We know that the Klein-Gordon operator is given by $(\partial^2+m^2)=(\partial_\mu\partial^{\mu}+m^2)$, which is used to describe the evolution of relativistic free particles.

How can we show that the relativistic propagator for the free particle $\langle x'|x\rangle$ is a solution to the Klein-Gordon equation? In other words, how do you apply the operators such that $(\partial^2+m^2)\langle x'|x\rangle=0$.

Additional information: the expression for the $\langle x'|x\rangle$ propagator can be written as $$\langle x'|x\rangle=\int\frac{d^3p}{(2\pi)^32\sqrt{\vec{p}^2+m^2}}e^{-\sqrt{\vec{p}^2+m^2}(t'-t)+ip\cdot(\vec{x}'-\vec{x})}=\int\frac{d^4}{(2\pi)^3}\delta(p^2-m^2)\theta(p^0)e^{-ip\cdot(x'-x)},$$ where the last expression is the Lorentz invariant form.

Edit: In a previous exercise I showed that the Feynman propagator in the form:


is a Green's function of the KG operator: $$(\partial^2+m^2)G_F(x'-x)=-i\delta(x'-x).$$

And now in this one I'm being asked to show that the propagator $\langle x'|x\rangle$ is a solution to the KG equation. However, the step function gives me a delta, so I'm not sure how to proceed. Another expression for the propagator in my notes is (after performing integration):

$$\langle x'|x\rangle=\frac{m}{(2\pi)^2}\frac{K_1(m\sqrt{-(x'-x)^2})}{\sqrt{-(x'-x)^2}}$$

Where $K_1$ is the modified Bessel function.

  • $\begingroup$ The propagator is a Green’s function of the KG equation, i.e it satisfies $$(\partial^2 + m^2)\langle x | x’\rangle = \delta(x-x’)$$ $\endgroup$ – bapowell Aug 20 '18 at 22:05
  • $\begingroup$ Thank you for your answer. I see where this comes, given the step function in the Lorentz invariant definition. However, for some reason the exercise I have says to show that $\langle x'|x\rangle$ is indeed a solution of the KG equation. In fact, I actually showed before that the Feynman propagator was a Green's function of KG, and now I must show that $\langle x'|x\rangle$ is the solution. $\endgroup$ – Charlie Aug 21 '18 at 0:00

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