# Squeezed gaussian expectation

1. Mar 7, 2008

### noospace

I'm trying to evaluate the expectation of position and momentum of

$\exp\left(\xi (\hat{a}^2 - \hat{a}^\dag^2)/2\right) e^{-|\alpha|^2} \sum_{n=0}^\infty \frac{\alpha^n}{\sqrt{n!}} |n\rangle}$

where $\hat{a},\hat{a}^\dag$ are the anihilation/creation operators respectively.

Recall $\hat{x} = \sqrt{\hbar/(2m\omega)}(\hat{a}^\dag + \hat{a})$.

The calculation of $\langle \hat{x} \rangle$ and $\langle \hat{p} \rangle$ are easy if one uses the Hadamard lemma

$e^X Y e^{-X} = Y+ [X,Y] + (1/2)[X,[X,Y]] + \cdots$.

I'm running into touble in the evaluation of $\langle\hat{x}^2\rangle,\langle\hat{p}^2\rangle$. In particular in the calculation of

$\exp\left(\xi (\hat{a}^\dag^2-\hat{a}^2)/2\right)(\hat{a}^\dag + \hat{a})^2\exp\left(-\xi (\hat{a}^\dag^2-\hat{a}^2)/2\right)$.

I was able to show that

$\exp\left(\xi (\hat{a}^\dag^2-\hat{a}^2)/2\right)(\hat{a}^\dag + \hat{a})^2\exp\left(-\xi (\hat{a}^\dag^2-\hat{a}^2)/2\right) = (\hat{a}^\dag + \hat{a})^2 + \xi([\hat{a}^2,\hat{a}^\dag^2]+[\hat{a}^\dag\hat{a},\hat{a}^\dag^2-\hat{a}^2]) + \frac{\xi}{2!}[(\hat{a}^\dag+\hat{a})^2,[\hat{a}^2,\hat{a}^\dag^2]+[\hat{a}^\dag\hat{a},\hat{a}^\dag^2-\hat{a}^2]]+\cdots$

but I can't figure out how to simplify this.

Last edited: Mar 7, 2008
2. Mar 8, 2008

### soarce

You may try to expand $(a^\dagger +a)^2=1+a^\dagger a + (a^\dagger)^2 + a^2$, then $a^\dagger a$ is the photon number operator...

Few years ago I did similar calculations and I was using the book of J.S.Peng and G.X. Li "introduction to modern quantum optics" where they give many results of some expectation values (however, you must check carefully their formulas against typing errors)
Also W.H. Louisell "Quantum statistical properties of radiation" may contain similar calculations.

but you may try to have a look directly at "the bible of quantum optics": Mandel & Wolf "Optical coherence and quantum optics"

3. Mar 8, 2008

### olgranpappy

oh?
it is also easy, if one knows how to do it.