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I am calculating normalized eigenstates for a problem with three particles, when I came across this problem:

what to do when $s_1=0$, I calculated the coefficients for

$|\downarrow\uparrow \rangle \otimes |\uparrow\rangle = \sqrt{\frac{1}{2}}|1,0\rangle\otimes |\frac{1}{2},\frac{1}{2}\rangle-\sqrt{\frac{1}{2}}|0,0\rangle \otimes |\frac{1}{2},\frac{1}{2} \rangle$ Since there is no table for $s_1 = 0$, does it mean that I have to look at the $1/2 \times 1/2$ table where $|s =0,m=0\rangle$?

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    $\begingroup$ Is the wrong state you are proposing normalized? $\endgroup$ – Cosmas Zachos Oct 12 '16 at 19:15
  • $\begingroup$ If you mean the last equation, I was trying to ask If I had calculated the combined state correct, I have not begun normalizing it yet, since I'm not sure if I calculated $|\downarrow \uparrow \uparrow\rangle$ the right way $\endgroup$ – hbar-gal Oct 12 '16 at 19:21
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    $\begingroup$ No you did not. What is the Kronecker product of a singlet with a doublet? $\endgroup$ – Cosmas Zachos Oct 12 '16 at 19:25
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    $\begingroup$ A spinless object and a doublet are just the doublet itself. Perhaps you'd review the C-G series before practicing examples. $\endgroup$ – Cosmas Zachos Oct 12 '16 at 19:32
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    $\begingroup$ $|0,0\rangle\otimes|s,m\rangle=|s,m\rangle$, right? $\endgroup$ – Cosmas Zachos Oct 12 '16 at 19:52
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Combining three doublets is well-answered in 29443, where you see that their 2x2x2=8 space is linearly rearranged into one quartet and two independent doublets, here dubbed a and b, with different symmetries w.r.t. the constituent doublets.

You may go ahead and use 3-j symbols, or different orders, etc... in your Clebsch table, but, at the end of the day, all you are doing is rewriting your 8 normalized states into a different orthonormal basis of states through a very sparse orthogonal matrix $O$,
$$ \begin{bmatrix} |\uparrow \uparrow\uparrow \rangle \\ |\downarrow \uparrow\uparrow \rangle \\ |\uparrow \downarrow\uparrow \rangle \\ |\uparrow \uparrow\downarrow \rangle \\ |\downarrow \downarrow\uparrow \rangle \\ |\downarrow \uparrow\downarrow \rangle \\ |\uparrow \downarrow\downarrow \rangle \\ |\downarrow \downarrow\downarrow \rangle \\ \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 & 0 & 0 &0 &0 &0 \\ 0 & 1/\sqrt{3} & -\sqrt{2/3} & 0 & 0 &0 &0 &0 \\ 0 & 1/\sqrt{3} & 1/\sqrt{6} & -1/\sqrt{2} & 0 &0 &0 &0 \\ 0 & 1/\sqrt{3} & 1/\sqrt{6} & 1/\sqrt{2} & 0 &0 &0 &0 \\ 0 & 0 & 0 & 0 & 1/\sqrt{2} &1/\sqrt{6} &1/\sqrt{3} &0 \\ 0 & 0 & 0 & 0 & -1/\sqrt{2} &1/\sqrt{6} &1/\sqrt{3} &0 \\ 0 & 0 & 0 & 0 & 0 &-\sqrt{2/3} &1/\sqrt{3} &0 \\ 0 & 0 & 0 & 0 & 0 &0 &0 &1 \\ \end{bmatrix} \begin{bmatrix} |3/2,3/2\rangle \\ |3/2,1/2\rangle \\ |1/2,1/2\rangle ^a \\ |1/2,1/2\rangle ^b \\ |1/2,-1/2\rangle ^b \\ |1/2,-1/2\rangle ^a \\ |3/2,-1/2\rangle \\ |3/2,-3/2\rangle \\ \end{bmatrix}. $$

That is, $O^T O=1 \!\! 1 $. The matrix is sparse just as your Clebsch table is, by selection rules. The essential part is the evident 3x3 orthogonal sub matrix and its mirror reflection.

So you may use this matrix to check your answers and normalizations, at the end of the day. For example, in these conventions, where a has been defined by an unconventional overall - sign, the addition of the trailing $\uparrow \uparrow$ first, as suggested, yields, $|\downarrow \uparrow\uparrow\rangle=|3/2,1/2\rangle/\sqrt{3}-\sqrt{2/3}|1/2,1/2\rangle^a$, as you might check.

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