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The question asks to find the Clebsch-Gordan Coefficients of $$\Big\vert \frac{3}{2} \frac{1}{2} \Big\rangle = C_{0, \frac{1}{2}} \Big\vert 1 0 \Big\rangle \Big\vert \frac{1}{2} \frac{1}{2} \Big\rangle + C_{1, -\frac{1}{2}} \Big\vert 1 0\Big\rangle \Big\vert \frac{1}{2}, -\frac{1}{2} \Big\rangle\qquad(1)$$ using ladder operators. I'm a little confused with how to do this. I know applying $$J_-\Big\vert \frac{3}{2} \frac{2}{2} \Big\rangle = (L_-+S_-) \Big\vert 1 1 \Big\rangle\Big\vert \frac{1}{2} \frac{1}{2} \Big\rangle\qquad (2)$$ gives me the right solutions of $C_{0, \frac{1}{2}} = \sqrt{\frac{2}{3}}$ and $C_{1, -\frac{1}{2}} = \sqrt{\frac{1}{3}}$ but I'm not sure how to get to $(2)$ from $(1)$ in the first place. What do I sub in? Any help is appreciated!

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You need to remember that, in your notation, $J_-$ acts on $\vert \textstyle\frac{3}{2}\frac{3}{2}\rangle$ on the left on your equality while on the right $L_-$ acts on the first state alone and $S_-$ acts on the second state alone: $$ \left(L_- +S_-\right)\vert 11\rangle\vert \textstyle\frac{1}{2}\frac{1}{2}\rangle =\left[L_-\vert 11\rangle \right]\vert \frac{1}{2}\frac{1}{2}\rangle + \vert 11\rangle\left[S_-\vert\frac{1}{2}\frac{1}{2}\rangle\right]\, . $$

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