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Addition of three angular momentum

  1. Aug 18, 2004 #1
    How can I couple three angular momentum?
    I am confused about this, in fact I don't understand how can I do this...
    I need to build eigenstates of total angular momentum for three particles J=j_1+j_2+j_3, someone can help me?

  2. jcsd
  3. Aug 18, 2004 #2
    If you can add two angular momentum, I don't see any problem. If you are able to deal with [tex]j_1[/tex] and [tex]j_2[/tex], say [tex]J_{dummy} = j_1 \otimes j_2[/tex] then the remaining is [tex] J= J_{dummy} \otimes j_3[/tex] :biggrin:

    Of course, it is easy to speak of the general [tex] j_1 \otimes j_2[/tex] without any specific value for them, whereas giving the general properties of [tex] j_1 \otimes j_2 \otimes j_3 \otimes j_4 \otimes j_5 \otimes j_6 \cdots[/tex] would not only be difficult, it would be useless.
  4. Aug 18, 2004 #3
    Ok, but is it the same way to couple j_12=j_1+j_2 and then J=j_12+j_3 or first j_23=j_2+j_3 and then J=j_1+j_23 ?
  5. Aug 18, 2004 #4
    Last edited by a moderator: Apr 21, 2017
  6. Aug 18, 2004 #5


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    Yes but, you will get different representations of states of total J. humanino is right, if you can couple 2, then you can couple 3 or ....... But, the algebra becomes horrific, and .... We are talking Clebsch-Gordan coefficients applied to Clebsch-Gordan coefficients, which really gets ugly for 6 or 12 individual particles. However, there's a lot of very elegant work , much due to G. Racah, that makes coupling of angular momenta much less formidable. (For coupling three angular momenta, one works with a 3-j symbol, a specially normalized and symmetrized set/product of CG coefficients.) For me the bible is Edmonds' Angular Momentum in Quantum Mechanics, but it is old.
    I'm sure a Google will produce lots on the subject.
    Reilly Atkinson
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