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Given its branching fractions how many Higgs can be detected

  • Thread starter Kara386
  • Start date
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1. Homework Statement
In 2011 the ATLAS experiment had a total integrated recorded luminosity of ##5.08fb^{-1}## at 7TeV, and in 2012 of ##21.3fb^{-1}## at 8TeV. Under these conditions Higgs bosons can be produced in several ways including gluon-gluon fusion and vector boson fusion, which have the cross sections ##\sigma_{ggf} = 19.0 \pm 7.5pb## and ##\sigma_{vbf} = 1.6 \pm 0.3 pb## respectively.

The branching ratio of ##H \rightarrow ZZ## is 2.87%, and for ##Z \rightarrow l^{+}l_{-}## is 3.37% where l is a lepton. How many Higgs, produced by the two processes listed above, could have been detected in the decay mode
##H \rightarrow ZZ^* \rightarrow 4l## where l is electrons or muons?

2. Homework Equations


3. The Attempt at a Solution
Integrated luminosity 2012 = ##L_{12}## and for 2011 = ##L_{11}##. I calculated the total number of Higgs produced while the detector was recording by multiplying integrated luminosities for every year by the cross sections for the processes, then adding:
##N = \sigma_{ggf}L_{12} + \sigma_{vbf}L_{12} + \sigma_{ggf}L_{11} + \sigma_{vbf}L_{11}##

##=1.9\times 10^4 fb \times 21.3 fb^{-1} +1.6\times 10^3 fb \times 21.3 fb^{-1} +1.9\times 10^4 fb \times 5.08 fb^{-1} +1.6\times 10^3 fb \times 5.08 fb^{-1}##

##= 543,748## Higgs produced in total.
I don't know how to find the number of detections. I don't have the total decay width and the only equation I know involving branching ratios is ##BF(i) = \frac{\Gamma_i}{\Gamma}##. I thought about just multiplying the probabilities but I think that's probably wrong, and anyway I'm given the branching ratio of Z to any leptons where I need specifically Z to electrons or muons, so I assume that has to be calculated somehow. Can anyone give me some pointers?

Thanks for any help!
 
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When I say multiplying probabilities I should clarify I mean multiplying the number of Higgs produced by the probability of the two decay modes i.e. ##543,748\times 0.0287 \times 0.0337 = 52.6##. I'm mainly wondering how to account for the fact that the branching ratio ##3.37%## is to any lepton and actually I want it for electrons or muons.
 
33,007
8,782
Multiplying probabilities works, but you have two Z decays instead of just 1, and you have two possible decays to consider for each Z. The branching fraction is nearly identical for leptons, see your other thread for details, so it is 3.37% for Z->ee and 3.37% for Z->µµ.

The cross section should be different for 7 and 8 TeV, but if the problem statement gives you one value for both energies, well...
 
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Multiplying probabilities works, but you have two Z decays instead of just 1, and you have two possible decays to consider for each Z. The branching fraction is nearly identical for leptons, see your other thread for details, so it is 3.37% for Z->ee and 3.37% for Z->µµ.

The cross section should be different for 7 and 8 TeV, but if the problem statement gives you one value for both energies, well...
Ok so it has to go ##H \rightarrow ZZ## so that means 15,606 pairs of Zs are produced. Then it can go ##Z \rightarrow e^+e^-\mu^+\mu^-##, ##Z \rightarrow e^+e^-e^+e^-##, ##Z \rightarrow \mu^+\mu^-\mu^+\mu^-##. Do I need to count ##Z \rightarrow \mu^+\mu^-e^+e^-## as a separate case as well?
 
33,007
8,782
Yes.
 
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If so, should the ##Z \rightarrow e^+e^- e^+e^-## decay and the decay to muon pairs be doubled? By the same kind of logic?
 
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How many permutations have to be considered? I presume ##e^+e^-## is equivalent to ##e^-e^+##? And on reflection I don't think the other decays have to be doubled, they're the same thing...
 
Last edited:
33,007
8,782
You can assign labels to the Z bosons: "Z number 1" and "Z number 2". One out of four options is "Z number 1 decays to electron+positron and Z number 2 decays to muon+antimuon". The other options are similar, just by changing the decays.

##e^+ e^-## and ##e^- e^+## is the same thing. You are overthinking this problem.
 
208
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You can assign labels to the Z bosons: "Z number 1" and "Z number 2". One out of four options is "Z number 1 decays to electron+positron and Z number 2 decays to muon+antimuon". The other options are similar, just by changing the decays.

##e^+ e^-## and ##e^- e^+## is the same thing. You are overthinking this problem.
Brilliant, thanks for your help!
 

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