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What the theta in schwinger function and what is theta formula?

is theta formula general solution of klein gordon equation? if so, what is its coefficient of $\exp\left(-ipx\right)$?

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I have a feeling that nobody looking at this has any idea of what the question your asking is, try rephrasing it? – Benjamin Horowitz Jul 3 '11 at 2:42
He is talking about the exact form of the propagator, with its Bessel functions. – Ron Maimon Sep 3 '11 at 23:55
It would make the question much better to clarify that, of course. – David Z Sep 6 '11 at 2:34

You are talking about the closed form formula for the massive propagator in 4 dimensions. It uses special functions which are of a Bessel function form, and it is extremely unenlightening.

The propagator itself is easily understood to be the Fourier transform of ${i\over k^2 -m^2 +i\epsilon}$, where the $\epsilon$ pole-presecription guarantees that it analytically continues to ${1\over k^2+m^2}$ if you make time purely imaginary. This form is the one that comes from the free field path integral, since the free field path integral for a scalar is just a simple statistical random field in imaginary time.

The best way to understand Schwinger's form is to first find the exact massless propagator. This can be done by using Gauss's law in 4 Euclidean dimensions, because the massless propagator satisfies the four dimensional Laplace's equation with a point source:

$$\nabla^2 \phi = \delta(x)$$

$$\phi(x) = {1\over 2\pi^2 x^2}$$

Where the pi factors come from the area of a 3d sphere in 4d. The analytic continuation of this to real time is:

$$2\pi^2 G(x,t) = {1\over t^2 - x^2 + i\epsilon}$$

And this has a singularity on the light cone which is smoothed over by the $\epsilon$. People often like to separate out this singularity, by replacing the $\epsilon$ with a symmetric prescription for the propagator

$$2\pi^2 G(x,t) = \mathrm{P.V} ({1\over t^2 - x^2 }) + \delta(x^2-t^2)$$

And this form appears in many places. That the limit is a delta function can be seen from the Cauchy-type delta function identity:

$$\delta(x) = {1\over x+i\epsilon} - {1\over x-i\epsilon} $$

I like the simpler form without the delta function and principle value, because in practice you evaluate all propagator integrals by Wick rotating anyway.

But you asked for the massive case. the massive propagator in Euclidean space has an illuminating representation, due to Schwinger:

$$ G(k) = \int d\tau e^{-\tau(k^2 + m^2)}$$

Which is a linear superposition in $\tau$ of spreading x-space Gaussians with normalization which shrinks by a factor of $e^{-m^2}$ every unit of proper time. This is the particle interpretation of the propagator--- the particle is doing Brownian motion in Euclidean space, disappearing with a steady rate.

The Fourier transform of the massive propagator in 3 dimensions is expressible in elementary functions:

$$G(r) = {e^{-mr} \over r}$$

But in 4 dimensions you need Bessel functions. There is nothing useful you can extract from this Bessel function form which goes beyond those properties which are obvious from The Schwinger representation and the elementary massless limit.

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