# Homework Help: Special H boson with very high cross section?

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1. Apr 20, 2015

### unscientific

1. The problem statement, all variables and given/known data
(a) Explain the results.
(b) Why is the cross section much higher? Suggest the dominant decay product pair.

2. Relevant equations

3. The attempt at a solution

Part(a)
For a centre of mass energy of $170GeV$, you produce pairs of oppositely charged $W^{\pm},W^{\mp}$ bosons. It is kinematically forbidden to produce $Z,Z$ pair and forbidden by charge to produce $Z,W^{\pm}$ pairs.
First group of 4 hadronic jets can be explained by production of 2 pairs of quark/anti-quark stream. Number of combinations = 3x4 = 12 because of 3 colours possible. ~45%

Second group can be explained by a quark/anti-quark stream and a lepton/neutrino stream. Number of possibilities = 3x2 + 2x3 = 12. (3 types of leptons, 2 charge types) ~45%

Third group can be explained by 2 pairs of lepton/neutrino stream. Number of possibilities = 3 (leptons must be oppositely charged) ~10%

Part (b)

This question stumbled me. Is there something special happening at 125 GeV? I suppose the heaviest quark-anti-quark pair it can decay to is the bottom/anti-bottom quark? Why is the cross section so much higher? Why is it that in part (a) only {u,d,c,s} quarks/anti-quarks are produced and not bottom (b) quarks??

2. Apr 20, 2015

### Staff: Mentor

The Z has a large width - you will get ZZ events with one Z a bit off-shell, but I guess you can ignore that here.
Why 4? And what about the other W?
I don't understand what you are adding here.

The LHC found a particle at this energy.
Right.
What does the Higgs couple to?
That is a mistake you have to fix in (a).

3. Apr 20, 2015

### unscientific

Ok, now thinking about it - each W boson can produce 5 possible types of quarks (u,d,s,c,b) so possibilities are 5x5 = 25? But 3 types of colour so 25x3 = 75.

For the second picture, 5 quark possibilities, 3 types of colour, 3 types of leptons, 2 charges of leptons so 5x3x3x2 = 90? Doesn't match the first one.

For the third picture, 3 types of leptons but the W bosons must be oppositely charged, so 3 types for the other W boson. So 3x3 = 9

The numbers are all wrong...

Right, should have noticed they were talking about the Higgs! The Higgs boson couples to mass, so it naturally has an extremely large cross section with the heaviest possible quark, bottom quark. Why is it 40,000 times as large though?

4. Apr 21, 2015

### Staff: Mentor

Sorry, I was thinking of Z decays with my comment about five quarks. No b in W decays as the top is too heavy.

You don't have so many options for the W. Let's consider the W+ boson: it either decays to up+antidown or charm+antistrange. Just two options, multiplied by three for colors => 6.
The W- has 6 options as well.

That is the dominant decay mode for the 125 GeV Higgs, right.
What is the electron to muon mass ratio? Do you see how those numbers could be related?

5. Apr 21, 2015

### unscientific

Why can't the W+ boson decay to charm/antidown? It is calibbo suppressed by $sin 13^o$ but it is possible? Also why can't W+ boson decay to $u\bar u, d\bar d, c\bar c$ pairs?

First picture
6 options for each W.
Total = 6x6=36 options

Second picture:
Quark producing W can either be W+ or w- = 2x6 = 12 options
For the lepton decay, 3 options.
Total = 12x3=36 options

Third Picture
3 options for each.
Total = 3x3 = 9 options

Mass of muon = 200 times Mass of electrons

I suppose $\sigma \propto m^2$?

6. Apr 22, 2015

### Staff: Mentor

It can, but the probability is small. Small enough to neglect it.
Quark + antiquark of the same type would violate charge conservation.

The new numbers look right.
Right.

7. Apr 22, 2015

### unscientific

Why is $\sigma \propto m^2$? I know that $\Gamma \propto m^5$. Cross section is also (Rate of reactions/Incident Flux) $\sigma = \frac{\Gamma}{J}$.

8. Apr 22, 2015

### Staff: Mentor

You have two massive particles in the interaction.
For a better description, see the Lagrangian and QFT.
You are mixing unrelated concepts here.

9. Apr 22, 2015

### unscientific

I thought lifetime $\tau \propto \frac{1}{\Gamma}$ and rate of reactions is decay width $\Gamma$?

10. Apr 22, 2015

### Staff: Mentor

The decay width is not the reaction rate. They are related if you take the partial width for the right process (here: electron/positron or muon/antimuon).
Also, the total width is not just the mass (the Higgs mass! Not the decay products) to the fifth power, it is a more complicated function as you have to take multiple processes with different thresholds into account.

11. Apr 25, 2015

### unscientific

True. Upon doing some reading I realized that there is indeed a relation between cross section and rates, using Fermi's golden rule:

$$\frac{d\sigma}{d\Omega} = \frac{\Gamma}{\nu} = \frac{1}{\nu}\frac{2\pi}{\hbar} |M_{fi}|^2\frac{dN}{dE}$$

where $\nu$ is incoming flux.

And for EM interactions like $e^+e^-\rightarrow \mu^+\mu^-$ or $e^+e^-\rightarrow \tau^+\tau^-$, it is $\sigma = \frac{4 \pi}{3}\left(\frac{\alpha \hbar c}{W} \right)^2$ where $\alpha = \frac{g_{EM}^2}{4\pi}$ and $W$ is the centre of mass energy.