2. The interaction between an electron gas and the potential field is given by

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1 68A. Solutions to Exercises II. April 5. Carry out a Hubbard Stratonovich transformation for the following interaction Hamiltonians, factorizing the terms in brackets, and write one or two sentences interpreting the effective Hamiltonian that it gives rise to a b c H I g d 3 x d 3 xψ xψ xψ xψ x xψ xψ x + ψ xψ x x + x x g. This is the local version of the BCS Hamiltonian introduced by Gorkov. It is useful if one wants to consider non-uniform such as a phase-twisted solutions. H I J H i, jσ i, j,σ,σ f iσ f jσ f jσ f iσ f iσ f jσ i j + i j f jσ f iσ + i j i j J H. This mean-field decoupling is often used to describe spin-liquids. The original Hamiltonian does not change the charge n f i at each site. The decoupled Hamiltonian develops a gauge symmetry f jσ e iφ j f jσ, i j e iφ i φ j i j which protects the charge conservation at each site. H I J σ ψ α b α b β ψ β α,β ψ σ b σ α + ᾱ b σ ψ σ + ᾱα J This is an unusual Hubbard Stratonovich transformation, because the decoupling field is a Grassman. This field never condenses, so that the above interaction has to be treated beyond mean-field theory at the level of Gaussian fluctuations. 3. The interaction between an electron gas and the potential field is given by β S EM dτ d 3 x eρxφx ɛ φ My apologies - there was a sign error infront of the eρx. 4

2 a Write down the path integral for a Coulomb gas of quantum electrons, involving an integral over the electron and potential fields. The path integral takes the form Z S D ψ, ψ, φe S dτd 3 x ψx τ m + eφx µ ψx ɛ φ b What equation does the potential φ satisfy at the saddle point of this path integral? The effective action is determined by the equation e S e f f φ D ψ, ψe S ψ,ψ,φ. 5 We can obtain the equation of motion for φ by differentiating this expression with respect to φ, as follows: δs e f f ln Z δ δφx δφx D ψ, ψ eρx + ɛ φx e S 7 Z so that the saddle point solution satisfies Gauss law or E eρ/ɛ. 6 φ eρx ɛ 8 c Why is the coefficient of the φ term negative? Give a physical interpretation. At first sight, this seems strange, because the last term in the action is minus one times the electromagnetic field energy: i.e minus one times the Coulomb energy! When we write the potential energy as eρφ we overcount the Coulomb energy by a factor of two. The negative correction term not only establishes Gauss s law as the saddle point, it also corrects for the overcounting. If you eliminate the particle density in terms of the potential, what is the energy density associated with the resulting electric field E φ? The answer to this question helps to explain the explanation just given above. If we insert the solution to Gauss s law into the total energy, replacing eρx ɛ φ, we obtain ɛ φ { }} { S EM d 3 x ɛ φ φ φ ɛ o d 3 φ x ɛ o 9 which is precisely the classical electric field energy, or Coulomb energy.

3 d Write down the effective action of the system when the fermions have been integrated out and interpret your result in terms of Feynman diagrams. The effective action is given by G { }} { F e f f T ln Det τ m µ +eφ d 3 φ xdτɛ β We can expand the logarithm as a power-series in the scattering potential as F TTr ln G + eφ d 3 E xdτɛ β TTr ln G TTr ln Geφ d 3 E xdτɛ β TTr ln G d 3 E xdτɛ β + n TTr Geφ n We can interpret the last term as the sum of loop diagrams for repeated scattering off the potential field φx. n TTr ln Geφ e Suppose the potential field acquires a mass term: β S EM dτ d 3 φ + κ φ x eρxφx ɛ What form does S EM take in real space when one integrates out the potential field? The electromagnetic part of the action can be written in the short-hand S EM eρ φ + φ M φ where M ɛ κ ɛ q κ in Fourier space. If we carry out the Gaussian integral over φ this gives or written out more explicitly S EM S EM eρ M eρ dτd 3 xd 3 x e ρx ρx 3 ɛ κ x x 3

4 Now ɛ κ ɛ q + κ e κr 4πɛ r so the final action corresponds to fermions interacting with a Yukawa potential. S dτ d 3 x ψ τ m µ ψ + d 3 xd 3 x Vx x : ρxρx : 4 where Vx x 4πɛ e κ x x x x. 3. Consider the Euclidean action for a single Harmonic oscillator S E β dτ ā τ + ω a a By re-writing the fields a and ā in terms of their momentum and position co-ordinates derive the action in terms of p and x. Hint: terms like dτx τ x dτ τ x vanish because they are perfect differentials of functions periodic in β. If we substitute then, a ā ω x + i p ω, ω x i p, 5 ω β ā τ + ωa ω β β x i p ω τ + ωx + i p ω ωxẋ + ω pṗ ipẋ xṗ + ω x + p { }} { ω 4 x + 4ω p + i β xp + β β dτ ipẋ + Hp, x dτ ipẋ + Hp, x 6 where the exact integrand appearing on line 3 is eliminated due to periodic boundary conditions, and Hp, x p + ω x is the Hamiltonian 4

5 b Carry out the Gaussian integral over the momentum field p and derive the corresponding action in terms of ẋ and x. If we integrate the action over momentum, then so the final action is dτp iẋ + p S β dτ iẋ dτ ẋ 7 ẋ dτ + ω x Mean field theory for antiferromagnetic Spin Density Wave Develop the mean-field theory for a three dimensional tight-binding cubic lattice with commensurate antiferromagnetic order parameter where Q π, π, π. M j Me iq R j 9 a Show that the mean-field free energy can be written in the form ɛk µ M σ M H MF ψ M σ ɛ k+q µ k + N s I k BZ ψ k where M M is the magnitude of the staggered mangetization, ψ k denotes the four-component spinor c k c ψ k k, c k+q c k+q ɛ k tc x +c y +c z, c l cos k l, l x, y, z is the kinetic part of the energy and the summation is restricted to the magnetic Brillouin zone, One half the original Brillouin zone. The action for the electrons moving in the Weiss field is given by S dτ c k τ + ɛ k µ c k + M j c j σc j + k j M j I from which we can read off the mean-field Hamiltonian H MF ɛ k µc kc k + M j c j σc j + k j M j I 3 5

6 If we now substitute M j Me i Q R j 4 where Q π, π, π, then H MF ɛ k µc kc k + k k M c k+q σc k + N s M I. 5 Dividing the Brillouin zone into two halves, we obtain H MF k BZ k BZ ψ k ɛk µc kc k + ɛ k+q µc k+qc k+q + M c k+q σc k + H.c + N s M ɛk µ M σ M ψ M σ ɛ k+q µ k + N s I. 6 I b On a tight binding lattice the kinetic energy has the nesting property that ɛ k+q ɛ k. Show that the energy eigenvalues of the mean-field Hamiltonian have a BCS form E k± ± ɛ k + M µ. 7 For a nested Fermi surface, the mean-field Hamiltonian becomes where the matrix hamiltonian hk H MF where the four dimensional Nambu matrices are τ 3, στ k BZ ψ khkψ k + N s M I. 8 ɛk µ M σ ɛ M σ ɛ k µ k τ 3 + M στ µ 9 σ. 3 σ The quasiparticle energies are determined by the eigenvalues of hk. We can determine these from DetE ɛ k τ 3 M στ + µ 3 We can determine the eigenvalues by squaring the determinant, to obtain Det E ɛ k τ 3 M στ + µ E + ɛ k τ 3 + M στ + µ Det E + µ ɛ k + M 6

7 where we have used the fact that τ 3 anticommutes with στ. From this, we see that E kp µ ± ɛ k + M 3 Note that in the half Brillouin zone, each of the corresponding eigenstates has a two-fold spin degeneracy. c Show that the mean-field free energy takes the form F k,p± T ln cosh βekp M + N s I µ From the energy eigenstates, we see that the mean-field Free energy is given by 33 F T k BZ,p± ln + e βekp + N s M I 34 where the factor of two is from the two-fold spin degeneracy of the states. We can expand the summation to the full Brillouin zone by dividing the first term by two, M F T ln + e βe kp + N s I k BZ,p± T ln + e βe k+ + e βe k M + N s I k BZ 35 Now with a bit of algebra, we deduce that + e βe k+ + e βe k + e β µ+ + e β µ e β µ+ e β µ 4 cosh e βµ βek+ βek 4 cosh cosh βek+ cosh βek 36 so we can write the mean field Free energy in the form F k,p± k,p± T ln cosh T ln cosh βekp µ M + N s I M + N s I µ βekp 37 7

8 Note the factor of two difference with the question - a typo in the original form. We can check that the above form is the correct one, by differentiating w.r.t µ to get the total particle number N F µ tanh βekp + N s k,p± βekp tanh f E kp 38 k,p± k,p± d Differentiating the mean-field Free energy w.r.t M, we obtain F N s M tanh k βekp Mp + ɛ k + M M I 39 from which we obtain k,p± ɛ k + M µp tanh T ɛ k + M I. 4 e Show that at half filling, the nesting guarantees that a transition to a spin-density wave will occur for arbitrarily small interaction strength I. The gap equation at T is given by sgnpe kp k,p ɛ k + I. 4 M If we set M, we obtain the critical I I c for an instability into the antiferromagnet, I c sgn ɛ k pµ ɛ k,p k Λ sgn ɛ pµ dɛnɛ p Λ ɛ Λ sgn ɛ pµ dɛnɛ ɛ p Λ µ dɛ ɛ N ln Λ µ 4 8

9 where we have replaced the integral over k by an integral over the density of states, introducing a cutoff Λ. To extract the infra-red divergence in the integral, we have replaced the density of states per spin by its value at ɛ, N. The critical value I c is then given by I c N ln 43 Λ µ a quantity that goes to zero as µ. In other words, the critical value of I c is zero at particlehole symmetry. f Calculate the zero temperature phase diagram assuming that the order remains commensurate at finite doping. A precise answer to this question would plot out the T c versus doping, determined from the two equations d 3 k + δ π f E k,p, 3 where δ is the doping away from half filling and d 3 k ɛ π tanh k + M µp 3 p± T ɛ k + I. M p± However, lets do it heuristically for the moment. Lets look first at the zero temperature phase diagram, plotting I c as a function of doping away from half-filling. If the doping is δ µn, then we have I c N ln ΛN δ Putting N /ɛ F, where ɛ F is the Fermi energy, we obtain We can estimate T c by writing or I c I c N ɛ F ln Λ/ɛ F δ. ln Λ T c T c Λe /I cn Its probably just as good to sketch the phase diagram. However, as an exercise I worked out the phase boundary in mathematica, as shown below. Notice that my finite grid size in integration has eliminated the region near I and δ. With some work this could have been fixed, but I drew the lines in by hand. 9

10 Figure : Phase boundary for antiferromagnet as a function of doping, calculated in mean field theory. All units with t.