Symmetry Properties for Wigner 3j-symbols

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

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

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

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

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

• You might check out "Quantum Theory of Angular Momentum" edited by L C Biedharn and H van Dam, published by Academic Press (1965). Jan 2 '19 at 16:32
• Check also "Quantum and Atom Optics" by Daniel Steck, available online. atomoptics-nas.uoregon.edu/~dsteck/teaching/quantum-optics Jun 24 '20 at 18:07

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.

• Thank you. I have one more question regarding 2: You mention the definition of the 3-j-symbol as the coupling of three angular momenta. Is this another (maybe explicit) definition not involving Clebsch-Gordan coefficients? Jan 3 '19 at 16:13
• @physicist23 not that I know of. AFAIR the $3j$-coefficient was introduced in the context of invariant theory; if anything I believe it predates the actual CG; Wigner at least studied and wrote in terms of $3j$’s, not CGs. Jan 3 '19 at 19:28
• In this case, how can I see the connection between the 3-j-symbols and the coupling of three angular momentum more obviously? I mean: The definition above only refers to CG and those are the coefficients for coupling two angular momenta, not three. Jan 3 '19 at 21:39
• If you want this interpretation in terms of CG you need $\langle j_1 m_!;j_2m_2\vert JM\rangle \langle JM; j_3 m_3\vert 00\rangle$ and use the fact that only $J=j_3$ and $M=-m_3$ can give you coupling to $0$. The CG $\langle JM; j_3 m_3\vert 00\rangle$ is basically the factor of $1/\sqrt{2 j_3+1}$. Jan 3 '19 at 22:06
• The $3j$ symbol is the invaraint tensor (Clebsch-Gordan) for coupling (if possible) three $SU(2)$ irreps to the identity irrep. Note that Clebsch-and Gordan wrote in the 19th century, long before Wigner. Wigner introduced it because $A\otimes B\otimes C=0$ is naturally more symmetric that the equivalent $A \otimes B= \bar C$, but it contains exactly the same information as the Clebshes for $A \otimes B= \bar C$ . Jun 24 '20 at 16:22

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