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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).

Integration contour 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!

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

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