I Higgs decay to Z and Z* at LHC

[Dhmht_Kr]

TL;DR Summary

What is the Z* and why it has different mass from Z ?

Goodmornig.I would like to explain me What is the Z* at higgs decay and why it has different mass from Z ?

 

[Orodruin]

It is an off-shell Z, ie, it is not a Z, but a set of particles produced by a Z-interaction.

The Higgs is too light to produce two Zs on-shell in a decay.

 

[malawi_glenn]

on-shell means that $E^2 = p^2 + m^2$ holds (in units of c=1).

you can read more here, section 2.2.2 https://arxiv.org/abs/hep-ph/0503172

With the same notation, we could write the muon decay as $\mu^- \to \nu_\mu + W^{-*}$

 

[malawi_glenn]

What is measured experimentally is 4 muons (2 positive & 2 negative).
Then you make a cut on your data to only keep events which has a muon pair (positive and negative) with an invariant mass = m_Z (+/- some treshold due to Z-width and experimental uncertainty).
After that, plot the invariant mass of the 4 muons, you should see a peak centered around m_H (which we now know is ≈ 125 GeV).

 

[Vanadium 50]

Here is a plot with more data:

 

[malawi_glenn]

Just one more flavor. If mH > 2mZ we would have the same signal 4 muons. But then one could make one more cut, namely that you would have two pair of muons with invariant mass = mZ

 

[Vanadium 50]

I'd turn it around - the fact that ZZ* is visible at all is remarkable. The Z* is off shell by 20-30Γ or so. Why is that decay large enough to see? The answer is that the longitudinal component of the Z is the Higgs.

 

[malawi_glenn]

Still the WW* Branching ratio is much higher for mH = 125 GeV
which is a much harder signal to measure, since the W decay is more "messy" than Z decays.

 

[Vanadium 50]

The WW will always be above ZZ. Don't blame me, Blame Clebsch and Gordon.