# What prevents this third-order QED scattering from having a non-zero amplitude?

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?

• Do you remember Furry theorem? Note that it works only for the sum of the diagrams not for individual ones. – OON Jan 18 '16 at 13:59
• @OON: Furry's theorem is usually stated as "diagrams with a fermion loop and an odd number of external photon legs vanish". It's not immediate to see how to apply it to this diagram. – ACuriousMind Jan 18 '16 at 14:14
• Bass, why do you think this diagram should vanish? It just looks like an $e^+ +e ^-\to e^+ + e^-$ concatenated with a bremsstrahlung diagram, and neither of those vanish. That the third-order contributions in total vanish doesn't mean the individual diagrams vanish - it might just destructively interfere with another third-order diagram. – ACuriousMind Jan 18 '16 at 14:16
• @ACuriousMind Yeah, that's true. I would still look at its sum with a similar diagram for the same process with not electron but positron radiating. – OON Jan 18 '16 at 14:32
• Your s-channel diagram and all the others where the photon is emitted from another leg, as well as all the t-channel diagrams definitively have a non-zero contribution. $e^+e^-\to e^+e^-\gamma$ was quite important to know at LEP (including a $Z^0$ exchange as well a photon), and a great deal of work has been poured into computing the cross-sections. The classic reference is Anthony C. Hearn, P. K. Kuo, and D. R. Yennie. Radiative corrections to an electron-positron scattering experiment. Phys. Rev., 187:1950–1963, Nov 1969. – user154997 Sep 15 '17 at 16:07