  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.
