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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?

Bob Riley
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2 Answers2

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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.

Ondřej Černotík
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Alternatively to Ondřej's answer, you can also see your density operator as a probabilistic mixture of number states, $$\rho=\sum_{n=0}^\infty p_n|n\rangle\langle n|\quad \text{with}\quad p_n=e^{-|\alpha|^2}\frac{|\alpha|^{2n}}{n!},$$ all of whose position averages are zero. Thus $$\text{Tr}(\rho X)=\sum_{n=0}^\infty p_n\text{Tr}\left(|n\rangle\langle n|X\right)=\sum_{n=0}^\infty p_n\langle n|X|n\rangle=0.$$ The take-home message in this is that convex decompositions of density operators are in general not unique for mixed states.

Emilio Pisanty
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