How would I go about evaluating expectation values like $\langle X \rangle$ and $\langle P \rangle$?
Work I've done:
I've done the integration over $\phi$ and rewrote $\rho$ as:
$\rho = e^{-|\alpha|^2} \sum_n \frac{|\alpha|^{2n}}{n !} \lvert n \rangle \langle n \rvert$, where $n$ is the number of states.
My intuition says to calculate expectation values using $\langle X \rangle = Tr(\rho X)$, but I'm having some difficulty with the calculation. Could someone help flesh out the details?
Since this is for a coherent state, is $\langle X \rangle$ going to be what you normally get for coherent states or will it be different since the state is a mixture?
Best Answer
The calculation $\langle \hat{X}\rangle = \mathrm{Tr}(\hat{\rho}\hat{X})$ is a good way to go. Now you just need to express the operator using creation and annihilation operators, $\hat{X} = \hat{a}+\hat{a}^\dagger$, $\hat{P} = i(\hat{a}^\dagger-\hat{a})$ (up to a constant), and plug it in. Then you just need to play with the expression a bit and, when doing the trace, sum the coefficients of the terms $|n\rangle\langle n|$, i.e., the diagonal elements.
If you had a single coherent state, you would obtain something like $\langle\hat{X}\rangle = \mathrm{Re}(\alpha)$, $\langle\hat{P}\rangle = \mathrm{Im}(\alpha)$ (again, modulo some constant depending on your definition of $\hat{X}$ and $\hat{P}$). But here you have a mixture of coherent states with the same amplitude $|\alpha|$ and different phases $\theta$. This means that in the phase space the states are placed on a circle around the origin with radius $|\alpha|$. Because all of them have the same weight in the mixture and are placed completely symetrically around the origin, the expectation values for both $\hat{X}$ and $\hat{P}$ should be zero.