5
$\begingroup$

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)?

$\endgroup$
3
  • 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

4
$\begingroup$

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 $\hat{a}^{\dagger}+\hat{a}$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.