I wrote out the bases states for 4 1/2-spin particles in $|S,m\rangle$ representation. I know I should have a 16-dimensional Hilbert space, but I only have 9: $$|2,2\rangle,\dots,|2,-2\rangle; |1,1\rangle,|1,0\rangle,|1,-1\rangle; |0,0\rangle$$ Where did the other 7 bases go?
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jng224
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Duplicate. – Cosmas Zachos Dec 22 '21 at 15:03
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@CosmasZachos, that post, with tensor powers, etc. is NOT helpful. The simplest explanations are the best. – aspiring_physicist Dec 22 '21 at 22:57
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But unless you appreciate addition of angular momentum, a simple explanation will fail you…. – Cosmas Zachos Dec 23 '21 at 03:48
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Some values of $S$ will be repeated. For instance $S=1$ occurs three times (that makes 6 more states you did not account for) and $S=0$ occurs twice (this is the last state unaccounted for).
Since $1/2\otimes 1/2=1\oplus 0$ you can decompose $(1\oplus 0)\otimes (1\oplus 0)$ to clearly see that $S=0$ occurs twice: once from $0\otimes 0$ and once from $1\otimes 1$. Likewise the three copies of $S=1$ occur in $1\otimes 1$, $1\otimes 0$ and $0\otimes 1$ respectively.
Finally be aware that, even if you have multiple copies of the same value of $S$, the states in each copy will be different and in fact can be made orthonormal.
ZeroTheHero
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Thank you. I refrained from using this, though, because in my mind I'm requiring that the "first two" are interacting with each other and the "last two" with each other. Does this argument satisfy the general, "any two interacting?" I hope that makes sense. – aspiring_physicist Dec 22 '21 at 22:59
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the coupling is transitive. If you first do “2nd + 3rd”, you will get different states but the same angular momentum contents. – ZeroTheHero Dec 22 '21 at 23:34
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nonetheless different sets of basis states. at here are three copies of the $S=1$ states you one can make an arbitrary but common transformation in each $m_s$ subspace to generate different sets. – ZeroTheHero Dec 23 '21 at 03:05