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What is the form of $$S^+px^2S$$? (where $S$ is the squeeze operator.) I know the form of $S^+pS$ and $S^+x^2S$ but no matter how I calculate it I cant find the form for $S^+px^2S$.

The main problem is that I cant find a neat form for the below equation. $$e^A B e^{-A}=B+[A,B]+(1/2!)[A,[A,B]]+...$$ (where $e^A=S$, $B=px^2$)

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If $S$ is the unitary squeeze opertor, then $S^\dagger x^2p S= (S^\dagger x^2 S)( S^\dagger p S)$. You say that you know both of these factors.

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mike-stone's answer is the most straight forward way if you know the forms of those factors. If not, it is straight forward to expand $px^2$ in terms of ladder operators to find your desired $e^ABe^{-A}$ forms.

$p =\frac{1}{\sqrt{2}}(\hat{a}^\dagger-\hat{a})$

$x =\frac{1}{\sqrt{2}}(\hat{a}^\dagger+\hat{a})$

You will have to find commutators of the form $[A,B^n]$, but this is fairly direct and outlined in most QM textbooks.

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