I am studying Scully and Zubiary's quantum optics book currently, and I ran across their definition of the P representation as:
$$P(\alpha,\alpha^*) = Tr[\rho \delta(\alpha^* - a^\dagger) \delta(\alpha - a)] $$$$P(\alpha,\alpha^*) = Tr[\rho \delta(\alpha^* - a^\dagger) \delta(\alpha - a)]. $$
They then go on to say that you can see that it is normalized by finding $\int P(\alpha,\alpha^*) d^2\alpha = 1$. In my attempt to do this, I inserted two sets of coherent states on either side of the delta functions such that (after applying the cyclical property of the trace to move $\rho$ to the middle):
$$ P(\alpha, \alpha^*) = \sum_n \frac{1}{\pi^2} \int \langle n | \beta \rangle \langle \gamma | n \rangle \langle \beta | \rho | \gamma \rangle \delta(\alpha - \beta) \delta(\alpha^* - \gamma^*) d^2 \beta \hspace{1mm} d^2 \gamma $$$$ P(\alpha, \alpha^*) = \sum_n \frac{1}{\pi^2} \int \langle n | \beta \rangle \langle \gamma | n \rangle \langle \beta | \rho | \gamma \rangle \delta(\alpha - \beta) \delta(\alpha^* - \gamma^*) d^2 \beta \hspace{1mm} d^2 \gamma. $$
Upon integrating over the delta functions, I get:
$$P(\alpha, \alpha^*) = \sum_n \frac{1}{\pi^2} \langle \alpha | n \rangle \langle n | \alpha \rangle \langle \alpha | \rho | \alpha \rangle = \frac{1}{\pi^2} \langle \alpha | \rho | \alpha \rangle $$
Which I don't believe is correct. So if anyone can give me any pointers or tips on how to properly demonstrate the normalization here, that would be wonderful. Thanks in advance!
