Baryons & addition of angular momentum

[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!

 

[dextercioby]

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!

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.

 

[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!

 

[dextercioby]

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!

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.

 

[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

 

[dextercioby]

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

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.