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Calculating Clebsch-Gordan coefficients

  1. Feb 9, 2017 #1
    1. The problem statement, all variables and given/known data
    The question asks me to calculate the non-zero Clebsch Gordan coefficients
    ##\langle j_1+j_2, j_s-2|j_1,m_1;j_2,m_2\rangle##
    Where ##j_s=j_1+j_2##.
    2. Relevant equations


    3. The attempt at a solution
    The ##j_s-2## part is the ##m## of a ##|j,m\rangle## and I know that m has to equal ##m_1+m_2## or that equation would just be zero. And I also know that I should apply the lowering operators ##J_-## to one of these two equations:

    ##\langle j_s, j_s-1|j_1,j_1-1;j_2,j_2\rangle = \sqrt{\frac{j_1}{j_s}}##
    Or
    ##\langle j_s, j_s-1|j_1,j_1;j_2,j_2-1\rangle = \sqrt{\frac{j_2}{j_s}}##

    I don't know which of these two I should be applying the lowering operator to. And then I don't know how to apply it because there are two ##j, m## sets in the ket, so when calculating the coefficient using ##\sqrt{j+m}\sqrt{j-m+1}## which m and j exactly is this thing supposed to change? How do you apply the operator to an inner product? And you can't apply ##J_-## to the RHS so I'm not sure what to do about that either.

    Essentially we're being taught via lots of examples, but it's hard to extract the rules from the examples without an explanation. Which is why my questions probably involve incredibly basic and fundamental concepts. Basically I have no idea what's going on, but it isn't for lack of reading or trying. I've definitely looked in lots of textbooks and on the internet, I just don't understand what they say! So any help is hugely appreciated, I'm trying to use this question to work out all the stuff I don't know. :)
     
  2. jcsd
  3. Feb 10, 2017 #2

    blue_leaf77

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    You can't, operators only act on kets.
    To start, we have the convention
    $$
    |j_1+j_2,j_1+j_2\rangle = |j_1,j_1;j_2,j_2\rangle
    $$
    In order to get ##|j_1+j_2,j_1+j_2-2\rangle##, you have to apply ##(J_-)^2## to the left side and hence also to the right side. This gives you
    $$
    (J_-)^2 |j_1+j_2,j_1+j_2\rangle = (J_-)^2 |j_1,j_1;j_2,j_2\rangle
    $$
    Then project both sides by ##\langle j_1,m_2;j_2,m_2|##
    $$
    \langle j_1,m_2;j_2,m_2|(J_-)^2 |j_1+j_2,j_1+j_2\rangle = \langle j_1,m_1;j_2,m_2|(J_-)^2 |j_1,j_1;j_2,j_2\rangle
    $$
    In the end, it looks like that you should get a piecewise answer.
     
  4. Feb 11, 2017 #3
    I don't know how to do the projecting thing. What I ended up with is:
    ##|j_s,j_s-2\rangle = \sqrt{\frac{2j_1(2j_1-1)}{j_s(4j_s-2)}} |j_1,j_1-2;j_2,j_2\rangle + \sqrt{\frac{2j_1j_2}{j_s(4j_s-2)}}|j_1,j_1-1;j_2,j_2-1\rangle + \sqrt{\frac{2j_2(2j_2-1)}{j_s(4j_s-2)}} |j_1,j_1;j_2,j_2-2\rangle + \sqrt{\frac{2j_1j_2}{j_s(4j_s-2)}}|j_1,j_1-1;j_2,j_2-1\rangle##

    ##=\sqrt{\frac{2j_1(2j_1-1)}{j_s(4j_s-2)}} |j_1,j_1-2;j_2,j_2\rangle + 2\sqrt{\frac{2j_1j_2}{j_s(4j_s-2)}}|j_1,j_1-1;j_2,j_2-1\rangle + \sqrt{\frac{2j_2(2j_2-1)}{j_s(4j_s-2)}} |j_1,j_1;j_2,j_2-2\rangle##

    So when I do the inner product, what's the rule there? On the LHS it's the inner product of something with itself so do I get the square of the LHS? As for the RHS I know you can move the constants out from the middle so you get inner products there too, and in this basis all these vectors are orthogonal I think - except if they were the RHS would just be zero.
     
  5. Feb 11, 2017 #4

    blue_leaf77

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    No, you are not projecting something on itself. You should project ##|j_s,j_s-2\rangle## onto ##\langle j_1,m_1;j_2m_2|##.
    For example for the first term in RHS you get
    $$
    \langle j_1,m_1;j_2m_2|j_1,j_1-2;j_2,j_2\rangle
    $$
    What's the condition on ##m_1## and ##m_2## so that the above inner product does not vanish?
     
  6. Feb 11, 2017 #5
    Oh, the inner product vanishes except for ##m=m_1+m_2##. Thanks! :)
     
  7. Feb 11, 2017 #6
    Ah, except which one is m? If ##m = j_s -2## then none of them vanish.
     
  8. Feb 11, 2017 #7

    blue_leaf77

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    I don't understand what you mean.
     
  9. Feb 11, 2017 #8
    For an inner product to exist it has to have ##m_1+m_2=m##. I can identify ##m_1## and ##m_2## in the kets, but is ##m=j_s-2##? From the ket on the RHS? Or actually in this case ##m_1,m_2## are in the bra, but I still don't know which value is m. Do you get it from the RHS or from the ket which is part of the inner product?
     
  10. Feb 11, 2017 #9

    vela

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    It seems a bit strange to claim that the inner product doesn't vanish only if ##m_1+m_2=m## when you have no idea what ##m## stands for. Anyway, you should rethink what the condition is for an inner product not to be 0.
     
  11. Feb 11, 2017 #10
    Well in my lecture notes there is ##\langle j,m|j_1,m_1;j_2,m_2\rangle = 0## unless ##m = m_1 + m_2##. It's just that I'm not entirely sure where to find the m in this scenario. I think it's #m = j_s-2##.
     
  12. Feb 11, 2017 #11

    vela

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    Yes, that's right. You have two bases you're dealing with here. One basis consists of eigenstates of total angular momentum ##\lvert j, m \rangle##; the other basis consists of states which are a product of the eigenstates of the individual angular momenta, i.e., ##\lvert j_1, m_1; j_2, m_2 \rangle##. Your notes say the state ##\lvert j, m \rangle## has no overlap with ##\lvert j_1, m_1; j_2, m_2 \rangle## if ##m \ne m_1+m_2##.

    Look back at post #4. Is that the kind of inner product you're calculating for terms on the RHS? What kind of state is on the left and right of each product?
     
  13. Feb 11, 2017 #12

    vela

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    You might notice that in your expression for ##\lvert j_s, j_s-2\rangle## in terms of the other basis, the requirement that ##m = m_1 + m_2## is met for each of the terms on the RHS.
     
  14. Feb 11, 2017 #13
    Ok, that isn't the right kind of inner product. On the way to my expression for the inner products I did find ##\langle j_s, j_s-1|j_1,j_1-1;j_2,j_2\rangle = \sqrt{\frac{j_1}{j_s}}## and ##\langle j_s, j_s-1|j_1,j_1;j_2,j_2-1\rangle = \sqrt{\frac{j_2}{j_s}}## and I do have the inner product
    ##\langle j_1, m_1; j_2, m_2|j_1, j_1-1; j_2,j_2-1\rangle## which looks a bit like that result combined, so could I apply that somehow? If these are bases then are the other two terms orthonormal, and so the inner product would be zero?
     
  15. Feb 11, 2017 #14

    blue_leaf77

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    In the RHS, you have this form of inner product ##\langle j_1,m_1;j_2,m_2|j_1,m_1';j_2,m_2'\rangle##. In order for this not to be zero, how must the ##m##'s be connected between those in bra and those in ket?
     
  16. Feb 11, 2017 #15
    Is it non-zero only for ##m_1 = m_1'## and ##m_2 = m_2'##?
     
  17. Feb 11, 2017 #16

    blue_leaf77

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  18. Feb 11, 2017 #17
    Ok, so then for the first inner product for example I'd need ##m_1 = j_1-2## and ##m_2 = j_2## for it to be non-zero, and I can write down similar things for the other inner products... should I know what ##m_1, m_2## are from somewhere? Because couldn't I just choose ##m_1 = j_1-2## and and ##m_2 = j_2##? That would make the other inner products zero, but that must be wrong because then if I chose to fulfil the conditions from another inner product I'd get a different answer.

    Does that even make sense? Sorry if it's a silly question!
     
  19. Feb 11, 2017 #18

    blue_leaf77

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    Yes.
    Indeed that should be the case, like I said in my first post you will have a piecewise solution. You can't express all three coefficients with only one common expression. There will be three conditions each with different answer for the inner product.
    You can check if your answer is correct from Clebsch-Gordan coefficients table.
     
  20. Feb 11, 2017 #19
    Taking the first inner product as an example, then. Set ##m_1 = j_1-2## and ##m_2 = j_2##. Then on the LHS I have

    ##\langle j_1, j_1-2; j_2, j_2|j_s, j_s-2\rangle##

    Now that works in the sense that ##j_1 + j_2 =j##, and ##m = m_1+m_2##, so is that equal to 1?
     
  21. Feb 11, 2017 #20
    No it wouldn't be, the inner product on the RHS is the one that's equal to 1.
     
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