It is often mentioned that a mean-field BdG Hamiltonian for a superconductor has the particle-hole symmetry. i.e., if $|\Psi_{\mathbf{k}}\rangle$ is an eigenstate of $H^{BdG}_{\mathbf{k}}$ with eigenvalue of $E_{\mathbf{k}}$, the state $\mathcal{C}|\Psi_{\mathbf{k}}\rangle=|\Psi^*_{\mathbf{-k}}\rangle$ is also an eigenstate of $H^{BdG}_{\mathbf{-k}}$, with an eigenvalue of $-E_{\mathbf{k}}$.
To be more precise, here I write down a simple BdG Hamiltonian of:
$ H^{BdG}=\sum_{k} \begin{pmatrix} c^{\dagger}_{k\uparrow} & c^{\dagger}_{k\downarrow} & c_{-k\uparrow} & c_{-k\downarrow} \end{pmatrix} \begin{pmatrix} \epsilon_{k}&0&0&\Delta\\ 0&\epsilon_{k}&-\Delta&0\\ 0&-\Delta&-\epsilon_{-k}&0\\ \Delta&0&0&-\epsilon_{-k} \end{pmatrix} \begin{pmatrix} c_{k\uparrow} \\ c_{k\downarrow} \\ c^{\dagger}_{-k\uparrow} \\ c^{\dagger}_{-k\downarrow} \end{pmatrix} $
At the limit of $\Delta\rightarrow0$, one can get eigenvectors like $\begin{pmatrix} 0&0&1&0 \end{pmatrix}^T$ (despite the mixing between degenerate states), which describe the excitations of single $c_{-k\uparrow}$, with energy $-\epsilon_{-k}$. And if we sum over $k$ to get the spectrum, we can see that the spectrum is symmetric for $E>0$ and $E<0$. The same result can also be shown without the BdG formula if we take:
$ H=\sum_{k}\epsilon_{k}c^{\dagger}_{k}c_{k}=\frac{1}{2}\sum_{k}[\epsilon_{k}c^{\dagger}_{k}c_{k}+(-\epsilon_{-k})c_{-k}c^{\dagger}_{-k}+Tr(H)] $
So here are my questions:
From the above-mentioned discussions, it seems that we can generalize the Hamiltonian for all systems to get a symmetric spectrum. However, this is not the case for most real solids.
So why does the above discussions invalid for real systems? What does the excitation of $c_{-k}$ mean for real condensed matter systems? Is the case different for superconducting and non-superconducting phases? And, what is the physical effect for such a symmetry?
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