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Feynman rules from Yang-Mills lagrangian

  1. May 8, 2013 #1
    In reading Ryder's book on quantum field theory he advocates reading off the Feynman rules directly from the Lagrangian in the path integral quantization method. I can sort of do this in phi-four theory, but it is not obvious in for example Yang-Mills theory, so I wondered if someone could explain to me why it is obvious that for example the term

    ##2gf^{abc}A^b_\mu A^c_\nu(\partial^\mu A^{\nu a}-\partial^\nu A^{\mu a})##

    corresponds to

    ##-2gf^{abc}[(r_\mu -q_\mu)g_{\nu \rho}-(p_\nu -r_\nu)g_{\mu \rho} + (q_\rho - p_\rho)g_{\mu \nu}].##

    where r,p and q are the momenta of the gauge bosons while a,b,c are their group indicies.
    I see that the derivatives goes over to momenta in momentum space, but what is the origin of the 'antisymmetry' between the momenta?

    How does one 'see' the exact structure?
     
  2. jcsd
  3. May 8, 2013 #2

    Bill_K

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    Since fabc is totally antisymmetric, there's six permutations of the indices, and what you need to do first is to write out (or at least imagine) all six of them:

    fabc(AbμAcνμAaν ± other 5 permutations of abc)

    (This automatically takes care of the two antisymmetric derivative terms.) Then make the momentum substitution.
     
  4. May 8, 2013 #3
    Alright. So that takes care of the antisymmetry of the momenta, but why is it that all the different permutations actually come into play? Why not just one permutation, - the permutation of abc in the lagrangian term above for example?

    Then there is the ##\rho## index. Where does that come from?

    Maybe you have a reference to somewhere this is explained in some detail?
     
    Last edited: May 8, 2013
  5. May 8, 2013 #4

    Bill_K

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    It's because a, b, c are dummy indices. take the first term fabcAbμAcνμAaν, and make the substitution a ↔ b. This is just bookkeeping, and doesn't change the term's value. But now it looks like fbacAaμAcνμAbν. Now use the fact that f is antisymmetric: - fabcAaμAcνμAbν

    Still the same value. Now the point is, do this all six possible ways, add them together and divide by 6. Still the same value! You've just rewritten it, to make explicit the antisymmetry.

    But now when you do the momentum substitution you'll get a sum of six terms, which is your second equation.
     
  6. May 8, 2013 #5
    Thank you! That makes a lot of sense! The only mystery left is the ##\rho## index. I guess that it is natural that every boson interacting at the vertex comes with it's own tensor index, thus giving rise to ##\mu, \nu## and ##\rho##, but in the term above there is only ##\mu## and ##\nu##, so how does this index arise?
     
  7. May 8, 2013 #6

    Bill_K

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    What you have written in the second equation is intended to be multiplied by AaμAbνAcρ, making a scalar.
     
  8. May 10, 2013 #7
    As you say one can make the antisymmetry explicit and divide by 6, since the six terms have the same value. But why then, does not this factor of 6 appear in the feynman rule? Well I've seen some include it and some not. Are these factors absorbed in the symmetry factor of the diagram or something?
     
  9. May 10, 2013 #8
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