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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?

Qmechanic
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Hamurabi
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2 Answers2

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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.

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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.