I offer three questions below, and you are welcome to think any or all of them over today and tomorrow. In two days I will give my answer, explain the underlying physics a bit, and comment your own answers, if you have been capable of typing them despite your skyrocketing glycemic index.

**1.**The charged pion is a particle made by a pair of quarks: for the positive one we have the composition , and for its antiparticle . The charged pion decays in about ten nanoseconds into a muon and its neutrino, as in . In the decay, the quark-antiquark pair "annihilate" into a negative W boson, which then materializes the muon-antineutrino pair. The question is: why does the pion prefer to decay to a muon-antineutrino pair instead than in an electron-antineutrino pair ? Note, weak interactions, which are responsible for the decay, are flavor-democratic: there is a thing called "lepton universality of the weak currents" which demands that the intensity of the couplings to all leptons be the same. Also note, the pion has a mass of 139 MeV, the muon a mass of 105 MeV, the electron a mass of 0.5 MeV, and the neutrino is virtually massless. How come the pion prefers the heavier muon, if there would be more "phase space" to yield an electron instead ?

**helicity is the key. Okay, you don't know what helicity is, so go to wikipedia, damn it.**

Hint:

Hint:

**2.**In the famous 1964 experiment that demonstrated the violation of CP invariance and won Cronin and Fitch their Nobel prize in Physics, the four experimenters (Christenson, Cronin, Fitch, and Turlay) allowed a beam of neutral kaons to decay inside a large bag filled with Helium. The observation of the decay of two charged pions of the neutral kaons constituted indisputable proof of the fact that these particles had violated the combined symmetry of charge-conjugation and parity. In fact, the beam of kaons had been treated in a way that only one of the two possible quantum states had survived.

The question is, why Helium ? What properties has this gas that were suitable for the experiment ? Mind you, this is a tricky one.

**Hint:**the answer has only in part something to do with Physics...

**3.**You have a b-tagging algorithm which is capable of properly identifying jets as being originated from b-quark hadronization. You measure two numbers: the probability P(btag|bjet)=0.5 as the probability of tagging a jet if it was originated by b-quark hadronization; and the probability P(btag|not bjet)=0.01 of incorrectly btagging a non-b-jet. From those, you then get P(no-btag|bjet) = 1-P(btag|bjet), and similarly P(no-btag|no-bjet) = 1-P(btag|no-bjet).

The question is: given a sample of 1000 jets tagged as b-jets by the algorithm, what fraction of them are b-jets ? In other words, what is P(bjet|btag) ?

**Hint:**think Bayesian!

Okay, now get your ass off the armchair and put down the glass of wine. It's time for some homework!

1. The spin-0 particle has to decay to two right-handed particles - because there's no left-handed antineutrino worth talking about, and because the total spin of the decay products (with opposite helicities) has to be zero.

But the W boson (and every gauge boson) only interacts with 1 right-handed particle and 1 left-handed antiparticle, as obvious from writing the cubic interaction term in the 2-component formalism (it couples to the 2-component stuff, and its conjugate, much like the kinetic term). Consequently, the amplitude would vanish for a massless charged lepton. The vanishing only disappears when the mass of the charged muon is taken into account. The decay rate will probably be proportional to mass(charged lepton)^2, so 206.8^2 smaller for the electron than the muon - while the differences between the phase spaces are less radical.

In the twistor space, the massless electron amplitude is "more than maximally helicity violating, MHV" amplitude and vanishes.

2. I suppose the purpose of helium, an inert gas, is to minimize chemical interactions with the pions (and photons): the electron shells are fully occupied and tight, and single-atomic helium is no greenhouse gas either. Everything has something to do with physics. Another reason why they used helium, also related to physics, is that Kripke could have had some fun with Sheldon's visit to Ira Flatow's show:

http://www.youtube.com/watch?v=d8F5eD-Qmec

3.

P(bjet|btag) = P(btag|bjet) P(bjet) / P (btag)

P(non-bjet|btag) = P(btag|non-bjet) P(non-bjet) / P (btag)

The sum of these two numbers must be 1. The first factor in the first numerator is 0.5. The first factor in the second-line numerator is just 0.01. This just tells me that the denominator is the standard normalization, so

P(bjet|btag) = P(btag|bjet) P(bjet) / [ P(btag|bjet) P(bjet) + P(btag|non-bjet) P(non-bjet) ] = 0.5 P(bjet) / [ 0.5 P(bjet) + 0.01 (1-P(bjet)) ]

I think that the dependence on the prior probability for P(bjet) doesn't disappear. The result above simplifies to

P(bjet|btag) = 0.5 p / (0.01 + 0.49 p)

where p = P(bjet), the prior probability. For example, for p=0.5 (and there is nothing special about this value, except that it may look "democratic" to some people), we get 0.9804, or 98.04% jets, or 980.4 jets. However, for a value of "p" going to zero, this ratio gets to zero, too. But for p=0.3, the result is still 95.5%, pretty close to 100%. I think it's clear that the p-dependence can't disappear: the desired percentage is a weighted average of known conditional probabilities but the weights depend on the relative mixture of jets with b and without b - among all jets.