Condensed Matter - HW3 :: Fermi Liquid Currents

PHSX 545

Problem 1 Various currents in Fermi liquid theory are given by the following expressions (spin-independent):

1 ∇ 0 j = ( pεp) δn¯p particle current V p X 0 1 ∂ε Π = p p δn¯ momentum current ij V i ∂p p p j 1 X q = ε0 (∇ ε0 ) δn¯ energy current (heat) V p p p p p X 0 where εp is energy in global equilibrium, andn ¯k is deviation of distribution function from local equilibrium, and that includes interactions between quasiparticles. (a) Calculate the particle current for a single excitation at momentum p. Hint: in the absence of quasiparticle interactions this would have been just the group velocity vp ≈ vf pˆ of the particle. With quasiparticle interactions it will be a different velocity u, that beside vp includes backflow currents from all other quasiparticles disturbed by the motion of the original one. Is it consistent with Galilean invariance principle (remember that we define co-moving reference frame by condition j′ = 0 - no particle current, in this case p′ = 0)? (b) Calculate momentum and energy currents. For momentum current use symmetry arguments to identify Fermi s,a liquid parameters Fℓ that enter Πij . Theoretically minded part of the class are strongly encouraged to do this calculation fully. The following expressions might help:

dΩˆ dΩˆ 1 dΩˆ dΩˆ 1 k kˆ = 0 k kˆ kˆ = δ k kˆ kˆ kˆ = 0 k kˆ kˆ kˆ kˆ = (δ δ + δ δ + δ δ ) 4π i 4π i j 3 ij 4π i j m 4π i j m n 15 ij mn im jn jm in Z Z Z Z 2

Answer of exercise 1 (a) The single particle is added into state p so that

δnk = δ(k − p) - this is change of the distribution function relative to the global(!) equilibrium. This particle perturbs other quasi- particles through interactions and we write

∂nk ∂nk jp = vk δnk − δεk ≈−δ(εk − µ) ∂εk ∂εk k X   where (sum over spins performed)

s ′ s δεk = 2 f (kˆ · kˆ )δnk′ = 2f (kˆ · pˆ) k′ X The integration over k now gives

s s dΩkˆ s f F1 jp ≈ v pˆ + N (v kˆ) 2f (kˆ · pˆ)= v pˆ + 2N (v ) pˆ = v pˆ 1+ f f 4π f f f f 3 f 3 Z   This expression we can re-write using definition of the effective mass

s pf F1 ∗ s pf j = 1+ = Galilean invariant system m = m(1 + F1 /3) = p m∗ 3 m   where m is bare mass of the particle. This is consistent with the Galilean invariance. Velocity u ≡ p/m defines a reference frame in which there is no particle current: ′ ′ j ∝ p = p − mu = 0 since the mass that enters this expression is the bare mass! NOTE: We can understand this expression as follows. Say we have a particle in state p, δnpσ = 1. This particle perturbs the energies of other states by shifting their energies and thus modifying the ‘ground state’ distribution: 0 0 ′ ′ ∂n ∂n ′ δεp′σ′ = fσ′σ(pˆ · pˆ)δnpσ = fσ′σ(pˆ · pˆ) δnp′σ′ = δεp′σ′ = fσ′σ(pˆ · pˆ) ∂εp′ ∂εp′ This change in the distribution can be seen as the “backflow” current, and we have to calculate the current due to the presence of the original particle relative to the “backflow” which is the new ‘ground state’: 0 0 ∂n ′ ∂n s ′ jpˆ = vp − ‘backflow’ = vp − vp′ δnp′σ′ = vp − vp′ fσ′σ(pˆ · pˆ)= vp − vp′ 2f (pˆ · pˆ) ∂εp′ ∂εp′ σ′pˆ′ σ′pˆ′ pˆ′ X X X

s F1 = vp + v pˆ f 3 (b) The energy current is exactly similar to the particle current: 1 q = ε0 (∇ ε0 ) δn¯ V k k k k k X where we take 0 − ∇ 0 εp = ξp + εf = vf (p pf )+ εf pεp = vf pˆ and so

0 0 ∂nk s ′ q = ε v kˆ δnk − 2f (kˆ · kˆ )δnk′ k f ∂ε0 k k k′ ! X X 3

∂nk − 0 − with the derivative = δ(εk εf ). In the second term we do integration over k first, ∂εk

0 dΩkˆ s ′ q = ε v kˆδnk + ε v kˆ F (kˆ · kˆ )δnk′ k f f f 4π k k′ X Z X After kˆ angle integration we replace k′ → k and obtain:

s s F1 F1 q = ε0 v kˆδn + ε (v kˆ)δn = v kˆ ε0 + ε δn k f k f 3 f k f k f 3 k k k k X X X   So for the energy current we have

s s F1 F1 q = v pˆ ε0 + ε = ξ0 v pˆ + ε 1+ v pˆ p f p f 3 p f f 3 f     For the momentum flux we have 0 1 ∂εp ∂nk s ′ Πij = pi δn¯p = pivf pˆj − kivf kˆj 2f (kˆ · kˆ )δnk′ V ∂pj ∂εk p k k′ X X X

dΩˆ = ppˆ v pˆ + p v k kˆ kˆ F s(kˆ · pˆ) i f j f f 4π i j Z s From the symmetry, the last integral is non zero when kˆikˆj overlaps with one of the harmonics of F (kˆ · pˆ). Since the product of two unit vectors can generally be expanded in spherical harmonics of ℓ = 0, 1, 2, one expects that there s will be contributions from interaction terms with F0,1,2. Explicitly we write for the interaction function

ˆ · 2 − s s s s s s s 3(k pˆ) 1 F (kˆ · pˆ)= F0 + F1 P1(kˆ · pˆ)+ F2 P2(kˆ · pˆ)= F0 + F1 (kˆ · pˆ)+ F2 2 Using the provided integral relations we can write for the quadrupolar term:

dΩˆ dΩˆ 1 1 2 k kˆ kˆ (kˆ · pˆ)2 =p ˆ pˆ k kˆ kˆ kˆ kˆ = (δ δ + δ δ + δ δ )ˆp pˆ = δ + pˆ pˆ 4π i j m n 4π i j m n 15 ij mn im jn jm in m n 15 ij 15 i j Z Z and for the momentum flux we have

s s s 1 F2 1 3F2 1 2 Π = pv pˆ pˆ + p v F0 δ − δ + δ + pˆ pˆ ij f i j f f 3 ij 2 3 ij 2 15 ij 15 i j   

s s F2 1 s F2 = v p + p pˆ pˆ + p v F0 − δ f f 5 i j f f 3 5 ij