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I'm quite unsatisfied with this answer, so I'm hoping to get an adequate one.

I'm trying to understand how to compute the reverse CB coefficents. I'll provide the simplest example.

If I have a $l = 1, s = 1/2$ particle and I'm trying to add the angular momentum. I can begin with the state. $$| 1,1,1/2,1/2 \rangle = |3/2, 3/2 \rangle$$ Now by applying the lowering operator ( where here I only write the subspace that the operator acts on) $$J_-|3/2, 3/2\rangle = L_- |1,1\rangle + S_- |1/2,1/2\rangle $$ and arrive at $$|3/2, 1/2\rangle = \sqrt{2/3}|1,0,1/2,1/2\rangle +\sqrt{1/3}|1,1,1/2,1/2\rangle$$ just by following the formulas for eigenvalues of lowering operators and normalizing. I don't know how to compute that $$|1/2, 1/2\rangle = \sqrt{2/3}|1,1,1/2,1/2\rangle - \sqrt{1/3}|1,0,1/2,1/2\rangle$$ though. Any help would be appreciated.

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  • $\begingroup$ Why did you fail to lower the spin of the doublet constituent? Why are you unsatisfied by the duplicate answer? How do you apply that answer here? $\endgroup$ – Cosmas Zachos Apr 26 at 21:33
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The $\vert 1/2,1/2\rangle$ state must be orthogonal to the $\vert 3/2,1/2\rangle$ state hence the relative sign and the switched coefficients: $$ \vert 1/2,1/2\rangle = -\sqrt{1/3}\vert 1,0;1/2,1/2\rangle +\sqrt{2/3} \vert 1,1;1/2,-1/2\rangle \tag{1} $$ (There’s a typo in the second component of your $\vert 3/2,1/2\rangle$ state.)

One can verify that the state of Eq.(1) is killed by $J_+$, which identifies this state has having maximum projection $m_s=1/2$.

Note that the overall sign is usually set by the Condon-Shortley phase convention, which states that the CG $\langle J,J\vert L,L; S,J-L\rangle$ is positive.

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