# 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?

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!
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 $P(\alpha)$. Commented Oct 11, 2022 at 11:12