Modified quantum harmonic oscillator: hamiltonians unitarily equivalent and energy spectrum

Question

Here is my Hamiltonian: $H_{\alpha, \beta} (q,p) = \frac{p^2}{2m} + \frac{1}{2}m \omega ^2q^2 + \alpha q + \frac{\beta}{2}(pq^2 + q^2p) + m \frac{\beta^2}{2}q^4$. How can I prove that $H_{\alpha, \beta}$ and $H_{\alpha, \beta‘}$ are unitarily equivalent and how I can compute the energy spectrum of the theory? It seems that the energy levels can be computed only perturbatively, because of the presence of term like $q$ or $q^4$, but in the text of the problem there no reference to perturbation theory.

Answer 1

The hamiltonian can be put in the form $H_{\alpha \beta} = \frac{1}{2m}(p + \beta m q^2)^2 + \frac{m \omega^2}{2}\left(q + \frac{\alpha}{m \omega^2}\right)^2 - \frac{\alpha^2}{m^2 \omega^4}$. Now define $P = p + \beta m q^2$ and $Q = q + \frac{\alpha}{m \omega^2}$. Since $[p,q] = -i \hbar$, it is easy to prove that also $[P,Q] = -i \hbar$, thus $P$ and $Q$ are canonically conjugate as well. In terms of these new variables the hamiltonian is $H_{\alpha \beta} = \frac{1}{2m}P^2 + \frac{m \omega^2}{2}Q^2 - \frac{\alpha^2}{m^2 \omega^4}$, which is a standard harmonic oscillator. This means that two hamiltonians with $\beta \neq \beta'$ are unitarily equivalent in the sense that they display the same spectrum (since they can always be rewritten as a harmonic oscillator). This also means that the spectrum of the theory is $E_n = \hbar \omega (n + 1/2) - \frac{\alpha^2}{m^2 \omega^4}$.

I hope I did not mess with the completion of squares, but in that case I hope the argument still holds.