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Feynman diagrams for spin 1/2 particles

  1. Mar 17, 2012 #1
    I think I have a pretty good handle on how scalar field scattering works in QFT, so now I'm trying to wrap my head around spin 1/2 particles, and I'm having a bit of trouble with it. For instance, in [itex]N + \phi \rightarrow N + \phi[/itex] scattering, an application of the Feynman rules leads to an expression like

    (2\pi)^4\delta(p + q - (p' + q'))\bar{u_{p'}}\frac{i}{-\not{p} -\not{q} - m + i\epsilon}u_{p} + crossing[/tex]

    So far so good--I see how the Wick expansion plus the definition of the fermion propagator lead to that. What I want to know (and I have a feeling this is going to turn out to be kind of a stupid question) is: how exactly do I go about evaluating this? I know that [itex]\bar{u_p}u_p = 2m[/itex], so that's a start on how to multiply the spinors, but a) that only works if the initial and final momenta are the same, and b) I still have to deal with the slashed momenta in the denominator.

    For part a), my first thought is that when the momenta are different, the result might be proportional to the dot product of the momenta--something like [itex]\bar{u_{p'}}u_p \propto p'\cdot p [/itex], but I don't know how to show that. Is that right?

    For part b), I know that the u's have been constructed such that [itex]\not{p}u_p = mu_p[/itex], so that would tell me how to evaluate the fraction if the slashed momentum were in the same direction as the spinor. But I'm not sure I understand how to handle the case where they're different, for instance [itex]\not{a}u_p[/itex]. Going along the same lines as above, my guess is that it'd work out to something like [itex]\not{a}u_p = (a\cdot p)u_p[/itex] but again I'm not really sure how to prove that. Am I on the right track here, or completely off-base?
  2. jcsd
  3. Mar 18, 2012 #2


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    Peskin and Schroeder has a section or 2 devoted to this. Basically, you want to get rid of slash operators on the denominator by multiplying top and bottom by the complex conjugate. And then you want to specify a frame in which to evaluate the expression (usually CoM frame), and denote, for example, p=(E,E,0,0) for particle coming in the x-direction. And express the outgoing particle in terms of scattering angles and whatnot.
  4. Mar 20, 2012 #3
    Thanks Matterwave, that makes sense. I managed to find out a bit more about this, too, which I thought I'd put here for posterity, in case anybody else stumbles across this thread.

    Things get simpler when you consider two things: a) usually once you get the S matrix, you're going to square it as part of getting a cross section, and b) usually you don't care about incoming or outgoing spin, so you have to average over incoming spins and sum across outgoing spins. Those two things mean that if your S matrix looks like [itex]\bar{u_{p'}}^{(r)}\not{q}u_p^{(s)}[/itex], what you're really interested is something more like [itex]\frac{1}{2}\sum_{rs}\bar{u_p}^{(s)}\not{q}u_{p'}^{(r)}\bar{u_{p'}}^{(r)}\not{q}u_p^{(s)}[/itex].

    That makes things simpler, due to some observations by Feynman. This quantity is a scalar, which is a 1x1 matrix, and 1x1 matrices are equal to their trace. So we really have:

    [tex]\frac{1}{2}\sum_{rs}\bar{u_p}^{(s)}\not{q}u_{p'}^{(r)}\bar{u_{p'}}^{(r)}\not{q}u_p^{(s)} = \frac{1}{2}Tr(\sum_{rs}\bar{u_p}^{(s)}\not{q}u_{p'}^{(r)} \bar{u_{p'}}^{(r)}\not{q}u_p^{(s)})[/tex]

    Next, we note that the trace is unaffected by cyclic permutations of its factors. So we can write:

    [tex]\frac{1}{2}Tr(\sum_{rs}\bar{u_p}^{(s)}\not{q}u_{p'}^{(r)}\bar{u_{p'}}^{(r)}\not{q}u_p^{(s)}) = \frac{1}{2}Tr(\sum_{rs}u_p^{(s)}\bar{u_p}^{(s)} \not{q}u_{p'}^{(r)} \bar{u_{p'}}^{(r)}\not{q})[/tex]

    Even though the LHS is a 1x1 matrix, and the RHS is a 4x4 matrix, their traces are equal. Now we have p and p' on the same side as each other, which in conjunction with the summations, allows us to use the identity [itex]\sum_{r}u_p^{(r)}\bar{u_p}^{(r)} = \not{p} - m[/itex]. So the above quantity reduces to:

    [tex]\frac{1}{2}Tr(\sum_{rs}u_p^{(s)} \bar{u_p}^{(s)}\not{q}u_{p'}^{(r)}\bar{u_{p'}}^{(r)}\not{q}) = \frac{1}{2}Tr((\not{p} - m)\not{q}(\not{p'}-m)\not{q})[/tex]

    At this point, you just have to distribute all the factors, and start beating on it with a bunch of gamma matrix trace identities to turn it into a real number, which does in fact involve a bunch of dot products like I was speculating above.
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