QED vertex as 2 equivalent matrix elements Consider just right handed fields $e_R,\bar{e}_R$ making up the standard electromagnetic current $A_\mu \bar{e}_R\gamma^\mu e_R$. Consider the matrix elements
$$\langle 0|J^\mu|e\bar{e}\rangle, \langle e|J^\mu|e\rangle$$
In words, the  first element describes the annihilation $e\bar{e}\to\gamma$ and the second describes $e\to e\gamma$. Both these vertices are exactly the same, however(note the helicity spinor crossing relations $v_L\sim u_R$); and the form factors must be related(the momenta must be carefully assigned though). This makes intuitive sense by draping around words like 'you can take the outgoing electron to be an incoming positron and therefore the matrix element is the same', but how exactly does this come about?
I suspect charge conjugation is involved, because only then can one convert between particle and antiparticle creation operators. Can someone clearly elucidate how these elements are equivalent, or in what sense?
A motivation for asking this is because people usually write down the hadron form factor as $\langle \pi^+|J^\mu|\pi^+\rangle=(p+p')^\mu F(q^2)$ and proceed to use it to evaluate cross sections where the final states are $\pi^+,\pi^-$ i.e. the matrix element $\langle 0|J^\mu|\pi^+\pi^-\rangle$, and I want to be sure how this correspondence originates.
 A: Consider some local operator $\phi(x)$, which may be a combination of elementary fields like quark fields, but which has a non-vanishing matrix element with a pion state,
$$\langle 0|\phi(x)|\pi^-_p\rangle=\sqrt{Z}e^{-ip\cdot x}.$$
Ignoring for simplicity any possible overall phase due to the charge conjugation, this implies
$$\langle \pi^+_p|\phi(x)|0\rangle=\langle \pi^+_p|C^{-1}C\phi(x)C^{-1}|0\rangle=\langle \pi^-_p|\phi^\dagger(x)|0\rangle=\sqrt{Z}e^{+ip\cdot x}.$$
So this is an example of the kind of crossing symmetry argument you are looking for but applied only to one-particle states.
More generally, this can be seen as a consequence of the LSZ reduction formula, which is treated in 7.2 of Peskin and Schroeder and from a different perspective ("polology") in 10.2 of Weinberg. The time-ordered correlation function
$$\int d^4x e^{ip\cdot x}\langle 0|T\phi(x)J(y)|\pi^+\rangle$$will have poles as $p$ goes on mass shell, both for positive and negative $p^0$, and the residue of the poles will be related to the matrix elements $\langle \pi^+|J(y)|\pi^+\rangle$ and $\langle 0|J(y)|\pi^-\pi^+\rangle$ respectively, so that gives a non-perturbative formulation of the crossing symmetry.
