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Clebsch-Gordan coefficients calculation

  1. Aug 30, 2016 #1
    1. The problem statement, all variables and given/known data
    So i have to calculate the Clebsch-gordan coefficients for the state j1=3/2 and j2=1/2



    2. Relevant equations
    Recursiom formula, lowering and uppering operator



    3. The attempt at a solution
    I have tried to calculate the first set of the coefficients stating that:
    L- l2,2> = L- l3/2,1/2>l1/2,1/2>

    For the left side I got that it is equal to 2hl2,1>. Is that correct? I can't get the right coefficients on the right side. In general, I have trouble understanding which combinations of j and m to use tobget what I have to do. Is there some procedure to get all of them in order? Right now I am trying to guess what to use and see if I get the right combination.
     
  2. jcsd
  3. Aug 30, 2016 #2

    blue_leaf77

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    Yes.

    You should start from the states in the two representations which are proportional to each other. One of such states are ##|3/2,3/2\rangle |1/2,1/2\rangle## and ##|2,2\rangle##. Therefore, ##|2,2\rangle = c|3/2,3/2\rangle |1/2,1/2\rangle##. The constant of proportionality ##c## is in general a unimodular complex number, but conventionally it is chosen to be ##1##. Thus ##|2,2\rangle = |3/2,3/2\rangle |1/2,1/2\rangle##. Then apply the lowering operator on both sides like you have done and in the RHS, use ##L_- = L_{1-}+L_{2-}##.
     
    Last edited: Aug 30, 2016
  4. Aug 30, 2016 #3

    DrClaude

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    Yes, assuming that h means ##\hbar##.

    Given that you are adding together ##j_1## and ##j_2##, then the allowed values of ##J## are
    $$
    J = j_1 + j_2, j_1+j_2 - 1, \ldots, \left| j_1 - j_2 \right|
    $$
    Second, the triangle rule has to be followed: ##M = m_1 + m_2##. So in the case you have, the result will have to be 0 since ##M = 2## while ##m_1+m_2 = 1##.
     
  5. Aug 30, 2016 #4
    Thank you! I got the first ones:
    I2,1> =√3/2 I1/2,1/2> + 1/2 I3/2,-1/2>

    Is that ok? I tried applying lowering again on this expression but didn't get the solution. How do I know where to use the operator next?
     
  6. Aug 30, 2016 #5

    blue_leaf77

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    If that square root applies only on the 3 in the numerator, then it's fine.
    Why won't it work? To get |2,0> you should indeed apply the lowering operator once more on both sides. Maybe you just made a mistake during calculation.
     
  7. Aug 30, 2016 #6

    DrClaude

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    That's not correct, check the values of ##j_!## and ##j_2##.
     
  8. Aug 30, 2016 #7

    blue_leaf77

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    Actually he is using ##|m_1,m_2\rangle## notation on the RHS while the values of ##j_1## and ##j_2## are only implied.
     
  9. Aug 30, 2016 #8

    DrClaude

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    Staff: Mentor

    Right o:) Nevermind....
     
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