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?


1 Answer 1


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.

  • $\begingroup$ $\Delta$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.