I have the retarded propagator for a free scalar field in 1+1 dimensions. Inside the light cone, this looks like $J_0(m \sqrt(t^2-x^2))$, $J$ being a Bessel function. When I take the massless limit, however, this goes to a constant.

In 3+1 dimensions, where I have a $J_1$ function as my retarded propagator, this goes to 0. Intuitively this makes sense to me, since a massless particle only travels at the speed of light. But what's different in two dimensions so that the propagator goes to a finite constant for $x \ne \pm t$?

Extra question: how do I obtain an expression for the propagator on the light cone ($t= \pm x$)? I know it's going to diverge and I'm going to find a delta function, but how do I exactly get it?


2 Answers 2


The propagators of massless Bosons are only zero inside the lightcone if the number of spatial dimensions is odd and larger than 1.

You can determine the exact form of the propagator on the light-cone, even though it is divergent, and do so for any number of spatial dimensions, in the following way:

Use this expression to derive the propagator in d spatial dimensions from the 1 dimensional case with,

$P_d{(t,r)}\ =\ \frac{1}{2\pi^a}\ \frac{\partial^a }{\partial (s^2)^a}\ P_1{(t,r)} $

where $~P{(t,r)}~$ is the propagator, with $~~a=(d-1)/2~$ and $~~s^2=t^2-r^2$

For the real valued Klein Gordon propagator this becomes:

$P_d^{KG}{(t,r)}\ =\ \frac{1}{2\pi^a}\ \frac{\partial^a }{\partial (s^2)^a} \left\{\ {\cal H}(s^2) J_o(ms)\ \right\}$

Where ${\cal H}(s^2)$ is the Heaviside step function which is 1 inside the light-cone and 0 outside the light-cone. In the mass-less case this becomes:

$P_d^{KG}{(t,r)}\ =\ \frac{1}{2\pi^a}\ \frac{\partial^a }{\partial (s^2)^a} \left\{\ {\cal H}(s^2)\ \right\}$

The following graphs showing the mass-less cases are from numerical simulations:

Simulated mass-less boson propagators in various dimensions

In the case with 1 spatial dimension you see the Heaviside step function. The 3d case is the first order derivative of the step function. The 5d case is the 2nd order derivative of the step-function and so on.

The even dimension cases are non-zero inside the light-cone because of the 1/2 order derivatives.

The operator which derives the d-dimensional propagator from the 1-dimensional propagator is derived in my paper here: http://www.physics-quest.org/Higher_dimensional_EM_radiation.pdf in section V.


Using the poisson representation of bessel function $J_n(z) = \frac{(z/2)^n}{\sqrt{\pi} \Gamma(n+1)} \int_0^{\pi} \cos{(z\cos{\theta})} \sin^{2n}{\theta} d\theta$, you can determine the constant..

In 3+1 d, for example, for a space-like distance, we can make a lorentz transformation such that $r = x-y$ is purely spatial, the amplitude is then $D(x-y) \sim \int dp \frac{p}{\sqrt{p^2 +m^2}} e^{ipr} \sim \delta{(r)}$..(I use Peskin's notation, c.f, Chap.2.4, Eqn(2.52))

  • $\begingroup$ I know what constant it goes to, since $J_0(0)=1$. What I don't understand is the physically what this means, since I expect it to go to 0. Why is $\int dk \frac{p}{\sqrt(p^2+m^2)} e^{ipr} $ a delta function? I know it diverges for $r=0$, but I would like to see how this is a delta function, since I normally see the delta function as $\delta(x)=\int \frac{dp}{2 \pi} e^{ipr}$ $\endgroup$
    – user22710
    Oct 14, 2014 at 9:01
  • $\begingroup$ I omitted the constant in the calculation for simplicity……Maybe you can consult Peskin to find the constant $\endgroup$ Oct 14, 2014 at 12:36
  • $\begingroup$ Sorry, I don't understand your answer. Inside the causal patch ($t>x$, strictly bigger) my propagator doesn't go to 0 and I find this wierd, unregarding on which constant you go to. Besides, I don't know how to find the delta function for $t=x$, which for light like separation. I have Peskin & Schroeder and I've read what you said, but I don't see any delta function, they're just considering the long distance behavior of the propagator for spacelike or timelike separation, I'm interested in the light like case $\endgroup$
    – user22710
    Oct 14, 2014 at 13:27

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