# Baryons & addition of angular momentum

1. Dec 24, 2004

### yxgao

I was wondering if someone could help me explain this problem.
In a baryon, two quarks are in l=0 and one quark is in the l=1 state. Quarks are spin-1/2 particles. What values can the total angular momentum take?

I know the answer is 1/2, 3/2, and 5/2, but I'm confused about how they arrive at this answer.

Thx for any help!

2. Dec 24, 2004

### dextercioby

Apply Clebsch-Gordan's theorem for the angular momentum states:
$$(l=0,s=\frac{1}{2}),(l=0,s=\frac{1}{2}),(l=1,s=\frac{1}{2})$$

Daniel.

3. Dec 24, 2004

### yxgao

That's what the solution said. I don't fully understand the theorem. Is there any way you can explain this better, in context of the problem?
Thanks a lot for help!

4. Dec 25, 2004

### dextercioby

Well,it's simple.U have three particles.Each of the paricles has two possible quantum states for the angular momentum.One for the angular,one for the spin.So that means 6 uniparticle states in all.U're interested in composing these 6 values for the angular momentum.So u apply the CG theorem and write:
$$0\otimes\frac{1}{2}\otimes 0\otimes\frac{1}{2}\otimes 1\otimes\frac{1}{2}$$
Begin to apply the the theorem from the right.You could do it from the left,as well.
$$0\otimes\frac{1}{2}\otimes 0\times\frac{1}{2}\otimes(\frac{1}{2}\oplus\frac{3}{2})=0\otimes\frac{1}{2}\otimes 0\otimes (0\oplus 1\oplus 1\oplus 2)=0\otimes\frac{1}{2}\otimes(0\oplus 1\oplus 1\oplus 2)$$

$$=0\otimes(\frac{1}{2}\oplus\frac{1}{2}\oplus\frac{3}{2}\oplus\frac{1}{2}\oplus\frac{3}{2}\oplus\frac{3}{2}\oplus\frac{5}{2})$$

$$=\frac{1}{2}\oplus\frac{1}{2}\oplus\frac{3}{2}\oplus\frac{1}{2}\oplus\frac{3}{2}\oplus\frac{3}{2}\oplus\frac{5}{2}$$

Daniel.

5. Dec 25, 2004

### yxgao

Thanks for your detailed explanation. How do you go from the last expression in the third to last line to the second to last line? Why do you only have 7 and not 8 terms in the parenthesis?
Thx

6. Dec 25, 2004

### dextercioby

I used the distributivity of the multiplication of representations towards the addition of the representations.The reason why in the last bracket there are 7 instead of 8 terms is that one of the 4 terms in the bracket is '0' and therefore the product of 1/2 with zero is still 1/2.The operator for the irreductible representation of ratio '0' is $\hat{1}$,and so,every operator of the irreductible representation "a" when getting multiplied with the unit operator remains th same.

Daniel.