Question about Glauber-Sudarshan $P$ representation

Question

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?

Answer 1

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!

Answer 2

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 $$ \chi_1(\beta) = \int \mathrm d^2\gamma\; P(\gamma)\, \mathrm e^{\beta \gamma^\ast - \beta^\ast \gamma} . $$