# Squeezed state expectation value form

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 $$e^A=S$$, $$B=px^2$$)

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.
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.