Lecture 12: Mar 1st 2012
Reading Griffiths chapter 6
Homework: 6.4, 6.6, 6.7, 6.8
0) C, P and CP Violation
The weak force violates both parity and charge conjugation.
The historical example is if you measure the electron decay direction in the weak decay
of the 60Co atom where the spin has been aligned upwards the electron is emitted
downward. The mirror image interaction where two coordinates are inverted does not
exists. i.e. if the electron is emitted along z then inverting x and y reverses the spin
vector but not the electron direction so that the electron is emitted along the direction of
the spin.
The electron has spin oriented oppositely to its momentum (helicity) and is classified as a
left-handed particle. The antineutrino has spin oriented along its direction of travel and is
classified as right handed. Similarly, positrons are generally right handed and all
neutrinos are left-handed. Note, electrons and positrons do not have a purely right or left
handed helicity since you can apply a velocity frame transformation that will reverse the
direction of the velocity and thus the helicity. If the neutrinos were mass-less they would
have an absolute helicity since any velocity transformation would still leave them moving
at the speed of light in the same direction. The fact that neutrinos are mass-less actually
breaks the standard model in which only (right) left handed (anti) neutrinos would exist
and interact via the weak force.
Interaction via weak force clearly violate charge conjugation. Charge conjugating an
interaction with a left-handed neutrino will give a left-handed anti-neutrino, which does
not exist.
The weak force carriers have a V-A, vector-axial(or pseudo)vector form. The axial part
of the force carrier has the opposite parity compared typical force carriers and thus
intrinsically flips the parity of the interaction leading to parity violation.
Though in most cases the weak force does obey the combination of charge conjugation
and party, CP. We expected the product of C and P to always be conserved for the weak
force. Charge conjugated weak force diagrams did not exist but the charge conjugated
parity flipped version did exist.
+  + 
happens as often as the charge conjugated and parity inverted
-  - u
Note also that this interaction does not preserve helicity for the muon but helicity is not
necessarily perfectly conserved for massive particle. However, this does lead to a
reduction in the probability of the interaction referred to as helicity suppression. This
suppression is much greater for the decay to electrons, where the electron is moving
closer to the speed of light and is less often in the correct helicity state.
However CP is violated in some common decays.
K  - e+ 
happens 3.3x10-3 more often than
K  + e- u
CP violation was surprising but welcome. Charge conjugation violation would not
explain the matter-anti-matter asymmetry of the universe because though charge
conjugated weak interactions didn’t exist the additionally parity flipped interactions did.
If there were interactions where matter particles decayed more often than antimatter
particles then it might explain the baryon asymmetry of the universe.
Research concentrates in two directions. Direct CP violation matter decays that happen.
more often than an anti-matter decays.
Bs(b s )  K+pi- 39% more than Bs( b s) -> K-pi+
Oscillation processes that convert matter to anti-matter.
1) Scattering and decay introduction. Note we are skipping chapter 5.
The key elements to understand in elementary particle physics are lifetimes, scattering
cross-sections and bound states. Bounds states are analyzed using the same methodology
used for the hydrogen atom, with different potentials. We will not study bounds states in
detail.
Calculating scattering cross-sections and lifetimes will have a components representing
the strength and potential of the force, including components representing the conserved
quantities associated with the force, and components for kinematics properties of the
interaction.
Typically we split the components into two sets.
The first set components we call the dynamical information and will be represented by
the amplitude, or matrix element, M. M is calculated by evaluating the sum of the
probabilities of each possible way an interaction can take place. The probably of each
interaction, represented by a Feynman diagram, can be calculated using the rules for
constructing M from those diagrams. There may be several diagrams involved and we
will sum over those diagrams. The initial state particles might be in a superposition of
different quantum numbers and the final state might have many possible quantum states.
We will average over the possible initial states (the particle must be in one of them or a
superposition of several with associated probabilities) and sum over the final states (all
that are possible).
The second set of components is known as the phase space and represents the kinematics
of the initial and final particles. Since the final state particles may have a range of
possible kinematics, large or small phase space, to get the total cross-section or evaluate
the lifetime you will have to integrate over the possibilities. A large phase space
typically means that the interaction is more probable
2) Classical billiard ball scattering cross section example.
The classical problem has no complex quantum states and happens with 100% probability
if the objects approach closely enough to collide so M is essentially one. However, it is
illustrative since it will teach is about the kinematical part of the scattering interaction
and will familiarize us with necessary concepts.
Consider a ball radius R and assume the particle scattering off the ball is small so we can
neglect its radius.
The classical problem can be represented by the impact parameter, the distance of closest
approach and theta, the angle the particle scatters to. Drawing a line from the center of
the ball to the point where the scattering particle strikes the ball the scattering angle will
be determined by reflecting the trajectory around that line by an angle alpha.
Then b = Rsin
And 2 +  = 
Where theta is 0 when there is no scattering.
Since we are interested in theta
b = Rsin() = Rsin(/2 - /2) = Rcos(/2)
We will then calculate the differential probability for the particle to scatter to a small area
d around  and .
d=sindd
From a small area d around b and .
d = bdbd
In classical scattering the object will scatter from the angle  to the same angle  so we
are primarily interested in the differential probability, d/d, as a function of .
d/d = b/sin db/d
for this problem specifically we find from b = Rcos(/2) that
db/ d = -R/2 sin(/2)
d/d = b/sin db/d = (Rb/2) sin(/2)/sin = (R2/2) sin(/2)cos(/2)/sin = R2/4
and integrate over the whole are to get the cross section
 = d =  d/d d = R2/4  d = R2/4 2 2 = R2
3) Decays
We are also interested in the probability of decay per unit time or the decay rate, .
dN = -Ndt
a solution to this differential equation is:
N(t) = N(0)e-t
testing our solution:
dN = -N(0) e-tdt = -Ndt
also of interest is the half-life which is indicative of the mean lifetime:
 = 1/
Note that if you have many decays then you just need to sum the probabilities 1 + 2 …
A sum over the final states. The lifetime will be the reciprocal of total probability and
 = 1/tot.
Often we are interested in measuring branching ratios, the ratio of a partial probability to
the total probability. i/.
 is also seen in another aspect of particles. Short-lived particles have a width. We see
them not as existing at a precise mass value but rather spread out over range of values
with a probability distribution. The width is related to the mean lifetime and the  by the
uncertainty principle.
If we can only see a particle for  time then its energy/or mass will have an uncertainty.
tE > /2
 ~ /2E
 ~ 2E/
Since  is proportional to the spread in energy of the particle we often call  the width
and give it units of energy. MeV. If you want to convert from the energy units of the
width to find the lifetime you just use =6.582x10-22 MeVs.
Note that a shorted lived particle will have a larger width or a larger spread in energies.
Measuring the width of a particle is often the only way to find the lifetime. For strongly
decaying particles they decay so fast that you can’t even measure the lifetime, but you
can measure the width. For instance the delta particle we were considering has a very
large width since it decayed strongly. The width large that there is even some small
probability that protons(940MeV) and pions(140MeV) can scatter through the
delta(1232MeV, width 120 MeV with a long tail) resonance and become
sigmas(1189MeV) and kaons(493.7MeV). See isospin example included with HW
solutions next time.
In fact the scattering probability is always enhanced if it takes place via one of these
resonances. Once the resonance is made it must decay sooner or later so the probability
of that part of the interaction is 1. Then the probability is just that to make the resonance,
which involves fewer vertexes than to make the resonance as an intermediate state and
then decay all in one interaction.
The bare W and Z bosons which decay almost instantaneously also have a width. This is
a very important property in understanding why there finite probability for interactions
via the weak decay to happen in low energy interactions like nuclear decay or protons
and neutrons.
4) The Golden Rule, for decays or scattering.
Transition rates (scattering cross sections or decay rates) have the general form:
transition rate ~ |M|2 x phase space
This expression is derived from perturbation theory. M is just the 1st order probability to
go from the initial state to the final state via some operator. The phase space is the
density of these states.
Note we are often going to be interested in the differential decay rate or scattering cross
section. For example the differential probability, d, to go to a momentum range p
around dp
5) Decay example
Example, how to calculate the transition rate: Simple case of two body decay. Calculate
the transition rate from 1 particle to 2 new particles. In the two body case the
momentums and energies of the outgoing particles are fixed to values given by energy
and momentum conservation. If the two outgoing particles have the same mass then the
absolute values of their energies and 3 momentums must also be equal. In this case the
phase space and differential rates are simple.
For a two body decay, dropping the c and
per unit time .
terms we will calculate the decay probability
d = 1/2E |M|2 S [2(d4p2/(2)4)2(p22-m22) 2(d4p3/(2)4)2(p32-m32)]
(2)44(p1-p2-p3)
E is the initial energy or in CM frame just m1. The initial two delta functions require that
the real outgoing particles have four momentums compatible with their mass values. The
last delta function enforces energy and momentum conservation.
Generally defining  =  d/dp dp where the integral runs over the ranges of 4
momentums. Then performing the integral over E2 and E3
where: 2(p22-m22) = 2(E22-( p22+m22))=(1/2E2) (E2-( p22+m22)1/2)
d = 1/2m1  |M|2 S [(d3p2/(2)32E2) (d3p3/(2)32E3)] (2)4 4(p1-p2-p3)
where
1) M, the matrix element, will quantify the dynamics of the interaction
2) There are differential momentum elements for each 3 momentum. No term for
the energy since once the three momentums and the mass of the particle are
known the energy is fixed and we integrated them out using the delta function.
3) m1, the mass of the original particle, is the natural scale of the problem. 1/2m1 is
known as a density of states term for the initial particle where E=m1. Note that
1/2E terms exist for the two final state particles as well where they came our of
the delta function where possibilities for the energy of the final state particles
were defined. Technically we could have defined a delta function for the initial
state which would have extracted the 1/2m term. The integral over the initial 3
momentum would be trivial since they are all 0.
4) The remaining delta function requires that momentum and energy be conserved in
all four dimensions.
5) All the other terms are to make the normalization and units come out correctly
after the integral
6) S is a statistical factor for when you have identical particle in the final state. 1/n!
for each set of n identical particles.
Example of two massless particles: E2 = E3 = c|p2| = c|p3|, -p2 = p3
 = (1/2m1)  |M|2 S [(d3p2/(2)32E2) (d3p3/(2)32E3)] (2)4 4(p1-p2-p3)
= (S/8m1)1/(2)2  |M|2/(|p2||p3|) d3p2 d3p3 3(-p2-p3) (m - E2 - E3)
Integrating over p3 and the 3D delta function which just fixes it’s value to -p2.
= (S/8m1) 1/(2)2  |M|2/|p2|2 d3p2 (m - E2 - E3)
Note that there are two cases here depending on weather the physics of M will result in
an angular distribution. 1) We expect no angular distribution and M has no dependence
on the phi or theta of particle 2. Therefore it’s only a function of |p2|. 2) There is angular
distribution. This will depend on the spins and the force involved. For instance we saw in
the decay of Co 60 there is a definite angular distribution. If there is an angular function
and we have to understand M before proceeding. Let’s consider case 1.
Often we will integrate over all the remaining momentum factors for a decay.
Switch to spherical coordinates and integrate of phi and theta, which will give us 4.
d3p2 = |p2|2 d|p2| sin d d
= (S/8m1) 1/(2)2 4 |M|2 d|p2| (m - E2 - E3)
= (S/8m1) 1/(2)2 4 |M|2 d|p2| (m - 2|p2|)
We need the delta function in terms of |p2|. Look up identities in appendix.
= (S/8m1)M|2 d|p2| (1/2)((1/2)m - |p2|)
= (S/16m1)|M|2 d|p2| ((1/2)m - |p2|)
Integrating over the last delta function.
 = (S/16m1)|M|2
where M is evaluated at the momentum given by the conservation laws. M should be
proportional to momentum squared.
General case with masses later