Particle-hole symmetry of Bogoliubov - de Gennes equation

Question

Bogoliubov - de Gennes equation reads, $$\left( \begin{array}{cc} H_{0} - E_{F} & -i\sigma_{y}\Delta \\ i\sigma_{y}\Delta^{*} & E_{F} - H_{0}^{*} \end{array}\right) \left( \begin{array}{c} \psi_{e}^{\uparrow} \\ \psi_{e}^{\downarrow} \\ \psi_{h}^{\uparrow} \\ \psi_{h}^{\downarrow} \end{array}\right) = \mathcal{E} \left( \begin{array}{c} \psi_{e}^{\uparrow} \\ \psi_{e}^{\downarrow} \\ \psi_{h}^{\uparrow} \\ \psi_{h}^{\downarrow} \end{array}\right) $$

The above Hamiltonian which is $4$x$4$ obeys particle-hole symmetry, $$H = -\mathcal{C}H\mathcal{C}^{-1}$$ where $\mathcal{C} = \tau_{x}\mathcal{K}$, with $\tau_{x} = \sigma_{x} \oplus \sigma_{x}$ and complex conjugation operator $\mathcal{K}$. This symmetry implies that if You have solution $\Psi$ with energy $\mathcal{E}$, then You also have solution $\mathcal{C}\Psi$ with energy $-\mathcal{E}$. I always thought that if Hamiltonian posseses some symmetry then You have solution with the same energy. I am confused, cause I also read that holes are time-reversed electrons, $$\psi_{h} = \mathcal{T}\psi_{e}$$ with $\mathcal{T} = i\sigma_{y}\mathcal{K}$, but I can't show this from above BdG equation.

Answer 1

Great question. One should view the diagonalization of quadratic fermion models as arising from the representation theory of the spin group and its Lie algebra. Because your model is quadratic, the Hamiltonian lies in $\mathfrak{spin}_4$, which can be identified with $\mathfrak{so}_4$ by pushing forward via the universal covering map $$\gamma_*: \underbrace{\mathfrak{spin}_{2n}}_\text{quadratic models}\xrightarrow{\simeq} \underbrace{\mathfrak{so}_{2n}}_\text{skew-symmetric matrices}$$

$\mathfrak{so}_{2n}$ is the algebra of real antisymmetric matrices, which is precisely the statement of particle-hole symmetry. Therefore, "particle-hole symmetry" is already built into your model, not because of anything physical, but because it is quadratic. I would study general characteristics of the eigenvalue problem for $\mathfrak{so}_{n}$, where all of the properties you mention will become clearer and less obfuscated by physical data (see https://en.wikipedia.org/wiki/Skew-symmetric_matrix#Spectral_theory).

Answer 2

Particle-hole symmetry is not a symmetry in the usual sense. Generally, we think of a symmetry as a (unitary) operator $A$ that commutes with the Hamiltonian. So, we have $HA=AH$. In this case, if we have an eigenstate $|\psi\rangle$ with energy $E$, it's simple to show that $A|\psi\rangle$ is also an eigenstate with the same energy:

\begin{align} HA|\psi\rangle = AH|\psi\rangle=AE|\psi\rangle = EA|\psi\rangle \end{align} In the case of "normal" symmetries, the state $A|\psi\rangle$ can either be the same state as $|\psi\rangle$, or a new eigenstate with the same energy.

Particle-hole symmetry is not like this. Instead of having an operator that commutes with the Hamiltonian, we have an operator that anticommutes with the Hamiltonian, $H\mathcal{C}=-\mathcal{C} H$. In this case, we can try to redo the proof above, but we pick up a minus sign.

\begin{align} H\mathcal{C}|\psi\rangle = -\mathcal{C}H|\psi\rangle=-\mathcal{C}E|\psi\rangle = -E\mathcal{C}|\psi\rangle \end{align} Thus, in this case, the state $\mathcal{C}|\psi\rangle$ is an eigenstate with eigenvale $-E$. Unlike the case of a "normal" symmetry, $\mathcal{C}|\psi\rangle$ is necessarily a distinct state from $|\psi\rangle$, since it has a different energy. The only time you can have $\mathcal{C}|\psi\rangle=|\psi\rangle$ is if $E=0$; these zero-energy modes are the Majorana modes.