This question is related to this one. I'm assuming that we're in or on the the light-cone $s \leq 0$ in what follows.

Suppose I'm interested in computing the following Fourier transform, in the massless $m=0$ case: $$ H(x,y;m) = \int \frac{d^{4}p}{(2\pi)^4} \frac{e^{-i p \cdot (x-y)}}{(p^2 + m^2 + i \epsilon)^2} $$

The $\epsilon$ is there in the Feynman-prescription sense. We know that the massive $m \neq 0$ propagator is given by; $$ G_0(x,y;m) \ = \ \int \frac{d^4p}{(2\pi)^4} \frac{e^{-i p \cdot (x-y)}}{p^2 - m^2 + i \epsilon} \ = \ - \frac{i}{4 \pi^2} \frac{m}{\sqrt{-s}} K_{1}\left( m\sqrt{-s} \right) - \frac{1}{4 \pi} \delta(s) $$

Where $s = (x^0 - y^0)^2 - (\mathbf{x} - \mathbf{y})^2$. Setting $\mu = m^2$, we notice that; $$ \frac{\partial G_0(x,y;\sqrt{\mu})}{\partial \mu} = \frac{\partial }{\partial \mu } \left( \int \frac{d^4p}{(2\pi)^4} \frac{e^{-i p \cdot (x-y)}}{p^2 - \mu + i \epsilon} \right) = \int \frac{d^{4}p}{(2\pi)^4} \frac{e^{-i p \cdot (x-y)}}{(p^2 - \mu + i \epsilon)^2} = H(x,y;\sqrt{\mu}) $$

So differentiating in the above manner, I have been able to find: $$ H(x,y;m) = \frac{i}{8 \pi^2} K_{0}\left( m \sqrt{-s} \right) $$

This seems like a good answer, however things go wrong when I set $m=0$. I find that: $$ H(x,y;0) \ = \ \lim_{m\to 0^+} \left[ \frac{i}{8 \pi^2} \log\left( \frac{m(-s)}{4} \right) \right] $$

Meaning the Fourier transformed function diverges like a log as I take the mass to zero. Why is this happening! Surely we can examine massless scalar theories? How can a rectify the above?

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    $\begingroup$ What makes you think that this Fourier transform is free from the IR divergence you are observing? You have $d^4 p/p^4$ for small $p$, which suggests a logarithmic IR divergence . I think the $i\epsilon$ prescription you are using is not guaranteed to define the propagator as a distribution near $p=0$ (not $p^2=0$ -- it does regulate the null cone singularity). $\endgroup$ – Peter Kravchuk Dec 14 '17 at 7:26
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    $\begingroup$ No, I am saying that $1/(p^2+m^2+i\epsilon)^2$ is well-defined as a distribution away from $p=0$. This is because the only problem can come from the on-shell momenta $p^2+m^2=0$, but away from $p=0$ your $i\epsilon$ prescription can be interpreted as adding $\pm i\epsilon$ to $p^0$. This means that you have a boundary value of a holomorphic function near such points, and I believe the conditions are satisfied that for it to give a good distribution. Near $p=0$ you cannot re-interpret your $i\epsilon$-prescription that way. $\endgroup$ – Peter Kravchuk Dec 17 '17 at 22:20
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    $\begingroup$ For $m\neq 0$ your function is integrable near $p=0$ anyway, so you get a fine Fourier transform. For $m=0$ it is not obvious that your $i\epsilon$ prescription helps at $p=0$. You can regulate $p=0$ otherwise, and then the distribution you get will be ambiguous up to terms localized at $p=0$, which are $\delta$-functions and its derivatives. Since if you integrate your momentum space function agains something which vanishes at $p=0$ the result is unambiguous, it means you can only have delta-function ambiguity. This exactly what you are seeing. $\endgroup$ – Peter Kravchuk Dec 17 '17 at 22:26
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    $\begingroup$ Other ways to phrase it: by dimensional analysis your result should be dimensionless, and if $m=0$ you don't have anything to multiply $s$ by to make it dimensionless. Or, if you differentiate your Fourier transform, then you get transform of $p/(p^2+i\epsilon)^2$, which is free of $p=0$ problem, and thus $\partial_x H(x,y;0)$ is unambiguous, but this only defines $H(x,y;0)$ up to a constant shift. $\endgroup$ – Peter Kravchuk Dec 17 '17 at 22:30
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    $\begingroup$ Yes, you can regulate the divergence in various ways, and they all are going to differ by an additive constant, because you know the derivatives of $H$ unambiguously. Regarding your second question, $G(x,y;m)$ is not analytic at $m=0$, as you basically have shown yourself. The expansion of $K_1(x)$ goes like $\frac{1}{x}+\frac{1}{4}x\log c x+o(x)$ for some $c$. $\endgroup$ – Peter Kravchuk Dec 18 '17 at 5:33

The problem is that the Green's function is not analytic on the light cone, so the order of limits matters.

One approach would be to go back to the differential equation that this chain is meant to solve. You can show that if $$\left[\frac{\partial^2}{\partial t^2} - \nabla^2\right] G(t,\mathbf{x};t',\mathbf{x}') = \delta(t-t')\,\delta(\mathbf{x}-\mathbf{x}')$$ and we want to find a function that satisfies $$\left[\frac{\partial^2}{\partial t^2} - \nabla^2\right]^2 G_2(t,\mathbf{x};t',\mathbf{x}') = \delta(t-t')\,\delta(\mathbf{x}-\mathbf{x}')$$ then the following works $$G_2(t,\mathbf{x};t',\mathbf{x}') = \int \operatorname{d}^4 x'' G(t,\mathbf{x};t'',\mathbf{x}'')\, G(t'',\mathbf{x}'';t',\mathbf{x}').$$

Though where to take this "back to the beginning" approach from here to solve the problem, I'm not sure. One idea is to Fourier transform the spatial coordinates, giving you a 4th order ODE you can solve per mode, though it isn't clear what boundary conditions you'd need to apply to that to get the Feynman propagator.

I think your best bet, then, would be to take what I'd describe as "the standard approach." You just take your original expression for $H(x,y;m)$ with $m=0$ and evaluate the $p^0$ integral using the residue theorem from complex analysis. The remaining integrals you can probably find on a table of Fourier transforms.

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  • $\begingroup$ Thanks for your suggestions. I may try to attempt to solve for this function in terms of a 4th order DE in time. I must ask though, what do you mean when you say put the delta in the integral? I don’t understand. $\endgroup$ – QuantumEyedea Dec 14 '17 at 4:20
  • $\begingroup$ PS Also, isn’t the massless propagator a delta function and $(4 pi s)^{-1}$? $\endgroup$ – QuantumEyedea Dec 14 '17 at 4:23
  • $\begingroup$ $$-\frac{1}{4\pi}\delta(s) = \frac{\delta(\tau^2)}{4\pi} = \frac{\delta(\Delta t^2 - |\Delta \mathbf{x}|^2)}{4\pi} = \frac{1}{4\pi} \left(\frac{\delta(\Delta t - |\Delta\mathbf{x}|)}{\Delta t +|\Delta\mathbf{x}|} + \frac{\delta(\Delta t + |\Delta\mathbf{x}|)}{\Delta t -|\Delta\mathbf{x}|}\right) = \frac{1}{8\pi}\left(\frac{\delta(\Delta t - |\Delta\mathbf{x}|)}{|\Delta\mathbf{x}|} + \frac{\delta(\Delta t + |\Delta\mathbf{x}|)}{|\Delta\mathbf{x}|}\right) $$ $\endgroup$ – Sean E. Lake Dec 14 '17 at 4:28
  • $\begingroup$ @Greg.Paul For "Put the delta function in the integral" I mean insert $$G(t,\mathbf{x};t',\mathbf{x}') = \frac{\delta([t-t']^2 - |\mathbf{x}-\mathbf{x}'|^2)}{4\pi}$$ into the integral for $G_2$ in my answer. The massless propagator is not $(4\pi s)^{-1}$. Its most general form is $$G(\tau^2) = \frac{\delta(\tau^2)}{2\pi} \Theta(t-t') + A + B \frac{\delta(\tau^2)}{2\pi} \left[\Theta(t-t') - \Theta(t'-t)\right]$$ with $\tau^2 \equiv [t-t']^2 - |\mathbf{x}-\mathbf{x}'|^2$. In other words, the zero mass limit on the $K_1$ term is a constant. $\endgroup$ – Sean E. Lake Dec 14 '17 at 4:40
  • $\begingroup$ And the Feynman propagator comes from choosing $B=-\frac{1}{2}$ to get a function that is time reflection symmetric. $\endgroup$ – Sean E. Lake Dec 14 '17 at 4:42

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