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Clebsch-Gordan coefficient

  1. Sep 11, 2012 #1
    Hi, for the deutron triplet state, I am trying to find the coefficient using the lowering operator but I kept getting 0 hope someone can help me.

    Using
    J-|j,m>=[(j+m)(j-m+1)]1/2|j,m-1>
    I can get from |++> to 1/[itex]\sqrt{2}[/itex](|-+> + |+->) but from here I am stuck since j+m=0 for |+->.

    What is wrong with my working? Thanks
     
  2. jcsd
  3. Sep 11, 2012 #2
    Perhaps you are not remembering that J- has two pieces?

    [tex]J_- = J_{1-} + J_{2-}[/tex]
     
  4. Sep 12, 2012 #3
    [tex]J_- |+-> + |-+> = J_{1-} + J_{2-}|+-> + |-+>[/tex] I am not sure about this part but i split the operators into
    [tex]J_- = J_{1-}|+-> + |-+> + J_{2-}|+-> + |-+>[/tex]
    so [tex]J_{1-}|+-> + |-+> [/tex] is where I am stuck since [tex]m_1=1/2 , m_2=-1/2 [/tex] so everything will be reduced to 0.
     
  5. Sep 12, 2012 #4
    Don't forget the brackets!

    [tex]\left( J_{1-} + J_{2-} \right) \left(\lvert +- \rangle + \lvert -+ \rangle \right)[/tex]

    There are 4 terms here, 2 of which survive the lowering operators.
     
  6. Sep 12, 2012 #5
    Yeah I am aware there are 4 terms but is this correct?
    [tex]J_- = J_{1-}|+-> + |-+> + J_{2-}|+-> + |-+>[/tex]
    If its correct then I should be able to evaluate this [tex]J_{1-}|+-> + |-+> [/tex]
    I am aware that when the lowering operator acts on the lowest eigenvalue it will disappear but I cant continue since the formula has (j+m) term which is 0 for different spins. By the way I am doing revision for this so it would be better if anyone can just tell me what I did wrong here. This is not homework. Thanks
     
  7. Sep 12, 2012 #6
    You're not using the distributive law correctly: (A+B)(x+y) = Ax + Ay + Bx + By. Every state in the sum should have a lowering operator acting on it.

    Two of the terms vanish but two survive because you're acting on a + with the lowering operator.
     
  8. Sep 12, 2012 #7
    I really still don't see what I did wrong there. I simply use (A+B)(x+y)=A(x+y)+B(x+y). I am having problem here because I am using the lowering operator on + but I am getting 0 which is clearly wrong but I just don't know why I got it wrong. I will show my working

    Using the lowering operator [tex]J_-|jm>=((j+m)(j-m+1))^{1/2}|jm-1>[/tex]
    on |-+> I got this [tex]J_-|-1/2, 1/2>=((-1/2+1/2)(-1/2-1/2+1))^{1/2}|-1/2,-1/2>[/tex]
    [tex](-1/2+1/2)(-1/2-1/2+1)=(0)(0)=0[/tex] and for |+-> the operator is also acting on the second entry which is - so it goes to 0 as well.
    Clearly I am making a stupid mistake but I really don't see it myself. Hopefully someone can tell me whats wrong
     
  9. Sep 12, 2012 #8
    eg.

    [tex]S_{1-} \lvert +- \rangle = \lvert -- \rangle[/tex]
     
  10. Sep 12, 2012 #9
    Sorry, we cross-posted...

    [tex]\sqrt{(j+m)(j-m+1)} = \sqrt{(1/2 + 1/2)(1/2 - 1/2 + 1)} = 1[/tex]
     
  11. Sep 12, 2012 #10
    Yes but I believe this is for the case of |++> since j=m=1/2?
     
  12. Sep 12, 2012 #11
    It's also the case for |+-> because J1- only acts on the +. It leaves the - alone.

    Maybe that's the point you're missing? J1- lowers the first index and leaves the second alone; J2- acts on the second index and leaves the first alone.
     
  13. Sep 12, 2012 #12
    Yes I am aware of that but in the working I wrote a few post ago, the coefficient is 0 since we have (j+m) which is 0.
     
  14. Sep 12, 2012 #13
    but j=1/2 and m=1/2.
     
  15. Sep 12, 2012 #14
    Think this is the part that I got confuse. For |+-> , does it mean that j=1/2 , m=-1/2?
     
  16. Sep 12, 2012 #15
    Aha! I think we've nailed the problem. What you have here is the tensor product of two spin-1/2 states. So j1 = j2= 1/2. The + and - refer to m1 and m2 respectively.

    For instance |+-> would have j1 = j2= 1/2, m1 = 1/2 and m2 = -1/2.
     
  17. Sep 12, 2012 #16
    Haha I get it now. Thank you so much for explaining this to me!
     
  18. Sep 12, 2012 #17
    No problem :cool:
     
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