Symmetry Properties for Wigner 3j-symbols

Question

I am learning about coupled angular momenta and came across the Wigner 3-j-symbols which are defined by \begin{equation} \left( { \begin{array}{ccc} j_1 & j_2 & j_3 \\ m_1 & m_2 & m_3 \\ \end{array} } \right) = \frac{(-1)^{j_1 - j_2 - m_3}}{\sqrt{2 j_3 + 1}} \langle j_1 \, m_1 \, j_2 \, m_2 | j_3 \, (-m_3) \rangle. \end{equation}

According to Wikipedia, various symmetries and selection rules hold for these symbols. For example, it is said that the exponent of the sign factor is always an integer and that the symbols have something to do with the addition of three angular momenta.

I have already taken a look into Quantum Mechanics by Tannoudji, Messiah and Schwabl, but could only find the properties and no proof for these claims. That is why I want to ask if someone can provide me with a reference for those proofs, especially why

1. The exponent of the sign is always an integer.

2. \begin{equation} \sum_{m_1=-j_1}^{j_1} \sum_{m_2=-j_2}^{j_2} \sum_{m_3=-j_3}^{j_3} |j_1 m_1\rangle |j_2 m_2\rangle |j_3 m_3\rangle \begin{pmatrix} j_1 & j_2 & j_3\\ m_1 & m_2 & m_3 \end{pmatrix} = |0\,0\rangle \end{equation}

3. \begin{equation} \begin{pmatrix} j_1 & j_2 & j_3\\ m_1 & m_2 & m_3 \end{pmatrix} = \begin{pmatrix} j_2 & j_3 & j_1\\ m_2 & m_3 & m_1 \end{pmatrix} = \begin{pmatrix} j_3 & j_1 & j_2\\ m_3 & m_1 & m_2 \end{pmatrix} \end{equation}

4. \begin{equation} \begin{pmatrix} j_1 & j_2 & j_3\\ m_1 & m_2 & m_3 \end{pmatrix} = (-1)^{j_1+j_2+j_3} \begin{pmatrix} j_2 & j_1 & j_3\\ m_2 & m_1 & m_3 \end{pmatrix} = (-1)^{j_1+j_2+j_3} \begin{pmatrix} j_1 & j_3 & j_2\\ m_1 & m_3 & m_2 \end{pmatrix} \end{equation}

5. \begin{equation} \begin{pmatrix} j_1 & j_2 & j_3\\ -m_1 & -m_2 & -m_3 \end{pmatrix} = (-1)^{j_1+j_2+j_3} \begin{pmatrix} j_1 & j_2 & j_3\\ m_1 & m_2 & m_3 \end{pmatrix} \end{equation}

Thank you in advance.

Answer 1

1. If $j_1$ and $j_2$ are half-integer, their product $j_1\otimes j_2$ can only contain integer values of $j_3$; for instance, coupling $j_1=1/2$ to $j_2=1/2$ yields $j_3=0,1$. In such a case clearly the sum is an integer. Likewise, if one of $j_1$ or $j_2$ is half-integer, then $j_3$ will also be half integer; for instance the coupling $j_1=1/2$ to $j_2=1$ yields $j_3=1/2,3/2$. Again the sum is an integer.

2. By definition the $3j$ symbol is defined as the coupling of $j_1, j_2,j_3$ to a scalar.

3., 4. and 5. actually follow from the definition the $3j$ via 2.

In the case of "vectors", where $j_1=j_2=j_3=1$, the $3j$ can be thought of as related to the invariant constructed from $\vec j_1\cdot \vec j_2\times \vec j_3$. This can be evaluated as a determinant, which is invariant under cyclic permutations of column as in (3) but picks up a sign under transposition of two rows, as in 4. Likewise, 5. can be seen as related to the parity operation that interchanges the $\hat z$ axis. I would not take these analogies very far but at least they give you a sense of why one would expect some sort of sign to occur in elementary situations.

Answer 2

Let ${\bf z}=(\xi, \eta)$, with $|z|^2= |\xi|^2+|\eta|^2$. Set $$ \phi_{jm}({\bf z})= \frac{\xi^{j+m}\eta^{j-m}}{[(j+m)!(j-m)!]^{1/2}}= \langle{{\bf z}}|j,m\rangle $$ with the usual conditions on $j,m$. These spin-$j$ basis functions are orthonormal under the Bargmann-Fock inner product $$ \langle{\psi}|{\chi}\rangle= \frac{1}{\pi^2} \int_{{\mathbb C}^2} e^{-|\xi|^2-|\eta|^2 } \overline{\psi(\xi, \eta)}\chi(\xi,\eta)\,d^2\!\xi d^2 \!\eta, $$ in which $$ \langle{\xi^n\eta^m}|{\xi^{n'} \eta^{m'}} \rangle= \delta_{nn'} \delta_{mm'} n!m!. $$ This inner product is equivalent to the Schwinger boson formalism with $$ a_+^\dagger \to \xi, \quad a_-^\dagger \to \eta, \quad a_+\to \frac{\partial}{\partial \xi},\quad a_-\to \frac{\partial}{\partial \eta}. $$

Now, for a triangular set $\{j_1,j_2,j_3\}$ define $$ \chi_{j_1j_2j_3}({\bf z}_1,{\bf z}_2,{\bf z}_3) =[(j_1+j_2+j_3+1)!]^{-1/2} \frac{[{\bf z}_2,{\bf z}_3]^{j_2+j_3-j_1}[{\bf z}_3,{\bf z}_1]^{j_3+j_1-j_2}[{\bf z}_1,{\bf z}_2]^{j_1+j_2-j_3}}{[(j_2+j_3-j_1)!(j_3+j_1-j_2)!(j_1+j_2-j_3)!]^{1/2}}\\ = \sum_{m_1,m_2,m_3}\phi_{j_1m_1}({\bf z}_1)\phi_{j_2m_2}({\bf z}_2)\phi_{j_3m_3}({\bf z}_3)\left(\matrix{j_1&j_2&j_3\cr m_1&m_2& m_3}\right). $$ Here triangular set means that the exponents on the RHS are all positive integers, the bracket $[{\bf z}_1, {\bf z}_2]$ denotes $ \xi_1\eta_2-\xi_2 \eta_1$, and the coefficients $$ \left(\matrix{j_1&j_2&j_3\cr m_1&m_2& m_3}\right) $$ are defined by the expansion of the second line. They are Wigner's $3j$-symbols. The RHS is invariant under the action of ${\rm SL}(2, {\mathbb C})$ (because the $[{\bf z}_1, {\bf z}_2]$ are) and, a fortiori, under ${\rm SU}(2)$. The function $\chi_{j_1j_2j_3}({\bf z}_1,{\bf z}_2,{\bf z}_3)$ is therefore a spin singlet.

From the generating function definition we see that under an odd permutation of the columns we have $$ \left(\matrix{j_2&j_1&j_3\cr m_2&m_1& m_3}\right) =(-1)^{j_1+j_2+j_3}\left(\matrix{j_1&j_2&j_3\cr m_1&m_2& m_3}\right). $$ Similarly, since $m\to -m$ interchanges $\xi$ and $\eta$ and hence takes $[{\bf z}_1,{\bf z_2}]$ to minus itself, we have $$ \left(\matrix{j_1&j_2&j_3\cr m_1&m_2& m_3}\right) =(-1)^{j_1+j_2+j_3}\left(\matrix{j_1&j_2&j_3\cr -m_1&-m_2& -m_3}\right) $$

The state $$ |{0,0}\rangle= \sum_{m_1,m_2,m_3} \left(\matrix{j_1&j_2&j_3\cr m_1&m_2& m_3}\right)|{j_1,m_1}\rangle \otimes |{j_2,m_2}\rangle\otimes |{j_3,m_3}\rangle $$ is now clearly invariant under $$ D(U)|{j,m}\rangle = \sum_{m'}|{j,m'}\rangle D^{j}_{m'm}(U). $$ It is also normalized, but that takes more work to show