10
$\begingroup$

The LHC discovered the Higgs using the following decay mode: $$ H^0 \rightarrow \gamma \gamma $$

This got me thinking: if we reverse this mode, will it be valid? In some annihilation/decay pairs, these will stay valid:

$$e^+e^- \rightarrow \gamma\gamma$$

$$\gamma\gamma \rightarrow e^+e^-$$

So if the Higgs will decay into two high energy photons, will the annihilation of two high energy photons create a Higgs by this mode?:

$$ \gamma \gamma \rightarrow H^0 $$

Improve this question
$\endgroup$
2
  • $\begingroup$ I want to add a follow up question: if there is scattering between photons via higs does this have any interesting or observable effects in a photon gas? $\endgroup$
    – Shane P Kelly
    Commented Aug 3, 2015 at 20:39
  • 3
    $\begingroup$ The fusion of a top quark-anti top quark pair can produce a Higgs boson, and they, in turn, are the byproducts of gluons. If you could produce such a pair with photons, then this would be feasible. Not likely, though. $\endgroup$
    – HDE 226868
    Commented Aug 3, 2015 at 20:41

1 Answer 1

Reset to default
17
$\begingroup$

In principle, yes. You can reverse any decay process and the corresponding synthesis will be valid - in this case, since $H_0\to\gamma\gamma$ happens, then $\gamma\gamma\to H_0$ will also happen, assuming the kinematics work out.

However, the corresponding probability is very small. Out of all the possible things that could happen when two photons cross paths, turning into a Higgs boson is a relatively unlikely one. In fact, by far the most common option is nothing, because photons don't directly interact with each other. In order to get an interaction, you need one of the photons to quantum-fluctuate into a pair of charged particles, so that the other photon can interact with one of them - this makes it a second-order interaction at best. And Higgs production is even more unlikely because photons don't interact with the Higgs either, so you need another fluctuation to go from the charged particles to the Higgs boson. All the Feynman diagrams are third-order or higher, which makes for a very unlikely process indeed.

Improve this answer
$\endgroup$
3
  • 1
    $\begingroup$ Are the cross-sections any larger in a strong (relevant) field? For instance vacuum pairs near an event horizon and charged pairs near a charged particle? If nonzero, how large can this be? $\endgroup$
    – Eric Towers
    Commented Aug 3, 2015 at 23:47
  • $\begingroup$ @EricTowers I'm not sure offhand, though in a strong field there are a lot more photons to interact with so the overall Higgs production rate should rise. $\endgroup$
    – David Z
    Commented Aug 4, 2015 at 1:53
  • 1
    $\begingroup$ I'll just add that for the same reasons of the coupling being indirect the two-photon decay channel for the Higgs is quite rare: it only represents 0.2% of all decays. This was a major challenge in trying to see the Higgs in this way, although the 'clarity' of the signal relative to jets of hadrons made it worth the trouble. $\endgroup$
    – Rococo
    Commented Aug 4, 2015 at 4:48

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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