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)?

  • 1
    $\begingroup$ 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. $\endgroup$ Commented Nov 24, 2019 at 6:43
  • $\begingroup$ $\hat{S}$ in your question is not a squeeze operator, rather it is a displacement operator. $\endgroup$
    – Sunyam
    Commented Nov 25, 2019 at 19:35
  • $\begingroup$ @Sunyam, you're right, I forgot to put the squares in the annihilation and creation operators. I already corrected this $\endgroup$ Commented Dec 5, 2019 at 4:21

1 Answer 1


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, $$ (\hat{a}^{\dagger}-\hat{a})\hat{S} = \hat{S}(\hat{a}^{\dagger}-\hat{a})+\frac{(\xi-\xi^*)}{2} \hat{S}. $$

Repetitive application will recursively produce the answer.

You can also play the same game with $\hat{a}^{\dagger}+\hat{a}$.


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