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Calculation of Clebsch-Gordan coefficients

  1. Feb 9, 2012 #1
    1. The problem statement, all variables and given/known data
    I have two spin-1/2 particles and I need to calculate their Clebsch-Gordan coefficients.


    2. Relevant equations



    3. The attempt at a solution

    I followed the procedure of applying [itex]J_- [/itex] to [itex]|{j,m}\rangle [/itex]
    and [itex]J_{1-}[/itex] and [itex]J_{2-}[/itex] to [itex]|{m_1,m_2}\rangle [/itex] and comparing them. I got correctly
    [itex]\langle{1,1}|{1/2,1/2}\rangle =1 [/itex],
    [itex]\langle{1,0}|{1/2,-1/2}\rangle=1/\sqrt{2} [/itex],
    [itex]\langle{1,0}|{-1/2,1/2}\rangle =1/\sqrt{2} [/itex],
    [itex]\langle{1,-1}|{-1/2,-1/2}\rangle=1 [/itex].
    Now I want to find [itex]\langle{0,0}|{1/2,-1/2}\rangle [/itex]
    and [itex]\langle{0,0}|{-1/2,1/2}\rangle [/itex].
    Therefore I denoted [itex]|{0,0}\rangle = \alpha|{1/2,-1/2}\rangle + \beta|{-1/2,1/2}\rangle [/itex]
    and used the normalization condition [itex]|\alpha|^2 + |\beta|^2 = 1[/itex] and orthogonality to the [itex]|{1,0}\rangle[/itex] state. I got the equation
    [itex]|\alpha|^2 = 1/2[/itex] from which there are 2 options:
    [itex]\alpha = 1/\sqrt{2}[/itex] and [itex]\alpha = -1/\sqrt{2}[/itex] (only real coefficients by convention). How do I know which is the right option out of the two?

    Thanks!
     
    Last edited: Feb 9, 2012
  2. jcsd
  3. Feb 9, 2012 #2

    vela

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    The overall sign of ##\alpha## doesn't really matter. What's important is the relative phase between ##\alpha## and ##\beta##. What combination of ##\alpha## and ##\beta## will yield a state orthogonal to ##|1, 0\rangle##?
     
  4. Feb 9, 2012 #3
    Ok. I agree. But the standard table has a certain convention. How do I know how to pick the sign to fit this convention?
     
  5. Feb 9, 2012 #4

    vela

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    According to this page, this is the convention:

    The Condon-Shortley convention is that the highest m-state of the larger component angular momentum is assigned a positive coefficient.

    Still seems a bit ambiguous in your case, though.
     
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