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I am reading this article: Edge superconducting correlation in the attractive-U Kane-Mele-Hubbard model.

Considering just the first part of the article, where a negative-U Hubbard model with the intrinsic Spin-Orbital term is considered on the infinite single layer graphene. In deriving the gap equation and the so-called number equation. The author assumed that the superconductivity order parameter in sub-lattice A and sub-lattice B is equal. which is $$\Delta_A=\frac{U}{N}\sum_k\langle a_{-k\downarrow}a_{k\uparrow}\rangle=\Delta_B=\frac{U}{N}\sum_k\langle b_{-k\downarrow}b_{k\uparrow}\rangle=\Delta$$

However, according to my calculation, using the mean-field Hamiltonian with $\Delta_A=\Delta_B=\Delta$ in the article, I found that $\Delta_A \neq \Delta_B$. The gap-equation in the paper can only obtained when I let $\Delta=\frac{\Delta_A+\Delta_B}{2}$; similar to number equation, I have to average $n_a$ and $n_b$.

Because this is a rather lengthy calculation, I am not 100 percent sure that my calculation is right. I confirm myself that the Spin-Orbital term is not invariant under exchange of A and B, so there is no A,B sub-lattice symmetry.

So my question is: am I right? If so, is it proper in the article to let the order parameter on the two sub-lattice equal?

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1 Answer 1

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In graphene, there are symmetry elements which connect sublattices. For instance, there are few 180 degrees rotations which exchange $A$ and $B$. In principle, applying this transform you should get a connection between $\Delta_A$ and $\Delta_B$. Naively, they should be equivalent. Probably, up to the sign, but the sign also depends on a particular choice of the basis. By the way, why you call $\Delta$ order parameter?

I must admit, it is hard to follow the article because authors a bit too freely mix lattice representation with k-space representation and it needs time to understand the exact meaning of connection between them in the cited paper.

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  • $\begingroup$ $\Delta$ is the superconductor order parameter. Is there anything improper about this statement? $\endgroup$ Mar 13, 2014 at 1:11
  • $\begingroup$ I think there is a reason why they do not use this term in the article. To some extent it is. Maybe. $\endgroup$
    – Misha
    Mar 13, 2014 at 9:13

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