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I want to generalize BdG equation in order to compute the conductance of a junction of graphene with a metal superconductor. The previous works done until now on this hetrojunction is devotted to use proximity induced superconductivity on the graphene sheet. So both side of junctions are governed by Dirac equation. The question arises is that: when a metal is conducted to a graphene sheet directly, in a normal graphene side, because of 2*2 Dirac Hamiltonian the resulted BdG equation have a 4 component spinor just like the Beenakker's result C. W. J. Beenakker, Phys. Rev. Lett. 97, 067007 (2006) / arXiv:cond-mat/0604594

But from the other side, where a superconductor with Schrodinger like Hamiltonian is placed, we have a two components spinor for quasi particle excitations. So I don't know what should I do about boundary conditions of this wave function in order to compute the conductance of junctions.

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  • $\begingroup$ The problem is well discussed by Beenakker in his paper : you inject the one-body Hamiltonian $H$ (eq.4 on the arXiv version, it's a $4\times 4$ matrix) in the BdG $2\times 2$-matrix (eq.1 of the arXiv version) such that you have a $8\times 8$ eigenvalues problem to resolve. Nevertheless, the valley can be switched by applying the time-reversal operator, so you need only $4$-components to resolve the problem. Since the differential equation is of first-order, you need only to match the wave-functions, not the derivative. $\endgroup$
    – FraSchelle
    Commented Nov 3, 2014 at 21:46
  • $\begingroup$ For two interesting papers on the relativistic version of BdG equation, see dx.doi.org/10.1103/PhysRevB.59.7140 and dx.doi.org/10.1103/PhysRevB.59.7155 by Capelle & Gross, PRB (1999). $\endgroup$
    – FraSchelle
    Commented Nov 3, 2014 at 21:49
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    $\begingroup$ Well, reading again your question, I realised that what you did (or want to do) is wrong, Beenakker is right. You should replace "conducted" by "contacted" in your question I guess. When a superconductor (S) is "contacted" to graphene (G), S becomes chiral in a thin layer at the interface which is the only one which matters, and G becomes particle-hole symmetric all over. (say differently, G is particle-hole redundant, whereas S is valley redundant, so everything is fine: make a global theory, with spin it's $8\times 8$ matrix) $\endgroup$
    – FraSchelle
    Commented Nov 3, 2014 at 22:00
  • $\begingroup$ Finally, I note that this question is the same as a previous one of yours: physics.stackexchange.com/q/144517/16689 which has been already perfectly answered by @danu ! $\endgroup$
    – FraSchelle
    Commented Nov 3, 2014 at 22:03
  • $\begingroup$ So thanks @FraSchelle for those interesting papers. I am going to read these. $\endgroup$
    – M Salehi
    Commented Nov 4, 2014 at 6:08

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