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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?

SRS
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    It goes through a loop with charged particles in it. – knzhou Jun 13 '18 at 12:53
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    $\pi^0$ also decays into photons.... –  Jun 13 '18 at 12:53
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    @marmot The $\pi^o$ is not elementary, but consists of a charged quark and anti-quark that annihilate. This is different from the Higgs. – safesphere Jun 13 '18 at 13:09
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    Although not the case for the Higgs: A neutral particle can interact with the em-field via a magnetic or electric dipole moment. In fact only neutral (composite) particles can decay to only photons due to the charge conservation. – Sebastian Riese Jun 13 '18 at 13:10
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    This seems rather easily Googlable – John Rennie Jun 13 '18 at 13:24
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    @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. – safesphere Jun 13 '18 at 13:47
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    Possible duplicate of Why don’t photons interact with the Higgs field? and Can two colliding photons create a Higgs Boson?. – AccidentalFourierTransform Jun 13 '18 at 15:29
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    @safesphere The point of John Rennie's comment is that questions should demonstrate some research effort. It is legitimate to down-vote if they do not. Users should explain why googling (or searching other questions on this site) has not answered their question. High rep users should be aware of what is expected of them. – sammy gerbil Jun 14 '18 at 10:50
  • @sammygerbil The goal of this site should be to promote knowledge, not to police user with totalitarian rules to scare them off so they never come back. I see your point that SRS is not new, but shouldn't all users be treated equally? And I do see a lot of newcomers downvoted for an honest, if naive, attempt to ask a question. Such a snobist brushing off creates a bad taste, users don't come back, gain no knowledge, the site gets a bad reputation. We have one moderator having done 20,000 downvotes. Tens of thousands users lost. This is the most unwelcome site on the web. Cut throat Wild West. – safesphere Jun 14 '18 at 19:29
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    @safesphere What is expected of users is stated in the Help Centre - eg How to ask good questions. If users don't like the totalitarian rules here there are many other sites to go to instead. How do you know newcomers are down-voted because of a naive question, rather than a lack of effort or because the question is not clear? If a down-vote encourages them to go away and do some research, and try harder to answer their own questions, that is a good thing IMO. – sammy gerbil Jun 14 '18 at 21:23
  • "If a down-vote encourages them to go away ... that is a good thing" - This is the saddest thing I've ever heard here. Negative reinforcements have never worked for any positive purpose in the history of the humankind. They are fossils of the past. "I have a dream that one day this site will rise up and live out the true meaning of its creed - we hold these truths to be self-evident: that all users are created equal." – safesphere Jun 14 '18 at 22:41

1 Answers1

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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.

FrodCube
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    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. – Cosmas Zachos Jun 13 '18 at 14:15
  • Could you define the terms $N_c$, $v$, and $h$? Thank you! – Rococo Jun 13 '18 at 14:53
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    @Rococo done. $h$ is the Higgs field. – FrodCube Jun 13 '18 at 14:55
  • @FrodCube So what is your conclusion? A neutral particle can have electromagnetic interaction? – SRS Jun 14 '18 at 08:47
  • @SRS yes if that particle is also coupled to something else that is electrically charged – FrodCube Jun 14 '18 at 13:31