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Ward identity from Ward-Takahashi identity?

  1. Jun 2, 2013 #1
    The Ward-Takahashi identity for the simplest QED vertex function states that

    $$q_\mu \Gamma^\mu (p + q, p) = S^{-1}(p+q) - S^(p)^{-1}.$$

    Often the 'Ward-identity' is stated as, if one have a physical process involving an external photon with the amplitude

    $$M = \epsilon_\mu M^\mu$$

    then

    $$q_\mu M^\mu = 0$$
    if q is the momentum of the external photon. One can argue on that the latter identity is true because of current conservation, but can one show that it follows from the Ward-Takahashi identity above? If so how?
     
  2. jcsd
  3. Jun 2, 2013 #2
    I agree with you on that one. Electron to a photon and an electron is not a physical prosess at all. However suppose that one of the the electrons are not on shell by being coupled to a subdiagram; one could for example have photon +electron coming in - vertex function - virtual electron - vertex function - photon +electron. Now one of the electrons are virtual with a momentum equal to the sum of the incoming photon and electron.
     
  4. Jun 2, 2013 #3
    (sorry, I deleted my previous post because I realized you were looking for something more general. I think that what I'm writing below is a better answer to your question.)

    Here's my understanding, which is based on a reading of Peskin and Schroeder section 7.4.

    You start with your statement of the Ward-Takahashi identity, which is true for the electron vertex, and proceed to show that it is also true for any physical process, not just the simple 3-point vertex. That can be done either order-by-order by examining the topology of Feynman diagrams, or more generally by using the functional integral.

    Next, you appeal to the argument in the LSZ reduction formula, which says that S-matrix elements are proportional to the residue of the pole of [itex]M[/itex] on the mass shell of the external particles. If the [itex]M[/itex] on the left is on-shell, then neither [itex]M(p+q)[/itex] or [itex]M(p)[/itex] on the right are on-shell, so neither have a pole in the right place to contribute to the S-matrix. Thus, the right-hand side is zero when you extract out the poles to compute the S-matrix.
     
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