Physics 535 lecture notes: - 12 Oct 11th, 2007
Homework: 6.6, 6.8, 6.15
1) Decays
Interested in the probability of decay per unit time. This is known 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 > /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 .
Note that short lived particles 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 fewer vertices than to make the resonance as an intermediate state and
then decay all in one interaction.
2) The Golden Rule, for decays or scattering.
transition rate = 2/
|M|2 x phase space
This expression is derived from perturbation theory. M is just the 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 decay rate probability d to go to a momentum range dp around
p.
3) 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
d = 2/
|M|2 S 1/4m1 [(cd3p2/(2)32E2) (cd3p3/(2)32E3)] (2)4 4(p1-p2-p3)
where
1) M is the matrix element

2) There are differential momentum elements for each final state 4 momentum. No
term for the energy since once the three momentums and the mass are known the
energy is fixed. These terms are the momentum density of state terms. They are
constructed to be Lorentz invariant and to integrate to a probability of 1. The
derivation of the normalization constant E comes from the relativistic wave
equation, the Klein Gordan equation. If you have a wave solution with e-iEt there
will be a normalization constant involving E.
3) 1/4m1 is the density term for the initial particle where E1=m1.
4) 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(pp1)dp is p1.
5) S is a statistical factor for when you have identical particles in the final state.
6) All the other constants are to make the normalization come out correctly after the
integral.
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 and instead keep the final expression as a differential lifetime.
Consider the case of two massless particles. E2 = E3 = c|p2| = c|p3|, -p2 = p3
 =  2/ m1 |M|2 S/4 [(cd3p2/(2)32E2) (cd3p3/(2)32E3)] (2)4 4(p1-p2-p3)
remove the c’s and

 =  2/m1 |M|2 S/4 [(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)
integrate 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. 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 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.
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|)
4) Two to Two scattering
d = 2 |M|2 S 1/4((p1p2)2 – (m1m2c2)2) [(cd3p3/(2)32E3) (cd3p4/(2)32E4)] (2)4
4(p1+p2-p3-p4)
 remove the c’s and
d = |M|2 S 1/4((p1p2)2 – (m1m2)2) [(d3p3/(2)32E3) (d3p4/(2)32E4)] (2)4 4(p1+p2-p3p4)

= |M|2 S /4 p1(E1+E2) [(d3p3/(2)32E3) (d3p4/(2)32E4)] (2)4 4(p1+p2-p3-p4)
where p1 is the magnitude of either of the incoming momentums
= (1/8)2 |M|2 S 1/ p1(E1+E2) [(d3p3/E3) (d3p4/E4)] 3(-p3-p4) (E1 + E2 - E3 - E4)
integrate over one delta function. p4 = -p3
= (1/8)2 |M|2 S 1/ p1(E1+E2) [(d3p3/E3 E4)] (E1 + E2 - E3 - E4)
where E3 =  (m32 + p32), E4 =  (m42 + p32),
and total E = E1+E2 = E3+E4
Separating the angular and radial momentum parts. p2dpd
= (1/8)2 |M|2 S 1/p1E p2/E3 E4 p2dpd (E - E3 - E4)
and noticing that
(E1+E2)/ E3 E4 = E/ (m32 + p2) (m32 + p2)
= ((m32 + p2)+(m32 + p2))/ (m32 + p2) (m32 + p2)
therefore
dE = Epdp/ (m32 + p2) (m32 + p2) = Epdp/E3E4
= (1/8)2 |M|2 S 1/p1 p/E2 dp d (E - E3 - E4)
and integrating over the delta function
= (1/8)2 |M|2 S p/ p1 1/E2 d
where p is the magnitude of either of the final momentums
5) This leaves understanding the matrix element. The matrix element must basically
represent the strength of the force and the internal particles. Let’s consider a simple
theory with no spins or other complications.
For a simple one vertex process.
-iM = [-ig] [1/(d4q/(2)4)] [(2)4 4(p1-p2-p3)]
This is not representative of any physics process but is instructive. Integrate over the
delta function to preserve momentum and energy.
-iM = -ig
M=g
g2
1/g2
Works for electromagnetic and strong decays where the mass of the internal photon or
gluon line is near the real mass of the photon and gluon, except for the fact that they are
two vertex processes. We remember that we need more to understand the weak force.
For a simple two vertex process such as scattering or decay.
-iM = [-ig]  [i/(q2-m2)] [-ig] [1/(d4q/(2)4)] [(2)4 4(p1+p2-q)] [(2)4 4(q-p3-p4)]
Integrate over one delta function
-iM = [-ig] [i/((p1+p2)2-m2)] [-ig] 4(p1+p2-p3-p4)]
where
1) –ig are the coupling strengths of the force at each vertex
2) [i/(q2-m2)] is the propagator for the internal particle. We can see that a factor of 1
over mass squared is going to come in for a massive particle such as a W in and
small initial momentums such as in a decay.
3) There is an integration over the possible momentums of the internal particle.
Note the 4D integration. The energy is not fixed by the mass of the particle in this
case. However, when the internal momentum is small and the mass of the internal
particle large the probability is lowered by the propagator term.
4) There are terms that conserve momentum and energy at each vertex.
In fact in this case
M ~ g2/m2