One of the suggested diagrams for the Higgs production is the following:enter image description here

so basically an electron-positron pair annihilates and forms an (excited?) Z boson, which then decays into another (less excited?) Z boson and a Higgs boson.

Why can't the electron-positron pair decay directly into a Higgs boson?

Charge and lepton number would be conserved anyway, and if the pair has enough energy to produce the $Z^*$ boson in the first place it should have enough energy to produce the Higgs boson... ?


2 Answers 2


The electron-positron pair can produce directly a Higgs boson, but this process is very suppressed, because the coupling between the leptons and the Higgs is proportional to the tiny mass $m_e$:

$$g_{\rm Hee}=-i\frac{ m_e}{v},$$ where $v\approx 246 \,\rm{GeV}.$

On the other hand, the process $e^+ e^-\to H Z$ is more likely to happen, because the coupling between $H$ and $Z$ is proportional to the $W$ mass:

$$ g_{\rm hZZ}^{\mu\nu}=i g \frac{M_W}{\cos^2\theta_W} g^{\mu\nu}. $$ In the latter case we also have to take in account the propagator of the $Z$ boson, which introduces a suppression factor of order $1/m_Z^2$, but at the end we still have a larger cross section.


The $^*$ notation does not mean excited in this case, it means "off shell" (i.e. virtual or having the "wrong" mass).

At the second vertex the $Z^0$ is put "on-shell" by the emission of a Higgs (note, however, that it will decay very quickly in any case).

The lepton pair can annihilate directly to the Higgs, but the event is experimentally identical to annihilation to photons or $Z^0$s (because the thing that makes a coupling possible is that both side have compatible quantum numbers, so that (at tree level) all three possibilities decay to very similar end states).

The reaction pictured is experimentally identifiable because the on-shell $Z$ decays to an lepton pair with a mass of 90 GeV and the Higgs decays to a limited choice of end states that are mostly reconstructable and add up to the Higgs mass.

A surprising amount of collider physics is not so much about what can happen as about what can be uniquely shown to have happened.

Finally, I would certainly not describe this reaction as a "decay".

  • $\begingroup$ so the weak force is responsible for the e-e annihilation to produce a Higgs boson, and what's responsible for the Z into H mechanism? $\endgroup$
    – SuperCiocia
    Commented Apr 27, 2014 at 17:50
  • $\begingroup$ It appears that I misspoke earlier, give me a moment. $\endgroup$ Commented Apr 27, 2014 at 18:17
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    $\begingroup$ keep in mind that the higgs couples proportional to the mass $\endgroup$
    – anna v
    Commented Apr 27, 2014 at 19:08
  • $\begingroup$ There are trilinear and quadrilinear interaction term in lagrangian involving Z boson and higgs boson. The term you are looking for has a vertex factor $\frac{igM_z}{cos\theta_w}$, where $\theta_w$ is weak mixing angle. $\endgroup$
    – user44895
    Commented Apr 27, 2014 at 19:33
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    $\begingroup$ @Melquíades From an experimental point of view $b$'s are reconstructable objects characterized by taggable jets. If you get both $b$'s you can reconstruct the Higgs by it's mass. $\endgroup$ Commented Apr 27, 2014 at 20:04

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