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Consider the Fermi theory:

$$ \mathcal{L} = \frac{G_{F}}{2\sqrt{2}}\bar{n}\gamma_{\mu}(1-\gamma_{5})p \bar{\nu}\gamma^{\mu}(1-\gamma_{5})e $$

The cross section of $2 \to 2$ scattering calculated within the leading order of the perturbation theory grows with energy: $$ \sigma_{\bar{p}n \to e\bar{\nu}} \sim G_{F}^{2}s $$ Thus, as is typically said, it violates the unitarity since it breaks down the "tree level unitarity" at high energies.

My question is the following: why the first statement follows from the second statement? Maybe the perturbation theory cannot be considered for energies $s\gtrsim G_{F}^{-1}$, since the effective coupling is $G_{F}s$?

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  • $\begingroup$ One assumes you are cool with this? $\endgroup$ – Cosmas Zachos Jun 25 at 19:30
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"Unitarity" here really just means that all probabilities have to sum to unity in the end.

Obviously a cross section (and hence a probability) that grows linearly with energy is not bounded by 1.

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