Download Physics 535 lecture notes: - 10 Oct 4th, 2007 Homework: 6.2, 6.3

Physics 535 lecture notes: - 10 Oct 4th, 2007
Homework: 6.2, 6.3, 6.4
1) Review CP and CP Violation
We expected the product of C and P to be conserved for the weak force. Charge
conjugated weak force diagrams did not exist but the charge conjugated parity flipped
version did exist. From the point of view of the weak force carriers having V-A, vectoraxial(pseudo)vector, structure this made sense since the axial part of the force carrier has
the opposite parity compared typical force carriers and thus flips the parity of the
interaction.
CP violation was surprising but welcome. Charge conjugation violation was not enough
to explain the matter-anti-matter asymmetry of the universe because though charge
conjugated weak interaction didn’t exist the additionally parity flipped interactions did.
CP violation is seen in small and large amounts, but not enough to explain the universe.
New physics contributions could lead to unexpectedly large CP violation, which could
explain the problem. Searches for such violations are a large endeavor involving 4
current HEP experiments and one future one at the LHC. These experiments concentrate
on B physics where SM CP violation is large and where contribution from new physics
involving the Higgs particle might be large since the Higgs couples more strongly to
more massive particles.
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+
 
Processes that convert matter to anti-matter. Oscillation processes, where CP violation
might also happen in the oscillation
process or subsequent decays.
 
2) 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! The symmetry of this system should lead to 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. Without even the quarks you could build an isospin system out of two nucleon
combination.
Analogous to spin
I = ½, I3 = -½,½
p: |½ ,½>, n: |½,-½>
I = ½ + ½ =1, I3 = -1, 0, 1
pp: |1,1> = |½ ,½>|½ ,½>
pn:|1,0> = (1/2)(|½ ,½>|½ ,-½> + |½ ,½>|½ ,-½>)
nn: |1-1> = |½ ,-½>|½ ,-½>
I = ½ - ½ = 0, I3 = -1, 0, 1
pn(deuteron):|1,0> = (1/2)(|½ ,½>|½ ,-½> - |½ ,½>|½ ,-½>)
It was seen that staring with a helium nucleus that that atoms with an two additional
isospin triplet nucleons nn, pn or nn all had similar mass, excited states, and strong
scattering interactions.
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 anti-quarks.
u: |½ ,½>, d: |½ ,-½>, u : |½ ,-½>, d : -|½ ,½>,
Don’t realy pay attention to u and d separately or d and u since they are the same
particle fromthe point of
view of isospin except to always make the mesons from particle
antiparticle pairs.


For the pions |I,I3> I = ½ + ½ = 1, I3 = -1, 0, 1
Pi+: u d :
|1,1> = |½ ,½>|½ ,½>
Pi0: u u or d d :|1,0> = (1/2)(|½ ,½>|½ ,-½> + |½ ,½>|½ ,-½>)
Pi-=|1,-1>: d u : |½ ,-½>|½ ,-½>

also be determined using I3=Q-1/2(A+S).
 I3 can

There
should
be a singlet as well.  |I,I3> I = ½ - ½ = 1, I3 = 0

: the system breaks down somewhat at this point since we should have been including
the strange quark. The  is part strange and the states are a bit complex since isospin
with the strange isn’t perfect. See chapter 5 if you are interested
For the proton and neutron |I,I3>, have to do a triple addition of spins. The proton and
neutrons are the states where I = ½ + ½ - ½, I3 = -½,½. Though the sum of the three
isospins may be complex with the -1/2 being any of the three quarks you don’t have to
worry about it since you can’t break the proton up and measure the isospin of the
individual quarks
There is also a system with I = 3/2, I3 = -3/2, -1/2, 1/2, 3/2.
++: uuu: |3/2,3/2>, +: uud |3/2,1/2>, 0: udd, |3/2,-1/2>, -: ddd |3/2,-3/2>
Note that the two middle particles are isospin excited states of the proton and neutron.
Isospin allows us to figure out what combinations of quarks are allowed, which will have
similar properties relative to the strong force, and whether they are expected to be more
massive. These particles can also be in ground or excited spin and angular momentum
states. To classify a particle and understand it’s interactions all these quantum numbers
need to be determined.
In addition, since isospin is conserved in strong interaction it will have dynamical
implication on strong scattering interactions. Isospin will have to be conserved which
can decrease the probability of certain interactions occurring.
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.
The isospin allowed interactions are clear. Isospin for the light quarks is determined by
charge and baryon number so you can’t get something that violates charge conservation
or creates new baryons without an anti-baryon.
To see non trivial isospin effects we need a process that takes place through a eigenstate
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.
3) 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. Calculating scattering crosssections and lifetimes will have a components representing the strength and potential of
the force, which include components representing the conserved quantities associated
with the force, and components for kinematics properties of the interaction.
The first set components we call the dynamical information and will be represented by
the amplitude, or matrix element, M calculated by evaluating the relevant Feynman
diagrams using the rules associated with constructing M from those diagrams. The
internal components, or propagators, might have a range of values and we will have to
integrate over those values. Also note that 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) and sum over the final states (all are possible).
The second set of components is known as the phase space and represents the kinematics
of the initial and final particles. Note 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.
4) Classical billiard ball scattering cross section example.
The classical problem has no complex quantum states or Feynman diagrams so M is
essentially one. However, it is illustrative since it will teach is about the kinematic part
of the scattering interaction.
Consider a ball radius R and assume the particle scattering off the ball is small so we
don’t have to consider it’s radius.
The classical problem can be represented by the impact parameter, the distance of closest
approach if the particle did not scatter beforehand 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 probability for the particle to scatter to any small segment of
d=sindd
then the chance to scatter from a small segment db around b
d = bdbd
to a small segment sindd around  and  is
d = d/d d = b/sin db/d d
for this problem specifically
db/ d = -R/2 sin(/2)
d = d/d d
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