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For my system I can write down the Hamiltonian in this form:

$$ H = \begin{pmatrix} \epsilon_{1\downarrow}-\mu_{B}B & 0 & 0 & 0 \\ 0 & \epsilon_{2\uparrow}+\mu_{B}B & 0 & 0 \\ 0 & 0 & \epsilon_{1\uparrow}+\mu_{B}B & 0 \\ 0 & 0 & 0 & \epsilon_{2\downarrow}-\mu_{B}B \end{pmatrix} $$

where $\mu_{B}B$ are the Zeeman split.

Now I want to bring this system in contact with a superconductor and want to write down the Bogoliubov - de Gennes Hamiltonian. In my opinion the Bogoluibov - de Gennes Hamiltonian has this form since the Zeeman energy does not dependent that I have quasiparticle or holes:

$$ \mathcal{H} = \begin{pmatrix} \epsilon_{1\downarrow}& 0 & 0 & 0 \\ 0 & \epsilon_{2\uparrow} & 0 & 0 \\ 0 & 0 & \epsilon_{1\uparrow} & 0 \\ 0 & 0 & 0 & \epsilon_{2\downarrow} \end{pmatrix}\tau_{z} + \begin{pmatrix} -\mu_{B}B & 0 & 0 & 0 \\ 0 & +\mu_{B}B & 0 & 0 \\ 0 & 0 & +\mu_{B}B & 0 \\ 0 & 0 & 0 & -\mu_{B}B \end{pmatrix}1_{\tau} + H_{\Delta}\tau_{x} $$

Without Zeeman term the problem is a standard textbook problem, since I have directly my Nambu spinor $\psi^{\dagger} = \left(c^{\dagger}_{1\downarrow},c^{\dagger}_{2\uparrow},c^{\dagger}_{1\uparrow},c^{\dagger}_{2\uparrow},c_{1\downarrow},c_{2\uparrow},c_{1\uparrow},c_{2\uparrow}\right)$ and can write down the Hamiltonian in the correct second quantization notation.

But if I add the Zeeman energy in my second quantization Hamiltonian vanish my Zeeman term with this Nambu spinor. If a write in my Bogoliubov - de Gennes

$$ \begin{pmatrix} -\mu_{B}B & 0 & 0 & 0 \\ 0 & +\mu_{B}B & 0 & 0 \\ 0 & 0 & +\mu_{B}B & 0 \\ 0 & 0 & 0 & -\mu_{B}B \end{pmatrix}\tau_{z} $$

I get the correct results but this means that I will have a sign flip in my Bogoliubov - de Gennes Hamiltonian for quasiparticles an holes, which is in my opinion wrong!

My question now is how I must changes my Nambu spinor so that my Zeeman term in my Hamiltonian will do not vanish or can I have a sign flip for quasiparticles and holes in the Zeeman energy?

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Sorry, I found my mistake. The sign in the Zeeman term must flip for the holes due to the particle-hole symmetry. –  user27964 Mar 4 at 13:44
    
The best is obviously to start from a normal form of the Hamiltonian, in your case something like $$H\sim c^{\dagger}\left(\dfrac{p^{2}}{2m}-\mu+h\sigma_{z}\right)c+\Delta c^{\dagger}c^{\dagger}+\Delta^{\ast}cc$$ and to reproduce the de Gennes calculation. Some of the details are there: physics.stackexchange.com/questions/77298 The point is -- for an unknown reason -- teacher tell to the students that the Bogoliubov-deGennes Hamiltonian describes superconductivity. This is merely wrong, the mean-field Hamiltonian I wrote above does it, the BdG is just a formal rewriting of it. –  FraSchelle Mar 5 at 8:54
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Note also that your Nambu spinor is twice too large. Since particle and holes are coupled in the BdG description, your $c_{1,2}$ notation seems redundant for me. Consider also giving us your detailed solution, it may help others having the same problem. Your previous comment is merely wrong: the BdG representation needs both the matrix and the vector to be defined. Particle-hole symmetry means nothing if you gave us only the matrix. Have fun. –  FraSchelle Mar 5 at 9:00
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No, Nambu spinor is correct. I consider the proximity-induced effect in Carbon nanotubes and the 1 and 2 means the different valleys. In my opinion the answer is, that if I do a transformation from particle to hole and vice versa than this does not change the spin of the particle/hole. Means particle-hole transformation preserved the spin. Now if the Zeeman splitting shift the particle energy down must then for the hole the Zeeman splitting must shift the energy up. I hope this is clear. –  user27964 Mar 5 at 17:59

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