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- Summary
- Request for clarification on the Ward identity

Hi I've been reading Peskin & Schroeder lately and I have some confusions over the ward identity.

So I think I understand how the identity works at a practical level but not exactly where it comes from. To illustrate my questions (which are difficult to state generally), I will make use the example of the three point function ##G^\mu (x_1,x_2,x_3)=\langle \Omega |T{ A^\mu (x_1)\psi (x_2) \bar \psi (x_3)}|\Omega\rangle##. The fourier transform of this function looks like

$$\tilde{G}^\mu(p_1,p_2,p_3) = (2\pi)^4 \delta^4 (p_1 - p_2 + p_3) D^{\mu\nu}(p_2-p_3)\tilde{S}(p_2)i\Gamma_{\nu} (p_2,p_3)\tilde{S}(p_3), $$ where ##\Gamma## are the 1PI graphs, ##D## and ##S## are the photon and fermion propagators respectively. The Ward identity says if we throw out the photon progator and the constant factors before it, and then contract the remaining part with the photon momentum ##q = p_2 - p_3##, we'll get

$$q_\mu \tilde{S}(p_2)i\Gamma^\mu(p_2,p_3)\tilde{S}(p_3) \sim e(\tilde{S}(p_2) - \tilde{S}(p_3)).$$

Okay this is all good but I don't understand how this relates to the version of the Ward-Takahashi identity obtained from the path integral. Specifically, how can the above be dervied from the relation $$i\partial_\mu \langle 0 |T{ j^\mu (x)\psi (x_2) \bar \psi (x_3)}|0\rangle = \mathsf{Contact\ terms}$$ when the perturbative terms in the greens functions take the form $$ \langle 0 |T{A^\nu(x_1)\psi (x_2) \bar \psi (x_3)}[\int d^4x A^\mu (x) j_\mu (x)]^n|0\rangle.$$ I can't see how you can relate the two together since in the latter case there is a photon field at the same position as the current.

So I think I understand how the identity works at a practical level but not exactly where it comes from. To illustrate my questions (which are difficult to state generally), I will make use the example of the three point function ##G^\mu (x_1,x_2,x_3)=\langle \Omega |T{ A^\mu (x_1)\psi (x_2) \bar \psi (x_3)}|\Omega\rangle##. The fourier transform of this function looks like

$$\tilde{G}^\mu(p_1,p_2,p_3) = (2\pi)^4 \delta^4 (p_1 - p_2 + p_3) D^{\mu\nu}(p_2-p_3)\tilde{S}(p_2)i\Gamma_{\nu} (p_2,p_3)\tilde{S}(p_3), $$ where ##\Gamma## are the 1PI graphs, ##D## and ##S## are the photon and fermion propagators respectively. The Ward identity says if we throw out the photon progator and the constant factors before it, and then contract the remaining part with the photon momentum ##q = p_2 - p_3##, we'll get

$$q_\mu \tilde{S}(p_2)i\Gamma^\mu(p_2,p_3)\tilde{S}(p_3) \sim e(\tilde{S}(p_2) - \tilde{S}(p_3)).$$

Okay this is all good but I don't understand how this relates to the version of the Ward-Takahashi identity obtained from the path integral. Specifically, how can the above be dervied from the relation $$i\partial_\mu \langle 0 |T{ j^\mu (x)\psi (x_2) \bar \psi (x_3)}|0\rangle = \mathsf{Contact\ terms}$$ when the perturbative terms in the greens functions take the form $$ \langle 0 |T{A^\nu(x_1)\psi (x_2) \bar \psi (x_3)}[\int d^4x A^\mu (x) j_\mu (x)]^n|0\rangle.$$ I can't see how you can relate the two together since in the latter case there is a photon field at the same position as the current.