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Vertex Feynman Rule Derivations

  1. Dec 26, 2012 #1
    Alright, this is a pretty low level / silly question but I am having some issues.

    I would like to get the vertex rule for an interaction that has a field tensor involved like this...
    [itex] \partial_\mu Z_\nu -\partial_\nu Z_\mu[/itex]
    do I treat the two Z's as separate fields?

    For a generic field term with a derivative [itex] \partial_\mu \varphi [/itex] we pull out a [itex] k_\mu [/itex] in momentum space...but I am just being stupid with these indices.

    Advice, please?
     
  2. jcsd
  3. Dec 26, 2012 #2

    Hepth

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    Gold Member

    The term is linear in Z's, so you have to include both.

    Imagine an in Z state, then this would give you (up to factors of i and -1 ):

    [tex]
    k_{\mu} \epsilon_{\nu} - k_{\nu} \epsilon_{\mu}
    [/tex]

    after contraction with each term.
     
  4. Dec 26, 2012 #3
    ...huh? After contraction? We still have our indices?



    So if I started with

    [itex] \sigma_{\mu \nu} F^{\mu \nu}_Z [/itex]
    I should end with
    [itex] \sigma_{\mu \nu} (k^{\mu} \epsilon^{\nu} - k^{\nu} \epsilon^{\mu})
    [/itex]
    where
    [tex]
    \sigma_{\mu \nu} = \frac{i}{2} (\gamma_{\mu}\gamma_{\nu}-\gamma_{\nu}\gamma_{\mu})
    [/tex]
    I'm not sure how to get rid of the polarization though :/

    Is there some simple relationship with the gamma matrices I should know?
     
  5. Dec 27, 2012 #4

    Hepth

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    Gold Member

    By "contraction" I meant contracting the field operators in the lagrangian with those in the in/out states.

    What you have is right (up to factors of I, -1) , and can be simplified further due to the antisymmetric nature of each:

    [tex]
    = \sigma_{\mu \nu} (k^{\mu} \epsilon^{\nu} - k^{\nu} \epsilon^{\mu}) \\
    = \sigma_{\mu \nu} k^{\mu} \epsilon^{\nu} - \sigma_{\mu \nu} k^{\nu} \epsilon^{\mu} \\
    = \sigma_{\mu \nu} k^{\mu} \epsilon^{\nu} + \sigma_{\nu \mu} k^{\nu} \epsilon^{\mu}
    [/tex]
    Now relabel the 2nd term's indices to match the first (its ok since theyre all contracted)
    [tex]
    = 2 \sigma_{\mu \nu} k^{\mu} \epsilon^{\nu}
    [/tex]

    I'm not sure what you mean by getting rid of the polarization. It stays in there for vertices/feynman rules. If you have one in the in/out state you have to do polarization sums/averaging (remember: average over initial stats, sum over final).

    So in short the answer is that you're probably just thinking about the indices wrong. Its OK to have a polarization (sorta required) if you have an ingoing/outgoing boson of spin>0.

    So when you get to say, a vertex times a Z propagator times another vertex you just connect the propagator to the vertices' polarization vectors.
    So:

    [tex]
    (2 \sigma_{\mu \nu} k^{\mu}) \frac{-i (g^{\nu \alpha} - k^{\nu} k^{\alpha}/M_Z^2)}{k^2-M_Z^2} (2 \sigma_{\beta \alpha} k^{\beta})
    [/tex]

    Up to factors of I and -1 that I may have missed (and sign convention on incoming/outgoign momenta) and your gauge choice.

    You'll notice that the polarization sums for photons/Z/vector bosons are very similar to their propagators.
     
  6. Dec 28, 2012 #5
    σμvFμvμv(∂μAv-∂vAμ)
    μvμAvμAv(interchanging indices in second term)
    =2σμvμAv
    now to get vertex factor,substitute plane wave form and take out the factors already taken care by normalization and external lines.it is done for iL(LAGRANGIAN multiplied by i) so write the lagrangian and do like above you will get something like
    -iσμvkv. Notice I don't differentiate between subscript and superscript.it is unnecessary for flat space.
    edit:there should not be any polarization associated with it.
     
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