Abstract
We examine the stability of mixed-symmetry superconducting states with broken time-reversal symmetry in spatial-symmetry-broken systems, including chiral states, on the basis of the free-energy functional derived in the weak-coupling theory. We consider a generic a_1 + i a_2 wave state, with a_1 and a_2 being different symmetry indices such as (a_1,a_2) = (d,s), (p_x,p_y), and (d,d').The time-reversal symmetry of the mixed-symmetry state with the a_1- and a_2-wave components is broken when the phases of these components differ, and such a state is called the time-reversal-symmetry breaking (TRSB) state. However, their phases are equated by Cooper-pair scattering between these components if it occurs; i.e., when the off-diagonal elements S_{a_1 a_2} = S_{a_2 a_1} of the scattering matrix are nonzero, they destabilize the TRSB state. Hence, it has often been believed that the TRSB state is stable only in systems with a spatial symmetry that guarantees S_{a_1 a_2}=0. We note that, contrary to this belief, the TRSB state can remain stable in systems without the spatial symmetry when the relative phase shifts so that S_{a_1 a_2} = 0 is restored, which results in a distorted TRSB (a_1 + a_2) + i a_2 wave state. Here, note that the restoration of S_{a_1 a_2} = 0 does not imply that the symmetry of the quasi-particle energy E_k is recovered. This study shows that such stabilization of the TRSB state occurs when the distortion is sufficiently small and \Delta_{a_1} \Delta_{a_2} is sufficiently large, where \Delta_a is the amplitude of the a-wave component in the TRSB state in the absence of the distortion. We clarify the manner in which the shift in the relative phase eliminates S_{a_1 a_2} and prove that such a state yields a free-energy minimum. We also propose a formula for the upper bound of the degree of lattice distortion, below which the TRSB state can be stable.
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