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

Question

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.

Answer 1

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 $C^{JJ}_{j_1j_1;j_2,J-j_2}\ge 0$.