Invariance Principles and Conservation Laws

Recall from last lecture:

What are Clebsch-Gordon coefficients.

Addition of Angular Momenta

Example: Decay of a pseudoscalar to a pseudoscalar and a vector

Let's determine the angular momentum states possible in the decay of a pseudoscalar meson to a pseudoscalr meson and a vector meson (abbreviated as P to PV). A pseudoscalar meson has JP = 0- and a vector meson has JP = 1-. The initial state is pseudoscalar, | JP = 0- >, and the final state is a combination of pseudoscalar and vector in a state of relative orbital angular momentum L, | 0- ⊗ 1-; L, M >.

Notice something odd about the above situation. When one particle decays (at rest, since we can always transform to the CM frame) to two lighter particles, the two lighter particles emerge back-to-back. Yet, we can asign them angular momentum. What does this mean? How can two particles, traveling in opposite directions from a parent particle have relative orbital angular momentum? And if they do, what does this imply?

They can have relative orbital angular momentum. The classical analogue is two particles traveling back-to-back with momenta p, but slightly displaced from each other by an amount b, called the impact parameter. These two particles have relative angular momentum of L=bp. In a quantum mechanical sense, having relative angular momentum implies that if a large number of identically prepared parent particles decay, and we measure the angular distribution of the emerging particles, the distribution will have a shape determined by the angular momentum.

Now back to our p to PV decay. What is the range of allowed values for L, determined from the rules for addition of angular momenta? The angular momentum of the pseudoscalar and vector particles is | j1=0⊗j2=1 > = | j3=1 >. (In general, the combination of a 0 angular momentum with any other angular momentum L yields a total angular momentum of L.) The addition of the orbital angular momentum must result in a state of total angular momentum equal to 0, the angular momentum of the initial state. This can only occur for L=1.

What about the m's? The sum of the m's must add to give 0, so the following combinations are allowed:

| j3,m3=1,1; l,ml=1,-1 > | j3,m3=1,0; l,ml=1,0 > | j3,m3=1,-1; l,ml=1,1 >
Consulting the table of Clebsch-Gordon coefficients we find that to yield a state with | J,M=0,0 >, these must be combined with the following coefficients:
| 0,0 > = 1/sqrt(3) ( | j3,m3=1,1; l,ml=1,-1 > - | j3,m3=1,0; l,ml=1,0 > + | j3,m3=1,-1; l,ml=1,1 > ).

Up to now we have ignored parity. Is this process parity conserving or parity violating? The initial state has parity -1. The final state has parity (-1)(-1)(-1)L = (-1)L = -1, and is therefore parity conserving. Decays of this type can occur via the strong, electromagnetic, or weak interactions. If the one of the daughter particles had the opposite parity (i.e. change the pseudoscalar to a scalar or the vector to an axial-vector) then the decay violates parity and can occur only via the weak interaction.

Finally, what if the vector particle is a (real) photon? Because the photon is massless, it occurs in only two of the three possible polarization states, | j,m=1,±1 >. These are the transverse polarization states. The state with m=0 is called the longitudinal polarization state, and it doesn't exist for massless photons. (This state is available to virtual (massive) photons.) Therefore, if the above decay is to a pseudoscalar and a photon then the angular state is:

| 0,0 > = 1/sqrt(3) ( | j3,m3=1,1; l,ml=1,-1 > + | j3,m3=1,-1; l,ml=1,1 > ).

Copyright © Robert Harr 2005