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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.

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  • $\begingroup$ It is useful to start from the rewriting hamiltonian in terms of creation-annihilation operators. This makes Hamiltonian more clear but consumes lot of time. It seems that you should use perturbation theory but if you provide the text of the problem may be it ll be useful. Also, if you rewrite Hamiltonian in terms of creation-annihilation operators, it seems that equivalence is obvious $\endgroup$ Jan 15 '20 at 17:32
  • $\begingroup$ I put the Hamiltonian in this form $H = (\frac{p}{\sqrt{2m}} + \sqrt{\frac{m}{2}} q^2)^2 + (\frac{m}{2} \omega q + \frac{\alpha + i \hbar \beta}{\omega \sqrt{2m}})^2 - constant^2$. Under an proper change of variables It seems that is the Hamiltonian of a quantum harmonic oscillator without any strange term. $\endgroup$
    – dfgoe55
    Jan 16 '20 at 16:31
  • $\begingroup$ What is $\hbar$? How does it appear? $\endgroup$ Jan 16 '20 at 17:04
  • $\begingroup$ I have used the canonical commutation relations between $q$ and $p$ to rewrite the term $(pq^2 + q^2p)$ but the problem is that the constant I have written is a complex number and therefore the energy of the system is shifted by a complex quantity, that is obviously wrong. $\endgroup$
    – dfgoe55
    Jan 16 '20 at 17:17
  • $\begingroup$ Why you said that in term of creation-annihilation operators the equivalence is obvious? How can I prove that? $\endgroup$
    – dfgoe55
    Jan 16 '20 at 19:58
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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.

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