# Diagonalizing Hamiltonian which is Linear in Bose Field Operators

Often times in mean field theories, I've run into Hamiltonians of the form $$H = \omega a^\dagger + \omega^* a + \epsilon_0$$ where $a$ and $a^\dagger$ are the regular Bose field operators and $\epsilon_0$ is some constant energy offset. Is there a simple way to diagonalizing Hamiltonians of this form? I've tried searching for this and haven't found any methods on diagonalizing single field operators.

Use the direct representation of $a$ and $a^{\dagger}$:

$$a \left| n \right> = \sqrt{n} \left| n-1 \right>,$$ $$a^{\dagger} \left| n \right> = \sqrt{n+1} \left| n+1 \right>.$$

Apply the eigenvalue equation $H \left| \Psi \right> = E \left| \Psi \right>$ to an arbitrary state

$$\left| \Psi \right> = \sum_n \Psi_n \left| n \right>.$$

You will get the following recurrent relation:

$$(E - \epsilon_0) \Psi_n = \omega \sqrt{n} \Psi_{n-1} + \omega^* \sqrt{n+1} \Psi_{n+1}.$$

This is linear in $\Psi$ for all $n$, so either $\Psi_0 \neq 0$, or the whole state is just zero (which is of course not allowed by the principles of QM). It doesn't matter which value of $\Psi_0$ you choose, it will only impact the overall normalization.

So choose an arbitrary $\Psi_0$ and recover the rest from the recurrent relation.

Bonus questions: are your eigenstates normalizable? What does it mean mathematically and physically?