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?


Short answer: The particle-hole symmetry of the superconductor BdG Hamiltonian is not a real hysical symmetry of the system, but a manufactured one, resulting from doubling the degrees of freedom. It has to be accounted for by either counting only half the degrees of freedom or dividing by half all results.

Long anwser: The symmetry is just not there in the original Hamiltonian. Let's try it, as you say, with a usual free-particle Hamiltonian $H = \sum_k \epsilon_k c^{\dagger}_k c_k$ and now artificially write it as $$H = \frac{1}{2}\sum_k ( \epsilon_k c^{\dagger}_k c_k - \epsilon_k c_k c^{\dagger}_k ) + {\rm const.}$$ where we just used the anti-commutation relations, paying the price of some constant energy which we don't care about. Now let's say we decide to treat creation and annihilation as independent degrees of freedom. This is not true, but we can decide to do so, and write the Hamiltonian in matrix form just like for the BdG $$ H = \frac{1}{2}\left( c^{\dagger}_k c_k \right) \left(\begin{array}{cc} \epsilon_k & 0 \\ 0 & -\epsilon_k\end{array}\right) \left( \begin{array}{c} c_k \\ c^{\dagger}_k \end{array}\right)$$ and suddenly we can talk about particle-hole symmetry etc. But this is patently wrong - the positive energy spectrum and the negative energy spectrum relate the same degrees of freedom! The only different is whether the state is filled or empty.

Normally we associate with each pair of states in Fock space, $|0\rangle$ and $|1\rangle$, reflecting filled and empty level, one pair of annihilation and creation operators $d, d^{\dagger}$. In the BdG Hamiltonian, we treat $dd^{\dagger}$ and $d^{\dagger}d$ as two independent entities. The symmetry is not really there.

The reason we do it is that due to the SC term $\Delta$ this is a convenient option that allows us to easily diagonalize the Hamiltonian. However, we must account for this, and therefore when we calculate expectation values we either count over all the BdG degrees of freedom and divide by half at the end, or only count half of the degrees of freedom (usually choosing just $E>0$ or something like that).

| cite | improve this answer | |
  • $\begingroup$ Thanks for your answer. Here I'd like to add a comment on "counting half of the degrees of freedom". If we solve the superconducting Hamiltonian with Bogoliubov transformation (instead of constructing a BdG one), we can diagonialize it with quasi-particle excitations (say, bogolons). And after diagonalization, we define a ground state which corresponds to the vacuum of bogolon (i.e. becomes zero if a bogolon annihilation operator acts on it). This is nothing but that equivalent to the choice of reserving only E>0 states in the BdG scheme. $\endgroup$ – Siqi Wu Oct 21 at 7:09

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.