The symmetry of clebsch-gordan coefficients $\left< j_1j_2;m_1m_2 \middle| j_1j_2;JM \right>$ under exchange of $j_1,m_1$ and $j_2,m_2$ is \begin{equation} \left< j_1j_2;m_1m_2 \middle| j_1j_2;JM \right>=(-1)^{j_1+j_2-J}\left< j_2j_1;m_2m_1 \middle| j_2j_1;JM \right> \end{equation} Now consider the case in which $j_1=2$, $j_2=1$, $J=2$, $m_1=m_2=1$ and $M=2$.

Applying the above formula to this we get \begin{equation} \left< 21;11 \middle| 21;22 \right>=-\left< 12;11 \middle| 12;22 \right> \end{equation} I think this should mean $\left< 21;11 \middle| 21;22 \right>=0$ as the order of $j_1$ and $j_2$ shouldn't matter and $\left< 21;11 \middle| 21;22 \right>$ and $\left< 12;11 \middle| 12;22 \right>$ must be the same state. But it is non-zero.

The same logic works for $m_1=m_2=0$ though and $\left< 21;00 \middle| 21;20 \right>=0$, but it might just be a coincidence. It seems I have not fully understood what the symmetry relation means. Why must there be a phase factor at all?

In many places the $j_1$ and $j_2$ are omitted and a blind application of the symmetry equation will give $\left<11 \middle| 22 \right>=-\left<11 \middle| 22 \right>$, and hence we might conclude that $\left<11 \middle| 22 \right>=0$.

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    $\begingroup$ "The order of $j_1$ and $j_2$ shouldn't matter"?? You want a demonstration that $A\otimes B \neq B\otimes A$ ? Have you done composition of a spin 1/2 with a spin 1 explicitly to appreciate your imagined equality does not hold? $\endgroup$ – Cosmas Zachos May 23 '18 at 15:01
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    $\begingroup$ Hint: if J=3, you'd have a symmetric composition of the two js and their order wouldn't matter. So the composition to J=2 must be orthogonal to it, so, then, antisymmetric. $\endgroup$ – Cosmas Zachos May 23 '18 at 15:44

The CG coefficients are usually determined by first constructing the highest state in each irrep, i.e. the state for which $m=j$. This state is determined by the requirement that it must be killed by $L_+$, i.e. if $$ \vert j,j\rangle = \sum_{m_1,_2}\vert \ell_1m_1\rangle \vert \ell_2m_2 \rangle \langle \ell_1 m_1; \ell_2 m_2\vert jj\rangle $$ with $ \langle \ell_1 m_1; \ell_2 m_2\vert jj\rangle$ a Clebsch-Gordan coefficient, then $$ L_+\vert j,j\rangle=0 $$ generates are recursion relation between coefficients, and this relation can be solved in terms of a "seed" coefficient.

The recursion alone is not enough to fix the CGs. In the well-known example of $\ell=1/2$ and $\ell_2=1/2$, the states $$ \vert 00\rangle_\pm=\pm \frac{1}{\sqrt{2}}\left( \vert \textstyle\frac{1}{2}, \frac{1}{2}\rangle \vert \frac{1}{2} ,-\frac{1}{2}\rangle -\vert \frac{1}{2} ,-\frac{1}{2}\rangle \vert \frac{1}{2}, \frac{1}{2}\rangle \right)\, . \tag{1} $$ are both killed by $L_+=L_+^{(1)}+L_+^{(2)}$. Both states differ by an overall sign and are legitimate singlet states.

The usual Condon-Shortley phase convention is to choose the sign of the coefficient $$ \langle \ell_1 \ell_1; \ell_2 m_2\vert jj\rangle > 0\, ,\qquad m_2=j-\ell_1\, . $$ Obviously this choice is not symmetric under the exchange of $\ell_1$ and $\ell_2$: there is no guarantee that if $\langle \ell_1 \ell_1; \ell_2 m_2\vert jj\rangle > 0$ then $\langle \ell_1 m_1; \ell_2 \ell_2\vert jj\rangle > 0$ . If indeed $\langle \ell_1 \ell_1; \ell_2 m_2\vert jj\rangle > 0$ and $\langle \ell_1 m_1; \ell_2 \ell_2\vert jj\rangle > 0$, then all CGs will not depend on the order of $\ell_1$ and $\ell_2$.

The recursion relations for CGs however are symmetric under the interchange of $\ell_1$ and $\ell_2$, with consequence that this interchange cannot affect the magnitude of the CGs. Note that the general relation is $$ \langle \ell_2 m_2; \ell_1 m_1\vert jm_j\rangle =(-1)^{\ell_1+\ell_2-j} \langle \ell_1 m_1; \ell_2 m_2\vert jm_j\rangle $$


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