# Prove relation for squeezed state - Quantum Information

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1. Apr 14, 2016

### Pentaquark5

1. The problem statement, all variables and given/known data
Prove the following relation for $\zeta:=r e^{i \theta}$:
$$S(\zeta)^\dagger a S(\zeta)= a \cosh r - a^\dagger e^{i \theta} \sinh r$$

with $S(\zeta)=e^{1/2[\zeta^\ast a^2-\zeta(a^\dagger)^2]}$ and $a$ being the annihilation operator with eigenvalue $\alpha$.

2. Relevant equations
See above

3. The attempt at a solution
I used the following identity: $e^{At}Be^{-At}=B+\frac{t}{1!}[A,B+\frac{t^2}{2!}\big[A,[A,B]\big]+\mathcal{O}$ to write

\begin{align*} S(\zeta)^\dagger a S(\zeta)&=\underbrace{\left(e^{1/2[\zeta^\ast a^2-\zeta(a^\dagger)^2]}\right)^\dagger}_{e^A} a \underbrace{\left(e^{1/2[\zeta^\ast a^2-\zeta(a^\dagger)^2]}\right)}_{e^{-A}} \\ &= e^A \; a \; e^{-A}\\ &=a+[A,a]+\frac{1}{2} \big[A,[A,a]\big] \end{align*}

Computing the commutator $[A,a]$ yields:
$$[A,a]=\frac{1}{1}re^{i\theta}[a^\dagger a^\dagger a-a a^\dagger a^\dagger]$$

I am not sure what to do with this expression, though. Is there any way to simplify it?

If not, what other way is there to prove the identity?

Consider substituting $a^\dagger a$ with the help of the commutation $[a,a^\dagger]=1$. Unfortunately, the above commutator expansion will have no constant or vanishing term so that you have to compute "all" terms (if you are clever enough, you should be ablo to find the pattern, though),