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Calculating Clebsch Gordon Coefficients

  1. Mar 28, 2010 #1
    1. The problem statement, all variables and given/known data

    Find the Clebsch Gordon coefficients of [tex] 1/2 \otimes 1 = 3/2 \oplus 1/2 [/tex]

    2. Relevant equations

    We have some properties of the CG coefficients which might be useful:

    1) they are nonzero only if j is between j1-j2 and j1+j2
    2) m = m1+m2 for nonzero coefficients
    3) they are real

    3. The attempt at a solution

    I am horribly confused. I know that the CG coefficients are given as the coefficients in the expansion

    [tex] |j m, j_1 j_2 \rangle = \sum_{m_1} \sum_{m_2} | j_1 m_1, j_2 m_2 \rangle \langle j_1 m_1, j_2 m_2| jm, j_1 j_2\rangle [/tex]

    or

    [tex] \langle j_1 m_1, j_2 m_2 | j m \rangle [/tex]

    and I know that the possible |j m j1 j2> states are for the product space with j1 = 1, j2 = 1/2:

    [tex] | \frac{3}{2} \frac{3}{2} ,1 \frac{1}{2} \rangle, | \frac{3}{2} \frac{1}{2},1 \frac{1}{2} \rangle, | \frac{3}{2} \frac{-1}{2} ,1 \frac{1}{2} \rangle, | \frac{3}{2}\frac{-3}{2} ,1 \frac{1}{2} \rangle, | \frac{1}{2} \frac{1}{2} ,1 \frac{1}{2} \rangle, | \frac{1}{2} \frac{-1}{2} ,1 \frac{1}{2} \rangle [/tex]

    but I don't understand what on earth to do or where to even start. Any help would be great.
     
  2. jcsd
  3. Mar 28, 2010 #2
    You usually just look them up in a table. I hope your assignment doesn't want you to actually derive them. Also, I am a bit confused on your notation in the first sentence.
     
  4. Mar 28, 2010 #3
    Yes, I think the notation is what's confusing me. I computed some of the coefficients a couple of years ago in an undergraduate class, but I've never seen this notation. I guess what the author means is to find the coefficients for all the basis kets in the product space where j1 = 1/2, j2 =1. There are (2*j1+1)*(2j2+1)=6 such kets.

    I actually have figured it out - but boy, I think Shankar has weird notation... :yuck:

    All one needs do is to start with the maximal possible z component, and apply lowering operators for each given j value, along with applying orthogonality and normalization conditions. A bit tricky but doable...
     
  5. Mar 28, 2010 #4

    vela

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    Start with the [itex]| 3/2\mbox{ }3/2 \rangle[/itex] state. You know that it has to correspond to [itex]|1/2\mbox{ }1/2; 1\mbox{ }1\rangle[/itex]. Then apply the lowering operator to get [itex]| 3/2\mbox{ }1/2 \rangle[/itex], and so on.


    Edit: Oh sure, figure it out right before I post! ;)
     
  6. Mar 28, 2010 #5
    Thanks anyway! It's good to know that my solution wasn't just pure nonsense :-)
     
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