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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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  • $\begingroup$ Duplicate. $\endgroup$ Dec 22, 2021 at 15:03
  • $\begingroup$ @CosmasZachos, that post, with tensor powers, etc. is NOT helpful. The simplest explanations are the best. $\endgroup$ Dec 22, 2021 at 22:57
  • $\begingroup$ But unless you appreciate addition of angular momentum, a simple explanation will fail you…. $\endgroup$ Dec 23, 2021 at 3:48

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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.

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  • $\begingroup$ 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. $\endgroup$ Dec 22, 2021 at 22:59
  • $\begingroup$ the coupling is transitive. If you first do “2nd + 3rd”, you will get different states but the same angular momentum contents. $\endgroup$ Dec 22, 2021 at 23:34
  • $\begingroup$ so different bases but bases nonetheless? $\endgroup$ Dec 23, 2021 at 1:46
  • $\begingroup$ 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. $\endgroup$ Dec 23, 2021 at 3:05

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