# Green's function for 1D Klein Gordon equation in position space

I want to derive the Green's function for the 1D Klein Gordon equation in position space. The Klein-Gordon equation in 1D: $$\begin{equation} (\partial_t^2-\partial_z^2+m^2)\phi=f(z,t) \end{equation}$$ where $$f$$ is an arbitrary source. The Green's function is defined as $$\begin{equation} (\partial_t^2-\partial_z^2+m^2)G(z,t)=\delta(z)\delta(t) \end{equation}$$ In Fourier space I get: $$\begin{equation} \hat{G}(k,\omega)=\frac{1}{(-\omega^2+k^2+m^2)}=-\frac{1}{2\sqrt{k^2+m^2}}\Big(\frac{1}{\omega-\sqrt{k^2+m^2}}-\frac{1}{\omega+\sqrt{k^2+m^2}}\Big). \end{equation}$$ The Back transformation is $$\begin{equation} G(z,t)=-\int\frac{dz}{2\pi}e^{ikz}\frac{1}{2\sqrt{k^2+m^2}}\int\frac{d\omega}{2\pi}e^{-i\omega t}\Big(\frac{1}{\omega-\sqrt{k^2+m^2}}-\frac{1}{\omega+\sqrt{k^2+m^2}}\Big) \end{equation}$$ To evaluate the $$\omega$$ integral I have to shift the poles. Here I want to calculate the \textbf{retarded} Greens function and therefore I shift both poles into the lower half plane (see figure). I therefore have to evaluate integrals of the form $$\begin{equation} F(t):=\int\frac{d\omega}{2\pi}e^{-i\omega t}\frac{1}{\omega-\omega_0+i\epsilon},~~~\omega_0\in\mathbb{R} \end{equation}$$ The integral over the circle in the lower half plane can be parameterized by $$\omega=Re^{i\phi}$$ with $$\phi\in (\pi,2\pi)$$. We therefore have for $$t>0:$$ that $$|e^{-i\omega t}|=|e^{-iRe^{i\phi}t}|=|e^{R\sin\phi t}|\overset{R\rightarrow\infty}{\rightarrow}0$$. Therefore in the limit $$R\rightarrow\infty$$ we have $$\begin{equation} F(t)=\oint t\frac{d\omega}{2\pi}e^{-i\omega t}\frac{1}{\omega-\omega_0+i\epsilon}=-i\text{Res}\big[\frac{e^{-i\omega t}}{\omega-\omega_0+i\epsilon}\big]_{|\omega=\omega_0-i\epsilon}=-ie^{-i\omega_0 t} e^{-\epsilon t}\overset{\epsilon\rightarrow0}{\rightarrow}-ie^{-i\omega_0 t} \end{equation}$$ where the minus is coming from running in the mathematical negative direction. Therefore the retarded Green's function is: $$\begin{equation} G(z,t)=i\Theta(t)\int\frac{dk}{2\pi}e^{ikz}\frac{1}{2\sqrt{k^2+m^2}}\Big(e^{-i\sqrt{k^2+m^2} t}-e^{i\sqrt{k^2+m^2} t}\Big) \end{equation}$$ where I have also introduced the $$\Theta(t)$$, since the integrals which I have evalueted with the residue theorem only converge if $$t>0$$.

The question is now how I can evaluate the $$k$$ integral? If I look up the endresult inn wikipedia it should be $$\begin{equation} G(z,t)=\frac{1}{2}\big(1-\sin(m t)\big)\big(\delta(t-z)-\delta(t+z)\big)+\frac{m}{2}\Theta(t-|z|)J_0(um), ~~~u=\sqrt{t^2-z^2} \end{equation}$$ I guess since I calculate the retarded solution, that I cannot get the term $$\delta(t+z)$$? Also I do not have a clue how to evaluate the integral to end up with the result from wikipedia? I have tried several things: Partial integration, substituting a $$\text{sinh}$$ to get rid off the square roots. Nothing has worked so far, so if you have a good idea this would be great!

Many thanks!

• This seems like a pure math question and as such might be better suited to Math.SE – Nephente Aug 2 at 12:27
• Hi, ok I will also post it there! – Jan SE Aug 2 at 12:55