5
$\begingroup$

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?

$\endgroup$
  • 4
    $\begingroup$ It goes through a loop with charged particles in it. $\endgroup$ – knzhou Jun 13 '18 at 12:53
  • 4
    $\begingroup$ $\pi^0$ also decays into photons.... $\endgroup$ – marmot Jun 13 '18 at 12:53
  • 4
    $\begingroup$ This seems rather easily Googlable $\endgroup$ – John Rennie Jun 13 '18 at 13:24
  • 6
    $\begingroup$ @JohnRennie Virtually everything on this site is easily "gooleable". This is a good question compared to tons of nonsense questions that don't get downvoted. The downvote is undeserved. For example, your link does not make it clear if the decays it mentions have been actually observed, but a good answer could clarify this and other relevant matters. I believe such an answer would be helpful and thus so is the question. $\endgroup$ – safesphere Jun 13 '18 at 13:47
  • 3
    $\begingroup$ Possible duplicate of Why don’t photons interact with the Higgs field? and Can two colliding photons create a Higgs Boson?. $\endgroup$ – AccidentalFourierTransform Jun 13 '18 at 15:29
8
$\begingroup$

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.

$\endgroup$
  • 3
    $\begingroup$ You might further point out that this type of effective action term is virtually exactly the one that allows $\pi^0\to \gamma \gamma$, with a suitable $v\to f_\pi$ and charges' replacement, to dispel canards about elementarity versus compositeness, which miss how effective lagrangians bridge the gap. $\endgroup$ – Cosmas Zachos Jun 13 '18 at 14:15
  • $\begingroup$ Could you define the terms $N_c$, $v$, and $h$? Thank you! $\endgroup$ – Rococo Jun 13 '18 at 14:53
  • 1
    $\begingroup$ @Rococo done. $h$ is the Higgs field. $\endgroup$ – FrodCube Jun 13 '18 at 14:55
  • $\begingroup$ @FrodCube So what is your conclusion? A neutral particle can have electromagnetic interaction? $\endgroup$ – SRS Jun 14 '18 at 8:47
  • $\begingroup$ @SRS yes if that particle is also coupled to something else that is electrically charged $\endgroup$ – FrodCube Jun 14 '18 at 13:31

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.