= Solution
A constant phase leaves the photon unchanged and makes every QED interaction vertex invariant: its one $\psi$ and one $\bar\psi$ acquire opposite phases. Thus the propagators and vertex factors in the <QED Feynman rules> are unchanged. A Green function with $n$ insertions of $\psi$ and $\bar n$ of $\bar\psi$ acquires the phase $e^{ie\chi(n-\bar n)}$. For an invariant vacuum it can be nonzero only when its total inserted charge is zero.
Equivalently, the conserved charge operator $Q$ commutes with the <S-matrix>. If initial and final states have charges $Q_i,Q_f$, then
$$
0=\langle f|[Q,S]|i\rangle=(Q_f-Q_i)\langle f|S|i\rangle.
$$
The <scattering charge selection rule from global gauge symmetry> is therefore
$$
\boxed{\langle f|S|i\rangle\ne0\quad\Longrightarrow\quad Q_f=Q_i}.
$$
In diagrams the oriented fermion lines preserve charge at every vertex, and a photon is neutral. Pair creation is allowed because a particle and its antiparticle have opposite charges. Global phase invariance supplies this selection rule; local <gauge invariance> supplies the stronger momentum-dependent <Ward-Takahashi identity>.
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