 (z z 1 ) a N (z z N 1 ) Here, the coefficients a i are to be determined so that (3.78) C N (z i ) = u i, i = 1,..., N (3.79) The coefficients are then given by the recursion a i = g i (z i ), g 1 (z i ) = u i, i = 1,..., N g p (z) = g p 1(z p 1 ) g p 1 (z), p 2 (3.80) (z z p 1 )g p 1 (z) This requires the following calculation: g 1 (z 1 ) = a 1 = u 1 g 1 (z 2 ) = u 2 g 1 (z 3 ) = u 3 g 1 (z 4 ) = u 4... g 2 (z) = g 1(z 1 ) g 1 (z) g 2 (z 2 ) = a 2 = a 1 g 1 (z 2 ) (z z 1 )g 1 (z) (z 2 z 1 )g 1 (z 2 ) = a 1 u 2 (z 2 z 1 )u 2 g 2 (z 3 ) = a 1 g 1 (z 3 ) (z 3 z 1 )g 1 (z 3 ) = a 1 u 3 g 2 (z 4 ) = a 1 u 4... (z 3 z 1 )u 3 (z 4 z 1 )u 4 g 3 (z) = g 2(z 2 ) g 2 (z) (z z 2 )g 2 (z) g 3 (z 4 ) = a 2 g 2 (z 4 ) (z 4 z 2 )g 2 (z 4 )... g 4 (z) = g 3(z 3 ) g 3 (z) (z z 3 )g 3 (z) g 3 (z 3 ) = a 3 = a 2 g 2 (z 3 ) (z 3 z 2 )g 2 (z 3 ) g 4 (z 4 ) = a 4 = a 3 g 3 (z 4 ) (z 4 z 3 )g 3 (z 4 )... (3.81) Thus, the following triangular matrix p i, j has to be calculated: j = 1 j = 2 j = 3 j = 4... i = 1 a 1 = u 1 u 2 u 3 u 4... i = 2 a 2 g 2 (z 3 ) g 2 (z 4 )... i = 3 a 3 g 3 (z 4 )... i = 4 a

15 This can be done as follows: p 1, j = u j, j = 1,..., N p i, j = p i 1, i 1 p i 1, j (z j z i 1 )p i 1, j, j = 2,..., N and i = 2,..., j (3.82) The diagonal of this matrix then contains the coefficients a i which are needed for the recursion formula for C N (z): with C N (z) = A N(z) B N (z) A n+1 (z) = A n (z) + (z z n ) a n+1 A n 1 (z) B n+1 (z) = B n (z) + (z z n ) a n+1 B n 1 (z) and A 0 = 0, A 1 = a 1, B 0 = B 1 = 1 (3.83) An important fact to keep in mind: the bigger you chose N, i.e. the more coefficients a i you calculate, the bigger the numbers A N (z) and B N (z) are going to be for a given z. N = 128 will already lead to values of order Comparing functions on real and imaginary axes For a first check if an analytic continuation makes sense, there are exact relations at zero frequency. If a complex function is given around z = 0 as f(z) = α + βz + γz (3.84) with complex coefficients α = α + iα etc., we can write it in terms of imaginary frequencies iω n as f(iω n ) = α + iα β ω n + iβ ω n γ ω 2 n iγ ω 2 n +... (3.85) or of real frequencies as f(ω) = α + iα + β ω + iβ ω + γ ω 2 + iγ ω (3.86) and thus by comparing coefficients we find for the function value at zero f(iω n ) ωn =0 = f(ω) ω=0 (3.87) For the derivatives of f we find d Re f(iω n ) dω = d n ωn =0 dω Im f(ω) ω=0 (3.88) 43

16 d Im f(iω n ) dω = d n ωn =0 dω Re f(ω) ω=0 (3.89) d 2 dω 2 Re f(iω n ) = d2 n ωn =0 dω 2 Im f(ω) ω=0 (3.90) d 2 dω 2 Im f(iω n ) = d2 n ωn =0 dω 2 Re f(ω) ω=0 (3.91) Kramers Kronig relations If the function we try to calculate is analytic in the upper complex plane as well as in the lower complex plane (this is the case for all response functions of physical systems) it must obey the Kramers-Kronig relation g (y) = P dx g (x) π y x (3.92) where the notation g(z) = g (z) + ig (z) is used. x, y are used for real, z and u for complex variables. This is the real-axis analog of the more general Cauchy integral formula g(z) = 1 2πi du g(u) u z (3.93) which says that a holomorphic function defined on a disk is completely determined by its values on the boundary of the disk. Response functions are not analytic on the real axis, but otherwise we can deform the contour arbitrarily on the complex plane. If g(z) falls off in infinity, we can write g(z) = which yields g(z) = dx [g(x + iδ) g(x iδ)] 2πi x z dx g (x) π z x The real part of this equation is equal to Eq. (3.92). (3.94) (3.95) The practical calculation of a Kramers Kronig transformation is straightforward but requires the subtraction of the divergent part. It will usually be implemented on a mesh from a to b large enough so that g (a) and g (b) are negligibly small. 44

17 Then g (y) = b a dx g (x) g (y) π x y + g (y) π Re [log(b y) log(a y)] = b a dx g (x) g (y) π x y + g (y) π ( b y ) log y a (3.96) (3.97) Now the integrand is smooth everywhere. At x = y it should be estimated using the derivative dg (y)/dy. Kramers Kronig relations can become very useful if real or imaginary part of a Greens function are easier than the full function. Typically, the imaginary part falls off abruptly at the band edges and is exactly zero beyond. Then the imaginary part is perfectly suited for calculation on a mesh, but the corresponding real part only falls off as 1/ω. 45