[Physics] Particle-Hole transformation in Superconductor

condensed-mattersuperconductivitysymmetry

The Bogoliubov – de Gennes equation has a emergent particle-hole symmetry:

$$
\mathcal{P}H\mathcal{P}^{\dagger} = -H\text{.}
$$

My question is now what happen to the Nambu spinor:

$$
\mathcal{P}\begin{pmatrix}
u_{\uparrow}\left(\vec{k}\right) \\
u_{\downarrow}\left(\vec{k}\right) \\
v_{\uparrow}\left(\vec{k}\right) \\
v_{\downarrow}\left(\vec{k}\right) \\
\end{pmatrix} = \text{?}
$$

and to the Bogoliubov quasiparticle operator:

$$
\mathcal{P}\gamma_{\vec{k}\sigma}\mathcal{P}^{\dagger} = {P}\left(u_{\vec{k}\sigma}c_{\vec{k}\sigma} + \sigma v_{-\vec{k}-\sigma}c_{-\vec{k}-\sigma}^{\dagger}\right){P}^{\dagger} = \text{?,}
$$

where $c_{\vec{k}\sigma}$ is a annihilation operator for a electron.

From QFT I know that charge conjugation transform a fermion with momenta $\vec{k}$ and spin $s$ to a antifermion with momenta $\vec{k}$ and spin $s$. But in the Dirac spinor the spin and momenta will flip. Is this in the case of a superconductor the same. i.e.

$$
\mathcal{P}\begin{pmatrix}
u_{\uparrow}\left(\vec{k}\right) \\
u_{\downarrow}\left(\vec{k}\right) \\
v_{\uparrow}\left(\vec{k}\right) \\
v_{\downarrow}\left(\vec{k}\right) \\
\end{pmatrix} = \begin{pmatrix}
v_{\downarrow}\left(-\vec{k}\right) \\
v_{\uparrow}\left(-\vec{k}\right) \\
u_{\downarrow}\left(-\vec{k}\right) \\
u_{\uparrow}\left(-\vec{k}\right) \\
\end{pmatrix}^{\star}
$$

but what is with the Bogoliubov operators?

Best Answer

You take the problem upside-down. Once you got a Hamiltonian, if it has a particle-hole symmetry $P$ which by definition verifies $\left\{ H,P\right\} =0$ with an anti-unitary operation $P$, you can construct it explicitly, and then you know how it applies on the operator basis and so on.

For instance, a so-called s-wave superconductor can be described by the Bogoliubov-deGennes Hamiltonian density $$H=\frac{\left(p^{2}-p_{F}^{2}\right)}{ 2m}\tau_{z}+\tau_{x}\Delta_{x}+\tau_{y}\Delta_{y}$$ with the gap function $\Delta=\Delta_{x}+\mathbf{i}\Delta_{y}$ a complex one, and $p_{F}$ the chemical potential of an otherwise free electron gas of band mass $m$. Importantly, the $\tau$'s are Pauli matrices, and $\tau_{y}$ is the imaginary one verifying $\tau_{y}^{\ast}=-\tau_{y}$ ($\ast$ signifies complex-conjugation, not dual). Then $H$ has a unique particle-hole symmetry (up to a unitary transform of no consequence) $P=K\mathbf{i}\tau_{y}$ which has the important property that $P^{2}=-1$ (this is important in order to use the Altland-Zirnbauer classification and its topological consequences).

The Bogoliubov-deGennes Hamiltonian then reads $\mathcal{H}=\int dx\left[\hat{c}^{\dagger}H\hat{c}\right]$ with the particle-hole conjugate $\bar{c}=P\hat{c}$, ... note that the choice for an explicit representation $\hat{c}$ is of no importance in this construction.

The three discrete symmetries are

  • the anti-unitary one which anti-commutes with the Hamiltonian : it's usually called particle-hole, or charge conjugation
  • the anti-unitary one which commutes with the Hamiltonian : it's usually called time-reversal symmetry
  • the unitary one which anti-commutes with the Hamiltonian : it's usually called chiral symmetry, or sub-basis, or valley, ...

They are called discrete because: i) the anti-unitary ones can have only $\pm 1$ square and ii) the chiral symmetry can be seen as the product of the two previous ones, to be either present or not.

In modern terminology, the discrete symmetries are more important than the representation. I believe that's the reason for the confusion with high-energy physics. In fact for the Dirac equation one does the same thing, but the Dirac representation is so used that people usually take for granted one representation of the discrete symmetry. Open the book by Itzykson and Zuber, and you will see how they define the charge conjugation for both the Dirac and Majorana representations for instance.

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