Recoupling and Wigner Symbols

Recoupling changes the order in which angular momenta are added while keeping the total state space fixed. Clebsch–Gordan coefficients couple two angular momenta. Wigner $3j$, $6j$, and $9j$ symbols package the same information in forms that are better adapted to symmetry, tensor operators, and multi-particle coupling schemes.

The practical message is:

• $3j$ symbols are Clebsch–Gordan coefficients in a notation with simple permutation rules;
• $6j$ symbols change the coupling order for three angular momenta;
• $9j$ symbols compare two binary coupling schemes for four angular momenta.

This page explains definitions and usage. The basic definition of the total generator is in Total Angular Momentum. Large tables, selection-rule summaries, and reference identities live in Wigner Symbols.

Why Recoupling Is Needed

For two angular momenta, there is only one binary coupling step:

$$\mathbf J = \mathbf J_1+\mathbf J_2.$$

For three angular momenta, there are different coupling orders. Choosing a physically useful order is discussed in Angular Momentum Coupling Schemes. Algebraically, one can first combine $j_1$ and $j_2$:

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

or first combine $j_2$ and $j_3$:

$$\left| j_1,(j_2j_3)j_{23};JM \right\rangle.$$

Both bases diagonalize the same total $J^2$ and $J_z$, but they diagonalize different intermediate angular momenta:

$$J_{12}^2 \quad \text{versus} \quad J_{23}^2.$$

The change of basis between these two coupling schemes is a recoupling coefficient. Wigner $6j$ symbols are the standard compact notation for it.

Wigner 3-j Symbols

The Wigner $3j$ symbol is a symmetric way to write Clebsch–Gordan coefficients. With the Condon–Shortley convention used in this volume,

$$\langle j_1m_1,j_2m_2|JM\rangle = (-1)^{j_1-j_2+M} \sqrt{2J+1} \begin{pmatrix} j_1&j_2&J\\ m_1&m_2&-M \end{pmatrix}.$$

The $3j$ symbol is useful because many symmetry relations become more compact. It also exposes the magnetic quantum number rule:

$$m_1+m_2-M=0.$$

Equivalently, in the $3j$ notation,

$$m_1+m_2+m_3=0.$$

The triangle condition is still present:

$$\lvert j_1-j_2\rvert \le J \le j_1+j_2.$$

Thus $3j$ symbols do not introduce new physics beyond Clebsch–Gordan coefficients. They reorganize the same coupling data in a notation that is better for identities. Table and phase-convention warnings are collected in Clebsch–Gordan Tables and Conventions.

Wigner 6-j Symbols

The $6j$ symbol recouples three angular momenta. The relation between the two bases above is

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

The braces denote the Wigner $6j$ symbol:

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

The $6j$ symbol is independent of $M$. It depends only on the angular-momentum labels, because rotational symmetry fixes the same recoupling for every member of a total-$J$ multiplet.

For the symbol to be nonzero, each triangular triple must be allowed:

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

Each allowed triple also has an integer sum and obeys the angular-momentum step rules; triangle inequalities alone are insufficient. These conditions are necessary but not always sufficient, since additional cancellations can occur.

Three Spin-One-Half Example

Take three spin-$1/2$ particles and focus on total $J=1/2$, $M=1/2$. One coupling scheme first combines spins $1$ and $2$:

$$\left| (12)j_{12},3;\frac12,\frac12 \right\rangle, \qquad j_{12}=0,1.$$

Another first combines spins $2$ and $3$:

$$\left| 1,(23)j_{23};\frac12,\frac12 \right\rangle, \qquad j_{23}=0,1.$$

With positive ladder coefficients and the highest-weight phase rule specified in Clebsch–Gordan Tables and Conventions, the recoupling matrix is

$$\begin{pmatrix} \lvert(12)0,3\rangle\\ \lvert(12)1,3\rangle \end{pmatrix} = \begin{pmatrix} -\frac12&\frac{\sqrt3}{2}\\ \frac{\sqrt3}{2}&\frac12 \end{pmatrix} \begin{pmatrix} \lvert1,(23)0\rangle\\ \lvert1,(23)1\rangle \end{pmatrix},$$

where the common total labels $J=1/2$, $M=1/2$ have been suppressed.

This example shows what recoupling means physically. The total spin is the same, but the question “which pair is in a singlet?” depends on the coupling scheme. A state that is a definite singlet of particles $1$ and $2$ is a superposition of singlet and triplet sectors for particles $2$ and $3$.

Wigner 9-j Symbols

The $9j$ symbol compares two binary coupling schemes for four angular momenta. One scheme is

$$((j_1j_2)j_{12},(j_3j_4)j_{34})J,$$

while another is

$$((j_1j_3)j_{13},(j_2j_4)j_{24})J.$$

The recoupling coefficient is

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

The common $M$ label is suppressed on both sides. Both states are embedded in the same tensor-factor order $1,2,3,4$: changing parentheses is not a physical permutation of particles. The $9j$ symbol is useful, for example, when converting two-electron LS and jj bases.

Matching Symbols to Coupling Orders

To construct the required coefficient:

1. State the coupling scheme in words.
2. Draw or write the order of parentheses.
3. Identify which intermediate angular momenta are diagonal.
4. Check triangle rules before looking up values.
5. Check the phase convention.
6. Only then insert the $3j$, $6j$, or $9j$ symbol.

This prevents a common error: copying a symbol with the right numbers in the wrong positions. Wigner symbols have many symmetries, but they are not permission to ignore ordering.

Relation to Tensor Operators

Wigner symbols also appear in tensor-operator matrix elements. The Wigner–Eckart theorem separates magnetic quantum numbers from reduced matrix elements using Clebsch–Gordan or $3j$ symbols. When an operator acts inside a space with several coupled angular momenta, $6j$ symbols often appear because one must recouple the angular momentum on which the operator acts.

This is why $6j$ symbols show up in:

• fine and hyperfine structure calculations;
• addition of orbital and spin angular momentum;
• multi-electron atoms;
• nuclear shell-model matrix elements;
• spin-network and angular-momentum diagrammatics.

The tensor-operator side begins with Irreducible Spherical Tensors and is then used in Wigner–Eckart Theorem.

Common Mistakes

• Treating $3j$ symbols as a different physical object from Clebsch–Gordan coefficients.
• Forgetting the phase factor relating $3j$ symbols and Clebsch–Gordan coefficients.
• Reading a $6j$ symbol without first specifying the two coupling schemes.
• Assuming triangle rules guarantee a nonzero value.
• Mixing phase conventions from different tables.
• Forgetting that $6j$ recoupling coefficients are independent of $M$.
• Using a $9j$ symbol when two successive $6j$ recouplings would make the calculation clearer.

Exercises

1. Derive the upper-left entry of the three-spin recoupling matrix directly from product states, and find the probability that pair $(23)$ is a singlet when pair $(12)$ starts in a singlet.

Solution

For $J=M=1/2$, the two pair-singlet states are

$$|A_0\rangle=\frac{|\uparrow\downarrow\uparrow\rangle-|\downarrow\uparrow\uparrow\rangle}{\sqrt2},$$ $$|B_0\rangle=\frac{|\uparrow\uparrow\downarrow\rangle-|\uparrow\downarrow\uparrow\rangle}{\sqrt2}.$$

Only the shared product state contributes to $\langle B_0|A_0\rangle$, giving $-1/2$. The singlet probability for pair $(23)$ is its squared magnitude, $1/4$; the complementary triplet probability is $3/4$. The total $J$ remains $1/2$ throughout this change of description.

2. Check a triangle condition.

For $j_1=1$, $j_2=1/2$, what intermediate values of $j_{12}$ are allowed?

Solution

The allowed values obey

$$\lvert j_1-j_2\rvert \le j_{12} \le j_1+j_2.$$

Thus

$$\frac12 \le j_{12} \le \frac32,$$

so

$$j_{12} = \frac12,\frac32.$$

3. Use recoupling to solve a three-spin interaction.

For three specified spin-$1/2$ factors, take the rotationally invariant Hamiltonian

$$H=A\frac{\mathbf S_1\cdot\mathbf S_2}{\hbar^2} +B\frac{\mathbf S_2\cdot\mathbf S_3}{\hbar^2},$$

where $A,B$ are real constants with energy units. In the ordered $j_{12}=0,1$ basis at fixed $J=1/2,M$, use

$$R = \begin{pmatrix} -\frac12&\frac{\sqrt3}{2}\\ \frac{\sqrt3}{2}&\frac12 \end{pmatrix}$$

to find the two-by-two Hamiltonian and its energies. Check $R^\dagger R=I$, explain the degeneracy of each energy, and compare with the $J=3/2$ sector. Is $j_{12}$ conserved when $B\ne0$?

Solution

Both columns have norm one and their inner product is zero, so $R^\dagger R=I$. A pair scalar product divided by $\hbar^2$ has eigenvalues $-3/4$ in its singlet and $1/4$ in its triplet. Write $D=\operatorname{diag}(-3/4,1/4)$. The first pair is already diagonal in this basis; the second pair requires recoupling:

$$H_{J=1/2}=AD+BRDR^{\mathsf T} =\begin{pmatrix} -3A/4&\sqrt3 B/4\\ \sqrt3 B/4&A/4-B/2 \end{pmatrix}.$$

Its eigenvalues are

$$E_\pm=-\frac{A+B}{4}\pm\frac12\sqrt{A^2-AB+B^2}.$$

The matrix is independent of $M$, so each energy forms a total-$J=1/2$ doublet. The term proportional to $B$ mixes the two $j_{12}$ values when $B\ne0$, even though every component of total angular momentum remains conserved.

In the $J=3/2$ quartet, both pairs are triplets and $E_{3/2}=(A+B)/4$. This accounts for all eight states. As checks, $B=0$ recovers the first-pair singlet/triplet energies; for $A=B>0$, the two doublets have energies $-A,0$ and the quartet has energy $A/2$. Recoupling thus makes a rotationally invariant interaction calculable when one intermediate pair label is insufficient to diagonalize it.

4. Why is a $6j$ coefficient independent of $M$?

Solution

The two coupling schemes are two bases for the same irreducible total-$J$ multiplet structure. Rotational symmetry treats all magnetic components $M=-J,\ldots,J$ in the same way. The recoupling changes intermediate angular-momentum labels, not the orientation label inside the final multiplet. Therefore the coefficient depends on $j_1,j_2,j_3,J,j_{12},j_{23}$ but not on $M$.

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.

• National Institute of Standards and Technology, Digital Library of Mathematical Functions, Chapter 34, 3j, 6j, 9j Symbols, accessed 2026-10-13.

• M. E. Rose, Elementary Theory of Angular Momentum, Wiley, 1957.

• J. J. Sakurai and J. Napolitano, Modern Quantum Mechanics, 3rd ed., Cambridge University Press, 2020.

• D. A. Varshalovich, A. N. Moskalev, and V. K. Khersonskii, Quantum Theory of Angular Momentum, World Scientific, 1988.