I am trying to find the eigenstates/eigenvalues of the following Hamiltonian $$ \hat{H} = \hbar \omega \Big(\hat{a}^{\dagger}\hat{a}+\frac{1}{2}\Big)+A\big(\hat{a}^{\dagger}\hat{a}^{\dagger}+\hat{a}\hat{a}\big)\equiv \hat{H}_1+\hat{H}_2, $$ where $\hat{a}$ and $\hat{a}^{\dagger}$ are the standard ladder operators, with $[\hat{a}, \hat{a}^{\dagger}] = 1$.

My attempt

(1) I tried to check whether the two terms were commuting, to see if the problem admitted a separable solution, but $\big[\hat{H}_1, \hat{H}_2\big]= \hat{a}^{\dagger}\hat{a}^{\dagger}-\hat{a}\hat{a}$.

(2) I tried the trial solution (even states) $$ |\phi\rangle = \sum_n c_{n2}|2n\rangle $$ which when substituted gives $$ \hat{H}|\phi\rangle = \sum_n \left[\frac{\hbar \omega}{2}c_{2n}+\hbar \omega (2n) c_{2n}+c_{2n-2}+c_{2n+2}\right]|2n\rangle $$ which gives the infinite system $$ c_{2n}= \left(\frac{\hbar \omega}{2}+2n\hbar \omega\right)c_{2n} +c_{2n-2}+c_{2n+2}. $$ I am not sure if this leads anywhere.


Is there a standard way to approach this problem? Any help?

  • $\begingroup$ This is a standard kind of problem in perturbation theory, where it is straightforward to get the second-order corrections to the energy. Getting the exact solution is much trickier. $\endgroup$
    – Buzz
    Commented Jun 7, 2022 at 12:24
  • $\begingroup$ $A$ is not small though, so in general I can't use perturbation theory I think. $\endgroup$
    – Andrea
    Commented Jun 7, 2022 at 12:31

1 Answer 1


The operators $\hat C=\hat a^\dagger \hat a + \hat a\hat a^\dagger$, $\hat A=\hat a^\dagger \hat a^\dagger$ and $\hat B=\hat a\hat a$ span an $\mathfrak{su}(1,1)$ Lie algebra.

The problem is solved using an $SU(1,1)$ transformation $T$ (in your case that's just a Bogoliubov transformation) that will map $\hat A$ to a linear combination of $\hat A,\hat B$ and $\hat C$. The same transformation will also map $\hat B$ and $\hat C$ to linear combinations of $\hat A,\hat B$ and $\hat C$. One then chooses the parameters in $T$ to eliminate the non-diagonal terms.

There's another equivalent approach which is to write your Hamiltonian in terms of $x$ and $p$. With some elementary insight you should see how this is actually the Hamiltonian of a harmonic oscillator with frequency $\omega^\prime$ which is a multiple of the original frequency $\omega$ of your problem. Once $\hat H$ is expressed in terms of $\omega^\prime$ you can just quantize that in terms of new raising operators $\hat b$ and $\hat b^\dagger$.

It shouldn't be too hard for you to find literature on Bogoliubov transformations, or $\mathfrak{su}(1,1)$ (or equivalently $\mathfrak{sp}(2,\mathbb{R})$).

In fact, the Wikipedia article on Bogoliubov transformations is not a bad place to start. You should be able to invert $$ \hat b=u\hat a+v\hat a^\dagger\, ,\qquad \hat b^\dagger=u^*\hat a^\dagger+v^*\hat a $$ and re-express your $H$ in terms of $\hat b$ and $\hat b^\dagger$, then choose $u$ and $v$ to eliminate the $\hat b\hat b$ and $\hat b^\dagger\hat b^\dagger$.


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