Higgs is neutral and therefore cannot have electromagnetic interactions. Then how can it decay into a pair of photons? Does it mean that particles need not be charged to have electromagnetic interaction?
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The fact that the Higgs boson is electrically neutral means that it is a singlet under the $U(1)$ electromagnetic gauge group. Since the whole Lagrangian must be a singlet as well, any coupling to the Higgs must happen through other gauge singlets.
The lowest dimension term that does this is the following $5$-dimensional term
$$\mathcal{L}_{h\gamma\gamma}\sim \left[ \frac{(2e/3)^2 N_c}{16\pi^2v} \right] h F_{\mu\nu}F^{\mu\nu}$$
with $N_c=3$ is the number of colors and $v = 246\, GeV$ is the Higgs vev. This term is generated by heavy particles loops. The top quark is the one that gives the biggest contribution to this decay since the coupling of the Higgs to fermions is proportional to their mass. The coefficent in front is given by dimensional analysis up to $O(1)$ numbers assuming only the top quark contribution and $m_t\gg m_h$. (NOTE: there is a really simple way of computing it exactly from the QED $\beta$-function)
Using this effective vertex you can find the decay rate of the Higgs into photons with a good accuracy. You can also do the full (but more complicated) calculation including also the W boson loop.
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