# Computing ${\rm tr}\left[\hat{a}^{\dagger}\hat{a}\hat{D}(\alpha)\right]$

I'm having some trouble computing $${\rm tr}\left[\hat{a}^{\dagger}\hat{a}\hat{D}(\alpha)\right]$$ where $$\hat{D}(\alpha)$$ is the displacement operator.

I tried doing this using the number state basis, which allows us to write the trace of an operator $$\hat{A}$$ as $${\rm tr}\left(\hat{A}\right) = \sum_{n \geq 0} \langle n |\hat{A} | n \rangle$$ Using that $$\hat{a}^{\dagger}\hat{a}$$ sends $$|n \rangle \mapsto n |n \rangle$$, I got that $${\rm tr}\left[\hat{a}^{\dagger}\hat{a}\hat{D}(\alpha)\right]= \sum_{n \geq 0} n \langle n |\hat{D}(\alpha) | n \rangle,$$ but am unable to continue from here. I was wondering if someone could direct me to another approach or a reference where this is worked out. (For context, I actually want to compute the Wigner function of $$\hat{a}^{\dagger}\hat{a}$$.)

• Commented Jul 19, 2021 at 22:34
• You know the resolution of the identity in terms of coherent states, and the action of D on them, and the projection of number states on them, so... Commented Jul 20, 2021 at 0:26
• Oh, I can maybe use the resolution of the identity to rewrite the number states in terms of coherent states. I'll try this and see if I can get something apparent. Commented Jul 20, 2021 at 17:23
• So I wrote $|n \rangle$ as $\frac{1}{\pi} \int d^{2}\alpha \langle \alpha| n \rangle |\alpha \rangle$. Doing so and using that $D(\alpha)|\alpha \rangle = \alpha |\alpha \rangle$, I obtained that $\langle n | D(\alpha) |n \rangle = \frac{1}{\pi{n!}} \int d^{2}\alpha ( \alpha^{2n} e^{-1/2|\alpha|^{2}})$. So I guess I need to evaluate the last expression Commented Jul 20, 2021 at 18:32
• Except D(α) shifts the eigenvalue of the coherent state it acts on!! Commented Jul 20, 2021 at 19:03