Clebsch–Gordan Quick Reference

This page is a quick reference for using Clebsch–Gordan coefficients without losing signs, labels, or basis conventions. It is meant for table lookup and basis transformations. The canonical explanation is Clebsch–Gordan Coefficients, the table-reading guide is Clebsch–Gordan Tables and Conventions, and the low-spin table is Clebsch–Gordan Coefficients.

Clebsch–Gordan Coefficients as Overlaps

The uncoupled basis is

$$\lvert j_1,m_1\rangle \lvert j_2,m_2\rangle,$$

and the coupled basis is

$$\lvert j_1,j_2;J,M\rangle.$$

The Clebsch–Gordan coefficient is the overlap

$$C^{JM}_{j_1m_1,j_2m_2} \equiv \langle j_1,m_1;j_2,m_2\vert J,M\rangle.$$

The coupled state expands as

$$\lvert j_1,j_2;J,M\rangle = \sum_{m_1,m_2} C^{JM}_{j_1m_1,j_2m_2} \lvert j_1,m_1\rangle \lvert j_2,m_2\rangle.$$

This volume uses the Condon–Shortley phase convention unless a page explicitly states otherwise. With real coefficients in this convention, signs are fixed by the ladder-operator convention and the chosen highest-weight phases.

Allowed Labels

For two angular momenta $j_1$ and $j_2$,

$$J = \lvert j_1-j_2\rvert, \lvert j_1-j_2\rvert+1, \ldots, j_1+j_2.$$

For each allowed $J$,

$$M=-J,-J+1,\ldots,J.$$

The magnetic quantum number rule is

$$M=m_1+m_2.$$

If this rule fails, the coefficient is zero. The dimension check is

$$(2j_1+1)(2j_2+1) = \sum_{J=\lvert j_1-j_2\rvert}^{j_1+j_2} (2J+1).$$

Use this check to catch missing total-$J$ sectors before consulting a table.

Orthogonality and Inverse Expansion

For fixed $j_1$ and $j_2$, the Clebsch–Gordan coefficients form a unitary change of basis:

$$\sum_{m_1,m_2} \left( C^{JM}_{j_1m_1,j_2m_2} \right)^* C^{J'M'}_{j_1m_1,j_2m_2} = \delta_{JJ'}\delta_{MM'}.$$

Completeness in the uncoupled basis gives

$$\sum_{J,M} C^{JM}_{j_1m_1,j_2m_2} \left( C^{JM}_{j_1m'_1,j_2m'_2} \right)^* = \delta_{m_1m'_1}\delta_{m_2m'_2}.$$

The inverse expansion is

$$\lvert j_1,m_1\rangle \lvert j_2,m_2\rangle = \sum_{J,M} \langle J,M\vert j_1,m_1;j_2,m_2\rangle \lvert j_1,j_2;J,M\rangle.$$

For real coefficients, the inverse uses the same numerical coefficients after matching labels. In a complex convention, use complex conjugates.

Table-Reading Workflow

Before using a coefficient table, check these items in order:

1. Identify whether the table lists Clebsch–Gordan coefficients or Wigner $3j$ symbols.
2. Verify the phase convention, usually Condon–Shortley in physics tables.
3. Check the order of $j_1$ and $j_2$.
4. Apply $M=m_1+m_2$, the triangle range, and the integrality condition $j_1+j_2+J\in\mathbb Z$. Passing these tests is necessary, not sufficient, for a nonzero coefficient.
5. Check normalization by summing the squared coefficients for a fixed coupled state.

The table home is Clebsch–Gordan Coefficients. For a compact formula card, use Addition of Angular Momentum.

Quick Examples

For two spin-$1/2$ systems,

$$\frac12\otimes\frac12 = 1\oplus0.$$

In the standard convention,

$$\lvert 1,0\rangle = \frac{1}{\sqrt2} \left( \lvert\uparrow\downarrow\rangle + \lvert\downarrow\uparrow\rangle \right),$$

while

$$\lvert 0,0\rangle = \frac{1}{\sqrt2} \left( \lvert\uparrow\downarrow\rangle - \lvert\downarrow\uparrow\rangle \right).$$

The relative sign distinguishes the triplet state from the singlet state. The physical interpretation is developed in Singlet and Triplet States.

For $j_1=1$ and $j_2=1/2$,

$$1\otimes\frac12 = \frac32\oplus\frac12.$$

The $M=3/2$ state has only one product-basis component:

$$\left\lvert \frac32,\frac32 \right\rangle = \lvert1,1\rangle \left\lvert\frac12,\frac12\right\rangle.$$

Lowering this state and normalizing generates the $J=3/2$ multiplet; orthogonality then fixes the $J=1/2$ states up to the convention-fixed signs.

Relation to Wigner 3-j Symbols

Clebsch–Gordan coefficients and Wigner $3j$ symbols contain the same coupling information. With the convention used here,

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

The $3j$ symbol is often better for identities and symmetry relations. The Clebsch–Gordan coefficient is often better for changing basis between uncoupled and coupled states. Do not use a $3j$ table as a Clebsch–Gordan table without the phase and normalization factor.

Coupling Order Warning

Swapping the factor order changes the standard coefficients by

$$C^{JM}_{j_2m_2,j_1m_1}=(-1)^{j_1+j_2-J}C^{JM}_{j_1m_1,j_2m_2}.$$

The singlet changes sign while the triplet does not for two spin-half factors. For three or more angular momenta, one must also specify which pair is coupled first.

For example,

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

and

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

are different coupled bases. Their change of basis involves Wigner $6j$ symbols, not merely a Clebsch–Gordan table. Use Recoupling and Wigner Symbols when the problem has more than two angular momenta.

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.
• 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.
• R. N. Zare, Angular Momentum: Understanding Spatial Aspects in Chemistry and Physics, Wiley, 1988.

Quick Checks

1. Which total angular momenta appear in $2\otimes1$?

Solution

The allowed total angular momenta run from

$$\lvert2-1\rvert=1$$

to

$$2+1=3$$

in integer steps. Therefore

$$2\otimes1=3\oplus2\oplus1.$$

2. Can $\langle1,1;1,-1\vert1,1\rangle$ be nonzero?

Solution

The magnetic quantum number rule requires

$$M=m_1+m_2.$$

Here $m_1=1$ and $m_2=-1$, so $m_1+m_2=0$, while the coupled state has $M=1$. The coefficient is therefore zero.

3. Why is a $3j$ table not automatically a Clebsch–Gordan table?

Solution

The two notations differ by a phase and a normalization factor:

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

Using a $3j$ value directly as a Clebsch–Gordan coefficient generally gives the wrong sign or magnitude.