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The Bogoliubov transformation picks a set of boson operators $\{a_{k},a^{\dagger}_{k}\}$ and transforms them into a new set of boson operators generally written as: \begin{equation} b_{k}=\sum_{l} u_{kl}a_{l}+\sum_{p}u^{\prime}_{kp}a^{\dagger}_{p} \end{equation} I was now thinking: is this the most general transformation possible or would it be possible to use the following transformation: \begin{equation} b^{\prime}_{k}=\sum_{l} u_{kl}a_{l}+\sum_{p}u^{\prime}_{kp}a^{\dagger}_{p}+c \end{equation} where c is a constant. I am asking this beacuse the commutator of a constant with something else is always zero so if $b_{k}$ and its complex conjugate $b^{\dagger}_{k}$ have bosonic commutation properties the same should be true also for $b^{\prime}_{k}$ and $b^{\prime\dagger}_{k}$. However, I can not find any place where this is reported so maybe there is some problem and I can't see it right now.

Edit. A very trivial example, consider the following Hamiltonian: \begin{equation} H= \sum_{l}\omega_{l}a^{\dagger}_{l}a_{l}+i\sum_{l}V_{l}\left(a_{l}-a^{\dagger}_{l}\right)+ \sum_{l}A_{l}(a^{2}_{l}+a^{\dagger\;2}_{l}) \end{equation} If I want to bosonize this Hamiltonian I have to add a constant term to the boson operator right?

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  • $\begingroup$ In the example: how is $a_{l\alpha}$ related to $a_l$? $\endgroup$ Nov 9 '20 at 23:49
  • $\begingroup$ It should be possible to "diagonalize" the Hamiltonian in the example without using the Bogoliubov transformation. Just complete the square. $\endgroup$ Nov 13 '20 at 11:10
  • $\begingroup$ Yes I agree on that, but if I wanted to see this as a Bogoliubov transformation this would mean that I added a constant in the guess of the bosonic field.... Another example might be this one $\endgroup$
    – Yepman
    Nov 14 '20 at 10:23
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The real question is: why would you do this tranformation or why would you define such an operator?

The Bogoliubov transformation is usually used to diagonalize Hamiltonians (see wiki). Off-setting the operator by a constant does not change its time dependence. $b'_k$ is also stationary if $b_k$ is stationary. So all the additional constant does is offset the operator expectation values. I can not imagine a physical reason to perform this operation.

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  • $\begingroup$ Well I was thinking that in order to find the new bosonic operators I have to ask for [b_{k},H]=\Omega b_{k} so if for example the Hamiltonian had some linear terms in it the photonic field like a_{l}+a^{\dagger}_{l} I would need constants in the definition of the new bosonic field right? $\endgroup$
    – Yepman
    Oct 9 '20 at 14:18
  • $\begingroup$ @Yepman It would be helpful if you could write out the example explicitly. I can’t imagine how the constant term would help in any example, but maybe I missed something. $\endgroup$ Oct 9 '20 at 14:48

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