Invariance Principles and Conservation Laws

Recall from last lecture:

Parity and C-parity are multiplicative quantities.

Parity and C-parity are conserved in strong and electromagnetic interactions.

The parity of a state of angular momentum L is (-1)L.

Addition of Angular Momenta

Review of Rotation Operators

There is a discussion of this in appendix C of Perkins.

Recall that in quantum mechanics, to the measurable angular momentum is associated an operator, just like any other measurable quantity. In fact, angular momentum is more complex because there is a group of operators associated with it. Conceptually, it is easiest to begin with the operators for the angular momentum about each of the x, y, and z axes; I'll call these three operators Jx, Jy, and Jz.

Earlier I wrote out Jz in terms of the coordinates:

Jz = -i hbar(x d/dy - y d/dx)
The three angular momentum operators don't commute:
[Jx, Jy] = i Jz
[Jy, Jz] = i Jx
[Jz, Jx] = i Jy
Notice the cyclic nature of this relation.

The fact that the operators don't commute means that a simultaneous measurement of more than one is impossible. Another operator that does commute with any of the three is the square of the total angular momentum:

J2 = Jx2 + Jy2 + Jz2
The usual choice for simultaneous measurement is J2 with eigenvalue J(J+1), and Jz with eigenvalue M.

Another pair of useful operators is

J± = Jx ± i Jy
These are known as raising (+) and lowering (-) operators of angular momentum because when acting on a state with angular momentum quantum numbers j and m:
J+ | j, m > = C+jm | j, m+1 > = sqrt{j(j+1)-m(m+1)} | j, m+1 >
J- | j, m > = C-jm | j, m-1 > = sqrt{j(j+1)-m(m-1)} | j, m-1 >
The C±jm are coefficients that are given by the square root expression. Notice that if m=j, then the coefficient C+jm = 0, and if m=-j, C-jm = 0. Thus the raising and lowering operators "obey the rules" for angular momentum quantum numbers and do not produce states with m>j or m<-j.

Example of Adding Angular Momenta States

Now suppose we have a system with two (or more) angular momenta given by the states | j1, m1 > and | j2, m2 >, where the j's and m's are the corresponding quantum numbers (m's measured for a common axis), and we want to determine the angular momentum of the whole system, | J, M >. Classically we would simply add the vectors to get J = J1 + J2. But quantum mechanically, we can't simultaneously know all the components of angular momentum, so we can't perform the vector addition. In fact, the rules of quantum mechanics require that the x and y components of angular momenta take on all possible allowed values.

Therefore, the addition of | j1, m1 > and | j2, m2 > can result in a state with | j1 - j2 | < J < j1 + j2. The z components are known though, and their addition yields the z component of the total angular momentum, M = m1 + m2. The result is that the whole system is a superposition of states with the same M and different J's:

| j1, m1; j2, m2 > = ΣJ=|j1-j2|j1+j2 | J, M > < J, M | j1 j2 m1 m2 >
where the quantities < J, M | j1 j2 m1 m2 > are numbers called Clebsch-Gordon coefficients.

Example: Combining q and qbar to form a meson

If we combine a quark and an anti-quark to form a meson, what are the possible states of total angular momentum (spin of the meson) that can be formed? Begin by assuming that the q and qbar have orbital angular momentum L=0, so that the problem is simplified to combining to spin 1/2 particles. The combination of two spin 1/2 particles can yield J = 0 or 1. Begin with both particles aligned with spins down. This combination has M = -1, and can only be in the J = 1 state:

| j1, m1; j2, m2 > = | 1/2, -1/2; 1/2, -1/2 > = | J, M > = | 1, -1 >
The Clebsch-Gordon coefficient for this combinatino is 1.

Now apply the raising operator to this relation:

J+ | j1=1/2, m1=-1/2; j2=1/2, m2=-1/2 > = | 1/2, 1/2; 1/2, -1/2 > + | 1/2, -1/2; 1/2, 1/2 > = J+ | J=1, M=-1 > = sqrt(2) | 1, 0 >

Applying the raising operator a second time yields:

J+ { | 1/2, 1/2; 1/2, -1/2 > + | 1/2, -1/2; 1/2, 1/2 > } = | 1/2, 1/2; 1/2, 1/2 > + | 1/2, 1/2; 1/2, 1/2 > = 2 | 1/2, 1/2; 1/2, 1/2 > = sqrt(2) J+ | J=1, M=0 > = 2 | 1, 1 >

Finally, the J=0, M=0 state is obtained by noting that there is a combination of momenta that is orthogonal to the J=1, M=0 state, namely:

1/sqrt(2) | 1/2, 1/2; 1/2, -1/2 > -1/sqrt(2) | 1/2, -1/2; 1/2, 1/2 > = | 0, 0 >

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

Copyright © Robert Harr 2005