Solution of the PDE II: finite temperature $T\ge 0\leadsto n_B\ge 0$

Since we know the solution for $n_B=0$, we perform a transformation of variables and seek the solution for $n_B>0$ in the form

$$P(z,z^*t)=F(u,u^*,s),\quad{u=ze^{+i\left[ \bar{\Omega} - i\kappa\right]t}, u^*=z^*e^{-i\left[ \bar{\Omega} + i\kappa\right]t},s=t},$$ (87)

which leads to

$$\partial_t P = \left(i \left[ \bar{\Omega} - i\kappa\right]ze^{+i\left[ \bar{\Om... ...ft[ \bar{\Omega} + i\kappa\right]t} \partial_{u^*} +\partial_s\right)F(u,u^*,s)$$

$$= \left(i \left[ \bar{\Omega} - i\kappa\right]z\partial_z - i \left... ...ga} + i\kappa\right]z^* \partial_{z^*}\right) P(z,z^*t) + \partial_s F(u,u^*,s)$$

$$\Check= \left(i \left[ \bar{\Omega} - i\kappa\right]z\partial_z - i \left... ... \partial_{z^*} +2\kappa+2\kappa n_B\partial_z \partial_{z^*}\right) P(z,z^*t),$$ (88)

where in the last line we compared with the original PDE. Therefore, one has

$$\partial_s F(u,u^*,s ) = 2\kappa F(u,u^*,s ) + 2\kappa n_B \partial_z \partial_{z^*}P(z,z^*t)$$

$$= 2\kappa F(u,u^*,s )+ 2\kappa n_B e^{2\kappa s}\partial_u \partial_{u^*}F(u,u^*,s ),$$ (89)

where we used $\partial_z \partial_{z^*}=e^{2\kappa s}\partial_u \partial_{u^*}$, cf. Eq.(7.88). The big advantage now is that we had got rid of the first order derivatives with the $z$,$z^*$-dependent coefficients. Eq.(7.90) is now a standard diffusion equation with time $(s=t)$-dependent coefficients, which can be solved by Fourier transformation:

Reminder: Complex Fourier Transformation, cf (4.141)

$$\tilde{f}(w) \equiv \int d^2z e^{i{\bf zw}} f(z),\quad f(z) = \int \frac{d^2w}{(2\pi)^2}e^{-i{\bf zw}} \tilde{f}(w)$$

$${\bf zw} \equiv \frac{1}{2}\left(zw^*+z^*w\right)=(z_1,z_2)\left( \begin{array}[h]{c} w_1\\ w_2 \end{array}\right)$$ (90)

Reminder: Gauß Integrals

$$\int_{-\infty}^{\infty}dx e^{-ax^2+bx} = \sqrt{\frac{\pi}{a}}e^{\frac{b^2}{4a}},\quad \Re a>0$$ (91)

$$\int \frac{d^2w}{(2\pi)^2} e^{-i{\bf zw}} e^{-\frac{a}{4}{\bf ww}} = \frac{1}{\pi a}e^{-\frac{\vert z\vert^2}{a}}.$$ (92)

We now Fourier-transform Eq.(7.90), $\partial_s F= (2\kappa + 2\kappa n_B e^{2\kappa s}\partial_u \partial_{u^*})F$, to obtain

$$\partial_s \tilde{F}(w,w^*,s) = \left(2\kappa + 2\kappa n_B e^{2\kappa s}\left(-\frac{1}{4}{\bf ww}\right)\right) \tilde{F}(w,w^*,s)$$ (93)

$$\leadsto \tilde{F}(w,w^*,s) = \exp\left\{ 2\kappa s -\frac{1}{4}n_B \left(e^{2\kappa s} -1\right) {\bf ww}\right\} \tilde{F}(w,w^*,s=0)$$

$$\leadsto F(u,u^*,s) = \int \frac{d^2 w}{(2\pi)^2} e^{-i{\bf uw}} \exp\left\{ 2\kappa s ... ...ac{1}{4}n_B \left(e^{2\kappa s} -1\right) {\bf ww}\right\} \tilde{F}(w,w^*,s=0)$$

$$= \int d^2 u' \int \frac{d^2 w}{(2\pi)^2} e^{-i{\bf (u-u')w}} e^{\l... ... -\frac{1}{4}n_B \left(e^{2\kappa s} -1\right) {\bf ww}\right\}} F(u',u'^*,s=0)$$

$$= \frac{e^{2\kappa s}}{\pi n_B \left(e^{2\kappa s} -1\right)} \int ... ...ac{\vert u-u'\vert^2}{n_B \left(e^{2\kappa s} -1\right)}\right\} F(u',u'^*,s=0)$$

Now we remember $u=ze^{+i\left[ \bar{\Omega} - i\kappa\right]t}$, $s=t$, and write $u'=z'$ in $F(u',u'^*,s=0)=P(z',z'^*,t=0)$, to find

$$P(z,z^*,t) = \int d^2 z'\frac{1}{\pi n_B \left(1-e^{2\kappa t}\right)} \exp\le... ...\right]t}-z'\vert^2} {n_B \left(e^{2\kappa t} -1\right)}\right\} P(z',z'^*,t=0)$$

$$= \int d^2 z'\frac{1}{\pi n_B \left(1-e^{-2\kappa t}\right)} \exp\l... ...pa\right]t}\vert^2} {n_B \left(1-e^{-2\kappa t} \right)}\right\} P(z',z'^*,t=0)$$

$$\equiv \int d^2 z' G(z,z';t) P(z',z'^*,t=0),$$

$$G(z,z';t) \equiv \frac{1}{\pi n_B \left(1-e^{-2\kappa t}\right)} \exp\left\{-\frac... ...Omega} - i\kappa\right]t}\vert^2} {n_B \left(1-e^{-2\kappa t} \right)}\right\}.$$ (94)

This is the solution of the initial value problem of the PDE: we have explicitely constructed the propagator $G(z,z';t)$ and expressed the solution of the PDE at times $t>0$ in terms of the initial $P$-distribution $P(z',z'^*,t=0)$.