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C-G coefficients by matrix method

  1. Mar 7, 2014 #1
    How to find the Glebsch Gordan coefficients by matrix form. Let's suppose we have a two spin 1/2 particles, we get that (check attachment 1). These are the corresponding coefficients the classic way. Now if I want find them the matrix way, where the matrix should be diagonalized. How do I proceed? So my question is how knowing the old basis and the new basis -how can we find the matrix containing the Glebsch Gordan coeficients?

    I would like to add the the matrix main operator should be either Jz or J^2, as far as I know. Please, correct me if I am wrong.
    Thanks!
     

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    Last edited: Mar 7, 2014
  2. jcsd
  3. Mar 7, 2014 #2

    DrDu

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    I don't see an attachment.
     
  4. Mar 7, 2014 #3
    Sorry, I forgot to attach it. There, I will right away.
     
  5. Mar 7, 2014 #4

    DrDu

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    So, either I don't understand your problem or it is trivial:
    From your formulas you can directly read off the matrix elements, e.g.,
    ##\left\langle 1/2,1/2;- 1/2,1/2 | 0, 0\right\rangle=-1/\sqrt{2}##
     
  6. Mar 11, 2014 #5
    Yes that's true. But that is another procedure. What I meant was, I want to find those coefficients (not in the trivial way that you just used) but in another procedure. What am asking for is a procedure that helps me find C-G coefficients using a matrix in which we have to diagonalize at the end of the day to get the coefficients as elements in this very matrix. (I don't know how to write the matrix down using this website's program but I can attach a paper of what I think is correct if you want)
     
  7. Mar 11, 2014 #6
    Oh and how did you write those matrix elements, I am having difficulties using the notation. Can you explain what you wrote in your last reply?
     
  8. Mar 12, 2014 #7

    DrDu

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    Maybe you want the following:
    You can set up the matrix with elements
    [tex]
    \langle j_1, m'_1, j_2,m'_2|J^2|j_1, m_1, j_2,m_2\rangle
    [/tex]
    and analogously the matrix for ##J_z##.
    You already know the eigenvalues of the first matrix, namely J(J+1) with ##J=|j_1-j_2|\ldots j_1+j_2## and hence you can solve for a system of eigenvectors corresponding to these eigenvalues. The matrix for J_z will be block diagonal with these basis states and you can diagonalize it similarly, knowing that M_z runs from -J to +J.
    The matrix of the eigenvectors are the CG coefficients you are after.
     
  9. Mar 12, 2014 #8
    Yes! That is exactly what am asking for. Thank you for understanding.
    Okay so what you are saying is that I should put the elements of the matrix as you just mentioned. But if I returned to the first attachment there are |0, 0> |1, 1> |1, -1> and |1, 0> states (In which I should related them to the expanded form, for example: in terms of |1/2, 1/2>|1/2, 1/2>. So, does solving the matrix require me to put all of this into the form b=Ax
    A: Matrix
    x:basis the one with j1, j2, m1, m2
    b: basis the one with J, m

    Just so I could relate and know which coefficient is for which.


    GREAT THANKS DrDu!
     
  10. Mar 12, 2014 #9

    DrDu

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    Yes, in principle, the matrix A is the matrix which diagonalizes both J^2 and J_z. Of course there may be some pitfalls. E.g. as you are transforming a basis, I guess the relation is rather ##b^T=x^TA##.
     
  11. Mar 12, 2014 #10
    I am having a hard time developing this. Is there any paper or an link that can illustrate this. I would really appreciate it, I am trying to solve by hand, let's say the first row of the 4x4 matrix is:
    ( <1/2, 1/2|<1/2, 1/2|J^2|1/2,1/2>|1/2,1/2> <1/2,1/2|<1/2, 1/2|J^2|1/2, 1/2>|1/2,-1/2> <1/2, 1/2|<1/2 , 1/2|J^2|1/2,-1/2>|1/2,1/2> <1/2, 1/2|<1/2, 1/2|J^2|1/2, -1/2>|1/2,-1/2>)
    When I want to solve how would I know which is my J in order to get J(J+1)? (Please notice that the elements are in terms of j_1, j_2 not in terms of J.

    And afterwards, what should I do? I only solve the elements and then diagonalize?If so where did the hbar^2 disappear? I really have to know this and I have been trying nonstop but I can't seem to relate.

    Thanks in advance!
     
  12. Mar 12, 2014 #11

    DrDu

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    You should really try to figure this out without an article. I set hbar=1. So in your case you only have two values of J: 0 and 1 and you know them already, although, if you really want so, you could find them from the determinant of the eigenvalue problem [itex] \hat{J^2} -J(J+1)\hat{1}[/itex]. As you know them, you have to solve [itex] (\hat{J^2} -J(J+1)\hat{1})x=0[/itex]. Here, [itex]\hat{J^2}[/itex] is the matrix you just wrote down the first row off. You could do this using Gaussian elimination. As the problem is underdetermined, the vector x will not be determined completely and you could find several independent vectors which span the sub-space (of dimension 2J+1). You can then use these vectors to set up an eigenequation for J_z.
    In practice it may be recomendable to solve first the equations for J_z where you also know the eigenvalues: -1, 0, 0, 1. The eigenvalues -1 and 1 belong to J=1 and you only have to diagonalize the matrix for J^2 in the basis of the two eigenvectors with M_z=0.
     
  13. Mar 12, 2014 #12
    Thank you, it is just the terminology that is not helping me. Ok. If you may tell me if am thinking correctly:

    1)I fill the matrix with three other columns analogous to the first row I wrote in my previous reply.
    2)Then I get a big 4x4 matrix
    3)After that I solve for the eigenvalues to get. Here I bump into the following problem (element by element)

    <1/2, 1/2|<1/2, 1/2|J^2|1/2,1/2>|1/2,1/2> = 2 if I consider J=1 (Is this correct?)
    <1/2,1/2|<1/2, 1/2|J^2|1/2, 1/2>|1/2,-1/2> = 0 (Orthogonal)
    <1/2, 1/2|<1/2 , 1/2|J^2|1/2,-1/2>|1/2,1/2>=0 (Orhtogonal) and so on.

    What is this matrix that am ending up with? A very trivial one?! Am I working this correctly?

    4) with the final simplified 4x4 matrix, I diagonalize it by finding eigenvectors and placing them side to side in a decent matrix.
     
  14. Mar 12, 2014 #13
    I am guessing the elements should be <j_1 j_2|<m_1 m_2|J^2|J, M>. What do you say?
     
  15. Mar 12, 2014 #14

    DrDu

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    I don't understand what you mean with " if I consider J=1", note that J^2 is an operator not a number.
    Nevertheless the result is correct and the matrix will be quite trivial, thinking about it, it only has two non-vanishing non-diagonal entries.
     
  16. Mar 12, 2014 #15

    DrDu

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    Btw, as J_1=J_2=1/2 you could drop the corresponding labels from your matrix elements, also, instead of m_z=1/2 or -1/2 you could simply write + or -, i.e. [itex]|1/2, 1/2; 1/2, -1/2\rangle=|+-\rangle[/itex]. Would stop us both from going mad ...
     
  17. Mar 12, 2014 #16
    Two non vanishing, please check the attachment. It doesn't sound like it.
     

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  18. Mar 12, 2014 #17

    DrDu

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    I was referring to your post #12, not 13.
     
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