As the first author of arXiv:1109.4155, my answer to this question is yes. The topological p-wave SC state is insensitive to the sign of the coupling. The argument provided in our paper is quite general, the time-reversal broken SC state is supported by the underlying topological order in the Kitaev spin liquid, as described by the particular spin-gauge locking PSG, which is not going to change when we reverse the sign of the Kitaev coupling.

As you might have known, that the Kitaev honeycomb model is exact solvable by introducing four Majorana fermions $\chi^{0,1,2,3}$ on each site, such that the spin operator can be represented as $\vec{S}=\frac{i}{2}(\chi^0\vec{\chi}-\frac{1}{2}\vec{\chi}\times\vec{\chi})$, under the gauge singlet constraint $\vec{K}=\frac{i}{2}(\chi^0\vec{\chi}+\frac{1}{2}\vec{\chi}\times\vec{\chi})=0$. The mean-field Hamiltonian reads
$$H=J_K\sum_{\langle ij\rangle}(u_{ij}^a i\chi_i^0\chi_j^0+u_{ij}^0 i\chi_i^a\chi_j^a-u_{ij}^au_{ij}^0),$$
from which we can see that the sign of $J_K$ does not affect the structure of the mean-field ansatz. Or put it explicitly, under $J_K\to-J_K$, one only needs to transform $u_{ij}^0\to u_{ij}^0$, $u_{ij}^a\to -u_{ij}^a$, $\chi_i^0\to \chi_i^0$, $\chi_i^a \to(-)^i\chi_i^a$ (here $(-)^i$ stands for the minus sign on one sublattice), then the Hamiltonian is invariant. Such a transformation only changes the global sign of one set of the mean-field ansatz, so it will not affect the PSG classification.

The most prominent character of the PSG for the Kitaev spin liquid is an effect we called the spin-gauge locking. Note that the four Majorana fermions can transform under the $O(4)$ group, which factorizes into $O(4)\simeq SU(2)_\text{spin}\times SU(2)_\text{gauge}$. In the mean-field Hamiltonian given above, one can see that the $\chi^0$ fermion has a band structure that is totally different from that of the rest fermions $\chi^{1,2,3}$, therefore the $O(4)$ structure is completely broken. Thus to preserve the mean-field ansatz, any $SU(2)_\text{spin}$ rotation must be followed by the same $SU(2)_\text{gauge}$ rotation, i.e. the spin-gauge locking, which is an effect independent to the sign of $J_K$ obviously.

If the gauge structure is not broken, then the spin rotation symmetry is also preserved, which is just the case of the Kitaev spin liquid ground state. But as we introduce doping into the system, the holons (in the $SU(2)$ slave boson language) carries the gauge charge. As they condense at low temperature (which means the system goes superconducting), they necessarily break the $SU(2)_\text{gauge}$ structure, and simultaneously break the $SU(2)_\text{spin}$ symmetry as well, due to the spin-gauge locking effect. Thus the resulting superconducting state must break the time reversal symmetry, can becomes a topological superconductor.

In conclusion, the topological SC state is a consequence of the topological order (the spin-gauge locking effect) hidden in the Kitaev spin liquid, reversing the sign of $J_K$ does not change the topological order at all, thus will not affect the resulting SC state.

This post imported from StackExchange Physics at 2014-03-09 08:46 (UCT), posted by SE-user Everett You