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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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  • Thank you this is very helpful. I'm having problems evaluating the trace and showing that it is zero. Would you be it be possible to provide some more guidance on this part? – Bob Riley May 07 '13 at 08:32
  • 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 07 '13 at 08:40
  • 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 07 '13 at 09:08
  • 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 07 '13 at 09:21
  • 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 07 '13 at 09:31
  • That's just a power series you need to sum, you should be able to find a formula somewhere on the internet or in some books. Note that although you have terms $|\alpha|^{2n}$, they are divided by $n!$ which guarantees that the sum will converge. – Ondřej Černotík May 07 '13 at 09:35
  • Yea, that's right, but doesn't $\sum_n \langle n \rvert N \rvert n \rangle = \sum_n n$, which does not converge? – Bob Riley May 07 '13 at 09:50
  • No, you still have the terms $|\alpha|^{2n}/n!$ that go with it. – Ondřej Černotík May 07 '13 at 09:51
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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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