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Feynman Diagram Question, Antiparticle

  1. Jan 14, 2010 #1


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    Lets say we have a process where:
    Particle B(P) -> v(p2) + k

    some particle effectively goes to an antifermion and a photon, with momentums P, p2, and k.

    And the diagram is just straightforward, in arrow to vertex, out photon and out antifermion from vertex.

    Now the arrow on the antifermion points inward, but thats for fermion/particle number right? Momentum conservation would still be P = p2+k correct? not P +p2 = k (Arrows).
  2. jcsd
  3. Jan 14, 2010 #2


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    Hi Hepth! :smile:

    Sorry, not following you :redface:

    you can't have e- -> e+ + γ because the charges don't add up,

    and you can't change species and have eg proton -> e+ + γ because lots of other things don't add up. :confused:
  4. Jan 14, 2010 #3


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    Its effective, its not e- > e+ + gamma

    its actually part of a much larger diagram. I might as well say it:

    M (meson) -> positron + electron + photon

    Its done through a penguin diagram, but can be approximated with an effective process.

    Doing so, the photon can be emitted from either of the two fermions (positron/electron)
    When doing the matrix element of the photon being emitted from the positron, the "propagator" between the meson vertex and the QED vertex is a positron. My REAL questions is does this propagator change somehow. Normally, for a fermion (like on the electron-emission side) the propagator would be :

    \frac{\not q + m}{q^2 -m^2}

    Where q is just the final electrons momentum minus the photons, or the initial mesons minus the positrons.
    When the POSITRON emits the photon, do the momentum conservation rules change, or is that all taken care of in the spinors? would the propagator change, since its a "different" dirac equation.
    I know I would get a sign change from the QED vertex.
    would the propagator change from:
    \frac{i}{\not q -m} \rightarrow \frac{i}{-\not q -m}

    if p1 = electron
    p2 = positron
    P = meson
    k = photon

    would the positron emission vertex have:
    q = p2 + k
    or because of arrows :
    q = p2 -k
  5. Jan 14, 2010 #4


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    It seems like I'm running into problems reading papers where everyone uses different notation. Some like to mark momenta on the graphs of antiparticles with a minus already, others take it into account when doing conservation, some take it into account in the propagators. I guess what would answer my question would be if someone could, from feynman rules, write down or show me somewhere (having a hard time finding it) the amplitude for the compton, uncrossed-photon processes. One for electron, one for positron.

    I know the probability will be the same, but its the sign convention I'm worrying about BEFORE squaring. Would they be:
    \bar{u}\left(- i e \gamma^{\mu}\right) \epsilon_{\mu} \frac{i \left(\not q + m\right)}{q^2-m^2} \epsilon^{*}_{\nu}\left(- i e \gamma^{\nu}\right) u
    with [tex]q = p_f + k_2 = p_i +k_i [/tex]
    v\left(+ i e \gamma^{\mu}\right) \epsilon_{\mu} \frac{i \left(\not q + m\right)}{q^2-m^2} \epsilon^{*}_{\nu}\left(+ i e \gamma^{\nu}\right) \bar{v}
    with [tex]q = p_i - k OR = p_i + k [/tex]

    or does the propagator change in sign on the top "q", BEFORE I fill in what q is, etc.
  6. Jan 14, 2010 #5


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    got it
  7. Jan 14, 2010 #6


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    The answer to this question, I believe, has to do with the fact that when using DIRAC spinors, you never need to worry about signs in the propagator or the vertex. So, for example:

    fermion-fermion-photon vertex is always [tex]\qquad ie\gamma^\mu[/tex]

    fermion propagator is always [tex]\qquad\frac{i(p\!\!\!\slash+m)}{p^2-m^2}[/tex]

    and the direction of the momentum vs the spinor arrow is irrelevant (so no need to distinguish between fermions and antifermions in the propagator). This would NOT be true if using Weyl spinors!! But as long as you use Dirac spinors, all the signs should automatically work themselves out.
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