How to write BdG Hamiltonian in graphene?

In Beenakker's paper:Specular Andreev Reflection in Graphene, the BdG Hamiltonian is written as:

$$H_{BdG}=\begin{pmatrix}H-E_F&\Delta\\ \Delta^*& E_F-H\end{pmatrix}$$ from equation (1).

Where $H$ is the Hamiltonian of pure graphene and it is:

$$H=\begin{pmatrix}H_+&0\\ 0& H_-\end{pmatrix}$$

Where $\pm$ denotes different valleys and:

$$H_{\pm}=-i\hbar v(\sigma_x\partial_x\pm\sigma_y\partial_y)$$

Moreover, $H$ is written in the basis of four dimensional spinor $(\Psi_{A+},\Psi_{B+},\Psi_{A-},\Psi_{B-})$

Question is, what basis is $H_{BdG}$ written in? What does the $4\times4$ matrix of $\Delta$ looks like?

Finnally, why the original $8\times8$ BdG equation can be valley decoupled like this (Equ.7 in the paper):

$$\begin{pmatrix}H_\pm-E_F&\Delta\\ \Delta^*& E_F-H_\pm\end{pmatrix}{u\choose v}=\epsilon {u\choose v}$$

It is a bit strange because paring is bewteen two valleys, how can we decouple the equation into two seperate valleys?

It is a very beautiful paper! But as all the old Physical Review Letters a bit cryptic, the supplementary material in the arXiv (http://arxiv.org/pdf/cond-mat/0604594v3.pdf) version is helping a bit.

In the full $8\times8$ Hamiltonian, electrons from valley K are coupled with holes from valley K' via the proximized superconducting coupling $\Delta$, the same is true for electrons from valley K' that are coupled with holes in K. However, there are no other mechanisms that are coupling the two valleys. It means that the $8\times8$ Hamiltonian is composed of two identical $4\times4$ blocks that are not coupled to each other. Here it is assumed that electrons and holes in opposite valleys have also opposite spin so to fulfill $s$-wave pairing.

Regarding your specific question, I think that the $4\times4$, coupling matrix should look like: $$\left(\begin{matrix} 0 & 0 & \Delta & 0 \\ 0 & 0 & 0 & \Delta \\ \Delta & 0 & 0 & 0 \\ 0& \Delta& 0 & 0 \end{matrix}\right).$$

In this way you are coupling an electron $\Psi_\text{A+}^\dagger$ with $\Psi_{A-}^\dagger$. This is what you would expect from a mean field $s$-wave pairing.

You can find a similar calculation done with all the details in the supplementary material of the experimental verification of specular Andreev reflection in bilayer graphene: D. K. Efetov et al., Specular interband Andreev reflections at van der Waals interfaces between graphene and NbSe2. Nat. Pays. (2015). http://doi.org/10.1038/nphys3583 or http://arxiv.org/abs/1505.04812.

• I don't think you are correct at the form of the matrix $\Delta$. In the article and the new article you cited, $\Delta$ is treated as a scarlar. Therefore, the matrix should be an identity matrix. Could you give the explicit basis and the corresponding form of $\Delta$ matrix to justify your results? – an offer can't refuse Mar 30 '16 at 10:57
• I think we are both correct, indeed the matrix of the $\Delta$ is diagonal in the subspace with only electrons from valley K and holes from the time-reversed valley K'. The basis in the BdG should read $(\Psi_\text{A+},\Psi_\text{B+},\Psi_\text{A-},\Psi_\text{B-}$,$\Psi_\text{A+}^{\dagger},\Psi_\text{B+}^\dagger,\Psi_\text{A-}^\dagger,\Psi_\text{B-}^\dagger)$ – Dario Bercioux Mar 30 '16 at 11:11
• Please see my answer. – an offer can't refuse Apr 2 '16 at 2:13

According to @Bercioux answer, if we choose the basis: $$\phi_1=(\Psi_{A+},\Psi_{B+},\Psi_{A-}，\Psi_{B-},\Psi_{A+}^\dagger,\Psi_{B+}^\dagger,\Psi_{A-}^\dagger,\Psi_{B-}^\dagger)$$

The BdG Hamiltonian should be written like this: $$H_{BdG}^1=\begin{pmatrix}H_+-E_F&0&0&\Delta_2\\0&H_--E_F&-\Delta_2&0\\ 0&-\Delta^*_2&E_F-H_+&0\\ \Delta_2^*&0&0&E_F-H_- \end{pmatrix}$$ where $\Delta_2=\Delta I_2$, and note the that the minus sign is crucial.

This gives two decoupled $4\times4$ Hamiltonian, however the valley is not decoupled as in the paper: $$\begin{pmatrix}H_+-E_F&\Delta_2\\ \Delta_2^*&E_F-H_-\end{pmatrix}\text{ and } \begin{pmatrix}H_--E_F&-\Delta_2\\ -\Delta_2^*&E_F-H_+\end{pmatrix}$$

If we choose another basis: $$\phi_2=(\Psi_{A+},\Psi_{B+},\Psi_{A-}，\Psi_{B-},\Psi_{A-}^\dagger,\Psi_{B-}^\dagger,-\Psi_{A+}^\dagger,-\Psi_{B+}^\dagger)$$ The BdG Hamiltonian can be simply written as:

$$H_{BdG}^2=\begin{pmatrix}H-E_F&\Delta_4\\ \Delta_4^*&E_F-H\end{pmatrix}$$

where $\Delta_4=\Delta I_4$, the Hamiltonian can be easily valley decoupled, resulting equation (7) in the main text.