# Is there any definition for $(\hat{a}^{\dagger} \pm \hat{a})^{n}\hat{S} \left | 0 \right>$

Specifically, I want to apply the following operators

$$\hat{P}^n = \left( \hat{a}^{\dagger} - \hat{a} \right)^n \tag{1},$$ $$\hat{X}^n = \left( \hat{a}^{\dagger} + \hat{a} \right)^n \tag{2},$$ (with the power $$n=1,2, \cdots$$ and $$\hat{a},\hat{a}^{\dagger}$$ the annihilation, creation operators) to a squeezed vacum state $$\left | 0 \right>$$, that is

$$\hat{P}^n \hat{S} \left | 0 \right> = \left( \hat{a}^{\dagger} - \hat{a} \right)^n \hat{S} \left | 0 \right> \tag{3},$$ $$\hat{X}^n \hat{S} \left | 0 \right> = \left( \hat{a}^{\dagger} + \hat{a} \right)^n \hat{S} \left | 0 \right> \tag{4},$$ where $$\hat{S}$$ is the well known squeezed operator, defined by $$\hat{S}=\exp\left[\frac{1}{2}(\xi^{\ast}\hat{a}^{2} - \xi\hat({a}^{\dagger})^2 )\right],$$ with $$\xi=r e^{i\theta}.$$ So, the question is: Is there any trick, identity or whatever, in order to expand the right-hand side of equations (3) and (4)?

• On approach is to write the commutator of $S$ with $X^n$ and $P^n$. If the commutator's action on $|0\rangle$ is simple enough, things will work. I guess using Baker Campbell Hausdorff you can do this. but I am not sure. You should check it out if you have never seen it before. – physicophilic Nov 24 '19 at 6:43
• $\hat{S}$ in your question is not a squeeze operator, rather it is a displacement operator. – Sunyam Nov 25 '19 at 19:35
• @Sunyam, you're right, I forgot to put the squares in the annihilation and creation operators. I already corrected this – Julio Abraham Mendoza Fierro Dec 5 '19 at 4:21

Try this: $$e^{A} B \ e^{-A} = B+[A,B]+\frac{[A,[A,B]]}{2!}+\dots$$
With $$A = - \frac{1}{2}(\xi^{\ast}\hat{a} - \xi\hat{a}^{\dagger} )$$, $$B = (a^{\dagger}-\hat{a})$$, the above formula yields, $$(\hat{a}^{\dagger}-\hat{a})\hat{S} = \hat{S}(\hat{a}^{\dagger}-\hat{a})+\frac{(\xi-\xi^*)}{2} \hat{S}.$$
You can also play the same game with $$\hat{a}^{\dagger}+\hat{a}$$.