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I am studying Scully and Zubiary's quantum optics book currently, and I ran across their definition of the P representation as:

$$P(\alpha,\alpha^*) = Tr[\rho \delta(\alpha^* - a^\dagger) \delta(\alpha - a)]. $$

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):

$$ P(\alpha, \alpha^*) = \sum_n \frac{1}{\pi^2} \int \langle n | \beta \rangle \langle \gamma | n \rangle \langle \beta | \rho | \gamma \rangle \delta(\alpha - \beta) \delta(\alpha^* - \gamma^*) d^2 \beta \hspace{1mm} d^2 \gamma. $$

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!

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  • $\begingroup$ @CosmasZachos Why would the delta functions only collapse 2 of the integrals and not 4? $\endgroup$
    – user132849
    Commented Apr 6, 2021 at 2:00
  • $\begingroup$ You have 2, not 4 such, at first. $\endgroup$
    – Cosmas Zachos
    Commented Apr 6, 2021 at 3:04
  • $\begingroup$ @CosmasZachos You're saying the initial integral $\int P d^2 \alpha$ is 2 integrals? If so, I agree, but when inserting the other 4 ($d^2 \beta$ and $d^2 \gamma$), I believe that the delta functions $\delta(\alpha^* - \gamma^*)$ and $\delta(\alpha - \beta)$ are each two dimensional delta functions. Thus, I should get, for example, $\int f(\beta) \delta(\alpha - \beta) d^2 \beta = f(\alpha)$, right? $\endgroup$
    – user132849
    Commented Apr 6, 2021 at 4:12
  • $\begingroup$ Might like this one. $\endgroup$
    – Cosmas Zachos
    Commented Apr 6, 2021 at 15:57

1 Answer 1

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$$\int P(\alpha,\alpha^*) d^2\alpha =\int \!\! d^2\alpha ~ \operatorname{Tr}\left ( \delta(\alpha - a)~\rho ~\delta(\alpha^* - a^\dagger) \right ) \\ = \frac{1}{\pi^2} \int \!\! d^2\alpha~ d^2 \beta ~ d^2 \gamma~ \operatorname{Tr}\left ( \delta(\alpha - \beta)\langle \gamma | \beta\rangle \langle \beta | \rho | \gamma \rangle \delta(\alpha^* - \gamma^*) \right ). $$

Now, glibly, the two integrals in α and cc and the two complex δ functions disappear, since no integrand depends on those, hence
$$ ... = \frac{1}{\pi^2} \int \!\! d^2 \beta ~ d^2 \gamma~ \operatorname{Tr}\left ( ~ |\gamma\rangle \langle \gamma | \rho | \beta \rangle\langle \beta| ~\right ) \\ = \operatorname{Tr} ( \rho )=1.$$

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  • $\begingroup$ Ah I see. So if you perform it in the other order, then it's not a problem. But what if I wanted to do it by first integrating over $\beta$ and $\gamma$? And thank you very much for the answer and the reference to the paper, by the way! $\endgroup$
    – user132849
    Commented Apr 6, 2021 at 16:30
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    $\begingroup$ Apologies for not clarifying this in my comment... You'd be stuck with partial variables for β,γ, and then the coherent state overlap, and Gaussians, deltas, Jacobians... and you'd need that reference... while α is just dross... $\endgroup$
    – Cosmas Zachos
    Commented Apr 6, 2021 at 16:38
  • $\begingroup$ You mean, for example, instead of having $d^2\beta$, I might just have $dx_{\beta}$? If so, I guess that's where my confusion comes in. I (correctly or incorrectly) am understanding the deltas as two dimensional deltas that require integration over both $dx_{\beta}$ and $dy_{\beta}$. $\endgroup$
    – user132849
    Commented Apr 6, 2021 at 16:42
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    $\begingroup$ They are not, even some mathematicians go there, as per JMP paper in the comments.... $\endgroup$
    – Cosmas Zachos
    Commented Apr 6, 2021 at 17:36
  • $\begingroup$ Hmm. Okay well I won't waste any more of your time. Thank you so much for a clear answer and all the help! $\endgroup$
    – user132849
    Commented Apr 6, 2021 at 19:19

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