# Can a constant term be added to the new operators in the Bogoliubov transform?

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: $$$$b_{k}=\sum_{l} u_{kl}a_{l}+\sum_{p}u^{\prime}_{kp}a^{\dagger}_{p}$$$$ I was now thinking: is this the most general transformation possible or would it be possible to use the following transformation: $$$$b^{\prime}_{k}=\sum_{l} u_{kl}a_{l}+\sum_{p}u^{\prime}_{kp}a^{\dagger}_{p}+c$$$$ where c is a constant. I am asking this beacuse the commutator of a constant with something else is always zero so if $$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: $$$$H= \sum_{l}\omega_{l}a^{\dagger}_{l}a_{l}+i\sum_{l}V_{l}\left(a_{l}-a^{\dagger}_{l}\right)+ \sum_{l}A_{l}(a^{2}_{l}+a^{\dagger\;2}_{l})$$$$ If I want to bosonize this Hamiltonian I have to add a constant term to the boson operator right?

• In the example: how is $a_{l\alpha}$ related to $a_l$? Nov 9 '20 at 23:49
• It should be possible to "diagonalize" the Hamiltonian in the example without using the Bogoliubov transformation. Just complete the square. Nov 13 '20 at 11:10
• 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 Nov 14 '20 at 10:23

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.