# Optical theorem for antinormal ordering

In this paper : https://arxiv.org/abs/1206.3405

They consider a density matrix :

$$\rho = \int P(\alpha) |\alpha\rangle \langle \alpha |$$

Where $|\alpha \rangle$ are coherent states.

Using it, we can easily prove their formula (5) :

$$\langle ({a^{\dagger}})^m a^n \rangle =\operatorname{Tr}[\rho ({a^{\dagger}})^m a^n]=\int P(\alpha) \operatorname{Tr}[({a^{\dagger}})^m a^n|\alpha\rangle \langle \alpha |]$$

And we use the circular permutation of the trace + the fact that $\operatorname{Tr}(|\alpha \rangle \langle \alpha|)=1$, and we end up with :

$$\langle ({a^{\dagger}})^m a^n \rangle=\int \alpha^n {\alpha^{*}}^m P(\alpha)$$ that is their formula (5).

But I don't understand how we can find their formula (7) :

Indeed, we would have :

$$\langle a^n ({a^{\dagger}})^m \rangle=\operatorname{Tr}[\rho a^n ({a^{\dagger}})^m ]=\int P(\alpha) \operatorname{Tr}[(a^n {a^{\dagger}})^m |\alpha\rangle \langle \alpha |]$$

But to continue I would need to either know the action :

$({a^{\dagger}})^m |\alpha\rangle$ (I don't remember why exactly, but I know it is not ${\alpha^{*}}^m |\alpha \rangle$)

Or I would need to do the big commutation of the power of creation/annihilation.

Thus I'm a little stuck : how can we prove the formula $(7)$ of the article ?

I have read the pages : https://en.wikipedia.org/wiki/Glauber%E2%80%93Sudarshan_P_representation https://en.wikipedia.org/wiki/Optical_equivalence_theorem and I am still stuck.

What I understand from the first page is that we can write the density matrix either :

$$\rho = \int P(\alpha) |\alpha\rangle \langle \alpha |$$

or

$$\rho_A = \sum_{jk} c_{jk} a^k (a^{\dagger})^k$$

And we have the relationship $P(\alpha)=\frac{1}{\pi} \rho_A(\alpha, \alpha^*)$

So I think the usefull thing to use is the formula with $\rho_A$ but I'm stuck when using it to try to prove (7).

Insert a unity operator in the $|\alpha\rangle$-basis between $a^n$ and $(a^\dagger)^m$. Following the notation in the paper we have $$\langle a^n (a^\dagger)^m \rangle = \operatorname{tr}[\rho_a a^n (a^\dagger)^m] = \int_\alpha \frac{1}{\pi} \langle\alpha|\rho_a a^n (a^\dagger)^m |\alpha\rangle = \int_{\alpha\beta} \frac{1}{\pi^2} \langle\alpha|\rho_a \underbrace{a^n |\beta\rangle}_{\beta^n|\beta\rangle}\underbrace{\langle\beta| (a^\dagger)^m}_{\langle\beta|{\beta^*}^m} |\alpha\rangle = \int_{\alpha\beta} \frac{1}{\pi^2} \langle\beta|\alpha\rangle\langle\alpha|\rho_a |\beta\rangle\beta^n{\beta^*}^m = \int_\beta \frac{1}{\pi}\langle\beta|\rho_a|\beta\rangle \beta^n {\beta^*}^m = \int_\alpha Q_a(\alpha) \alpha^n {\alpha^*}^m.$$
We have used the fact that $\mathbb{I} = \int_\beta \frac{1}{\pi} |\beta\rangle\langle\beta|$ (as also given here).
• I am not sure to understand. Indeed you wrote $\langle \beta | \alpha \rangle = \delta_{\alpha \beta}$ but our base is not orthonormal as we work with coherent states. What's more I'm not sure if we can, to compute the trace write $\int_{\alpha} \langle \alpha \rho_a a^n (a^{\dagger})^m | \alpha \rangle$. Indeed, the $|\alpha\rangle$ covers more than the all space (they span the hilbert space with some recovering). Could you explain please ? – StarBucK Jan 15 '18 at 14:06
• I indeed missed a factor of $1/\pi$ for the trace. The rest was sloppy notation on my part, sorry about that. Hope it is clearer now. – noah Jan 15 '18 at 14:39
• Your last statement as noted in the above comment doesn't seem correct. Instead why don't you just bring $\langle \beta | \alpha \rangle$ to the left and remove identity. – Sunyam Jan 15 '18 at 16:18
• I meant that you can conclude : $\int_{\alpha\beta} \frac{1}{\pi^2} \langle\alpha|\rho_a |\beta\rangle\beta^n{\beta^*}^m \langle\beta|\alpha\rangle = \int_{\alpha\beta} \frac{1}{\pi^2} \langle\beta|\alpha\rangle\langle\alpha|\rho_a |\beta\rangle\beta^n{\beta^*}^m = \int_{\beta} \frac{1}{\pi} \underbrace{\langle\beta|\rho_a |\beta\rangle}_{Humisi-Q-function}\beta^n{\beta^*}^m$. – Sunyam Jan 16 '18 at 10:51