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Let $|0\rangle,...$ be the states of the harmonic oscillator. Then a squeezed state was defined as $|\xi\rangle =S(\xi)|0\rangle $, where $S(\xi):=e^{\frac{1}{2}( \xi (a^{ \dagger ^2}-a^2))}$, where $a$ and $a^{\dagger}$ are the canonical operators linked with the harmonic oscillator problem in QM.

Now we were supposed to show that $S^{\dagger}(\xi) X S(\xi)=Xe^{-\xi}$.

A hint says that we are supposed to differentiate $F(\xi):=S^{\dagger}(\xi) X S(\xi)$ and reexpress this expression in terms of $F$. Finally, there shall be a differential equation that does it.

The problem is: In my opinion the derivative is just: $F'(\xi):=\frac{1}{2}S^{\dagger}(\xi)(a^{ \dagger ^2}-a^2) X S(\xi)+\frac{1}{2}S^{\dagger}(\xi) X S(\xi)(a^2-a^{ \dagger ^2}).$

Now I do not see how to proceed.

If something is unclear, please let me know.

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up vote 3 down vote accepted

One option is to note that $S^{\dagger}(\xi)$ commutes with ${a^ \dagger }^2-a^2$, and that therefore the derivative of the former can be written as $$\frac d{d\xi}S^{\dagger}(\xi)=(a^{ \dagger ^2}-a^2)S^{\dagger}(\xi).$$ This will allow you to write $F'(\xi)$ in terms of $F(\xi)$ and $a^{ \dagger ^2}-a^2$.

However, this is not very tractable as a differential equation. Instead, you should play a similar game with the one on the right, to obtain $$F'(\xi):=\frac{1}{2}S^{\dagger}(\xi)\left[(a^{ \dagger ^2}-a^2) X + X(a^2-a^{ \dagger ^2})\right] S(\xi).$$ The part in square brackets is a commutator that you can and should evaluate; it reduces as it must to a simple expression, which leaves only a function of $F(\xi)$ in the right-hand side.

Finally, note that if you already know that $F(\xi)$ is expected to be $e^{-\xi}X$, then you know what you can expect $F'(\xi)$ to be in terms of $\xi$ and $X$ and therefore of $F(\xi)$.

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