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Consider a spin-1 particle that is a bound state of two particles of spin-1/2. How do you write the state $\left|\alpha\right\rangle$ if

  • the two particles are distinguishable?
  • the two particles are identical?

Since the two particles can combine in $S=0,1$ states, orbital angular momentum can take the values $$S=0 \; \longrightarrow \; \ell=1,$$ $$S=1 \; \longrightarrow \; \ell=0,1,2.$$ If the two particles are identical, the state ket must be antisymmetric, so it's only possible to have $$S=1, \;\;\; \ell=1.$$ But I can't understand how to write $\left|\alpha\right\rangle$.

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  • $\begingroup$ My previous answer was wrong. I don't know what I was thinking. Also, (1) I think you mean $m$ instead of $l$, and (2) with $S=0$ you can only have $m=0$, not $m=1$. $\endgroup$ – Arturo don Juan Feb 10 '18 at 21:25
  • $\begingroup$ @Arturo, no I mean $L$, relative orbital angular momentum $\endgroup$ – Stig Feb 11 '18 at 22:50

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