According to Standard model, the partial width of the decay of Higgs into dimuon (up to tree level) is:
$$\Gamma\approx\frac{m_H}{8\pi} \left(\frac{m_{\mu}}{\nu}\right)^2$$
with the Higgs mass $m_H=125 GeV$, muon mass $m_{\mu}=0.106 GeV$, and the vacuum expectation value of the Higgs field $\nu=246 GeV$, apparently the decay width is extremely small. Then why is the width of the resonance peak in the plot from ATLAS so wide? If it's due to experimental errors then is there any meaning in comparing it with the theoretical result? I'm having trouble understanding this. Could somebody please explain it for me?
[Physics] Width of the decay of Higgs boson into dimuon
higgsparticle-physicsstandard-model
Best Answer
It is mainly measurement and detector errors that make up the width in the plots you show. The Monte Carlo simulates the detector resolution and folds in the theoretical values when it says that the width agrees. The real width is expected to be much smaller.
In this we see that the real width is only given as a bound by the experiments
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So you were correct to be puzzled. The partial widths add up to the total width , that was how the width of the invisible neutrino decays of the Z have been found, by doing the sum and subtracting form the total. Leptonic machines have much better accuracies than hadronic. That is why the next collider will be a leptonic one, to study the Higgs accurately and nail down discrepancies to the standard model. Hadronic machines are just discovery machines.