Wigner Symbols Quick Reference

This page is a compact reference for Wigner $3j$ , $6j$ , and $9j$ symbols. Use it to identify which symbol belongs in a calculation, check selection rules, and avoid importing signs from the wrong convention.

For the conceptual treatment, use Recoupling and Wigner Symbols. For a table-style summary, use Wigner Symbols. For the representation-theory formulation, use Wigner 3j, 6j, and 9j Symbols.

Symbol Roles

Symbol	What it answers	Typical use
$3j$	How a two-angular-momentum coupling is written in symmetric notation	Clebsch–Gordan conversion, tensor operators, angular integrals
$6j$	How two coupling orders for three angular momenta are related	recoupling, spin–orbit and hyperfine matrix elements, intermediate-coupling bases
$9j$	How two binary pairings of four angular momenta are related	multi-particle angular momentum, atomic and nuclear recoupling

These symbols are kinematic angular-momentum coefficients. They do not by themselves determine energies, rates, or dynamics.

Convention

This volume uses the Condon–Shortley convention unless a page explicitly declares otherwise. The $3j$ relation to Clebsch–Gordan coefficients is

\langle j_1,m_1;j_2,m_2\vert J,M\rangle = (-1)^{j_1-j_2+M} \sqrt{2J+1} \begin{pmatrix} j_1&j_2&J\\ m_1&m_2&-M \end{pmatrix}.

Equivalently,

\begin{pmatrix} j_1&j_2&J\\ m_1&m_2&-M \end{pmatrix} = \frac{ (-1)^{j_1-j_2+M} }{\sqrt{2J+1}} \langle j_1,m_1;j_2,m_2\vert J,M\rangle.

The phase factor is part of the convention. Dropping it is the most common way to turn a correct table lookup into a wrong coefficient.

Three-j Selection Rules

A $3j$ symbol is written

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

It vanishes unless all of the following are satisfied:

m_1+m_2+m_3=0,

\lvert j_1-j_2\rvert \leq j_3 \leq j_1+j_2,

with cyclic variants of the triangle condition, and

\lvert m_i\rvert\leq j_i, \qquad j_1+j_2+j_3\in\mathbb Z.

Also require $j_i\ge0$ , $2j_i\in\mathbb Z$ , and $j_i-m_i\in\mathbb Z$ for every column. A useful special case is

\begin{pmatrix} j&j&0\\ m&-m&0 \end{pmatrix} = \frac{(-1)^{j-m}}{\sqrt{2j+1}}.

Selection rules are necessary conditions. They are not a guarantee that an allowed-looking symbol is nonzero.

Six-j Recoupling

A $6j$ symbol changes the coupling order for three angular momenta. The two bases

\left\lvert (j_1j_2)j_{12},j_3;JM \right\rangle

and

\left\lvert j_1,(j_2j_3)j_{23};JM \right\rangle

are related by

\begin{aligned} & \left\lvert (j_1j_2)j_{12},j_3;JM \right\rangle \\ &= \sum_{j_{23}} (-1)^{j_1+j_2+j_3+J} \sqrt{(2j_{12}+1)(2j_{23}+1)} \times \left\{ \begin{array}{ccc} j_1&j_2&j_{12}\\ j_3&J&j_{23} \end{array} \right\} \left\lvert j_1,(j_2j_3)j_{23};JM \right\rangle. \end{aligned}

The $6j$ symbol is independent of $M$ . It depends only on the angular-momentum labels because it compares two ways of organizing the same total- $J$ irreducible space.

For

\left\{ \begin{array}{ccc} j_1&j_2&j_{12}\\ j_3&J&j_{23} \end{array} \right\},

each of the following triples must obey the triangle inequalities and have an integer sum:

(j_1,j_2,j_{12}), \quad (j_{12},j_3,J), \quad (j_2,j_3,j_{23}), \quad (j_1,j_{23},J).

Nine-j Recoupling

A $9j$ symbol compares two binary coupling schemes for four angular momenta. Both states below have the same $J,M$ ; $M$ is suppressed. Use the fixed tensor-factor order $1,2,3,4$ , without an additional physical exchange operation. Then

\begin{aligned} & \left\langle (j_1j_2)j_{12},(j_3j_4)j_{34};J \right. \\ &\quad \left. \vert (j_1j_3)j_{13},(j_2j_4)j_{24};J \right\rangle \\ &= \sqrt{ (2j_{12}+1)(2j_{34}+1) (2j_{13}+1)(2j_{24}+1) } \times \left\{ \begin{array}{ccc} j_1&j_2&j_{12}\\ j_3&j_4&j_{34}\\ j_{13}&j_{24}&J \end{array} \right\}. \end{aligned}

For a $9j$ symbol, rows and columns encode allowed triangular couplings in the displayed pairing scheme. A $9j$ symbol can also be evaluated as a finite sum of products of three $6j$ symbols. Intermediate angular-momentum labels and dimension factors must be included.

Relation to Tensor Operators

The Wigner–Eckart theorem commonly uses a $3j$ symbol:

\langle j' m'\vert T_q^{(k)}\vert j m\rangle = (-1)^{j'-m'} \begin{pmatrix} j'&k&j\\ -m'&q&m \end{pmatrix} \langle j'\lVert T^{(k)}\rVert j\rangle.

This form separates magnetic quantum-number algebra from the reduced matrix element. When the state has several coupled angular momenta and the operator acts on one part of the system, $6j$ symbols often appear because one must recouple to expose the angular momentum being acted on.

Use Irreducible Spherical Tensors for tensor conventions and Wigner–Eckart Theorem for the theorem.

References

D. M. Brink and G. R. Satchler, Angular Momentum, 3rd ed., Oxford University Press, 1993.
A. R. Edmonds, Angular Momentum in Quantum Mechanics, Princeton University Press, 1957.
NIST Digital Library of Mathematical Functions, Chapter 34, 3j, 6j, and 9j symbols.
D. A. Varshalovich, A. N. Moskalev, and V. K. Khersonskii, Quantum Theory of Angular Momentum, World Scientific, 1988.
E. P. Wigner, Group Theory and Its Application to the Quantum Mechanics of Atomic Spectra, Academic Press, 1959.
R. N. Zare, Angular Momentum: Understanding Spatial Aspects in Chemistry and Physics, Wiley, 1988.

Quick Checks

Why must

\begin{pmatrix} 1&1&1\\ 1&0&1 \end{pmatrix}

vanish?

Solution

The lower labels of a $3j$ symbol must sum to zero. Here

1+0+1=2,

so the symbol vanishes.

Which symbol changes from $\left\lvert(j_1j_2)j_{12},j_3;JM\right\rangle$ to $\left\lvert j_1,(j_2j_3)j_{23};JM\right\rangle$ ?

Solution

This is a change of coupling order for three angular momenta, so the natural object is a Wigner $6j$ symbol:

\left\{ \begin{array}{ccc} j_1&j_2&j_{12}\\ j_3&J&j_{23} \end{array} \right\}.

Why do $6j$ symbols not depend on $M$ ?

Solution

A $6j$ symbol compares two ways of coupling the same angular momenta to the same total $J$ . Rotational symmetry makes the recoupling coefficient the same for every member of the $M=-J,\ldots,J$ multiplet. The dependence on magnetic labels is already handled by Clebsch–Gordan coefficients or $3j$ symbols.