Solid State Theory Solutions Sheet 9. Exercise 1. FS 2014 Prof. Manfred Sigrist Polarization of a neutral Fermi liquid Consider a system of neutral spin-1/2 particles each carrying a magnetic moment µ = field E couples to the particles by the relativistic spin-orbit interaction µ v HSO = × E · σ, 2 c µ 2 σ. An electric (1) where σ = (σ x , σ y , σ z ) is the vector of Pauli spin matrices. In the following we want to calculate the linear response function χ for the uniform polarization P = χE. (2) In the presence of the spin-orbit interaction we have to consider a more general situation of a distribution of quasiparticles with variable spin quantization axis. In such a case we must treat the quasiparticle distribution function and the energy as a 2 × 2 matrix, (ˆ np )αβ and (ˆ p )αβ , respectively. Furthermore, we require f to be a scalar under spin rotations. In this case f must be of the form fˆαβ,α0 β 0 (p, p0 ) = f s (p, p0 )δαβ δα0 β 0 + f a (p, p0 )σ αβ · σ α0 β 0 . (3) (a) Expand n ˆ p , ˆp , and fˆσσ0 (p, p0 ) in terms of the unit matrix σ 0 = 1 and the Pauli spin matrices 1 x σ = σ , σ 2 = σ y , σ 3 = σ z and write down Landau’s energy functional E. (b) Assume that the electric field is directed along the z direction, E = Ez zˆ. Show that the polarization of such a system is given by Pz = Here, δnip = function. 1 2 tr µ X ∂E = ∗ py δn1p − px δn2p . ∂Ez m c p (4) δˆ np σ i and δˆ np is the deviation from the equilibrium (Ez = 0) distribution (c) The application of the electric field changes the quasiparticle energy in linear response according to δ˜ ip = δip + 2 X ii f (p, p0 )δnip0 V 0 with δnip = p ∂n0p i δ˜ = −δD (0p − F )δ˜ ip , ∂p p (5) where δD denotes the Dirac delta function. Use the ansatz δ˜ ip = αδip and show that α = 1/(1 + F1a /3). (d) Compute χ according to Eq. (2). Solution. We introduce the notation σ 0 = 1, σ 1 = σ x , σ 2 = σ y and σ 3 = σ z . (a) We expand δˆ np = 3 X δnip σ i , ˆp = i=0 3 X ip σ i (S.1) i=0 and fˆσσ0 (p, p0 ) = 3 X σ i f ij (p, p0 )σ j where i,j=0 1 f ij  s i = j = 0;  f , a f , i = j = 1, 2, 3; =  0, i 6= j. (S.2) The energy functional is then given by E = E0 + X 1 X X (δˆ np )βα fˆαβ,α0 β 0 (p, p0 )(δˆ np0 )β 0 α0 2V 0 0 0 (ˆ p )αβ (δˆ np )βα + p,αβ = E0 + X pp αβα β tr [ˆ p δˆ np ] + p = E0 + 2 X p,i 1 2V X tr σ i δˆ np f ij (p, p0 ) tr σ j δˆ np0 pp0 ,ij 2 X i ii ip δnip + δnp f (p, p0 ) δnip0 , V 0 (S.3) pp ,i where E0 denotes the energy of the ground state. In the last line we used the fact that tr[σ i σ j ] = 2δij . (b) For an electric field along the z-direction the quasiparticle (QP) energy matrices have the form µEz ˆp = 0p σ 0 + py σ 1 − px σ 2 . (S.4) ∗ 2m c Thus, together with (S.3), the polarization in the z-direction is obtained as Pz = µ X ∂E = ∗ py δn1p − px δn2p . ∂Ez m c p (S.5) (c) We now look how the QP energies and the distribution of the quasiparticles are changed in linear response to the application of the electrical field and additionally consider the QP interaction. The bare QP energy changes according to ˆp = ˆ0p + δˆ p (S.6) where ˆ0p = 0p σ 0 is the bare quasiparticle energy in the absence of the electric field and δˆ p = µEz 1 2 p σ − p σ . y x 2m∗ c (S.7) The components of the dressed QP energy are different from the bare values due to the QP interaction and will change according to δ˜ ip = δip + 2 X ii f (p, p0 )δnip0 . V 0 (S.8) p We can relate the change in the distribution function to the change in the QP energies in linear response ∂n0 i δnip = 0 δ˜ = −δD (0p − F )δ˜ ip . (S.9) ∂p p The solution of the coupled Eqs. (S.8,S.9) can be found by using the ansatz δ˜ ip = αδip where α contains all the contributions from the QP interaction. Because  i = 1;  py , 1 µE µE z z 1 2 i −p , i = 2; δip = tr p σ − p σ σ = × x y x 2 2m∗ c 2m∗ c  0, i = 0, 3; 2 (S.10) (S.11) it is clear that only the i = 1, 2 components of δ˜ ip will not vanish. Next we note that (as in the lecture) the dependence of fˆσσ0 (p, p0 ) on p and p0 can be reduced to the relative angle θpˆ ,ˆp0 by assuming spherical symmetry of the system. For the momenta pj appearing in δip we also only have to consider the dependence on the solid angle, |p| ≈ ~kF . Therefore, 2 X a 2 X a fpp0 δn1,2 f (cos θpˆ ,ˆp0 )δD (0p0 − F )δ1,2 0 = −α p p0 V 0 V 0 p p Z 02 0 Z p dp 0 = −2α δD (p0 − F ) dΩ0 f a (cos θpˆ ,ˆp0 )δp1,2 0 (2π~)3 Z dΩ0 a f (cos θpˆ ,ˆp0 )δ1,2 . = −αN (F ) p0 4π where N (F ) = m∗ kF π 2 ~2 (S.12) (S.13) (S.14) is the density of states at the Fermi energy. Now we use again the expansion in spherical harmonics. We remember from the lecture f a (cos θpˆ ,ˆp0 ) = X fla Pl (cos θpˆ ,ˆp0 ) = X l fla l 4π X ∗ 0 0 Y (θ , φ )Ylm (θ, φ). 2l + 1 m lm (S.15) We also note that, since the momenta only depend on the solid angle, P which can be written in terms of spherical harmonics of degree l = 1, we have δip = m bm Y1m . Using the orthogonality relation, we then find that Z X Ylm (θ, φ) Z X X dΩ0 a a ∗ 0 b = f (θ0 , φ0 )Y1m0 (θ0 , φ0 ) (S.16) f (cos θpˆ ,ˆp0 )δ1,2 dΩ0 Ylm m l p0 4π 2l + 1 m l m0 X Ylm (θ, φ) X X bm0 = fla δ1l δmm0 (S.17) 2l + 1 m l m0 fa X fa = 1 bm Y1m (θ, φ) = 1 δip . (S.18) 3 m 3 Collecting all results and plugging them into (S.8), we finally find α=1−α F1a 3 ⇒ α= 1 1+ F1a 3 . (S.19) (d) Using the expression (S.5) we find for the polarization in a similar way Z µ N (F )V dΩ Pz = − ∗ α py δ1p − px δ2p m c 2 4π µ 2 E N ( )V Z dΩ z F 2 2 = − α p + p y x m∗ c 2 2 4π | {z } (S.20) (S.21) 2 sin2 θ =~2 kF Z π µ 2 E N ( )V z F 2 2 2π = − α~ k dθ sin3 θ F m∗ c 2 2 4π 0 | {z } (S.22) = 23 = − µ2 N a Ez , ∗ 2 2m c 1 + F1 3 3 (S.23) where N V = 3 kF 3π 2 with N the total number of particles. The susceptibility is then given by χ=− F V N (F ) µ2 µ2 N . a = − F Fa ∗ 2 2m c 1 + 1 m∗ c2 3 1+ 1 3 3 The response is dielectric. 4 (S.24)