# Squeeze state in quantum SHO, uncertainty

#### Pengwuino

Gold Member
SOLVED: Squeeze state in quantum SHO, uncertainty

1. Homework Statement

So we're working with what are called squeeze states defined as |nu> = S(nu)|0> where nu is a real number, |0> the ground state of the oscillator, and S(nu) = exp[nu(a^2 - a*^2)/2] with a* = hermitian conjugate of a (unless someone can point me towards a dagger on my keyboard :) ). The creation and annihilation operators are the usual a* and a respectively which unfortunately I can't write because i've lost whatever program use to make pretty equations in html :(. I am attempting to find the uncertainties in position and momentum.

3. The Attempt at a Solution

Now I know the usual e^(A)Be^(-A) = B + [A,B] + [A,[A,B]]/2 + [A,[A,[A,B]]]/6 + ...

Finding <x^2> for example, we do <0|S*(nu)x^(2)S*(nu)|0>. We take our above identity and attempt to solve for what this little mess will be.

[A,B] = constant * (aX + Xa - a*X - Xa*) (1)

which is a problem in of itself

Further down the line it gets even worse, so far I have

[A,[A,B]] = constant * (a[X,a^2] - [X,a^2] + a*[X,a^2] - 2aX - 2Xa + [X,a^2]a* + a[X,a*^2] + 2a*X + 2Xa* + [X,a*^2]a - a*[X,a*^2] - [X,a*^2]a*) (2)

It only gets worse from here... Now this can't be the right way to do it or I'm missing something fairly obvious.

The question is, in (1), the <x^2> will have things like <0|aX|0> which really doesn't seem to help considering X, the position operator, doesn't commute with the operator N which gave the states of the harmonic oscillator, thus it will not give me the eigenvalue of X. Am I missing something?

In (2), i can see that equation reducing down into simply a's and a*'s which upon doing [A,[A,[A,B]]], will finally result in constants thus terminating the series. However what's left before that seems to be a deal breaker for the solution. Anyone have any suggestions?

Last edited:
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#### Pengwuino

Gold Member
Ok nevermind, I found my mistake :)

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