Let us assume that we have a Hamiltonian of the form:

$$ H = \sum_{k,\sigma,s}\epsilon_{\sigma s}\left(k\right)c_{k\sigma s}^{\dagger}c_{k\sigma s} + \sum_{k,s}\Delta_{0}\left(k\right)c_{k\uparrow s}^{\dagger}c_{-k\downarrow s}^{\dagger} + \Delta_{1}\left(k\right)c_{k\uparrow s}^{\dagger}c_{-k\downarrow -s}^{\dagger} + \text{ h.c.} $$

This is, for example, the case in graphene with a $d$-wave pairing term, where s stands for conduction or valence band electrons (see arXiv:1406.0101).

In the case of BCS Hamiltonian

$$ H_{\text{BCS}} = \sum_{k,\sigma}\epsilon_{\sigma}\left(k\right)c_{k\sigma}^{\dagger}c_{k\sigma} + \sum_{k,s}\Delta\left(k\right)c_{k\uparrow}^{\dagger}c_{-k\downarrow}^{\dagger} + \text{ h.c.} $$

We have a Bogoliubov transformation $U\in\text{SU}\left(2\right)$ of the form

$$ U = \begin{pmatrix} u & -v \\ v^{\star} & u^{\star} \end{pmatrix} $$ with $\left|u\right|^{2} + \left|v\right|^{2} = 1$, which brings the BCS Hamiltonian $H_{\text{BCS}}$ in a diagonal form. Further, the relation $\left|u\right|^{2} + \left|v\right|^{2} = 1$ can be derived by the fact that the new quasiparticles must obey anticommutation relations.

My question is: How I must modify the Bogoliubov transformation to diagonalize the Hamiltonian $H$ with the two pairing terms? I know one way is to write the Hamiltonian as a Bogoliubov - de Gennes Hamiltonian and then diagonalize the matrix. However, the problem is that the transformation must belongs to $U\left(4\right)$ and I am very interested in the transformation itself and not only the eigenenergies. Is there somebody how had some experience how to diagonalize the Hamiltonian $H$?

  • $\begingroup$ You just wrote the mean-field BCS Hamiltonian in term of spin, but your $u$ and $v$ do not contain any spin. In the general expression, $u$ and $v$ are spin matrices, that's the way to do it for higher order angular momentum of the Cooper pair. Accordingly, you do not really have two pairing term, but your $\Delta$ matrix gets a spin structure. For pure s-wave, one can take a scalar for $\Delta$ in the spin space. $\endgroup$ – FraSchelle Apr 20 '16 at 13:03
  • $\begingroup$ Oh, sorry, I misread your Hamiltonian. I do not know what $s$ is in your case, but replace spin in my previous comment by s, whatever degree of freedom it is. It is not clear to me whether this degree of freedom has a finite number of realisation (like 2 for a 1/2 spin), but you have to attach a s meaning to your $u$ and $v$ for sure. $\endgroup$ – FraSchelle Apr 20 '16 at 13:05
  • $\begingroup$ Yes, $s$ has a finite number of realisation ($s = \pm 1$). In the case of graphene $s = +1$ is the case where $c$ is a fermionic operator for conduction band and $s = -1$ then a fermionic operator for valence band. $\endgroup$ – Lars Milz Apr 20 '16 at 13:10
  • $\begingroup$ So it goes exactly like for the spin, say you will have a $\sigma$ su(2) algebra for spin, a $\tau$ su(2) algebra for the particle-hole and a $\Sigma$ (for your s) su(2) algebra for conduction and valence band. But it is not completely clear whether $s$ has some particle hole meaning as well. It's strange to me to keep both particle-hole and conduction-valence, but why not ? $\endgroup$ – FraSchelle Apr 20 '16 at 13:14
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    $\begingroup$ @LarsMilz, the paper by Suhl et al., Phys. Rev. Lett. 3, 552 (1959) has the solution to the two-band BCS theory. Not sure if this is the same as your problem. $\endgroup$ – leongz Apr 21 '16 at 3:16

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