I have the Hamiltonian

$H=\sum\limits_k [Ab^{\dagger}_{k}b_{k} + B(b^{\dagger}_kb^{\dagger}_{-k}+b_{k}b_{-k})]$,

where $b^{\dagger}_k$ and $b_k$ are fermionic creation and annihilation operators.

I know that diagonalized form of this Hamiltonian has spectrum of the following form $E_k=\sqrt{A^2+4B^2}$

I want to check it by myself, so I use Bogoliubov transformation



where $u^2_k+v^2_k=1$ and $\left\{ {b_k,b_{k'}} \right\}=\delta_{kk'}$.

Condition for diagonalized Hamiltonian is $[a_k,H]=[u_kb_k-v_kb^{\dagger}_{-k},H]$, so I start to calculate commutators $[b_k,b^{\dagger}_{k'}b_{k'}]$, $[b_k,b^{\dagger}_{k'}b^{\dagger}_{-k'}]$, $[b^{\dagger}_{-k},b^{\dagger}_{k'}b_{k'}]$, $[b^{\dagger}_{-k},b_{k'}b_{-k'}]$, after that I have to equate the coefficients with the same operators $b_k$, $b^{\dagger}_{-k}$ on the right and on the left.

But there I meet some difficulties, because I have the third order terms like $-2b^{\dagger}_{k'}b_{k}b_{k'}$ and $-2b^{\dagger}_{k'}b_{k'}b^{\dagger}_{-k}$ from commutators $[b_k,b^{\dagger}_{k'}b_{k'}]$, $[b_k,b^{\dagger}_{k'}b^{\dagger}_{-k'}]$ respectively.

1) Is Bogoliubov transformation that I use correct?

2) What to do with third order terms? Maybe it's just my mistake and I calculate these commutators wrong.

Thank you.

  • $\begingroup$ You probably calculate commutators wrong, $[a_k, H]$ contains no third order terms. Besides, $v_k$ has different sign in expressions for $a_k$ and $a_k^\dagger$. Are your operators related by hermitian conjugation? $\endgroup$
    – Gec
    Mar 21, 2019 at 20:41
  • $\begingroup$ @Gec, you are right, I did a mistake in calculation. I suppose that coefficients $v_k$ and $u_k$ have to be complex and operators $a_{k}$ and $a^{\dagger}_{k}$ are hermitian conjugated, but I'm not sure where comes that minus sign for $a_{k}$. $\endgroup$
    – C-Roux
    Mar 21, 2019 at 21:14

1 Answer 1


$$ [b_k,b^{\dagger}_{k'}b_{k'}] \\ = b_k b^{\dagger}_{k'}b_{k'} - b^{\dagger}_{k'}b_{k'}b_k \\ = b_k b^{\dagger}_{k'}b_{k'} + b^{\dagger}_{k'}b_kb_{k'} \\ = \{b_k, b^{\dagger}_{k'}\}b_{k'} \\ = \delta_{k, k'}b_{k'}. $$

Note that the third line stems from the fact that $b_k$ and $b_{k'}$ anti-commute.

  • $\begingroup$ Oh, now I see where was the problem, I permuted operators $b_{k}b_{k'}$ without changing sign due to anti-commutation relation. Thank you for your answer. $\endgroup$
    – C-Roux
    Mar 21, 2019 at 21:05

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