Lecture 26: April 26th 2012
Reading 9.1-9.5
Reading for next time 9.5-9.6
Homework
1) 9.5, 9.29
When Comparing to the experimental result, remember that the decay to a muon/electron is only part of the total decay width of the tau. Is solving the problem using the assumption muon mass is small supported by the experimental data?
2) What quarks or anti-quarks does not interact with.
3) 9.16
1) Consider two scattering interactions for electrons at medium energy:
t channel neutrino electron scattering:
In CM frame
We can express this this in terms of s, t and u
s channel anti-neutrino electron scattering via annihilation:
In CM frame
In electron (anti)neutrino scattering s channel has the angular dependence and t is isotropic.
Also the s channel diagram can be obtained from the t channel diagram by an s to t crossing.
2) Scattering cross sections
usingd/d = (1/8)2 |M|2 S |pf|/|pi| 1/E2, where E is the total energy of both particles
t channel neutrino electron scattering:
s channel anti-neutrino electron scattering:
There is also a difference in the total cross section!
Why is anti-neutrino scattering one third as likely? We can make a helicity agreement. The backward scattering interaction for the anti-neutrino basically violates helicity conservation for the antineutrino. This relationship would have been different if the W+ didn’t interact with neutrinos in a pure V-A way.
Helicity conservation is an important factor in how neutrinos interact that has to be considered when detecting neutrinos and anti-neutrinos.
3) Pion decay
Since the anti u and d quarks in the pion interact via the strong force it is incorrect to consider this as an annihilation diagram. However, it is instructive.
,
The terms to first order in the quark mass are will cancel our since they have an odd number of gamma matrices. The term to second order in the quark masses turns out to be zero uses eq 7.66 and and chapter 7 trace 15.
In CM the quark momentums will be approximately 0 and we will gets terms with the quark mass times the energy of the outgoing particles.
However, this is not quite a correct treatment. The distribution of momentums of the quarks, and thus the possible angular distributions of terms like , will be constrained by the strong force and we should treat them in a more generalized way.
Where F will be a function of the pion or quark momentums times a scalar function of the initial momentum vector,.
The last is anti-symmetric under interchange of p1 and p3, which are equal/opposite.
using
The decay width is:
Where using homework 3.16
(mA,0,0,0) = (EB + EC,0,0,0)
(mA,0,0,0) = (EB + (pB2 +mC2)1/2,0,0,0)
(mA,0,0,0) = (EB + (EB2 -mB2 +mC2)1/2,0,0,0)
m A = EB + (EB2 -mB2 +mC2)1/2
(m A - EB)2 = EB2 -mB2 +mC2
m A2 -2 m AEB + EB2 = EB2 -mB2 +mC2
EB = (m A2 + mB2 - mC2)/2 m A
If we compare the decay width to muons and electrons
Again we can make a helicity argument. The electron or muon is forced to be in the wrong helicitystate which can happen more often the muon.
2) Quark generations
Consider the decay:
The Width should be:
Since for and involve the same type of strong interactions between quarks by isospin symmetry we don’t expect them to be different.
However is we compare and we find:
We expected the kaon decay width to be larger!
To solution to this is to have a suppression factor when decaying across generations in the matrix element. For the standard d to u decay there is a factor of and for the suppressed s to u decay there is a factor of and .
It was also observed that s to d transitions only occurred at extremely small rates. There had to be a mechanism that almost completely suppressed them.
The solution was the introduction of weak quark eigenstates as the states that couple to the weak force and express the strong eigenstates as superposition’s of the weak eigenstates and viceversa.
Weak states expressed as super-positions of strong states.
and strong states expressed as super-positions of weak states.
or represented as a rotation matrix
The weak force will only relate quarks of the same weak generation
Then for weak quark decay.
For the interaction within the generation d u:
Then only the contribution within the weak generation contributes.
The amplitude will have a factor of
For interactions across the generation s u:
Only allow interactions within the weak generation contributes
The amplitude will have a factor of