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Suppose we are given a particle-hole symmetric Hamiltonian H.
This type of Hamiltonian is usually diagonalized by a Bogoliubov transformation. In this context in the literature the local electron operator is either expanded in terms of eigenmodes at energies E as $$ \Psi_{\sigma}(x)=\sum_{E>0}u_{\sigma, E}(x) \ \gamma_{E} + v^{*}_{\sigma, E}(x) \ \gamma^{\dagger}_{E} $$ with $\sigma=\uparrow,\downarrow$ and $v^{*}_{\sigma, E}=u_{\sigma, -E}$ or as $$ \Psi_{\uparrow}(x)=\sum_{E>0}u_{E}(x) \ \gamma_{\uparrow, E} + v^{*}_{E}(x) \ \gamma^{\dagger}_{\downarrow, E} $$ $$ \Psi_{\downarrow}(x)=\sum_{E>0}-v^{*}_{E}(x) \ \gamma_{\uparrow, E} + u_{E}(x) \ \gamma^{\dagger}_{\downarrow, E} $$ with $v^{*}_{E}=u_{-E}$. So in one case the spin is included in the coherence factors and in the other case it is include in the operators. I am trying to understand why these two formulations are equivalent.

First both expansions seem reasonable to me since for some state $|\Psi\rangle$ and with $\mathcal{P}$ being the particle hole symmetry operator $$ \mathcal{P}\gamma^{\dagger}_{E}|\Psi\rangle=\gamma^{\dagger}_{E}|\Psi\rangle=e=\gamma_{-E}|\Psi\rangle $$ and also $$ \mathcal{P}\gamma^{\dagger}_{\uparrow, E}|\Psi\rangle=\gamma^{\dagger}_{\downarrow,E}|\Psi\rangle=e=\gamma_{\downarrow,-E}|\Psi\rangle $$ $$ \mathcal{P}\gamma^{\dagger}_{\downarrow, E}|\Psi\rangle=-\gamma^{\dagger}_{\uparrow,E}|\Psi\rangle=e=-\gamma_{\uparrow,-E}|\Psi\rangle $$ Nevertheless I am still somewhat confused why the two formulations are equivalent. In particular because the extra "spin"-degree of freedom in the second formulations has doubled the size of the Hilbert space. I suppose this redundancy is removed by identifying two states that differ by the application of the particle hole symmetry operator?

I would be very happy if someone could clarify the equivalence for me.

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  • $\begingroup$ I do not understand this question. The representations are not equivalent, they are the same, aren't they ? A Bogoliubov transformation is a canonical transformation (i.e. preserving the anti-commutation relations between the $\gamma$'s) which diagonalises the Bogoliubov-deGennes Hamiltonian. Please do that once and you will see there is no conceptual difficulty, only rhetorical ones : many presentations forget that the s-wave superconductivity is spin-degenerate. $\endgroup$ – FraSchelle Oct 6 '15 at 21:44

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