Question about Glauber-Sudarshan $P$ representation I'm reading Scully's 'Quantum Optics'. I've got some question about the Glauber-Sudarshan $P$ representation.
It's straight forward that
$$
P(\alpha) = \frac{e^{\vert \alpha \vert ^2}}{\pi^2} \int \left<- \beta\right|\rho \left| \beta \right> e^{\vert \beta \vert ^2 -\beta \alpha ^* +\beta ^* \alpha} d^2 \beta
$$
while $P(\alpha)$ can also be represented using characteristic function
$$
P(\alpha) = \frac{1}{\pi^2} \int e^{-\beta \alpha ^* +\beta ^* \alpha} \chi _1(\beta) d^2 \beta
$$
in which
$$
\chi_1 (\beta) = Tr[\hat{D}_\beta \rho]e^{\frac{\vert \beta \vert ^2}{2}}
$$
$\hat{D}_\beta$ is the displacement operator.
Comparing these two formulas, I just cannot get them equivalent. Can anybody give me any tips on how to demonstrate these two formulas are equivalent?
 A: I'll assume that we already know that a $P$-representation exists, i.e., that we can always write
$$\rho = \int \mathrm d^2\gamma\; P(\gamma)\, |\gamma \rangle\langle \gamma| \tag{1}$$ for some function or distribution $P(\gamma)$. Then it suffices to plug in (1) into both expressions
$$ P_1(\alpha) = \frac{\mathrm e^{\vert \alpha \vert ^2}}{\pi^2} \int \mathrm d^2\beta\; \left\langle -\beta \right| \rho \left| \beta \right\rangle\, \mathrm e^{\vert \beta \vert ^2 -\beta \alpha ^* +\beta ^* \alpha} $$
and
$$ P_2(\alpha) = \frac{1}{\pi^2} \int \mathrm d^2\beta\; \mathrm e^{-\beta \alpha ^* +\beta ^* \alpha}\, \chi _1(\beta) $$
and find that $P_1(\alpha) = P(\alpha)$ and $P_2(\alpha) = P(\alpha)$. [NB: You have a sign wrong in the exponent of $P_1$!]
Both calculations are straightforward if you use $\int \mathrm d^2 z\; \mathrm e^{zw^\ast - z^\ast w} = \pi^2 \delta^{(2)}(w)$. I will not write those calculations out here but give the intermediate step
$$ \chi_1(\beta) = \int \mathrm d^2\gamma\; P(\gamma)\, \mathrm e^{\beta \gamma^\ast - \beta^\ast \gamma} . $$
A: There are many useful properties that you can take advantage of: one is
$$\rm{Tr}[D(\alpha)D(-\beta)]=\pi \delta^2(\alpha-\beta).$$ This lets you write, for any operator $F$ and any bounded operator $G$ (e.g., Cahill+Glauber 1969),
$$\rm{Tr}[FG]=\frac{1}{\pi}\int d^2\xi \rm{Tr}[F D(\xi)]\rm{Tr}[GD(-\xi)].$$
Applying this to the first expression finds
\begin{align}
P(\alpha)&=\frac{e^{|\alpha|^2}}{\pi}\int d^2\beta\langle -\beta|\rho|\beta\rangle e^{|\beta|^2+\beta^*\alpha-\beta\alpha^*}\\
&=\frac{e^{|\alpha|^2}}{\pi^2}\int d^2\beta d^2\xi\rm{Tr}[D(-\xi)|\beta\rangle\langle -\beta|]\rm{Tr}[D(\xi)\rho] e^{|\beta|^2+\beta^*\alpha-\beta\alpha^*}.
\end{align} You should be able to work out the first trace from properties of displacement operators and/or coherent states. The second trace starts to look like the characteristic function. Then, if you can do the integral over $\beta$ and change the name of the variable $\xi$, you'll have your answer, without ever assuming that the $P$-function takes a particular diagonal form!
