3 III. GYROTROPY CONDUCTIVITY IN THE NORMAL STATE Substituting the Green function in the eqn. (20) and performing summation over the Matsubara frequencies, we obtain can produce only the terms proportional to even powers of the product kq being not responsible for the gyrotropy properties. The off diagonal elements of ˆ τ ( k ) have the mixed par- ity. So, the dispersive terms proportional to the odd powers of the product kq can arise only from the part of eqn. (12) containing the off diagonal elements of ˆ τ ( k ) matrices. Hence, for the calculation of gyrotropy of con- ductivity, only the last two lines in eqn. (12) consisting of interband terms are important. They are equal to Here f ( ξ$_{±}$ ( k$_{±}$ )) is the Fermi distribution function and k$_{±}$ = k ± q / 2. By changing the sign of momentum k → − k in the first term under integral and making use that ξ$_{λ}$ ( k ) is even function of k , we come to Using the identity where ˆ γ$_{l}$ = γ$_{l}$/ | γ | , one can rewrite the above expression as Analytical continuation of this expression from the dis- crete set of Matsubara frequencies into entire half-plane ω > 0 is performed by the usual substitution iω$_{n}$ → ω + i/τ . I$_{l}$ = ∫ d $^{3}$k (2 π ¯ h ) $_{3}$ˆ k$_{l}$ { f ( ξ$_{−}$ ( k$_{−}$ )) − f ( ξ$_{+}$ ( k$_{+}$ )) i ¯ hω$_{n}$ + ξ$_{−}$ ( k$_{−}$ ) − ξ$_{+}$ ( k$_{+}$ ) − − f ( ξ$_{+}$ ( k$_{−}$ )) − f ( ξ$_{−}$ ( k$_{+}$ )) i ¯ hω$_{n}$ + ξ$_{+}$ ( k$_{−}$ ) − ξ$_{−}$ ( k$_{+}$ ) } . (21) I$_{l}$ = 2 ∫ d $^{3}$k (2 π ¯ h ) $_{3}$ˆ k$_{l}$ [ ξ$_{+}$ ( k$_{+}$ ) − ξ$_{−}$ ( k$_{−}$ )][ f ( ξ$_{+}$ ( k$_{+}$ )) − f ( ξ$_{−}$ ( k$_{−}$ ))] ( ξ$_{+}$ ( k$_{+}$ ) − ξ$_{−}$ ( k$_{−}$ )) 2 − ( i ¯ hω$_{n}$ ) 2 . (22) ∂γ$_{l}$ ∂k$_{i}$ ∂γ$_{m}$ ∂k$_{j}$ { τ$_{l,}$$_{+}$$_{−}$τ ∗ m, + $_{−}$[ G$_{−}$G$_{+}$ − F$_{−}$F † $_{+}$] + τ$_{l,}$$_{−}$$_{+}$τ ∗ m, − $_{+}$[ G$_{+}$G$_{−}$ − F$_{+}$F † $_{−}$] } . (15) τ$_{l,}$$_{+}$$_{−}$τ ∗ m, + − = τ ∗ l, − $_{+}$τ$_{m,}$$_{−}$$_{+}$ = δ$_{lm}$ − ˆ γ$_{l}$ ˆ γ$_{m}$ + ie$_{lmn}$ ˆ γ$_{n}$, ∂γ$_{l}$ ∂k$_{i}$ ∂γ$_{m}$ ∂k$_{j}$ { ( δ$_{lm}$ − ˆ γ$_{l}$ ˆ γ$_{m}$ )[ G$_{−}$G$_{+}$ + G$_{+}$G$_{−}$ − F$_{−}$F † + (16) − F$_{+}$F † $_{−}$] + ie$_{lmn}$ ˆ γ$_{n}$ [ G$_{−}$G$_{+}$ − G$_{+}$G$_{−}$ − F$_{−}$F † + + F$_{+}$F † $_{−}$] } . I$_{l}$ = 2 ∫ d $^{3}$k (2 π ¯ h ) $_{3}$ˆ k$_{l}$ [ f ( ξ$_{+}$ ( k$_{+}$ )) − f ( ξ$_{−}$ ( k$_{−}$ ))] × { 1 ξ$_{+}$ ( k$_{+}$ ) − ξ$_{−}$ ( k$_{−}$ ) + (¯ hω ) 2 ( ξ$_{+}$ ( k$_{+}$ ) − ξ$_{−}$ ( k$_{−}$ )) 3 } , (23) γ ( k ) = γ$_{0}$ k , (17) γ ( k ) = γ$_{⊥}$ ( k$_{y}$ ˆ x − k$_{x}$ ˆ y ) + γ$_{‖}$k$_{x}$k$_{y}$k$_{z}$ ( k 2 x − k 2 $_{y}$)ˆ z. (18) We shall work at frequencies smaller when the band splitting ¯ hω < γ$_{0}$k$_{F}$ far from the resonance region ¯ hω ≈ γ$_{0}$k$_{F}$ but still in the collisionless limit ωτ > 1 where one can decompose the integrand in powers of ω $^{2}$: Starting this point we need the explicit form of spin- orbit coupling vector γ ( k ). Its momentum dependence is determined by the crystal symmetry. 9,17 For the cu- bic group G = O , which describes the point symmetry of Li$_{2}$(Pd$_{1}$$_{−}$$_{x}$,Pt$_{x}$)$_{3}$B, the simplest form compatible with the symmetry requirements is where γ$_{0}$ is a constant. For the tetragonal group G = C$_{4}$$_{v}$ , which is relevant for CePt$_{3}$Si, CeRhSi$_{3}$ and CeIrSi$_{3}$, the spin-orbit coupling is given by The frequency independent term in eqn. (23) corre- sponds to the current density ν B introduced in eqn. (4). We are interested in linear in q part of density of cur- rent. Expanding the integrand up to the first order in ∂ξ ± ∂ k q one can prove by direct calculation that this term vanishes. Thus, in the normal state ν = 0 as it should be in gauge invariant theory (see Section V and $^{14}$). The frequency dependent term determines the current The gyrotropy current j$_{g}$ , which is linear with respect to the wave vector q , originates from the last term in the eqn. (16). One can show that for the tetragonal crystal with the symmetry group G = C$_{4}$$_{v}$ , for the electric field lying in the basal plane the linear in the component of wave vector q part of conductivity is absent. In that follows we continue calculation for the metal with cubic symmetry where ˆ γ = ˆ k signγ$_{0}$ . We put ˆ γ = ˆ k taking γ$_{0}$ as a positive constant. Thus, we obtain for gyrotropy current 21 Let us find first the gyrotropy conductivity in the nor- mal state. After substitution of the Fourier component of the vec- tor potential by the Fourier component of an electric field A = c E /iω , we obtain where j g i ( ω, q ) = ie$_{ijl}$ e $^{2}$γ2 0 c ¯ hq$_{m}$ (¯ hω ) $^{2}$I$_{lm}$ A$_{j}$ ( ω, q ) , (24) I$_{lm}$ = ∫ d $^{3}$k (2 π ¯ h ) $_{3}$ˆ k$_{l}$ [ − 3 f ( ξ$_{+}$ ) − f ( ξ$_{−}$ ) ( ξ$_{+}$ − ξ$_{−}$ ) 4 ( ∂ξ$_{+}$ ∂k$_{m}$ + ∂ξ$_{−}$ ∂k$_{m}$ ) + + 1 ( ξ$_{+}$ − ξ$_{−}$ ) 3 ( ∂f ( ξ$_{+}$ ) ∂ξ$_{+}$ ∂ξ$_{+}$ ∂k$_{m}$ + ∂f ( ξ$_{−}$ ) ∂ξ$_{−}$ ∂ξ$_{−}$ ∂k$_{m}$ )] . (25) j$_{gi}$ ( ω$_{n}$, q ) = ie$_{ijl}$ e $^{2}$γ2 0 c I$_{l}$A$_{j}$ ( ω$_{n}$, q ) , (19) I$_{l}$ = ∫ d $^{3}$k (2 π ¯ h ) $_{3}$ˆ k$_{l}$ × T ∞ ∑ m = −∞ [ G$_{+}$ ( K$_{+}$ ) G$_{−}$ ( K$_{−}$ ) − F$_{+}$ ( K$_{+}$ ) F † $_{−}$( K$_{−}$ ) − − G$_{−}$ ( K$_{+}$ ) G$_{+}$ ( K$_{−}$ ) + F$_{−}$ ( K$_{+}$ ) F † $_{+}$( K$_{−}$ )] . (20) j g i ( ω, q ) = e$_{ijl}$e $^{2}$¯ h $^{3}$γ 2 $_{0}$ωq$_{m}$ I$_{lm}$E$_{j}$ ( ω, q ) . (26)