# How does the exchange symmetry of the Clebsch-Gordan coefficients arise from the phase convention?

In the book by Zettili, he explains how the sign of the Clebsch-Gordan coefficients is determined by the phase convention that $$\langle j_1,j_2; j_1, (j-j_1)\vert j, j \rangle$$ is real and positive. He then immediately says that this implies the exchange symmetry relation

$$\langle j_1,j_2;m_1,m_2\vert j,m\rangle=(-1)^{j-j_1-j_2}\langle j_2, j_1; m_2, m_1 \vert j, m\rangle.$$

Is it obvious how this follows directly from the convention? I don't understand how one can so easily find this general result even though the convention only considers one specific state.

Once you have fixed the phase of the highest state $$\vert JJ\rangle$$, the phase of all other $$\vert JM\rangle$$ follows from this as the action of the lowering operator multiplies by $$\sqrt{(J+M)(J-M+1)}$$, which is a positive factor.
Consider for instance the $$J=2,M=2$$ state constructed in the decomposition of $$(j_1=2)\otimes (j_2=1)$$. Its expression is $$\vert 2 2\rangle_1 = C_{22;10}^{22} \vert 22\rangle \vert 10\rangle + C_{21;11}^{22} \vert 21\rangle \vert 11\rangle\, ,\\ = \sqrt{\frac{2}{3}}\vert 22\rangle \vert 10\rangle - \frac{1}{\sqrt{3}} \vert 21\rangle \vert 11\rangle\, . \tag{1}$$ The action of $$J_-=J_-^{(1)}+J_-^{(2)}$$ yields (after normalization) $$\vert 21\rangle_1= \frac{1}{\sqrt{6}}\vert 21\rangle \vert 10\rangle + \frac{1}{\sqrt{3}}\vert 22\rangle \vert 1,-1\rangle - \frac{1}{\sqrt{2}} \vert 20\rangle \vert 11\rangle\, . \tag{2}$$ Clearly here the sign of the last term is related to the sign $$\vert 21\rangle \vert 11\rangle$$ in (1), as the action of $$J_-$$ on $$\vert 21\rangle \vert 11\rangle$$ will give a piece proportional to $$\vert 20\rangle \vert 11\rangle$$ (there is also a piece proportional with $$\vert 21\rangle \vert 10\rangle$$ that will combine with something else from $$J_-\vert 22\rangle\vert 10\rangle$$, but it doesn't enter in the argument.)
Now, reverse the order of $$j_1,j_2$$ so that $$\vert 2 2\rangle_2 = C_{11;21}^{22} \vert 11\rangle \vert 21\rangle+ C_{11;21}^{22} \vert 11\rangle\vert 21\rangle\, ,\\ = \frac{1}{\sqrt{3}} \vert 11\rangle\vert 21\rangle - \sqrt{\frac{2}{3}}\vert 10\rangle\vert 22\rangle \, .\tag{3}$$ The action of $$J_-$$ on $$\vert 11\rangle\vert 21\rangle$$ will now produce a term proportional to $$\vert 11\rangle\vert 20\rangle$$ with positive cofficient rather than negative (as in Eq.(2)) because the sign of term in $$\vert 11\rangle\vert 21\rangle$$ in (3) is now positive, whereas in (1) it is negative: $$\vert 21\rangle_2= \frac{1}{\sqrt{2}}\vert 11\rangle\vert 20\rangle -\frac{1}{\sqrt{6}}\vert 10\rangle\vert 21\rangle -\frac{1}{\sqrt{3}}\vert 1,-1\rangle\vert 22\rangle$$
Hence you see how the initial relative phases of $$\vert 22\rangle_1$$ in (1) and $$\vert 22\rangle_2$$ in (3) actually trickle down to all the CGs. This phase difference is forced by the Condon-Shortly phase convention, which distinguishes between the orderings since it is such that $$C^{JJ}_{j_1j_1;j_2,J-j_2}\ge 0$$.