Prove relation for squeezed state - Quantum Information

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SUMMARY

The discussion focuses on proving the relation for squeezed states in quantum information, specifically the equation S(ζ)† a S(ζ) = a cosh r - a† e^{i θ} sinh r, where S(ζ) = e^{1/2[ζ* a² - ζ (a†)²]}. The annihilation operator a is utilized, and the commutation relation [a, a†] = 1 is essential for simplification. Participants suggest using the Taylor expansion of hyperbolic functions and computing the commutator [A, a] to progress in the proof.

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  • Understanding of quantum mechanics, specifically squeezed states
  • Familiarity with the annihilation operator and its properties
  • Knowledge of commutation relations in quantum mechanics
  • Ability to apply Taylor expansions of hyperbolic functions
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  • Explore the Taylor expansion of hyperbolic functions and their significance in quantum mechanics
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Pentaquark5
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Homework Statement


Prove the following relation for ##\zeta:=r e^{i \theta}##:
<br /> S(\zeta)^\dagger a S(\zeta)= a \cosh r - a^\dagger e^{i \theta} \sinh r<br />

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

<br /> \begin{align*}<br /> S(\zeta)^\dagger a S(\zeta)&amp;=\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}} \\<br /> &amp;= e^A \; a \; e^{-A}\\<br /> &amp;=a+[A,a]+\frac{1}{2} \big[A,[A,a]\big]<br /> \end{align*}<br />

Computing the commutator ##[A,a]## yields:
<br /> [A,a]=\frac{1}{1}re^{i\theta}[a^\dagger a^\dagger a-a a^\dagger a^\dagger]<br />

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!
 
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Pentaquark5 said:
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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