Tree-level cross sections for processes described by Fermi theory behave like $\sigma $ $\sim$ $G_{F}^2 \cdot s$, where $G_{F}$ is the Fermi constant and $\sqrt s$ is the energy entering in the process.
Then it is clear that for high energies Fermi theory violates unitarity, since as $s$ increases also $\sigma$ does. Starting from here, I was wondering if it was possible to obtain the Fermi EFT cutoff just by imposing unitarity.
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$\begingroup$ Haven't you already done it? $\endgroup$– AndrewCommented Jun 6 at 19:32
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$\begingroup$ Thank you @Andrew, I was just missing one step and now maybe I have found the solution. I will write it as a reply to the question. $\endgroup$– onibakuCommented Jun 7 at 9:20
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Here is what I came up with, I will leave it here since maybe it could be useful to someone.
Following this answer https://physics.stackexchange.com/a/139242/378284 it can be seen that unitarity of the S-matrix imposes the behaviour $\sigma \sim s^{-1}$ to the total cross section of a scattering process.
Hence the unitarity bound applied to the cross section of Fermi theory results in the following inequality
\begin{equation}
\sigma_{Fermi} \sim G^2_{F}s \lesssim \frac{1}{s}
\end{equation} which gives
\begin{equation}
\sqrt{s} \lesssim \frac{1}{\sqrt{G_{F}}} \sim 300 GeV
\end{equation}
(where $G_{F} = 1.1663787 \cdot 10^{-5} GeV^{-2}$). This is indeed what I was expecting as a cutoff to the Fermi theory, since Fermi EFT operators $\frac{G_{F}}{\sqrt{2}}\bar{\psi}\psi\bar{\psi}\psi$ carry a dimensionful coefficient dependent on the Higgs vacum expectation value scale:
\begin{equation}
\frac{G_{F}}{\sqrt{2}}=\frac{1}{2v^2_{ew}}; \
v_{ew} \simeq 250 GeV \sim \sqrt{\frac{1}{G_{F}}}
\end{equation}
So, even if the infrared description breakes down at energies $E \sim M_{W} \sim 80 GeV$, the coefficients of the irrelevant operators are suppressed by a larger effective scale $v_{ew} \sim 250 GeV$.