Physics 535 Lecture Notes: - 11 Oct 9Th, 2007

Physics 535 lecture notes: - 11 Oct 9th, 2007

Homework: 6.2, 6.3, 6.4

1) Isospin example with strangeness

I3 can also be determined using I3=Q-1/2(A+S). Transition between p- and 0K0,-K+,+K- via a 0

This transition is not impossible since as we see in the book the delta has a very wide probability distribution. We will discuss this again today’s section on decays.

First figure out isospins, I and I3, for each particle. I3=Q-1/2(A+S), where S is -1 for strange quarks.

p uud |1/2,1/2>, I3 = +1 -1/2(1+0) u udd |1/2,-1/2>

+ u d |1,1> 0 u u |1,0> - u d |1,-1>, I3 = -1 -1/2(0+0)

+ uus |1,1>, I = +1 -1/2(1-1) 3 0 uds |1,0>

- dds |1,-1>

K+ u s |1/2,1/2> I3 = +1 -1/2(0+1) K- u s |1/2,-1/2> I3 = -1 -1/2(0-1)

K0 d s |1/2,-1/2> I3 = 0 -1/2(0+1)

K 0 d s |1/2,1/2> I3 = 0 -1/2(0-1)

0 udd |3/2,-1/2> I3 = 0 -1/2(1)

First the transition, 0  +K- does not conserve isospin projection, I3. It’s forbidden! Note that this state also creates two 2 quarks!

The other two transition are allowed transition from I =1+1/2 to an I = 3/2 state and back to a I=1+1/2 state. However, when transitioning from the initial I = 1+1/2 state to the I=3/2 state the initial state can be expressed as a superposition of I=3/2 and I=1/2 isospin eigenvectors. The rules for adding spin or isospin say that starting from states I =1/2 and 1 the combined state could be either I=1-1/2 or I=1+1/2. In this case we know the final state will be I=3/2 so we can only have that transition at a reduced probability corresponding to the percentage of the original state that is in the total 3/2 state. In this case the transition through a |1/2,-1/2> state, which would be a neutron, is kinematically forbidden Note that though that the total I for the first to particles may be misaligned on aligned the I3 projection is fixed and we know it’s value.

11,-1>|1/2,1/2> = 1/3|3/2,-1/2> -  (2/3)|1/2,-1/2>

The initial transition, which is the same for both allowed interactions, will happen with probability 33%.

Similarly for the transition from the delta to the final states the probabilities can be just read from the same constants solving the eigenvalue problem in reverse

|3/2,-1/2> = 1/3|1,-1>|1/2,1/2> +  (2/3)|1,0>|1/2,-1/2>

The first one is for the - decay and the second for the 0 decay.

Therefore the ratio is 0K0,-K+,+K- 2/9:1/9:0 2:1:0

Let’s try p+  ++ -> +K+ The delta is in the |3/2,3/2> = |1,1>|1/2,1/2> Compared to the others 1: 2/9 : 1/9 : 0

2) 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 cross- sections and lifetimes will have a components representing the strength and potential of the force, which includes 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. There also 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) 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. 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

3) 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 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 don’t have to consider its 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

4) Decays

Interested in the probability of decay per unit time. This is know as the decay rate, . dN = -Ndt and N(t) = N(0)e-t then dN = -N(0) e-tdt = -Ndt also if interest is the mean lifetime is  = 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.

Often we are interested in measuring branching ratios. 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 > h /2  ~ h /2E

 ~ 2E/ h

Since  is proportional to the spread in energy of the particle we often call  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 h .

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 width. For instance the delta particle we were considering has a very large width since it is made and decays strongly. So large that there is even some small probability that protons and pions can scatter through the delta resonance and become sigmas and kaons.

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 less vertices than to make the resonance as an intermediate state and then decay all in one interaction.

5) The Golden Rule, for decays or scattering.

transition rate = 2/ h |M|2 x phase space

Note we are often going to be interested in the differential decay rate or scattering cross section. For example the probability d to go to a momentum range p around dp

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 momentums must also be equal. In this case the phase space and differential rates are simple.

For a two body decay

2 3 3 3 3 4 4 d = 2/ h m1 |M| S/4 [(cd p2/(2) 2E2) (cd p3/(2) 2E3)] (2)  (p1-p2-p3)

where 1) M is the matrix element

2) There are differential momentum elements for each 3 momentum. No term for the energy since once the three momentums are known the energy is fixed. 3) The E is coming from the probability density of a free particle. Think of a particle with momentums dp then with fixed mass E2 = p2+m2, 2EdE = 2pdp, then density = dE/E = pdp/E2, and integrated at p the density = p/E2 ~ 1/E. 1/E is the a normalization factor. 4) All the other terms are to make the normalization come out correctly after the integral. 5) The delta function requires that momentum and energy be conserved in all four

dimensions. Note that the delta function is defined such that the integral of p(p1) is p1. 6) S is a statistical factor for when you have identical particle in the final state.

Typically we will integrate over all these factors for a decay. However in cases where there is an angular distribution involved in the matrix element we may wish to not integrate over that variable.

For two massless particles. E2 = E3 = c|p2| = c|p3|, -p2 = p3

2 3 3 3 3 4 4  =  2/ h m1 |M| S/4 [(cd p2/(2) 2E2) (cd p3/(2) 2E3)] (2)  (p1-p2-p3)

2 2 3 3 3 2 = S/8 h m1 1/(2)  |M| /|p2||p3| d p2 d p3  (-p2-p3) (mc - E2 - E3)

integrate over p3 and the 3D delta function which just fixes it’s value to -p2.

2 2 2 3 2 = S/8 h m1 1/(2)  |M| /|p2| d p2 (mc - E2 - E3)

Note that there are two cases here. 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 |p |. 2) 2 There is an angular function and we have to understand M before proceeding. Let’s consider case 1.

Switch to spherical coordinates and integrate of phi and theta, which will give us 4. 3 2 d p2 = |p2| d|p2| sin d d

2 2 2 = S/8 h m1 1/(2) 4 |M| d|p2| (mc - E2 - E3)

2 2 = S/8 h m1 1/(2) 4 |M| d|p2| (mc - 2|p2|)

We need the delta function in terms of |p2|. Look up identities in appendix.

2 = S/8 h m1 M| d|p2| (1/2)((1/2)mc - |p2|)

2 = S/16 h m1 M| d|p2| (1/2)((1/2)mc - |p2|)