I have a rather trivial question. I am looking for the decomposition of $1/2\otimes 1/2\otimes 1/2$. It should give, $0,1/2$ and $3/2$. I thought one must get as the overall dimension of this space 8, but counting, I just get 7. Does one have 2 singlets?
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1$\begingroup$ Possible duplicate: Adding 3 electron spins. $\endgroup$– Emilio PisantyCommented Aug 31, 2012 at 13:50
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$\begingroup$ The question is similar, true but I am more interested in the formal mathematical expression of this problem. $\endgroup$– HamurabiCommented Aug 31, 2012 at 15:19
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2 Answers
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From where did you get the idea that one can get a spin zero representation? The product of an even/odd number of Fermion representations always gives a Boson/Fermion representation.
In your particular case, repeated use of $$1/2 \otimes s = (s-1/2) \oplus (s+1/2)$$ gives $$1/2 \otimes 1/2 \otimes 1/2=(0\oplus 1) \otimes 1/2 = (0 \otimes 1/2) \oplus (1 \otimes 1/2) =1/2 \oplus (1/2 \oplus 3/2).$$ Thus one gets two spin 1/2 and one spin 3/2 representations.
answered Aug 31, 2012 at 16:54
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With the caveat that I'm 30+ years rusty at this stuff: Two 1/2's and one 3/2 (and no singlet) does have dimension 8. And Mathematica's Clebsch-Gordon routines affirm this decomposition.
