- #1

- 17

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## Homework Statement

Prove the following relation for ##\zeta:=r e^{i \theta}##:

[tex]

S(\zeta)^\dagger a S(\zeta)= a \cosh r - a^\dagger e^{i \theta} \sinh r

[/tex]

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

## Homework Equations

See above

## 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

[tex]

\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*}

[/tex]

Computing the commutator ##[A,a]## yields:

[tex]

[A,a]=\frac{1}{1}re^{i\theta}[a^\dagger a^\dagger a-a a^\dagger a^\dagger]

[/tex]

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?

Thanks in advance!