# Derivation of symmetry properties of Clebsch-Gordan coefficients

According to wikipedia and some books, Clebsch-Gordan coefficients follow certain symmetry relations listed below (from Wikipedia)

I am interested in proving (or at least know the method for deriving) the first two expressions. Wikipedia mentions Wigner 3j symbols but I would like to find a derivation without using them. I have found a derivation for the first one in a book, but the starting point of it is

"Under change of parity $$\textbf{r}\to -\textbf{r}$$, a wave function behaves as $$\Psi_{jm} \: \to \: (-1)^{j-m} \Psi_{j,-m}"$$

Which I don't understand, since only change in wave function under parity I have ever seen before is by a factor of $$(-1)^\ell$$ using spherical harmonics.

I would like to know why that change under parity holds and the way to proceed in order to derive the second symmetry property, which I guess is related to Pauli's exclusion principle.

The relative sign of CGs is fixed by the recursion relation but the overall sign of all CGs is determined by the Condon-Shortley convention. To see this clearly, consider the singlet state constructed from two spin-1/2 states: \begin{align} \vert 00\rangle=\textstyle\frac{1}{\sqrt{2}}\vert \textstyle\frac{1}{2}\frac{1}{2}\rangle\vert \textstyle\frac{1}{2},-\frac{1}{2}\rangle - \frac{1}{\sqrt{2}} \vert \textstyle\frac{1}{2},-\frac{1}{2}\rangle \vert \textstyle\frac{1}{2}\frac{1}{2}\rangle \tag{1} \end{align} is just equally valid as a singlet state as \begin{align} \vert 00\rangle’=-\textstyle\frac{1}{\sqrt{2}}\vert \textstyle\frac{1}{2}\frac{1}{2}\rangle\vert \textstyle\frac{1}{2},-\frac{1}{2}\rangle + \frac{1}{\sqrt{2}} \vert \textstyle\frac{1}{2},-\frac{1}{2}\rangle \vert \textstyle\frac{1}{2}\frac{1}{2}\rangle\, , \end{align} since the state $$\vert 00\rangle’$$ is killed by $$L_+$$, and is orthogonal to $$\vert 10\rangle$$.
More generally, the choice $$-\vert JJ\rangle$$ is as valid as $$\vert JJ\rangle$$, as far as being killed by $$L_+$$ and orthogonality with other states.
Hence, to determine the symmetry under exchange $$(j_1m_1)\leftrightarrow (j_2m_2)$$, it is sufficient to construct the compare the sign of the CGs $$C^{JJ}_{j_1j_1;j_2,J-j_1}$$ and $$C^{JJ}_{j_1,J-j_2;j_2j_2}$$. It’s not hard to see that there are $$j_1+j_2-J$$ steps between these values when recursively solving $$L_+\vert JJ\rangle=0$$, and the Clebsch’s alternate in sign, so $$(-1)^{j_1+j_2-J}$$ comes out of that.
To get those CGs with $$J$$ interchanged with $$j_1$$ or $$j_2$$, we need to consider the integral relations between group functions: \begin{align} \int d\beta \, \sin\beta \, d^{J}_{JM’}(\beta) d^{j_1}_{j_1m_1’}(\beta) d^{j_2}_{J-j_1;m_2’}(\beta)&=C^{JJ}_{j_1j_1;j_2,J-j_1}C^{JM’}_{j_1m_1’;j_2m_2’}\frac{2}{2J+1} \tag{2}\\ &=C^{j_1j_1}_{JJ;j_2,J-j_1}C^{j_1m_1}_{JM’;j_2m_2’}\frac{2}{2j_1+1} \tag{3} \end{align} Since the $$d^j$$ functions are real, one can readily interchange the role of $$d^{J}_{JM’}(\beta)$$ and $$d^{j_1}_{j_1m_1’}(\beta)$$ in (2) without changing the left hand side of (2), but changing the CGs and the factor $$2/(2J+1)$$ on the right to become (3). That’s the origin of the $$\sqrt{2J+1/2j_1+1}$$ factor in the interchange of argument of the CGs. Since the CGs of the type $$C^{JJ}_{j_1j_1;j_2m_2}>0$$ by the Condon-Shortley phase convention, the overall sign connecting the CGs with $$J$$ and $$j_1$$ exchanged comes out of evaluating directly the integral on the LHS of (2). The integral is somewhat simplified because \begin{align} d^{J}_{JM}(\beta)= \sqrt{\frac{(2J)!}{(J+M)!(J-M)!}}\cos(\beta/2)^{J+M} \left(-\sin(\beta/2)\right)^{J-M} \end{align} although the general form of $$d^{j_2}_{m_2m_2’}(\beta)$$ is quite messy. Intuitively, it’s clear that the phase can only depend on $$j_2$$ and $$m_2$$, i.e. on the values of angular momenta that are not interchanged.