Classical massless QED has axial current conservation. When quantizing the theory, we expect that suddenly $\partial_\mu \hat{j}^{\mu5}\neq0$ (as an operator equality).

I have two questions regarding this:

Why is it then, that when trying to prove indeed that $\partial_\mu \hat{j}^{\mu5}\neq0$ in the quantum version, we go on to try to prove that $$\partial_\mu\langle\Omega\left|\hat{j}^{5 mu} (x) \hat{j}^\nu (y) \hat{j}^\lambda (z)\right|\Omega\rangle\neq0$$? I understand that the above relation would imply the violation of a naive Ward-identity, but what's the connection between axial current conservation and Ward-identities? Is it the case that if the Ward identity is violated then *necessarily* the current cannot be conserved?

When calculating $\langle\Omega\left|\hat{j}^{5 mu} (x) \hat{j}^\nu (y) \hat{j}^\lambda (z)\right|\Omega\rangle$, it appears that the $\hat{j}$'s are treated as external photon lines, as in this image: why is that? I would naively calculate such a transition amplitude by plugging in functional derivatives w.r.t. to the Fermion and anti-Fermion sources, for example, like so: $$\langle\Omega\left|\hat{j}^{5 mu} (z) \hat{j}^\nu (x) \hat{j}^\lambda (y)\right|\Omega\rangle = \gamma_{\mu\alpha_{1}\alpha_{2}}\gamma_{5\alpha_{2}\alpha_{3}}\gamma_{\nu\alpha_{4}\alpha_{5}}\gamma_{\lambda\alpha_{6}\alpha_{6}}\left.\left(\frac{\left(-\frac{1}{i}\frac{\delta}{\delta\eta_{\alpha_{1}}\left(z\right)}\right)\left(\frac{1}{i}\frac{\delta}{\delta\bar{\eta}_{\alpha_{3}}\left(z\right)}\right)\left(-\frac{1}{i}\frac{\delta}{\delta\eta_{\alpha_{4}}\left(x\right)}\right)\left(\frac{1}{i}\frac{\delta}{\delta\bar{\eta}_{\alpha_{5}}\left(x\right)}\right)\left(-\frac{1}{i}\frac{\delta}{\delta\eta_{\alpha_{6}}\left(y\right)}\right)\left(\frac{1}{i}\frac{\delta}{\delta\bar{\eta}_{\alpha_{7}}\left(y\right)}\right)\mathcal{Z}_{QED}}{\mathcal{Z}_{QED}}\right)\right|_{\mbox{sources}=0}$$

This post imported from StackExchange Physics at 2014-08-07 15:35 (UCT), posted by SE-user PPR