Physics 535 lecture notes: - 13 Oct 16th, 2007
Homework: 6.6, 6.8, 6.15
No homework for next week.
Midterm Exam Thursday Oct 25th. Covering Griffiths 1-4, 6 Perkins 1, Lecture Notes
1) The Golden Rule, for decays or scattering.
transition rate = 2p/ |M|2 x phase space
For a two body decay
dG = 2p/ |M|2 S 1/4pm1 [(cd3p2/(2p)32E2) (cd3p3/(2p)32E3)] (2p)4 d4(p1-p2-p3)
Two to Two scattering
ds = 2 |M|2 S 1/4Ö((p1p2)2 – (m1m2c2)2) [(cd3p3/(2p)32E3) (cd3p4/(2p)32E4)] (2p)4 d4(p1+p2-p3-p4)
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) 1/4pm1 or 1/4Ö((p1p2)2 – (m1m2c2)2) is the density term for the initial particles.
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 pd(p-p1) 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.
One Dirac delta function can always be used to integrate one of the final state 3 momentums. This just enforces that in CM frame once the 3 momentum of one particle is known the other must be equal and opposite.
The second Dirac delta function can be decomposed into an angular component and a momentum magnitude component. If the matrix element depends on angles then the angular part cannot be integrated until the matrix element is calculated. We should be able to integrate the momentum magnitude part, though in some cases a change in variables to energy might be useful in some cases.
Many decay and scattering cases are covered by a few instances of the above equations.
Two body decay of m1 particles with m2 and m3.
G = S|M|2p0/8pm12
p0 = (1/2m1)Ö(m14 + m24 + m34 - m12m22 – m12m32 - m22m32)
which reduced for m2 = m3 = 0
G = S|M|2/16pm12
Two to two scattering
ds/dW = (1/8p)2 |M|2 S |pf|/|pi| 1/E2 dW
Note we will find that the two to two scattering typically does have an angular distribution even for spinless particles.
2) 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 two vertex process such as scattering or decay. In this case A+B à C à A+B
-iM = [-ig] ò [i/(q2-m2)] [-ig] [1/(d4q/(2p)4)] [(2p)4 d4(p1+p2-q)] [(2p)4 d4(q-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 particles such as a W 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. The internal, or virtual, particle can have a mass that is far from it’s default value. 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.
Integrate over one delta function
-iM = [-ig] [i/((p1+p2)2-m2)] [-ig] d4(p1+p2-p3-p4)]
3) Consider the case of the decay of a composite particle of A and B where m>mA, mB
M ~ g2/m2
Compare this to a simple one vertex process.
-iM = [-ig] ò[1/(d4q/(2p)4)] [(2p)4 d4(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
Gµg2
tµ1/g2
We need more to understand the weak force. For instance for a meson to decay it must be considered as a two to two process with the quark and anti-quark of the meson annihilating each other and then the weak boson pair producing a lepton and antineutrino.
For this process the phase space will just be the phase space for one to two decay, meson to lepton and antineutrino but the Matrix element must be calculated as a two to two process.
G = S|M|2p0/8pm12
p0 = (1/2m1)Ö(m14 + m24 + m34 - m12m22 – m12m32 - m22m32)
for a small electron mass
G ~ |M|2/16pm1
G ~ (g2/mW2)2/16pm1
g can be identified with a, g = Ö(4pa)
G ~ p(a/mW2)2/m1
G ~ paW2/m1
Note that the units don’t work out in our toy model unless you give our coupling constant the dimensions of momentum.
Units: (MeV/c) 2/(MeV/c2) = MeV: The Expected units of width
Divide by = 6.582x10-22 MeVs: Width in 1/s or lifetime in seconds
aW = 10-5
m1 = mp = 139.6 MeV
Therefore
G = 2.250x10-12 MeV
t = /G = 2.925x10-10 sec
Fairly similar to what we will eventually get for the weak force for pion decay to electron and anti electron neutrino, though not surprising that it’s a few orders of magnitude off the real value.
G µ aW2ml2/m
In the real theory the coupling constant is dimensionless
Units MeV/c2
Multiply out the c2 to give MeV: Expected units of width
Divide by : Width in 1/s or lifetime in seconds
Instead consider this as an annihilation two to two scattering process A + B à A + B. Where still m> mA, mB and small momentums
ds/dW = (1/8p)2 |M|2 S |pf|/|pi| 1/E2 dW
s = 4p (1/8p)2 (g2/mW2)2 |pf|/|pi| 1/E2
= 1/4 (a/mW2)2 |pf|/|pi| 1/E2
= 1/4 (aW2)2 1/E2
Multiply by (c)2 c = 197.3 MeV fm will give units of area
Consider the collision at E = 197.3MeV
= 2.5x10-11 fm2 = 2.5x10-41 m2
again very similar to what we will eventually get for the weak force!
s µ aW2/E2
4) Renormalization
If we compute higher order diagrams in our toy theory or in real quantum electrodynamics we find that there are divergent term proportional to the logarithm of the momentum of the propagator particle as you integrate to infinity.
One way to fix this is to assume a new propagator term into the matrix element computation.
-M2/(q2-M2), with a large mass M
This is equivalent to saying that at large energy the particles interact via a new force with a force carrying bosom of mass M. In fact this is now just what we think happens. We had many problems that the phenomenology of the standard model did not answer such as dark matter. Introduction a new set of particles and interactions as can solve most of these problems. Performing the new integral you find that for finite N you find new terms that modify the masses and couplings that then multiple the same factors that we previously calculated. Therefore, in low energy reactions we see these as small corrections to the standard interactions and masses. In our experiments we have just been measuring these modified masses constants. One can even calculate the mass M which you will find to be about 1TeV, the scale at which we expect to find new physics at the LHC. All the quantum field theories are renormalizable in this way and all are consistent assuming we find new particles and interactions at a higher scale.