Addition of Angular Momentum

Purpose

For two angular momenta acting on distinct factors, total angular momentum is

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

In full tensor-product notation,

$$J_i = J_{1i}\otimes I_2 + I_1\otimes J_{2i}.$$

The uncoupled basis diagonalizes $J_1^2,J_{1z},J_2^2,J_{2z}$; the coupled basis diagonalizes $J_1^2,J_2^2,J^2,J_z$. Clebsch–Gordan coefficients are the unitary change-of-basis amplitudes between them.

This card collects the rules needed to choose allowed sectors, convert bases, check coefficients, and evaluate scalar interactions. The operator meaning of the sum belongs at Total Angular Momentum, and the coefficient construction belongs at Clebsch–Gordan Coefficients.

At a glance

Task	Formula
Total operator	$\mathbf J=\mathbf J_1+\mathbf J_2$
Projection	$M=m_1+m_2$
Allowed total values	$J=\lvert j_1-j_2\rvert,\ldots,j_1+j_2$ in steps of one
Coupled expansion	$\lvert JM\rangle=\sum_{m_1m_2}C_{m_1m_2}^{JM}\lvert j_1m_1;j_2m_2\rangle$
Dimension check	$(2j_1+1)(2j_2+1)=\sum_J(2J+1)$
Scalar product	$\mathbf J_1\cdot\mathbf J_2=(J^2-J_1^2-J_2^2)/2$
Two spin halves	$\tfrac12\otimes\tfrac12=1\oplus0$

The labels $j_1,j_2$ are held fixed in the compact notation below unless they must be displayed explicitly.

Allowed total angular momenta

The tensor product decomposes as

$$j_1\otimes j_2 = \bigoplus_{J=\lvert j_1-j_2\rvert}^{j_1+j_2}J,$$

where $J$ increases in integer steps. For each allowed $J$,

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

The triangle conditions can be stated as

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

with $j_1+j_2+J$ an integer. For two irreducible factors, each allowed $J$ appears once. With three or more angular momenta, the same final $J$ can appear with multiplicity and additional intermediate-coupling labels are needed.

Dimension counting is a fast completeness check:

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

If this equality fails after listing sectors, an allowed $J$ value or multiplet has been omitted.

Coupled and uncoupled bases

Write an uncoupled product state as

$$\lvert j_1m_1;j_2m_2\rangle \equiv \lvert j_1,m_1\rangle \otimes \lvert j_2,m_2\rangle.$$

It has a definite total projection because

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

so

$$M=m_1+m_2.$$

It is not generally an eigenstate of $J^2$, because

$$J^2 = J_1^2+J_2^2 + 2\mathbf J_1\cdot\mathbf J_2.$$

The coupled state is

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

with eigenvalues

$$J^2\lvert j_1,j_2;J,M\rangle = \hbar^2J(J+1) \lvert j_1,j_2;J,M\rangle$$

and

$$J_z\lvert j_1,j_2;J,M\rangle = \hbar M \lvert j_1,j_2;J,M\rangle.$$

Use the uncoupled basis when separate projections dominate the Hamiltonian or measurement. Use the coupled basis when rotationally invariant scalar couplings or total-$J$ measurements dominate. Neither basis is intrinsically more physical.

Clebsch–Gordan transformation

Define

$$C_{m_1m_2}^{JM} \equiv \langle j_1,m_1;j_2,m_2 \vert J,M\rangle.$$

Then

$$\begin{aligned} \lvert j_1,j_2;J,M\rangle &= \sum_{m_1,m_2} C_{m_1m_2}^{JM} \\ &\qquad\times \lvert j_1m_1;j_2m_2\rangle. \end{aligned}$$

The coefficient vanishes unless

$$M=m_1+m_2.$$

The inverse transformation is

$$\begin{aligned} \lvert j_1m_1;j_2m_2\rangle &= \sum_{J,M} \left(C_{m_1m_2}^{JM}\right)^* \\ &\qquad\times \lvert j_1,j_2;J,M\rangle. \end{aligned}$$

Standard Condon–Shortley Clebsch–Gordan coefficients are real, but writing the complex conjugate makes the unitary structure explicit and remains correct under more general rephasings.

Orthogonality and completeness

Unitarity of the basis transformation gives

$$\sum_{m_1,m_2} C_{m_1m_2}^{JM} \left(C_{m_1m_2}^{J'M'}\right)^* = \delta_{JJ'}\delta_{MM'},$$

and

$$\sum_{J,M} C_{m_1m_2}^{JM} \left(C_{m_1'm_2'}^{JM}\right)^* = \delta_{m_1m_1'}\delta_{m_2m_2'}.$$

For one fixed coupled state,

$$\sum_{m_1,m_2} \lvert C_{m_1m_2}^{JM}\rvert^2 =1.$$

These relations are stronger checks than verifying a few individual signs. Within each fixed-$M$ block, the coefficient matrix must be unitary.

Phase and symmetry rules

Clebsch–Gordan signs depend on basis phases. This card uses the standard Condon–Shortley convention. In that convention, interchanging the two coupled factors gives

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

Reversing all projections gives

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

For identical angular momenta $j_1=j_2=j$, exchange acts on a coupled state as

$$P_{12}\lvert j,j;J,M\rangle = (-1)^{2j-J} \lvert j,j;J,M\rangle.$$

This is the exchange symmetry of the angular-momentum factor. The full state of identical particles must also include spatial and any other internal degrees of freedom before bosonic or fermionic exchange symmetry is assessed.

Relation to Wigner 3j symbols

With the same phase convention,

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

The $3j$ symbol is not numerically identical to the Clebsch–Gordan coefficient; the phase, square-root factor, and sign of $M$ matter. Use Clebsch–Gordan Tables and Conventions before mixing sources.

Scalar products and coupled interactions

The identity

$$\mathbf J_1\cdot\mathbf J_2 = \frac12 \left( J^2-J_1^2-J_2^2 \right)$$

makes the coupled basis especially useful. On a state with fixed $j_1,j_2,J$,

$$\begin{aligned} \mathbf J_1\cdot\mathbf J_2 &\longrightarrow \frac{\hbar^2}{2} \big[ J(J+1) \\ &\qquad -j_1(j_1+1) -j_2(j_2+1) \big]. \end{aligned}$$

For an isotropic coupling

$$H_{\mathrm{int}} = A\mathbf J_1\cdot\mathbf J_2,$$

the energy in total-$J$ sector is

$$\begin{aligned} E_J &= \frac{A\hbar^2}{2} \big[ J(J+1) \\ &\qquad -j_1(j_1+1) -j_2(j_2+1) \big]. \end{aligned}$$

This formula underlies exchange, spin–orbit, and hyperfine splittings, but the coefficient $A$ and additional Hamiltonian terms are system dependent.

Within the fixed $j_1,j_2$ product space, the projector onto one allowed total-$J$ sector can be written as

$$P_J = \prod_{J'\ne J} \frac{ J^2-\hbar^2J'(J'+1)I }{ \hbar^2 \left[ J(J+1)-J'(J'+1) \right] },$$

where the product runs over the other allowed total values. This spectral projector is useful when a basis-independent sector decomposition is needed.

Two spin-one-half particles

For two spin halves,

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

The triplet is

$$\lvert1,1\rangle = \lvert\uparrow\uparrow\rangle,$$ $$\lvert1,0\rangle = \frac{1}{\sqrt2} \left( \lvert\uparrow\downarrow\rangle + \lvert\downarrow\uparrow\rangle \right),$$ $$\lvert1,-1\rangle = \lvert\downarrow\downarrow\rangle.$$

The singlet is

$$\lvert0,0\rangle = \frac{1}{\sqrt2} \left( \lvert\uparrow\downarrow\rangle - \lvert\downarrow\uparrow\rangle \right).$$

The triplet states are symmetric under factor exchange, while the singlet is antisymmetric. Let $I_4$ be the identity on the two-spin space and let $\boldsymbol\sigma_1,\boldsymbol\sigma_2$ act on the first and second factors. The sector projectors are

$$P_{J=1} = \frac14 \left( 3I_4+ \boldsymbol\sigma_1\cdot\boldsymbol\sigma_2 \right),$$ $$P_{J=0} = \frac14 \left( I_4- \boldsymbol\sigma_1\cdot\boldsymbol\sigma_2 \right).$$

They follow from the eigenvalues $+1$ and $-3$ of $\boldsymbol\sigma_1\cdot\boldsymbol\sigma_2$ in the triplet and singlet sectors. The full derivation and entanglement interpretation are at Two Spin-Half Particles and Singlet and Triplet States.

Frequent coupling with spin one-half

When angular momentum $j$ is coupled to a spin half, the possible totals are $J=j\pm1/2$ for $j>0$. In the displayed phase convention,

$$\begin{aligned} \lvert j+\tfrac12,M\rangle &= \sqrt{ \frac{j+M+1/2}{2j+1} } \lvert j,M-\tfrac12\rangle \lvert\tfrac12,\tfrac12\rangle \\ &\quad+ \sqrt{ \frac{j-M+1/2}{2j+1} } \lvert j,M+\tfrac12\rangle \lvert\tfrac12,-\tfrac12\rangle, \end{aligned}$$

and

$$\begin{aligned} \lvert j-\tfrac12,M\rangle &= -\sqrt{ \frac{j-M+1/2}{2j+1} } \lvert j,M-\tfrac12\rangle \lvert\tfrac12,\tfrac12\rangle \\ &\quad+ \sqrt{ \frac{j+M+1/2}{2j+1} } \lvert j,M+\tfrac12\rangle \lvert\tfrac12,-\tfrac12\rangle. \end{aligned}$$

Terms with an out-of-range constituent projection vanish at the endpoints. These formulas are common in spin–orbit coupling and spinor spherical harmonics; verify the factor order and phase convention before importing them into a table-based calculation.

Three or more angular momenta

For three factors, one may couple $j_1$ and $j_2$ first to $j_{12}$ and then couple $j_{12}$ with $j_3$, or choose a different intermediate pair. The final total-$J$ space is the same, but the basis labels differ. Wigner $6j$ symbols perform the recoupling transformation, and $9j$ symbols organize related four-angular-momentum changes of scheme.

Use Recoupling and Wigner Symbols rather than treating the two-factor coefficient as sufficient when multiplicities occur.

Calculation workflow

1. State the factor order and the Condon–Shortley convention.
2. List allowed $J$ values from the triangle rule and check dimensions.
3. For a specified $M$, retain only product states with $m_1+m_2=M$.
4. Choose the basis adapted to the dominant commuting operators in the Hamiltonian.
5. Obtain coefficients from a convention-matched table or by highest-weight lowering and orthogonality.
6. Check normalization and, when several states share $M$, mutual orthogonality.
7. For scalar couplings, use the $J^2-J_1^2-J_2^2$ identity before constructing large matrices.

Common mistakes

• Assuming $J=j_1+j_2$ is the only allowed total.
• Assuming definite $m_1,m_2$ implies definite $J$ rather than only definite $M$.
• Forgetting the rule $M=m_1+m_2$.
• Omitting tensor-factor identities in operator expressions and then mixing subsystem actions.
• Reading a Wigner $3j$ table as a Clebsch–Gordan table.
• Mixing coefficient signs from different phase or factor-order conventions.
• Treating direct-sum representation labels as energy levels before a Hamiltonian is specified.
• Adding spin or exchange degeneracy a second time after coupled sectors have already been counted.
• Applying two-factor uniqueness to a many-factor problem with multiplicities.
• Enforcing identical-particle symmetry on the spin factor without considering the rest of the wavefunction.

Canonical links

• Coupled and Uncoupled Bases identifies the compatible operator sets.
• Clebsch–Gordan Quick Reference gives convention-aware starter coefficients.
• Clebsch–Gordan Tables provides fixed low-angular-momentum entries.
• Spin–Orbit Coupling applies the scalar-product identity.
• Hyperfine Structure uses the same coupling algebra for nuclear and electronic angular momenta.
• Wigner Symbols gives mathematical conventions for $3j$, $6j$, and $9j$ symbols.

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.

Exercises

1. For $j_1=3/2$ and $j_2=1$, list the allowed total values and verify the dimension count.

Solution

The allowed values are

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

The product-space dimension is

$$(2j_1+1)(2j_2+1) = 4\cdot3 =12.$$

The coupled dimensions sum to

$$(2)+(4)+(6)=12.$$

Thus the listed multiplets exhaust the tensor product.

2. Expand $\lvert\uparrow\downarrow\rangle$ in the coupled basis and find the probabilities of total spin $J=1$ and $J=0$.

Solution

Adding the definitions of $\lvert1,0\rangle$ and $\lvert0,0\rangle$ gives

$$\lvert\uparrow\downarrow\rangle = \frac{1}{\sqrt2} \left( \lvert1,0\rangle + \lvert0,0\rangle \right).$$

Therefore

$$\Pr(J=1)=\frac12, \qquad \Pr(J=0)=\frac12.$$

The total projection is $M=0$ in both sectors.

3. Two spin halves interact through $H=A\mathbf S_1\cdot\mathbf S_2$. Find the triplet and singlet energies.

Solution

For $s_1=s_2=1/2$,

$$E_J = \frac{A\hbar^2}{2} \left[ J(J+1)-\frac34-\frac34 \right].$$

For the triplet $J=1$,

$$E_{1} = \frac{A\hbar^2}{4}.$$

For the singlet $J=0$,

$$E_{0} = -\frac{3A\hbar^2}{4}.$$

The splitting is $E_1-E_0=A\hbar^2$. Its ordering depends on the sign of $A$.

4. Use the spin-half coupling formula to construct the $J=1/2$, $M=1/2$ state obtained from $j=1$ and spin $1/2$.

Solution

Use the $J=j-1/2$ formula with $j=1$ and $M=1/2$:

$$\begin{aligned} \lvert\tfrac12,\tfrac12\rangle &= -\sqrt{\frac{1}{3}} \lvert1,0\rangle \lvert\tfrac12,\tfrac12\rangle \\ &\quad+ \sqrt{\frac{2}{3}} \lvert1,1\rangle \lvert\tfrac12,-\tfrac12\rangle. \end{aligned}$$

Both product terms have total projection $M=1/2$, and the squared coefficients sum to one. The overall sign could be changed by rephasing the entire coupled multiplet, but relative signs must remain convention consistent.