Ward identity from Ward-Takahashi identity?

1. Jun 2, 2013

center o bass

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. Jun 2, 2013

center o bass

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.

3. Jun 2, 2013

Chopin

(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 $M$ on the mass shell of the external particles. If the $M$ on the left is on-shell, then neither $M(p+q)$ or $M(p)$ 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.