A question on the doped Kitaev-Heisenberg model?

Question

Recently, some groups have studied the effects of doping the Kitaev model on honeycomb lattice(e.g.,http://arxiv.org/abs/1109.6681 and http://arxiv.org/abs/1109.4155) and their calculations show the existence of a topological p-wave superconducting phase.

My question is as follows: As we know,one of the amazing properties of the Kitaev model is that the low energy physics(ground states properties and excitations above it) of the model does not depend on the signs of the coupling constants in the Hamiltonian. So for the (hole)doped case, is the existence of topological p-wave superconducting ground states still insensitive to the signs of the couplings we choose?

Thanks in advance.

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

Answer 1

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

Comments

:Haha, it's really you(or You), I know you, thanks very much, I have learned much from your comment.

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

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In Kitaev's original paper, from the point of view of exactly solvability, the insensitivity to the signs can be derived from the local $Z_2$ transformations of Majorana operators. So I want to know whether this sign independent phenomena is general for the (class of exactly soluble) quantum compass models ? And what is the underlying mechanism for the phenomena of sign insensitivity ?

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

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But in the other paper arXiv:1109.6681, under the $U(1)$ slave-boson approach, the authors present a time-reversal invariant(TRI) topological SC, so whether it's TRI or not depends on what method we use?

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

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@K-boy In our paper, we have mentioned that in the low doping limit, the physics is governed by the spin liquid, while for large enough doping, the topological order will be destroyed, and give way to the fermi liquid physics. The difference between our work and arXiv:1109.6681 is that we focus on the spin liquid side, and their approach is from the fermi liquid side. Without the protection by the topological order, the resulting SC can be TRI or TRB depending on the details.

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

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@K-boy I have not thought of the reason for the sign insensitivity. I tended to think that it is an accident. But anyway your question is good and worth thinking about.

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

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@ Everett You Ok, I see. I will keep it as a question in my mind.

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