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Gauge Invariance (QED)

  1. Mar 8, 2015 #1
    My book says that in this case $$e^+e^- \rightarrow \gamma \gamma $$ gauge invariance requires that $$k_{1\nu}(A^{\mu\nu} + \tilde{A}^{\mu\nu})=0=k_{2\mu}(A^{\mu\nu} + \tilde{A}^{\mu\nu})$$ Please see attachment. My question is how does this statement hold? Capture.PNG Capture.PNG
     
    Last edited: Mar 8, 2015
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  3. Mar 9, 2015 #2

    vanhees71

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    This is a Ward-Takahashi identity. It comes from current conservation, ##\partial_{\mu} j^{\mu}=0##, which is a necessary condition for gauge invariance.
     
  4. Mar 9, 2015 #3
    But the author insisted that these conditions are met although the quantities in the equations k1ν(Aμν+A~μν)=0=k2μ(Aμν+A~μν) each separately are all different from zero. Why would he say that if it is already a consequence of ward identity? @vanhees71
     
  5. Mar 10, 2015 #4

    vanhees71

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    The Ward-Takahashi identities refer to the proper vertex function and is valid order by order in ##\hbar## (number of loops), i.e., it's valid only for the complete set of diagrams of a given order.
     
  6. Mar 10, 2015 #5
    How does this have to do with my question? I can't relate.
     
  7. Mar 10, 2015 #6

    vanhees71

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    It has all to do with your last question, why it doesn't apply to a single diagram but only to the sum of the two diagrams relevant for the N-point function at the given loop order (which here is tree-level).
     
  8. Mar 10, 2015 #7

    Avodyne

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    If you write out the expressions for ##A^{\mu\nu}## and ##\tilde{A}^{\mu\nu}## that you get from computing these diagrams, you will find that ##k_{1\nu}(A^{\mu\nu} + \tilde{A}^{\mu\nu})=0## and ##k_{2\mu}(A^{\mu\nu} + \tilde{A}^{\mu\nu})=0##. So, while this can be predicted from gauge invariance, it is also the result of doing the explicit calculation.
     
  9. Mar 10, 2015 #8
    Are you telling me that Ward Takashi holds for the amplitude and not necessarily for each of the Feynman diagrams whose sum is the amplitude? @vanhees71
     
  10. Mar 11, 2015 #9

    vanhees71

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    Yes, that's it. The same is true for other symmetries, like charge conjugation symmetry, which implies that n-photon vertices with an odd number of photons must vanish. This also holds true only for the sum at a given loop order. Take, e.g., the one-loop triangle diagrams. You need to add both contibutions (which are different only by the orientation of the electron-positron loop making up the triangle).
     
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