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I'm doing independent studies on electron-positron scattering, specifically the annihilation diagram contribution to the M matrix in Bhabha scattering, and this is the equation I recovered with the following initial conditions: An electron, associated with external momentum p_1, and a positron, associated with p_2, annihilate to produce a virtual photon, associated with internal momentum q, that produces a positron-electron pair (p_4 and p_3, respectively). I'm just going to assume you have a running understanding of the Feynman Rules.

These are the associated spinors for the particles involved in the interaction.

$$u\to e^- enters\\\overline u\to e^- exits \\ v \to e^+ exits\\ \overline v \to e^+ enter$$

This is the term added when a vertex is reached, it is a 4 x 4 matrix for fermions. $$-ig_e\gamma^\mu$$

The recovered equation:

$$ (2\pi)^4\int{[\overline u ^{(S3)} (p_3)(-ig_e\gamma^\mu)v^{(S4)}(p_4)}](\frac{-ig_{\mu \nu}}{q^2})[u ^{(S1)} (p_1)(-ig_e\gamma^\nu) \overline v^{(S2)}(p_2)]\ \delta^4(q-p_3 -p_4) \\ \times \delta^4(p_1+p_2 -q)\ d^4q $$

where $g_e$ is the quantum-electrodynamic coupling constant.

How do I reduce this down (I see that I can contract the index on $\gamma^\nu$) and integrate the derivatives of the Dirac delta function with arguments $q, p_1,p_2,p_3,$ and $p_4$? I've never come across having to integrate derivatives of the delta function.

Is it possible to use Eq. (17) on http://mathworld.wolfram.com/DeltaFunction.html ?

Any help is appreciated.

Qmechanic
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Jeesubmunu
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1 Answers1

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There are no derivatives of the Delta function in your integral. Performing the integration with one of the delta functions you recover

  • energy-momentum convervation in the vertices and
  • overall energy-momentum conservation in the form of a leftover $$ \delta^4(p_1 + p_2 - p_3 - p_4)$$

For further simplification you need to consider the absolute matrix element squared with sums over the particle spins (you average over the incoming and sum over the outgoing): $$\frac{1}{4} \sum_{s_1, s_2 s_3 s_4} \vert \mathcal M \vert^2 = \sum_{s_1, s_2 s_3 s_4}\vert \text{your integral} \vert^2 $$ and use completeness relations like $$ \sum_s u^s(p) \bar u^s(p) = p_\mu \gamma^\mu - m $$ $$ \sum_s v^s(p) \bar v^s(p) = p_\mu \gamma^\mu + m $$

Good luck!

Neuneck
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  • I have found that your answer, as well as the one posted http://physics.stackexchange.com/questions/61500/getting-rid-of-double-delta-function-in-feynman-rules?rq=1, both essentially cleared up any misconception I had before. – Jeesubmunu Sep 09 '13 at 10:29