Relationship between vertex function and current

[geoduck]

Is there a proof or obvious reason why the vertex function is related to the current via:

$$S(p')\Gamma^\mu(p',p)S(p)=-\int \int dxdy e^{ipx}e^{-ipy} \langle T \psi(x) j^\mu(0) \bar{\psi}(y)\rangle$$

In the free-field case one can see it's true. But the interacting case?

 

[king vitamin]

I would use that expression as a definition of the vertex function in the interacting case. The right hand side becomes an infinite perturbation series for the measurable/physical quantity on the left.

Actually, I notice it implies using bare propagators on the LHS, so the expression does not contain external leg corrections. I believe this definition is used to include only the irreducible vertex corrections, and one defines irreducible propagator corrections separately (self-energy), so that when building higher order diagrams one can easily keep track of divergences. (Also, if you're using LSZ to get S-matrix elements, you should leave off the legs anyways)

 

[geoduck]

I would use that expression as a definition of the vertex function in the interacting case. The right hand side becomes an infinite perturbation series for the measurable/physical quantity on the left.

Actually, I notice it implies using bare propagators on the LHS, so the expression does not contain external leg corrections. I believe this definition is used to include only the irreducible vertex corrections, and one defines irreducible propagator corrections separately (self-energy), so that when building higher order diagrams one can easily keep track of divergences. (Also, if you're using LSZ to get S-matrix elements, you should leave off the legs anyways)

I can't see how the RHS of the equation is irreducible. I'm assuming $$j^\mu(0)=\psi(0)\gamma^\mu \bar{\psi}(0)$$, and the entire RHS side, when done perturbatively, seems to sum all diagrams with two electrons and a photon, with the photon polarization left off. This would include the reducible diagrams too.

This is actually comes from a proof of the Ward identity using the operator formalism, so we're not trying to get S-matrix elements

 

[king vitamin]

Actually, you're right, this isn't the irreducible vertex function. Also, I think the S's are the exact electron propagators to be consistent with the RHS. In that case, gamma is the amputated three-point function. It's consistent with the standard definition of gamma in its relation to the S-matrix for the interaction of electrons with an external field, and it's use in defining the electric charge and magnetic moment.