# Density Operator, Expectation Value, Coherent States

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?

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.
• For both $\hat{X}$ and $\hat{P}$, you get two terms -- $\hat{a}|n\rangle\langle n|$ and $\hat{a}^\dagger|n\rangle\langle n|$. Therefore, you get terms such as $|n-1\rangle\langle n|$ or $|n+1\rangle\langle n|$ but none of these contain any diagonal terms $|n\rangle\langle n|$ so taking the trace gives you zero. If this short remark is not enough, I'll edit it into the answer in more detail. – Ondřej Černotík May 7 '13 at 8:40
• So for example: You are saying $\langle X \rangle = Tr(\rho X) = e^{-|\alpha|^2} \sum_n \frac{|\alpha|^{2n}}{n !} \langle n \rvert X \rvert n \rangle = e^{-|\alpha|^2} \sum_n \frac{|\alpha|^{2n}}{n !} (\sqrt{n}\langle n \rvert n-1 \rangle+\sqrt{n+1}\langle n \rvert n+1 \rangle) = 0$. Is this how to evaluate the trace correctly? What would happen if I tried to evaluate $\langle N \rangle$, which is equal to $|\alpha|^2$ normally. Is there a good way to evaluate this expectation value using the trace? – Bob Riley May 7 '13 at 9:08
• Yes, that's the way to go. If you wanted to evaluate $\langle\hat{N}\rangle$, you would use the same approach, calculating terms such as $\langle n|\hat{N}|n\rangle$ = n, and should indeed get $|\alpha|^2$. – Ondřej Černotík May 7 '13 at 9:21
• Thank you again for your clear guidance! One more thing... I'm not sure I see how $\langle N \rangle$ reduces to $|\alpha|^2$. You have: $\langle N \rangle = Tr(\rho N) = e^{-|\alpha|^2} \sum_n \frac{|\alpha|^{2n}}{n !} \langle n \rvert N \rvert n \rangle$, but there are still exponential terms of powers of $|\alpha|^{2n}$. How does this all reduce? You are saying it will still be $\langle N \rangle = |\alpha|^2$ still even in this mixed state? I could think of writing $N = a^\dagger a$ and evaluating this way, but I don't see how it simplifies. – Bob Riley May 7 '13 at 9:31
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.