# Bogoliubov transformation for fermionic Hamiltonian

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

$$a_k=u_kb_k-v_kb^{\dagger}_{-k}$$,

$$a^{\dagger}_k=u_kb^{\dagger}_k+v_kb_{-k}$$,

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.

• 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?
– Gec
Mar 21, 2019 at 20:41
• @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 $a_{k}$. Mar 21, 2019 at 21:14

$$[b_k,b^{\dagger}_{k'}b_{k'}] \\ = b_k b^{\dagger}_{k'}b_{k'} - b^{\dagger}_{k'}b_{k'}b_k \\ = b_k b^{\dagger}_{k'}b_{k'} + b^{\dagger}_{k'}b_kb_{k'} \\ = \{b_k, b^{\dagger}_{k'}\}b_{k'} \\ = \delta_{k, k'}b_{k'}.$$
Note that the third line stems from the fact that $$b_k$$ and $$b_{k'}$$ anti-commute.
• 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. Mar 21, 2019 at 21:05