condensed matter - Why does a superconductor obey particle-hole symmetry?

We normally solve the Bogoliubov-de Gennes (BdG) equations in order to compute the energy spectrum of a superconductor. The Nambu spinor is a common object that is used in formulating these equations. In other words, the problem is formulated with the a priori assumption that the superconductors obeys particle-hole symmetry. But how do we know that it will obey particle-hole symmetry? I'm sure that by assuming particle-hole symmetry, computing the spectrum by solving the BdG equations, and comparing with experiments we get a good agreement. So we know it's true. I just don't understand why that is the case. Is there any physical reason behind it?

Answer

The particle-hole symmetry is not exactly a property of the superconductors. It's rather a property of the free electron gas, with the Fermi level -- separating all the empty states above from all the occupied below it -- cutting a parabola. The property to have the spectrum symmetry $\varepsilon_{\mathbf{k}}=\varepsilon_{-\mathbf{k}}$ (see the quotation below for the notations meaning) is mandatory to have a particle-hole symmetry also.

I quote below the book by C. Kittel, Quantum theory of solids (1963), chapter 5.

In treating the electron system it is particularly convenient to redefine the vacuum state as the filled Fermi sea, rather than the state with no particles present. With the filled Fermi sea as the vacuum we must provide separate fermion operators for processes which occur above or below the Fermi level. The removal of an electron below the Fermi level is described in the new scheme as the creation of a hole. We consider first a system of $N$ free non-interacting fermions having the Hamiltonian $$H_{0}=\sum_{\mathbf{k}}\varepsilon_{\mathbf{k}}c_{\mathbf{k}}^{+}c_{\mathbf{k}}$$ where $\varepsilon_{\mathbf{k}}$ is the energy of a single particle having $\varepsilon_{\mathbf{k}}=\varepsilon_{-\mathbf{k}}$ . We agree to measure $\varepsilon_{\mathbf{k}}$ from the Fermi level $\varepsilon_{F}$ .

In the ground state [... $\Phi_{0}$ ...] all one-particle states are filled to the energy $\varepsilon_{F}$ and above $\varepsilon_{F}$ all states are empty. We regard the state $\Phi_{0}$ as the vacuum of the problem: it is then convenient to repreent the annihilation of an electron in the Fermi sea as the creation of a hole. Thus we deal only with electrons (for states $k > k_{F}$) and holes (for states $k < k_{F}$). The act of taking an electron from $\mathbf{k}'$ within the sea to $\mathbf{k}''$ outside the sea involves the creation of an electron-hole pair. The language of the theory has a similarity to positron theory, and there is a complete formal similarity between particles and holes.

We introduce the electron operators $\alpha^{+}$ , $\alpha$ by $$\alpha_{\mathbf{k}}^{+}=c_{\mathbf{k}}^{+}\;;\;\alpha_{\mathbf{k}}=c_{\mathbf{k}}\;\;\text{for}\;\;\varepsilon_{\mathbf{k}}>\varepsilon_{F}$$ and the hole operators $\beta^{+}$ , $\beta$ by $$\beta_{\mathbf{k}}^{+}=c_{-\mathbf{k}}\;;\;\beta_{\mathbf{k}}=c_{-\mathbf{k}}^{+}\;\;\text{for}\;\;\varepsilon_{\mathbf{k}}<\varepsilon_{F}$$ The $-\mathbf{k}$ introduced for the holes is a convention which gives correctly the net change of wave-vector or momentum: the annihilation $c_{-\mathbf{k}}$ of an electron at $-\mathbf{k}$ leaves the Fermi sea with a momentum $\mathbf{k}$ . Thus $\beta_{\mathbf{k}}^{+}\equiv c_{-\mathbf{k}}$ creates a hole of momentum $\mathbf{k}$ .

So it is essentially a matter of convention in defining the operators. Obviously, it is no more strange than calling all the excitations of a solid a particle I believe.

For superconductor, though, the particle-hole symmetry plays a more prominent role than just a redefinition of what is called quasi-electron and quasi-hole. Since the phonons (as in the BCS model) couple electrons at $\mathbf{k}$ and $-\mathbf{k}$, they couple electrons and holes in the prescription cited above. Hence the crucial role played by the Nambu spinor.

This is nevertheless not an exact symmetry as Lubos Motl mentioned, but it can hardly be contradict. Indeed, the ratio $\Delta / \varepsilon_{F}$ is so small ($10^{-3} -10^{-5}$ for BCS superconductors) that the linearisation of the spectrum close to the Fermi energy is a really good approximation, and so the relation $\varepsilon_{\mathbf{k}}=\varepsilon_{-\mathbf{k}}$ is well verified.