I have learned that in the Dyson-Wick expansion of the QED scattering operator
$$ S=e^{-i\int_{t_i}^{t_f}H\mathrm{d}t} $$
with the QED interaction Lagrangian
$$ H=e\bar\psi\gamma^\mu A_\mu\psi $$
in the limits $t_i\to-\infty$, $t_f\to\infty$, the first-order and third-order terms all vanish. For the first-order terms (single-vertex Feynman diagrams) this seems clear to me, since they correspond to an $e^-e^+$ pair that annihilates into a real photon, which is prevented by energy-momentum conservation at the single vertex.
The same argument holds for many third-order diagrams.
However, there are also three-order diagrams where I cannot see why their amplitudes should vanish, for example this one:
Here, time flows from left to right.
If I understand it correctly, this diagram corresponds to the term (where $\mathcal N$ means the normal ordering "operator"):
$$ \int\int\int dx_1dx_2dx_3 \,\mathcal N\left\{ (\overline\psi^-\gamma^\mu {A_\mu}^-\psi^+)_{x_3}, (\overline\psi^-\gamma^\nu {A_\nu}^+\psi^-)_{x_2}, (\overline\psi^+\gamma^\rho {A_\rho}^-\psi^+)_{x_1}\right\} $$
The process is
- destruction of real $e^+e^-$ pair and creation of virtual photon at the (left-most) point $x_1$.
- destruction of the virtual photon and creation of a virtual electron and a real positron at the (middle) point $x_2$.
- destruction of the virtual electron and creation of an electron-photon pair at the (right-most) point $x_3$.
I don't see why this process should have a zero amplitude. I tried to evaluate it and did not see a delta function that prevents it (like in energy-momentum conservation).
What prevents this process from being physical, i.e. from having a non-zero amplitude?