Addition of three angular momentum

[stefano]

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?

Thank's

 

[humanino]

If you can add two angular momentum, I don't see any problem. If you are able to deal with $$j_1$$ and $$j_2$$, say $$J_{dummy} = j_1 \otimes j_2$$ then the remaining is $$J= J_{dummy} \otimes j_3$$

Of course, it is easy to speak of the general $$j_1 \otimes j_2$$ without any specific value for them, whereas giving the general properties of $$j_1 \otimes j_2 \otimes j_3 \otimes j_4 \otimes j_5 \otimes j_6 \cdots$$ would not only be difficult, it would be useless.

 

[stefano]

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 ?

 

[humanino]

yes ! The tensor product is associative.
http://www.wordiq.com/definition/Tensor_algebra
Hey, welcome in PF by the way !

 

[reilly]

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 ?

....
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.
Regards,
Reilly Atkinson