- #1
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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!