Clebsch–Gordan Coefficients

Clebsch–Gordan coefficients are the numbers that convert between two natural bases for a composite angular momentum system.

The uncoupled basis labels the two angular momenta separately:

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

The coupled basis labels the total angular momentum:

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

The Clebsch–Gordan coefficients are the amplitudes in the expansion

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

They are change-of-basis coefficients, not new dynamical assumptions.

For the operator meaning of the total generator and the two bases before computing coefficients, see Total Angular Momentum and Coupled and Uncoupled Bases.

Allowed Total Angular Momenta

For angular momenta $j_1$ and $j_2$, the allowed total $J$ values are

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

For each $J$,

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

Dimension counting gives the consistency check

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

Magnetic Quantum Number Selection Rule

The coefficient

$$\langle j_1,m_1;j_2,m_2|J,M\rangle$$

vanishes unless

$$M=m_1+m_2.$$

This follows because the total $z$ component is

$$J_z=J_{1z}+J_{2z}.$$

The selection rule is often the fastest way to see which product states can appear in a coupled state.

Normalization and Orthogonality

For fixed $j_1,j_2,J,M$, the coefficients are normalized:

$$\sum_{m_1,m_2} \left| \langle j_1,m_1;j_2,m_2|J,M\rangle \right|^2 =1.$$

The full transformation between uncoupled and coupled bases is unitary. Thus the inverse expansion is

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

With real Condon–Shortley coefficients, the inverse uses the same numbers. In other phase conventions, complex conjugation must be tracked carefully.

Writing

$$C^{JM}_{m_1m_2} := \langle j_1,m_1;j_2,m_2\mid J,M\rangle$$

for fixed $j_1,j_2$, the two full unitarity relations are

$$\sum_{m_1,m_2} C^{JM*}_{m_1m_2} C^{J'M'}_{m_1m_2} = \delta_{JJ'}\delta_{MM'},$$

and

$$\sum_{J,M} C^{JM}_{m_1m_2} C^{JM*}_{m'_1m'_2} = \delta_{m_1m'_1}\delta_{m_2m'_2}.$$

The first states that coupled basis vectors are orthonormal; the second states that they span the full tensor-product space.

Phase Convention

Clebsch–Gordan coefficients depend on phase conventions. This volume uses the Condon–Shortley convention unless explicitly stated otherwise.

The convention fixes the relative phases of angular momentum states through the standard ladder-operator action and chooses the highest-weight coupled state with positive leading coefficient. Tables from different sources may differ by signs if they use different state phases.

The signs are not arbitrary once a convention is chosen. Mixing conventions inside one calculation is a common source of wrong answers.

Two Spin-One-Half Example

For $j_1=j_2=1/2$,

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

The coupled states are

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

Thus, for example,

$$\langle \tfrac12,\tfrac12;\tfrac12,-\tfrac12|1,0\rangle = \frac{1}{\sqrt2},$$

and

$$\langle \tfrac12,-\tfrac12;\tfrac12,\tfrac12|0,0\rangle = -\frac{1}{\sqrt2}.$$

These signs assume the convention used above. The symmetry interpretation of the resulting scalar singlet and vector triplet is developed in Singlet and Triplet States.

How Small Coefficients Are Computed

For small angular momenta, a practical method is:

1. Start with the highest-weight state $\lvert J,J\rangle$ for the largest $J=j_1+j_2$.
2. Apply the total lowering operator $J_-=J_{1-}+J_{2-}$.
3. Normalize the resulting state.
4. Use orthogonality to find states with the same $M$ but different $J$.
5. Continue lowering and normalizing.

This method is often more transparent than memorizing a formula. It also exposes the phase convention.

How to Read Tables

A table of Clebsch–Gordan coefficients usually fixes $j_1,j_2,J$ and lists coefficients for allowed $m_1,m_2,M$. Before using a table, check:

• whether it lists $\langle j_1,m_1;j_2,m_2\vert J,M\rangle$ or a related Wigner $3j$ symbol,
• which phase convention it uses,
• whether $j_1$ and $j_2$ have been interchanged,
• whether the table omits entries that vanish by $M=m_1+m_2$.

For table conventions, sign checks, and $3j$ conversion rules, see Clebsch–Gordan Tables and Conventions. For $3j$ notation and the $6j$ or $9j$ symbols used when three or more angular momenta can be coupled in different orders, see Recoupling and Wigner Symbols.

Wigner $3j$ symbols and Clebsch–Gordan coefficients contain the same information but differ by phase and normalization factors. Do not identify them without the conversion formula.

In the Condon–Shortley convention used here,

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

The coefficients organize much more than coupled-state notation. They enter spherical-tensor matrix elements, multipole expansions, orbital–spin coupling, atomic term construction, two-particle partial waves, selection rules, and the recoupling networks generalized by $6j$ and $9j$ symbols.

Common Mistakes

• Forgetting the rule $M=m_1+m_2$.
• Treating coefficient signs as convention-free facts.
• Confusing the labels $j_1,j_2$ with the total label $J$.
• Reading Wigner $3j$ symbol tables as Clebsch–Gordan tables.
• Assuming the uncoupled product basis is less physical than the coupled basis. The better basis depends on the Hamiltonian and measurement being considered.

Cross-Links

• Clebsch–Gordan Quick Reference
• Clebsch–Gordan Tables and Conventions
• Tensor Product Representations in Angular Momentum
• Total Angular Momentum
• Two Spin-1/2 Particles
• Coupled and Uncoupled Bases
• Singlet and Triplet States
• Spin–Orbit Coupling
• Angular Momentum Algebra
• Ladder Operators
• Spin Rotations
• Clebsch–Gordan Coefficients as Representation Coefficients
• Tensor Product Representations
• Tensor Products
• SU(2)

References

• A. R. Edmonds, Angular Momentum in Quantum Mechanics, Princeton University Press, 1957.
• D. A. Varshalovich, A. N. Moskalev, and V. K. Khersonskii, Quantum Theory of Angular Momentum, World Scientific, 1988.
• J. J. Sakurai and J. Napolitano, Modern Quantum Mechanics, 3rd ed., Cambridge University Press, 2020.
• R. N. Zare, Angular Momentum: Understanding Spatial Aspects in Chemistry and Physics, Wiley, 1988.
• B. C. Hall, Quantum Theory for Mathematicians, Springer, 2013.

Exercises

1. For $j_1=1$ and $j_2=1/2$, list the allowed values of $J$.

Solution

The allowed values are

$$J=\left|1-\frac12\right|,\ldots,1+\frac12.$$

Therefore

$$J=\frac12,\frac32.$$

2. Explain why the coefficient $\langle 1,0;1,0\vert 1,1\rangle$ must vanish.

Solution

The magnetic quantum number rule requires

$$M=m_1+m_2.$$

Here $m_1=0$ and $m_2=0$, so $m_1+m_2=0$. But the coupled state has $M=1$. Therefore the coefficient vanishes.

3. For $j_1=3/2$ and $j_2=1$, list the allowed $J$ values and verify the tensor-product dimension by summing the coupled multiplet dimensions.

Solution

The triangle rule gives

$$J=\frac12,\frac32,\frac52.$$

The uncoupled dimension is $(2j_1+1)(2j_2+1)=4\times3=12$, while the coupled dimensions sum to

$$(2\cdot\tfrac12+1) +(2\cdot\tfrac32+1) +(2\cdot\tfrac52+1) =2+4+6=12.$$

4. Suppose every uncoupled basis vector is rephased as $\lvert j_1,m_1;j_2,m_2\rangle\mapsto e^{i(\alpha_{m_1}+\beta_{m_2})}\lvert j_1,m_1;j_2,m_2\rangle$ and every coupled vector as $\lvert J,M\rangle\mapsto e^{i\gamma_{JM}}\lvert J,M\rangle$. How do the coefficients change, and which displayed identities remain invariant?

Solution

The transformed coefficient is

$$C^{JM}_{m_1m_2} \mapsto e^{i(\gamma_{JM}-\alpha_{m_1}-\beta_{m_2})} C^{JM}_{m_1m_2}.$$

The selection rule and both unitarity relations are unchanged because the phases cancel between a coefficient and its complex conjugate. Individual signs or phases are therefore convention dependent, while probabilities and complete basis identities are not.