- #1
Norman
- 895
- 4
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
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: