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I'm following Peskin & Schroeder and I'm trying to calculate the momentum space representation for the following diagram, Q4 in this link. Paper

The loop is what's causing me problems. I'm not sure how to write loops in scattering amplitudes.

In general how to electron loops fit into the amplitude?

For the specific question I think the answer would be something like

$\bar{u}(k')v(k'_+)(-ie\gamma^\mu)(-ie\gamma^\nu)(-1)\int\frac{d^4l}{(2\pi)^4}tr\left[electron-propagator\right]\bar{v}(k_+)u(k)$

I've left out the photon propagator but the part inside the trace is what I'm unsure of.

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    $\begingroup$ Loops are not treated differently from any other internal lines. What is your confusion? $\endgroup$
    – ACuriousMind
    Commented Feb 8, 2015 at 13:43
  • $\begingroup$ I've updated my question @ACuriousMind $\endgroup$
    – Okazaki
    Commented Feb 8, 2015 at 14:39

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First, note that you don't have a trace yet. You get a trace when you do the spin sum for the external fermion lines for the square of the amplitude; c.f. Peskin and Schroeder p. 132.

Second, you need two photon propagators; one for each line in the diagram. One will couple to your $\gamma^\mu$ and to a new vertex with some $\gamma^\rho$ on the loop, and the other will couple to your $\gamma^\nu$ and to a new vertex with some $\gamma^\sigma$ on the loop.

As for the loop itself, it's just the product of the propagators (placed in the appropriate order with the gamma matrices): $$\frac{i(l \cdot \gamma + m)}{l^2-m^2+i\varepsilon} (-i e\gamma^\rho) \frac{i((l-q)\cdot \gamma + m)}{(l-q)^2 - m^2 + i \varepsilon} (-i e \gamma^\sigma)$$ wherein $q$ is the momentum transferred through the loop, as in the problem prompt.

In general, loops work just that way: go around the loop one full time. Each time you hit a propagator, put a factor in the integrand with appropriate momentum. Each time you hit a vertex, put in the appropriate vertex factor, with the appropriate momentum if your theory has momentum-dependent vertices. When you get all the way around, you're done with that loop and ready to integrate.

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