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Total Nuclear Angular Momentum

  1. Oct 1, 2016 #1
    1. The problem statement, all variables and given/known data
    The nucleons in a nucleus have a net orbital angular momentum ##L = 1## and spin angular momentum ##S = \frac{1}{2}##. What are the possible values of the total nuclear spin ##J##, i.e. the total angular momentum of the nucleus, and what are the possible values of ##J_z## in each case?

    2. Relevant equations


    3. The attempt at a solution

    I am very confused about how to solve this question.

    I know that the total angular momentum ##\vec{J} = \vec{L} + \vec{S}## and that the lengths of these vectors are ##|\vec{L}| = \sqrt{l(l+1)}## and ##\vec{S} = \sqrt{s(s+1)}## respectively (in units of ##\hbar##).

    At a guess, I would say that the total nuclear spin can be 1/2 or 3/2, but I don't have a particularly convincing reason.

    I'd really appreciate some help on what I need to understand to solve questions like this.

    Thank you!
     
  2. jcsd
  3. Oct 1, 2016 #2

    kuruman

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    That is correct, that's how angular momenta add in QM. What about the possible values of Jz?
     
  4. Oct 2, 2016 #3
    I think ##J_z = m_j \hbar## so the possible values should be ##j_z = \frac{3}{2}, \frac{1}{2}, -\frac{3}{2}, -\frac{1}{2}##
     
  5. Oct 2, 2016 #4

    kuruman

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    That is correct for J = 3/2. For J = 1/2 you have another pair of +1/2 and -1/2 values for a total of 6. Note that the number of states is
    (2J1+1) + (2J2 + 1) = (2×3/2+ 1) + (2×1/2+1) = 4 + 2 = 6.
    Counting the other (L + S) way, you get the same number of states
    (2L + 1) (2S + 1) = (2×1 + 1)×(2×1/2 + 1) = 3×2 = 6
    which is as it should be.
     
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