Prove relation for squeezed state - Quantum Information

  • #1

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!
 

Answers and Replies

  • #2
I am not sure what to do with this expression, though. Is there any way to simplify it?
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),
You will also need to know the Taylor expansion of hyperbolic functions.
 

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