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sbh77
Dec4-09, 07:56 PM
1. The problem statement, all variables and given/known data
V(x) = \int \frac{d^3q}{(2\pi)^3} \frac{-g^2}{|\vec{q}|^2 + m^2} \exp^{i \vec{q} \cdot \vec{x}}

= -\frac{g^2}{4\pi^2} \int_0^{\infty} dq q^2 \frac{exp^{iqr}-exp^{-iqr}}{iqr} \frac{1}{q^2+m^2}

= \frac{-g^2}{4\pi^2 i r} \int_{-\infty}^{\infty} dq \frac{q exp^{iqr}}{q^2+m^2}


2. Relevant equations

none

3. The attempt at a solution

I understand how the measure changes to a single integral with a factor of 4\pi (integrating out the angles leaving the radial one) and I also know that a vector in spherical coordinates is represented just by a quantity in the radial direction. The problem I am having is I don't know where the term \frac{exp^{iqr}-exp^{-iqr}}{iqr} came from. It almost looks like sin but the "i" is in the denominator.

My last question is why does the second line become the third? I see the limits of integration are changed but why did the second exponential term disappear? Is it a symmetry thing?

Thanks!
1. The problem statement, all variables and given/known data



2. Relevant equations



3. The attempt at a solution

lanedance
Dec4-09, 09:19 PM
here it is in tex, just need the tags either side
V(x) = \int \frac{d^3q}{(2\pi)^3} \frac{-g^2}{|\vec{q}|^2 + m^2} e^{i \vec{q} \cdot \vec{x}}

= -\frac{g^2}{4\pi^2} \int_0^{\infty} dq q^2 \frac{e^{iqr}-e^{-iqr}}{iqr} \frac{1}{q^2+m^2}

= \frac{-g^2}{4\pi^2 i r} \int_{-\infty}^{\infty} dq \frac{q e^{iqr}}{q^2+m^2}

lanedance
Dec4-09, 09:25 PM
note 1/i = -i

can you decribe the problem a bit more, looks like an integral over momentum space?
what are x & r this problem?

lanedance
Dec4-09, 09:27 PM
also the 3rd line probably comes about by splitting the second line into a sum of 2 intergals and changing variables in one, q -> -q then recombining...

sbh77
Dec5-09, 04:57 AM
ah, my mistake, x and r are the same thing. I am just comparing the first Born approximation with that of the transition matrix element for two fermions scattering off of each other. By making this comparison it can be seen that the potential (in momentum space) is

V(q) = \frac{-g^2}{q^2+m^2} (sorry, I don't know what these tags are)

where "g" is the coupling constant in the Yukawa theory and m is the mass of the fermion. So I am just inverting this back to coordinate space. "q" is the momentum difference between the inbound and outbound fermion - along the propagator in a Feynman diagram.

Fourier transforms have never been my strong point! :)

Thanks again!