The point of the question is to ask what is the function given by the following integral: $$ H(x,y) \ \equiv \ \int \frac{d^{4}p}{(2\pi)^{4}} \frac{e^{-i p \cdot (x-y)}}{(p^{2}+m^{2}-i\epsilon)^{2}} $$

This is closely related to the propagator (for $(x-y)^{2} < 0$): $$ G(x,y) \ = \ \int \frac{d^{4}p}{(2\pi)^{4}} \frac{e^{-i p \cdot (x-y)}}{p^{2}+m^{2}-i\epsilon} \ = \ - \frac{ i m }{ 4 \pi^{2} \sqrt{ - (x-y)^{2} }} K_{1}\left( m \sqrt{ - (x-y)^{2} } \right) $$

The reason I ask this question, is because in calculations like for the following Feynman diagram:

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I would have in position space the following: $$ \propto \int d^{4}u\ G_{0}(x,u) G_{0}(u,u) G_{0}(u,y)\\ = \int d^{4}u \left[ \int \frac{d^{4}p}{(2\pi)^{4}} \frac{e^{-i p \cdot (x-u)}}{p^{2}+m^{2}-i\epsilon} \right] \left[ \int \frac{d^{4}k}{(2\pi)^{4}} \frac{1}{k^{2}+m^{2}-i\epsilon} \right] \left[ \int \frac{d^{4}q}{(2\pi)^{4}} \frac{e^{-i p \cdot (u-y)}}{q^{2}+m^{2}-i\epsilon} \right] \\ = \int \frac{d^{4}p}{(2\pi)^{4}} \frac{e^{-i p \cdot (x-y)}}{(p^{2}+m^{2}-i\epsilon)^{2}} \ \int \frac{d^{4}k}{(2\pi)^{4}} \frac{1}{k^{2}+m^{2}-i\epsilon} \\ = H(x,y) \int \frac{d^{4}k}{(2\pi)^{4}} \frac{1}{k^{2}+m^{2}-i\epsilon} $$

You can regulate the integral over $k$ however you like, but what to do with $H(x,y)$ here?

It seems that $H(x,y)$ pops up a lot when you do calculations like this. Is there a way to evaluate $H(x,y)$? Surely this must have been done somewhere?


1 Answer 1


Note that you don't need to go through the Fourier transform to get the result you're seeking. All you need is translation invariance of $G$ (i.e. $G(x,y) = G(x+w, y+w) = G(|x-y|)$) to say that the first integral gives: \begin{align} I & \equiv \int \operatorname{d}^4 u \ G_0(x,u)\, G_0(u,u)\, G_0(u,y) \\ & = G(0) \int \operatorname{d}^4 u \ G_0(x,u)\, G_0(u,y), \end{align} as long as the Green's function $G$ has been suitably regularized to make $G(0)$ finite. Note that the convolution chain can grow infinitely long. Consider an exactly solvable perturbation of the form $\Delta \mathcal{L} = - \frac{1}{2}(\Delta m^2) \phi^2$, you'll get just such an infinite chain in producing the net propagator with shifted mass.

One way to calculate $H$ is to note that it can be produced by the limit: \begin{align} H(x,y) &= \lim_{\mu^2 \rightarrow 0} \int \operatorname{d}^4 u \ G_0(x,u; m)\, G_0(u,y; \sqrt{m^2 + \mu^2}) \\ & = \lim_{\mu^2 \rightarrow 0} \int \frac{\operatorname{d}^4 p}{(2\pi)^4} \frac{\mathrm{e}^{-ip\cdot(x-y)}}{ \left(p^2 + m^2 - i\epsilon \right) \left(p^2 + m^2 + \mu^2 - i\epsilon \right) } \\ & = \lim_{\mu^2 \rightarrow 0} \int \frac{\operatorname{d}^4 p}{(2\pi)^4} \frac{\mathrm{e}^{-ip\cdot(x-y)}}{\mu^2} \left[\frac{1}{p^2 + m^2 - i\epsilon} - \frac{1}{p^2 + m^2 + \mu^2 - i\epsilon}\right] \\ & = \lim_{\mu^2 \rightarrow 0} \frac{1}{\mu^2} \left[G_0(m|x-y|) - G_0\left(\sqrt{m^2 + \mu^2}|x-y|\right)\right], \end{align} where I used partial fraction decomposition in the intermediate step. Working out the final result, which can likely be done using the Bessel function derivative identities, is left as an exercise for the reader.

  • $\begingroup$ @ Sean_E._Lake This is very interesting, thanks for the insight. Are you saying that I can actually take the limit $\mu^{2} \to 0$, or is this as far as I can take the computation? (it seems to me that the first term will blow up as $\mu^{2} \to 0$) $\endgroup$ Commented Jun 14, 2017 at 18:19
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    $\begingroup$ I believe you can take that limit - it's set up exactly like a derivative, which is why I suggested the dlmf identities. Note how the solvable $\Delta m^2$ perturbation produces an infinite series that shifts the squared mass, so you can think of what form a translation operator in squared mass would take, and it would involve derivatives with respect to squared mass, as we produced here. $\endgroup$ Commented Jun 14, 2017 at 18:23
  • $\begingroup$ @Sean_E._Lake I think I understand what you're saying; If I define a function $f(\mu^{2}) = G_{0}(x,y;\sqrt{m^{2}+\mu^{2}})$ such that $f(0) = G_{0}(x,y;m)$, then your above limit corresponds exactly to $- f^{\prime}(0) = - \frac{d G(x,y;\sqrt{m^{2}+\mu^{2}})}{d(\mu^{2})} \big|_{\mu^{2} = 0}$. Is this correct? $\endgroup$ Commented Jun 15, 2017 at 3:59
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    $\begingroup$ Yep, although you don't need to define a separate function. Consider what $$-\frac{\partial G_0}{\partial m^2}$$ looks like. For more fun, you can do a chain of $n$ convolutions with $$\left(\frac{[-1]^n}{n!}\right) \frac{\partial^n G_0}{\partial [m^2]^n}.$$ $\endgroup$ Commented Jun 15, 2017 at 14:29

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