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It is well known that the Feynman propagator can be expressed as a Green function of the Klein-Gordon operator:

$$G_F(x'-x)=i\int\frac{d^4p}{(2\pi)^4}\frac{e^{-ip\cdot(x'-x)}}{p^2-m^2+i\epsilon}$$

On the other hand, we can define the retarded and advanced propagators such as the former exists only if $$(t'-t)>0$$ and the second exists only if $$(t'-t)<0$$. For example, the retarded propagator has the expression:

$$G_R(x'-x)=i\int\frac{d^4p}{(2\pi)^4}\frac{e^{-ip\cdot(x'-x)}}{(p^0-E_p+i\delta)(p^0+E_p+i\delta)}$$

Or performing the temporal integral:

$$G_R(x'-x)=\theta(x'^{0}-x^0)\int\frac{d^3p}{(2\pi)^3}\frac{e^{-i\vec{p}\cdot(\vec{x'}-\vec{x})}}{2E_p}(e^{iE_p(x'^{0}-x^0)}-e^{-iE_p(x'^{0}-x^0)})$$

It is well known that the Feynman propagator is a Green function of the Klein-Gordon operator. However, how can we verify that the retarded propagator is a Green function of the Klein-Gordon operator?

From what I've read, the Feynman and the retarded propagator differ by a solution for the Klein-Gordon equation, but all the books and articles I've seen skip this demonstration.

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\begin{equation} G_F(x'-x)=i\int\frac{d^4p}{(2\pi)^4}\frac{e^{-ip\cdot(x'-x)}}{p^2-m^2+i\epsilon} \end{equation}

we can in fact, do the one integral to get,

\begin{equation} G_F(x'-x)=\int\frac{d^3p}{(2\pi)^3}\frac{e^{-i\vec{p}\cdot(\vec{x'}-\vec{x})}}{2E_p}(-e^{iE_p(x'^{0}-x^0)}\theta(x'^{0}-x^0)-e^{-iE_p(x'^{0}-x^0)}\theta(x^0 - x'^{0})) \end{equation}

\begin{equation} G_R(x'-x)=\theta(x'^{0}-x^0)\int\frac{d^3p}{(2\pi)^3}\frac{e^{-i\vec{p}\cdot(\vec{x'}-\vec{x})}}{2E_p}(e^{iE_p(x'^{0}-x^0)}-e^{-iE_p(x'^{0}-x^0)}) \end{equation}

Then, $G_{F} - G_{R}$ is given by (also taking $x^{0} = 0$, $x^{\prime 0} = t$ )

\begin{equation} G_{F}(x') - G_{R}(x')= \int\frac{d^3p}{(2\pi)^3} \frac{e^{-i(\vec{p}\cdot\vec{x'} + E_{p}t)}}{2E_p} \end{equation}

which is indeed a solution of the homogeneous Klein Gordon equation i.e $ (\Box + m^{2})[G_{F}(x') - G_{R}(x')] = 0$

I might be little sloppy with i's and other things but this is the basic idea.

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  • $\begingroup$ Thank you very much, I think I was misreading the question, but I now understand better the concepts. I redid the calculations and they do agree, and also verified something similar to practice with the advanced propagator. $\endgroup$ – Charlie Aug 26 '18 at 15:38

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