Glauber P function of Schrödinger cat state I am trying to find the Glauber P function for a superposition of Coherent States:
$$\psi=\frac{1}{\sqrt{2}}(|\beta\rangle +|-\beta\rangle)$$ where $\beta$ is real.
The P function is defined as:
$$P(\alpha)=\frac{e^{|\alpha|^2}}{\pi^2}\int d^2u \langle-u|\rho|u\rangle e^{|u|^2}e^{\alpha u^*-\alpha^*u}$$
Using the orthogonality relation $$\int d\alpha^2 e^{\alpha (z-z')^*-\alpha^* (z-z')}=\pi^2 \delta^{(2)}(z-z')$$ one can evaluate two of the terms very easily:
$$\int d^2u \langle-u|\beta\rangle\langle \beta|u\rangle e^{|u|^2}e^{\alpha u^*-\alpha^*u}=\pi^2 e^{-|\beta^2|}\delta^{(2)}($$
$$\int d^2u \langle-u|-\beta\rangle\langle -\beta|u\rangle e^{|u|^2}e^{\alpha u^*-\alpha^*u}=\pi^2 e^{-|\beta^2|}\delta^{(2)}(\alpha-\beta)$$
However the mixed terms are not so easy to evaluate:
I get:
$$\int d^2u \langle-u|-\beta\rangle\langle \beta|u\rangle e^{|u|^2}e^{\alpha u^*-\alpha^*u}=\int d^2u e^{-|\beta|^2}e^{u(\beta-\alpha)^*}e^{u^*(\beta+\alpha)}$$
This one is not so simple to evaluate, because one can't use the orthogonality relation so easily.
I tried expression $u$ and $\alpha$ as complex numbers, but then I get the Fourier Transform of an exponential, which blows up. There must be something wrong or something simple is eluding me.
I found this publication https://www.researchgate.net/publication/230943507_Non-Classical_States_of_the_Electromagnetic_Field and also the wikipedia article https://en.wikipedia.org/wiki/Glauber%E2%80%93Sudarshan_P_representation contains in the last chapter something for this.
 A: There is a small trick that can be employed to calculate this term:
Starting from:
$$\int d^2u \langle -u|\beta\rangle\langle -\beta|u\rangle e^{|u|^2}e^{\alpha u*-\alpha^*u}=  \int d^2u e^{-|\beta|^2}e^{-\beta u^*}e^{-\beta^* u}e^{\alpha u*-\alpha^*u}$$
Now Taylor expanding the middle two exponentials we get:
$$\int d^2u \sum_{n,m}e^{-|\beta|^2}\frac{(-u^*\beta)^n}{n!}\frac{(-u\beta^*)^m}{m!}e^{\alpha u*-\alpha^*u}$$
However, $u^*$ is an additiona factor one gets if one differentiates the last exponential with respect to $\alpha$ and similarly for $u$ and $\alpha^*$ Therefore one can write this as:
$$\int d^2u \sum_{n,m}e^{-|\beta|^2}\frac{(-\frac{\partial}{\partial \alpha}\beta)^n}{n!}\frac{(\frac{\partial}{\partial \alpha^*}\beta^*)^m}{m!}e^{\alpha u*-\alpha^*u}$$
Now one can pull out the factors containing the derivatices from the integral due to linearity and one gets:
$$\sum_{n,m}\frac{(-\frac{\partial}{\partial \alpha}\beta)^n}{n!}\frac{(\frac{\partial}{\partial \alpha^*}\beta^*)^m}{m!}e^{-|\beta|^2}\int d^2u e^{\alpha u*-\alpha^*u}$$
The term inside the integral is just the two dimensional Dirac Delta leaving us with:
$$\pi^2e^{-|\beta|^2}\sum_{n,m}\frac{(-\frac{\partial}{\partial \alpha}\beta)^n}{n!}\frac{(\frac{\partial}{\partial \alpha^*}\beta^*)^m}{m!}\delta^{(2)}(\alpha)$$
And now reversing the Taylor expansion we get and invoking that $\beta$ is real
$$\pi^2e^{-|\beta|^2}e^{-\frac{\partial}{\partial \alpha}\beta+\frac{\partial}{\partial \alpha^*}\beta}\delta^{(2)}(\alpha)$$
