Inverting squeezing and displacement operators: how do I turn $D(\alpha)S(\xi)$ into $S(\xi')D(\alpha')$? This question is about inverting the product of squeezing operator and a displacement operator in the following way:
I have $D(\alpha)S(\xi)$ and I'd like to turn it into $S(\xi')D(\alpha')$,
where
$$D(\alpha)=e^{\alpha a^\dagger-\alpha^* a} \qquad\text{ and }\qquad S(\xi)=e^{\frac{1}{2}(\xi^* a^2-\xi a^{\dagger2})}.$$
Can anyone explain how this is done (i.e. how to find $\xi',\alpha'$ as a function of $\xi, \alpha$)?
 A: Start from the definitions
$D(\alpha)\equiv\exp(\alpha a^\dagger - \alpha^* a)$
and
$S(\xi)\equiv\exp[\frac12(\xi a^{\dagger 2}-\xi^* a^2)]$.
Remember the identity
$$e^A B e^{-A}=e^{\operatorname{ad}_A}B\equiv \sum_{k=0}^\infty\frac{1}{k!}[\underbrace{A\cdots A}_k,B],$$
with $[\underbrace{A\cdots A}_k,B]\equiv \operatorname{ad}_A^k B$ denoting the iterated commutator of $A$ with $B$ (e.g. $[AA,B]\equiv[A,[A,B]]$).
We then have
$$S(\xi)D(\alpha)S^\dagger(\xi)=\sum_{k=0}^\infty\frac{1}{k!} \operatorname{ad}_A^k D(\alpha)$$
with $A\equiv \frac12(\xi a^{\dagger 2}-\xi^* a^2)$. Observe that
\begin{align}
  [A,\alpha a^\dagger -\alpha^* a] &=
    (\xi\alpha^*) a^\dagger - (\xi\alpha^*)^* a =
    (\xi\alpha^*) a^\dagger - (\xi^*\alpha) a, \\
  [A,[A,\alpha a^\dagger -\alpha^* a]] &=
    |\xi|^2(\alpha a^\dagger - \alpha^* a).
\end{align}
From these, we can deduce the more general rules:
\begin{align}
  \operatorname{ad}_A^{2k}(\alpha a^\dagger-\alpha^* a) &=
    |\xi|^{2k} (\alpha a^\dagger-\alpha^* a), \\
  \operatorname{ad}_A^{2k+1}(\alpha a^\dagger-\alpha^* a) &=
    |\xi|^{2k} [(\xi\alpha^*) a^\dagger-(\xi\alpha^*)^* a],
\end{align}
and therefore
$$
S(\xi)D(\alpha)S^\dagger(\xi) =
\sum_{k=0}^\infty |\xi|^{2k} \left[
  \frac{1}{(2k)!} (\alpha a^\dagger -\alpha^* a) +
  \frac{1}{(2k+1)!} (\xi\alpha^* a^\dagger -\xi^*\alpha a) 
\right] \\
= (\alpha a^\dagger -\alpha^* a) \cosh(|\xi|) +
(e^{i\theta}\alpha^* a^\dagger -e^{-i\theta}\alpha a) \sinh(|\xi|)
\\= (\alpha \cosh(|\xi|) + e^{i\theta}\alpha^* \sinh(|\xi|))a^\dagger - c.c. \\
= D(\alpha \cosh(|\xi|) + e^{i\theta}\alpha^* \sinh(|\xi|)).
$$
We conclude that
$$S(\xi)D(\alpha) = D(\alpha \cosh(|\xi|) + e^{i\theta}\alpha^* \sinh(|\xi|)) S(\xi).$$
A: I found how to find it again [this is not a proof].
Here $\xi=re^{i\theta}$. We have:
$a^\dagger D(\alpha) = D(\alpha) a^\dagger + \alpha^*D(\alpha)$
$a D(\alpha) = D(\alpha) a + \alpha D(\alpha)$
$a^\dagger S(\xi) = S(\xi) a^\dagger ch(r) - S(\xi)ae^{-i\theta}sh(r)$
$a S(\xi) = S(\xi) a ch(r) - S(\xi)a^\dagger e^{i\theta}sh(r)$
Then, applying $a$ to "$D(\alpha)S(\xi)=S(\xi)D(\beta)$", we find that we must have:
$\alpha=\beta ch(r)- \beta^*e^{i\theta}sh(r)$ ie $\beta=\alpha ch(r)+\alpha^*e^{i\theta}sh(r)$
