Download Physics 535 lecture notes: - 9 Oct 2nd, 2007 Homework: Griffiths: 4.8

Physics 535 lecture notes: - 9 Oct 2nd, 2007
Homework: Griffiths: 4.8, 4.10, 4.11, 4.29
Reading for next time: Griffiths Chapter 6
1) Review Parity and Charge Conjugation
Parity, P: Discrete symmetry involving inversion through all axis’s.
Each particle has two possible parity eigenvalues. +1 -1
Mesons have parity -1*(-1)l
Baryons have parity 1*(-1)l
Anti-baryons have parity -1*(-1)l
The photon, a vector particle, has parity -1
This comes from the quarks having positive parity and the anti-quarks negative parity
with an extra factor for the orbital angular momentum. The spin component has already
been taken care of by the intrinsic parity of the quarks.
Mesons in the l=0, j=0 state with anti-parallel spins have parity -1 and are called
pseudoscalars
This differentiates them from a true scalar particle such as the Higgs.
Mesons in the l=0, j=1 state with parallel spins have parity -1 and are called vectors –
they have the same spin and parity as the photon.
Mesons in the l=1, j=1 states with anit-parallel spins have parity 1 and called are
pseudovectors.
The total product of the parities will be conserved in an interaction. The final particles
can have an angular distribution characterized by an orbital angular momentum quantum
number to conserve parity.
Also parity will be conserved in that the inverted interaction should happen at the same
rate. From this the parity violation of the weak interaction can be seen since the inverted
interaction does not occur.
Charge conjugation, C: Discrete symmetry involving converting particles to antiparticles
Only particles that are their own anti-particles conserve are eigenstates of charge
conjugation and must preserve charge conjugation number. These particles will have
charge conjugation eigenvalues of 1 and -1.
The neutral mesons have charge conjugation number (-1)s+l
The photon has charge conjugation number -1
The total product of the charge conjugation eigenvalues will be conserved in an
interaction if the interaction involves eigenstates of parity.
Charge conjugated interactions should occur with the same rate and distributions as the
base interaction.
The weak force also clearly violates charge conjugation. Left handed neutrinos exist but
left handed antineutrinos do not exist.
Example rho0
Parity
rho0  pi0 pi0, pi+ pi-1 = -1 * -1 (-1)l=1
This interaction will have an angular distribution
Charge Conjugation
rho0  pi0 pi0
rho0 is s=1
-1  +1*+1, forbidden
rho0  pi+ pi-, not forbidden since the charged pions are not C eigenstates.
pi-  e- 
Note that pion decay to electrons, pi-  e-  , is suppressed. The antineutrino is in the
right-handed state. The right-handed antineutrino and the right-handed positron couples
to the weak force and the similarly the left-handed neutrino and left-handed electron

couple to the weak force. This decay forces the electron into the right handed state to

conserved spin. However, the muon being heavier and moving more slowly has more of
a chance of being force into the “wrong” state, actually being in the correct state in the
frame where the weak interaction took place. This is called helicity suppression.
2) CP and CP violation
We expected the product of C and P to be conserved.
+  + 
happens as often as the charge conjugated and parity inverted
-  - 
Note that the parity inversion flips the neutrino from left to right handed.
However it was eventually seen that

K  - e+ 
happens 3.3x10-3 more often than
K  + e- 
This has also been seen in the B sector in the past few years. For instance
B
s  K+pi- 39% more than Bs -> K-pi+


There is one more interesting effect.
K 0 and K 0 are not eigenstates of C or CP.
However a combination of the two is. This combination is possible because there is a
diagram that converts one to the other.


The CP eigenstate combinations are:
|KS> = (1/2)( K 0 - K 0 ), CP even
|KL> = (1/2)( K 0 + K 0 ), CP odd
by parity
KS can decay into two P odd pions and KL into three P odd pions.
 
The three body decay happened with lower probability. Therefore the KL has a longer
 
lifetime. However, in this case also CP violation decays of the pion have been observed.
In addition, consider this system as a system of two states with a coupling that can
translate one state to the other. This is like a spring system of two springs with third
spring the couple the two together. In the spring system the two eigenstates are where the
spring move together in the same direction or in opposite directions. If you start in a
states with one spring moving you will evolve into a state with the other spring moving at
a rate dependent on the coupling spring. This is identical to the phenomenology of the
two Kaon system and they are governed by the same equations.
In the spring system there is a small energy difference between the two states. Similarly
there is an energy difference between the kaon states that can be seen as a mass
difference. Also there will be a rate of oscillation between the two states, which can be
measured.
This oscillation system exists for K, B and D mesons and has been seen in all three, the
later two 1 year ago and this year respectively. For the B meson the oscillation rate is on
the terahertz rate.
3) Isospin
Another such spin like conservation principle comes from the similarity between down
and up quarks. They both have approximately the same mass and both interact with the
strong force the same way! Therefore there should be a conserved quantity, isospin, that
will govern what sort of interactions are allowed and at what probabilities. Isospin
conservation was noticed before it was even understood that the proton and neutron are
made up of quarks from just noting that the proton and neutron were very similar and
treating them as two of the same type of particle with different isospins. It was seen that
staring with a helium nucleus that that atoms with an two additional nucleons nn, pn or
nn all had similar mass, excited states, and strong scattering interactions. Though these
systems are different from the point of view of the
This can even be extended to strange particles since the strange mass is not that different.
However, the predictions start to be slightly less accurate. Note, since the masses if the u
and d quarks are not exactly the same there is even some very small amount of isospin
violation in that case.
Isospin:
Define I and I3
I is going to lead to 2I+1 states delineated by I3=Q-1/2(A+S) which goes from –I to I in
integer steps
Q = charge, A = baryon number and S=strangeness
For the light quarks this comes from assigning the quarks isospin, I3, quantum numbers u,
½ and d -½ and the inverse for the antiquarks.
For the proton and neutron |I,I3>
P=|1/2,1/2>
N=|1/2,-1/2>
and for the pions
Pi+=|1,1>
Pi0=|1,0>
Pi-=|1,-1>
Example: Consider the pion and nucleon colliding via the strong force. At first glance
these all such processes happen at the same rate, especially for the elastic processes. To
see the isospin effects we need a process that takes place through a eigenstatea of the
combined isospin quantum number, ++, uuu, 0, udd.
For pi+p: pi+p -> ++ -> pi+p
For pi-p: pi-p -> 0 -> pi-p
or : pi-p -> 0 -> pi0n
pi+p=|1,1>|1/2,1/2>= ++ =|3/2,3/2>
pi-p=|1,-1>|1/2>|1/2>=(1/3)|3/2,-1/2> - (2/3)|1/2,-1/2>
0=|3/2,-1/2>
The pi+p process has an amplitude three times as larger than the pi-p process at the delta
particle energy. The cross section actually goes as the square of the amplitude so this
process happens 9 times as often - you go from pi+p to the delta particle and then back
again. This is representative of the quarks involved in that in the later case there are two
udd combinations that could be made and the probability has to be divided up between
them. However there is also a charge exchange process pi-p to pi0n that happens twice
as often so total rate is only 3 times as large.
pi0n=|1,0>|1/2>|-1/2>=(2/3)|3/2,-1/2> + (1/3)|1/2,-1/2>
You get a factor of 1 from the pi-p to delta and 2 from the delta to pi0n.