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Baryons & addition of angular momentum

  1. Dec 24, 2004 #1
    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. jcsd
  3. Dec 24, 2004 #2

    dextercioby

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

    Daniel.
     
  4. Dec 24, 2004 #3
    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!
     
  5. Dec 25, 2004 #4

    dextercioby

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    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:
    [tex] 0\otimes\frac{1}{2}\otimes 0\otimes\frac{1}{2}\otimes 1\otimes\frac{1}{2} [/tex]
    Begin to apply the the theorem from the right.You could do it from the left,as well.
    [tex] 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)[/tex]

    [tex]=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})[/tex]

    [tex]=\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} [/tex]

    Daniel.
     
  6. Dec 25, 2004 #5
    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
     
  7. Dec 25, 2004 #6

    dextercioby

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    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 [itex] \hat{1} [/itex],and so,every operator of the irreductible representation "a" when getting multiplied with the unit operator remains th same.

    Daniel.
     
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