Bogoliubov-de-Gennes (BdG) formalism

Question

Suppose you treat the mean-field BCS superconductor Hamiltonian $H$ in "BdG style" by re-writing it as

$H = \frac{1}{2} \sum_k \psi_k^{\dagger} H_{BdG} \psi_k$

where, in terms of original annhiliation and creation operators appearing in $H$, $\psi_k = ( a_{k, u}, a_{k, d} , a_{-k, d}^{\dagger} , -a_{-k, u}^{\dagger} )^T$. Here $u$ and $d$ suggest up and down spin.

The eigenvalues of $H$ can be found by finding eigenvalues of $H_{BdG}$.

But, how to find the eigenvectors of original Hamiltonian $H$ from knowledge of $H_{BdG}$? What is general prescription?

Answer 1

When you diagonalize $H_\text{BdG}$, you find the eigenvalues and the eigenvectors. Let us assemble these (column) vectors into a unitary matrix $U$ (this is always possible since $H_\text{BdG}$ is Hermitian), so that $U^\dagger H_\text{BdG} U=E$ is diagonal (i.e. the eigenvalues). Define $\Gamma_k = U^\dagger \psi_k$, then

$ H=\frac{1}{2} \Gamma_k^\dagger E \Gamma_k $

Therefore $\Gamma_k$ gives you the correct "single-particle" eigenstates of $H$ and can be used to build any excitations of the Hamiltonian. To be more explicit, the $i$-th row of $\Gamma_k$ is $\sum_j(U^\dagger)_{ij}\psi_{kj}$, with energy $E_i$.

Lastly, because the particle-hole symmetry of the Hamiltonian, the spectrum of $H_\text{BdG}$ always comes in $\pm E$ pairs, and correspondingly $\gamma_{E}$ and $\gamma_{-E}$ are actually related by hermitian conjugation: $\gamma_{-E}=\gamma_E^\dagger$.