Clebsch–Gordan Coefficients and Tensor-Product Decomposition

Clebsch–Gordan coefficients tell us how to express a state with definite total angular momentum in a basis of separate angular momenta. They form a unitary change of basis: the vector space does not change, but its basis is reorganized into irreducible rotation multiplets.

Helpful background. Tensor-Product Representations explains why combining two spins gives a tensor product that can split into a direct sum. Clebsch–Gordan Coefficients develops the construction of the coefficients and their physical applications.

For irreducible $SU(2)$ representations,

$$V_{j_1}\otimes V_{j_2} \cong \bigoplus_{J=\lvert j_1-j_2\rvert}^{j_1+j_2}V_J.$$

The Clebsch–Gordan coefficients are the entries of the unitary basis transformation

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

Orthogonality and the magnetic selection rule

The weight selection rule is

$$C^{JM}_{j_1m_1,j_2m_2}=0 \quad\text{unless}\quad M=m_1+m_2.$$

Unitarity gives orthonormality and completeness:

$$\sum_{m_1,m_2} C^{JM*}_{j_1m_1,j_2m_2} C^{J'M'}_{j_1m_1,j_2m_2} = \delta_{JJ'}\delta_{MM'},$$ $$\sum_{J,M} C^{JM}_{j_1m_1,j_2m_2} C^{JM*}_{j_1m'_1,j_2m'_2} = \delta_{m_1m'_1}\delta_{m_2m'_2}.$$

These identities are convention independent. Individual coefficient signs are not.

Phase conventions

The commonly used Condon–Shortley convention chooses real coefficients and fixes the highest-weight phase. Other internally consistent phase choices describe the same basis change. Never combine coefficients or $3j$ symbols from different tables without checking conventions.

In a standard convention,

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

A dimension and selection-rule check

For $j_1=1$ and $j_2=1/2$, the product space has $3\times2=6$ dimensions. The allowed total spins are $J=3/2$ and $J=1/2$, whose dimensions add to $4+2=6$. This check counts basis vectors; it does not yet determine their coefficients.

At $M=3/2$, only $(m_1,m_2)=(1,1/2)$ is possible. Normalization therefore fixes the magnitude of that coefficient to one. At $M=1/2$, two product vectors are available, so the $J=3/2$ and $J=1/2$ states are two orthonormal combinations of them. The ladder algebra determines the combinations, and the phase convention fixes their remaining signs.

Exercise

For $j_1=j_2=1/2$ and $M=0$, write the two allowed product basis vectors. Suppose one normalized coupled vector is $(|+ -\rangle+|- +\rangle)/\sqrt2$. Find a normalized real vector orthogonal to it, and explain the remaining freedom.

Solution

The selection rule allows $|+ -\rangle$ and $|- +\rangle$, where the signs abbreviate $m=\pm1/2$. An orthogonal normalized vector is

$$\frac{|+ -\rangle-|- +\rangle}{\sqrt2}.$$

Its negative is equally normalized and orthogonal. A consistent phase convention fixes that sign when one constructs the full coupled basis. Orthogonality alone does not fix it.

References

• A. R. Edmonds, Angular Momentum in Quantum Mechanics, Princeton University Press, 1957 — coupling coefficients and phase conventions.
• NIST Digital Library of Mathematical Functions, §34.1, Special Notation — the relation between Clebsch–Gordan coefficients and $3j$ symbols.