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  QED proper vertex Ward identity derived from global symmetry and Schwinger-Dyson Equations?

+ 1 like - 0 dislike

In QED, according to Schwinger-Dyson equation, $$\left(\eta^{\mu\nu}(\partial ^2)-(1-\frac{1}{\xi})\partial^{\mu}\partial^{\nu}\right)\langle 0|\mathcal{T}A_{\nu}(x)...|0\rangle = e\,\langle 0|\mathcal{T}j^{\mu}(x)...|0\rangle + \text{contact terms}$$ And the term $\left(\eta^{\mu\nu}(\partial ^2)-(1-\frac{1}{\xi})\partial^{\mu}\partial^{\nu}\right)$ is just the inverse bare photon propagator, so if we put the photon on shell, then the l.h.s will yield the complete n-point Green function with the complete photon propagator removed and also multiplied by a factor $Z_3$, the vector field renormalization constant.
But the r.h.s gives $$\partial_{\mu}\, \langle 0|\mathcal{T}j^{\mu}(x)...|0\rangle = \text{contact terms}$$ which is the common complete (n-1)-point complete Green function.

So if we truncate all the n-1 external complete propagators, then we are left with the proper vertex Ward identity.

The problem is, now the constant $Z_3$ appeared.
But the well known Ward identity, e.g. $$p_\mu\Gamma^\mu_P(k,l)=H(p^2)[iS^{-1}(k)-iS^{-1}(l)]$$ doesn't contain $Z_3$.

Where went wrong? Please help.

This post imported from StackExchange Physics at 2014-09-30 06:47 (UTC), posted by SE-user LYg
asked Sep 28, 2014 in Theoretical Physics by LYg (15 points) [ no revision ]

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