# Reverse Clesbch Gordan coefficients

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.

• 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? – Cosmas Zachos Apr 26 '19 at 21:33

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.