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):(adsbygoogle = window.adsbygoogle || []).push({});

$$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)?

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# How does a virtual particle becomes a point interaction?

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