1. Jun 15, 2004

### Norman

I am working through some old particle physics notes of mine and when reading Griffiths "Introduction to Elementary Particles" I stumbled across a very perplexing problem:

Problem 4.9, pg 138:

"When you are adding angular momentum... it is useful to check your results by counting the number of states before and after the addition. For instance in Example 4.1 we had two quarks to begin with, each would have $m_s=+\frac{1}{2}$ or $m_s=-\frac{1}{2}$ so there were four possibilities in all. After adding the spins, we had one combination with spin one (hence $m_s=1,0,-1$ ) and one with spin 0 ($m_s=0$ )-- again, four states in all."

Griffiths seems to imply that you should always have the same number of possible states before and after addition of their spins. I am fairly confident this is wrong. But it seems so odd that such a horribly incorrect statement would be published.

My counter example to Griffiths counting argument:
addition of Spin 1/2 and Spin 1 particles:
Before the addition there are 5 possible states, 2 from the Spin 1/2 and 3 from Spin 1.
i.e.:
$$S_1=\frac{1}{2}, m_{s,1}=+\frac{1}{2}, -\frac{1}{2}$$
$$S_2=1, m_{s,2}=1,0,-1$$

After addition, the total spin would either be 1/2 or 3/2, with two states coming from the 1/2 state again but now four states coming from the 3/2 total spin state. This is a total of 6 final states.
i.e.:
$$S=\frac{1}{2}, m_{s}=+\frac{1}{2}, -\frac{1}{2}$$
and
$$S=\frac{3}{2}, m_{s}=+\frac{3}{2}, +\frac{1}{2}, -\frac{1}{2}, -\frac{3}{2}$$

Am I wrong here? Or did I just mis-understand the implication of the problem. I know he did not out right state it, but I think it is clearly implied in the problem. I believe the argument only holds when you add the angular momentum of particles with the same angular momentum. i.e. 3 quarks or 2 mediators (spin 1), etc.
Thanks,
Norm

Last edited: Jun 15, 2004
2. Jun 16, 2004

### reilly

With a spin 1/2 state and a spin 1 state there are 6 states. Three have 1/2 spin up with each of the three spin 1 states. The other three come from spin -1/2 and the various spin 1 states. This follows from the fact that state vectors for two independent systems are product states -- you count unique combinations.
Regards,
Reilly Atkinson

3. Jun 16, 2004

### Norman

Reilly,

I guess I don't understand your comment. Griffiths (I think) is saying you need to count the number of possible states before and after you make the addition. Are you talking about before you add them, or after? Because before you add them, you only have 5 possible values of the z component of spin- 2 from spin 1/2 and 3 from spin 1.
Cheers,
Norman

4. Jun 16, 2004

### reilly

Norman -- The rule for counting states is

Number of states =(2*j1+1)(2*j2+1)(2*j3+1)....
when combining angular momentum j1, j2, j3, ...

Again, any particular j1z sate can be combined with all j2 states, and in turn with j3 states, and so forth. Two coins provide four states, a pair of dice (spin 5/2)provides 36 states. In QM, the number of states must be independent of the representation used -- so before and after must be the same. This is an enormously important concept -- a great deal of physics rests upon it(as in the Standard Theory). Most any book on QM, or angular momentum, or on the rotation group will discuss the point.

Just adding the collective number of individual angular momentum states won't cut it.

Regards,
Reilly Atkinson

5. Jun 16, 2004

### Tom Mattson

Staff Emeritus
Right. You can't compare the number of single-particle states to the number of states after addition. If you let |MJ1,MJ2> represent the state in which the spin-1 particle has magnetic quantum number MJ1 and the spin-1/2 particle has magnetic quantum number MJ2, then you have six of those states. That is what you compare with the six final states.

6. Jun 24, 2004