content_image/439295.jpg content_image/439296.jpg content_image/439297.jpg content_image/439300.jpg content_image/439312.jpg content_image/439311.jpg content_image/439299.jpg content_image/439293.jpg content_image/439294.jpg content_image/439298.jpg content_image/439309.jpg content_image/439310.jpg content_image/439313.jpg
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 

<img src='content_image/439295.jpg'>

Using the identity 

<img src='content_image/439296.jpg'>

where ˆ γ$_{l}$ = γ$_{l}$/ | γ | , one can rewrite the above expression as 

<img src='content_image/439297.jpg'>

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 

<img src='content_image/439300.jpg'>

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 

<img src='content_image/439312.jpg'>

Let us find first the gyrotropy conductivity in the nor- mal state. 

<img src='content_image/439311.jpg'>

<img src='content_image/439299.jpg'>

## III. GYROTROPY CONDUCTIVITY IN THE NORMAL STATE 

Substituting the Green function in the eqn. (20) and performing summation over the Matsubara frequencies, we obtain 

<img src='content_image/439293.jpg'>

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 

<img src='content_image/439294.jpg'>

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 . 

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 $^{2}$: 

<img src='content_image/439298.jpg'>

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 

<img src='content_image/439309.jpg'>

where 

<img src='content_image/439310.jpg'>

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 

<img src='content_image/439313.jpg'>

