How does a virtual particle becomes a point interaction?

1. Feb 5, 2016

jim burns

If you want to show that a propagator of a heavy particle reduces to a point interaction at distances large compared to the inverse mass of the particle by Taylor expanding the propagator (for simplicity take 1-dimension):

$$G(x)=\int^\infty_{-\infty} \frac{dk}{2\pi} \frac{e^{-ikx}}{k^2+M^2}= \int^\infty_{-\infty} \frac{dk}{2\pi} \frac{e^{-ikx}}{M^2(1+\frac{k^2}{M^2})}= \int^\infty_{-\infty} \frac{dk}{2\pi} \frac{e^{-ikx}}{M^2} (1-\frac{k^2}{M^2}+...(-1)^n\left(\frac{k^2}{M^2}\right)^n)$$

then how does one justify this expansion when k>M? The first term gives the δ(x) point interaction, and the rest give derivatives of δ(x). But how is the Taylor expansion in the integrand justified?

It seems that the answer must have to do with the fact that x>M-1 in the exponential?

Also, a related question: if you have an arbitrary function

$$f(x)=\int^\infty_{-\infty} \frac{dk}{2\pi} e^{-ikx} f(k)$$

and specify that x>M-1, does that tell you anything about the contribution of f(k) for k>M to f(x)?

2. Feb 6, 2016

vanhees71

Why are you doing the expansion in the first place? You can just calculate the propagator in position representation. You find it explicitly done in older textbooks on quantum field theory like Schweber or in the Pauli Lectures.

3. Feb 6, 2016

jim burns

I was following some lectures (McGreevy's notes - if anyone's interested it's page 16 of http://physics.ucsd.edu/~mcgreevy/s15/215C-2015-lectures.pdf ) and they showed the expansion so I thought it was important to understand why it could be done mathematically.

I was thinking that physically high energy virtual modes don't contribute to the propagator at long time scales, so that you should be able to ignore the integral at high k when your time argument of the propagator in position space is long (>M-1). However, I don't see the justification mathematically.