6
$\begingroup$

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?

$\endgroup$
3
$\begingroup$

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: $\gamma_{-E}=\gamma_E^\dagger$.

$\endgroup$

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.