I encounter a problem which is related to Bogoliubov-de Gennes(BdG) Hamiltonian. Suppose I use BdG theory to transform my many body Hamiltonian into the following form:

\begin{equation} \hat{H} = \sum_{k \sigma} \xi_{k} c^{\dagger}_{k \sigma} c_{k \sigma} - \sum_{k} ( \Delta^{*} c_{-k \downarrow} c_{k \uparrow} + \Delta c^{\dagger}_{k \uparrow} c^{\dagger}_{k \downarrow}) \end{equation}

Then, we can rewrite our Hamiltonian into matrix form by introducing a vector of creation and annihilation operator $\phi_k = ( c_{k \uparrow} ~~ c^{\dagger}_{-k \downarrow})^{T}$:

\begin{equation} \hat{H} = \sum_k ( c^{\dagger}_{k \uparrow} ~~ c_{-k \downarrow}) \begin{pmatrix} \xi_{k} & - \Delta \\ -\Delta^{*} & -\xi_{k} \end{pmatrix} \begin{pmatrix} c_{k \uparrow} \\ c^{\dagger}_{-k \downarrow} \end{pmatrix} \end{equation}

However, when I expand matrix form Hamiltonian to see whether I can recover the original one, I encounter a problem in the $-\xi_{k}$ term. Firstly, the expansion of such matrix form Hamiltonian is as follow:

\begin{equation} \hat{H} = \sum_{k} \xi_{k} c^{\dagger}_{k \uparrow} c_{k \uparrow} -(\Delta^{*} c_{-k \downarrow}c_{k \uparrow} + \Delta c^{\dagger}_{k \uparrow}c^{\dagger}_{-k \downarrow}) - \xi_{k} c_{-k \downarrow} c^{\dagger}_{-k \downarrow} \end{equation}

Since $k$ and $-k$ are dummy variables, I can re-label the term $c_{-k \downarrow}c^{\dagger}_{-k \downarrow} \rightarrow c_{k \downarrow}c^{\dagger}_{k \downarrow}$. But when I use the anti-commutation rules to normal order the $ c_{k \downarrow}c^{\dagger}_{k \downarrow}$, I got a trouble since $\{c_{k \downarrow}, c^{\dagger}_{k \downarrow} \} = c_{k \downarrow} c^{\dagger}_{k \downarrow} + c^{\dagger}_{k \downarrow} c_{k \downarrow} = \delta_{k,k} \delta_{\downarrow, \downarrow} = 1$, implying that $ c_{k \downarrow} c^{\dagger}_{k \downarrow} = 1 - c^{\dagger}_{k \downarrow} c_{k \downarrow}$. Therefore, my $- \xi_{k}$ term would become like the following:

\begin{equation} \begin{split} -\xi_{k} c_{-k \downarrow} c^{\dagger}_{-k \downarrow} &\rightarrow -\xi_{k}(1 - c^{\dagger}_{k \downarrow} c_{k \downarrow}) \\ &= \xi_{k} c^{\dagger}_{k \downarrow} c_{k \downarrow} - \xi_{k} \end{split} \end{equation}

This means that the expansion of the matrix form Hamiltonian will acquire a constant energy shift which is inconsistent with the original one. Could anyone help me to find out the mistake that I had made in the whole derivation? Thank you.


1 Answer 1


You are correct. There is a $c$-number energy shift: $$ \hat H_{\rm Bogoliubov}= a^\dagger_i H_{ij}a_j +\frac 12 \Delta_{ij} a^\dagger_i a^\dagger_j +\frac 12 \Delta^{\dagger}_{ij} a_i a_j\\ = \frac12 \left(\matrix{ a^\dagger _i &a_i}\right)\left(\matrix{ H_{ij}& \phantom {-}\Delta_{ij}\cr \Delta^{\dagger}_{ij}& -H^T_{ij}}\right) \left(\matrix{ a_j\cr a^\dagger_j}\right) +\frac 12 {\rm tr}(H). $$ Keeping track of the ${\rm tr} (H)$ is not important for quasiparticle energies, and so it commonly omitted, but it is essential when you are interested in the dynamics of the condensate.

  • $\begingroup$ Thank you for your comment, Mike. I think you are right. The matrix form Hamiltonian I wrote is copied from a textbook and I did not realised the author dropped out all the constant energy shift, since the author only wanted to show the quasi-particle spectrum of BdG Hamiltonian. $\endgroup$
    – Ricky Pang
    Commented Mar 28, 2021 at 1:02

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.