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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 $\xi, \alpha$)?

glS
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MarcO
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2 Answers2

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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 $$S(\xi)D(\alpha) = D(\alpha \cosh(|\xi|) + e^{i\theta}\alpha^* \sinh(|\xi|)) S(\xi).$$

glS
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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 $\beta=\alpha ch(r)+\alpha^*e^{i\theta}sh(r)$

MarcO
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