# 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?

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 $$\chi_1(\beta) = \int \mathrm d^2\gamma\; P(\gamma)\, \mathrm e^{\beta \gamma^\ast - \beta^\ast \gamma} .$$
• Thanks for your solution! I understand your solution by inserting the relation between $\rho$ and $P(\alpha)$. I have one more question. Is it possible to prove the equivalence without inserting the relation between $\rho$ and $P(\alpha)$. Oct 11, 2022 at 11:12
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!