advanced30 min studyLesson 404 in Physics

Adding angular momentum with Clebsch–Gordan coefficients

Resolve coupled spin states into product amplitudes and use total spin to compute exchange energies.

Before you begin

On this page

The question for today

What you’ll understand

Distinguish product and coupled bases of a tensor-product space.
Construct two-spin triplet and singlet states by lowering and orthogonality.
Calculate probabilities and exchange energies without treating angular momenta as classical arrows.

Two spins have separate measurement labels, but an interaction often cares about their total angular momentum. Clebsch–Gordan coefficients translate between these two useful descriptions. They are amplitudes in a unitary basis change, not additional forces or probabilities assigned independently. The simplest pair of spin-one-half systems already reveals an important distinction: zero total z component can belong either to a triplet with nonzero total spin or to a singlet with zero total spin.

Tensor products and the coupled basis

For distinguishable subsystems, the total generator is $\mathbf J=\mathbf J_1\otimes I+I\otimes\mathbf J_2$ . Operators acting on different factors commute. The product basis labels the two individual z components, whereas the coupled basis labels total squared angular momentum and its z component. The individual squared magnitudes remain fixed when each subsystem is restricted to a selected multiplet. Switching bases therefore reorganizes the same tensor-product space without changing its dimension or adding physical states.

The allowed total labels run in unit steps from $|j_1-j_2|$ to $j_1+j_2$ . Counting their dimensions reproduces the product dimension. A nonzero coefficient requires $M=m_1+m_2$ , because both descriptions must have the same total z eigenvalue. This condition is necessary but does not determine a coefficient's magnitude or sign. Relative signs are fixed by phase conventions and orthogonality, and can affect interference when product states are coherently compared.

|JM\rangle=\sum_{m_1,m_2} C^{JM}_{j_1m_1,j_2m_2}|j_1m_1\rangle|j_2m_2\rangle,\qquad \sum_{m_1,m_2}|C^{JM}_{j_1m_1,j_2m_2}|^2=1

The dimensionless C values are Clebsch–Gordan amplitudes for fixed j1 and j2. Their squared magnitudes are product-basis probabilities in a normalized coupled state.↔ Scroll wide equations horizontally.

Constructing a triplet and a singlet

For two spins one half, the highest total magnetic state is $|\uparrow\uparrow\rangle$ . It has total spin one. Lowering the total spin means lowering either factor and adding the results coherently, which produces the normalized symmetric combination of the two opposite-spin products. A second lowering yields the two-down state. These three states make one triplet. Their construction uses the individual ladder factors and the total ladder factor, so normalization is fixed rather than guessed.

The remaining opposite-spin combination must be orthogonal to the triplet's middle state. Choose its relative minus sign and normalize to obtain the singlet. The total raising and lowering operators annihilate it, consistent with total spin zero. Both middle states give half probability for each opposite-spin product, so a local z measurement alone does not distinguish them. Their relative sign instead changes joint measurements along other axes and their response to an exchange interaction.

|1,0\rangle=\frac{|\uparrow\downarrow\rangle+|\downarrow\uparrow\rangle}{\sqrt2},\qquad |0,0\rangle=\frac{|\uparrow\downarrow\rangle-|\downarrow\uparrow\rangle}{\sqrt2}

Arrows denote individual z-spin eigenstates with eigenvalues plus or minus hbar/2. The displayed singlet phase is a convention; its relative minus sign is required by orthogonality.↔ Scroll wide equations horizontally.

A basis change can diagonalize an interaction

The identity obtained by squaring the total generator expresses the scalar product of the two spins in terms of three squared angular momenta. An isotropic exchange Hamiltonian therefore becomes diagonal in the coupled basis. Its singlet and triplet energies can differ even though the interaction preserves rotations. Rotational invariance enforces equal energies among the three magnetic triplet components, not equality between the singlet and triplet, since those are distinct irreducible representations.

The sign of the exchange coefficient decides which total spin is lower. This statement uses the explicit Hamiltonian convention; condensed-matter texts sometimes absorb a minus sign or a numerical factor into their exchange parameter. For identical particles, the spin state must also be combined with a spatial state of the appropriate exchange symmetry. The two-spin calculation by itself does not authorize every spin and spatial combination for identical fermions. Here the spins can be associated with distinct, localized subsystem labels.

H_{\rm ex}=\frac{A}{\hbar^2}\mathbf S_1\cdot\mathbf S_2=\frac{A}{2\hbar^2}(J^2-S_1^2-S_2^2),\qquad E_{J=1}=\frac A4,\quad E_{J=0}=-\frac{3A}{4}

A is an energy, and each constituent has spin one half. The isotropic interaction is the complete Hamiltonian in this illustrative model; external fields and additional orbital energies are omitted.↔ Scroll wide equations horizontally.

Work it through

Resolve a product preparation into exchange levels

Take $A=4\,\mathrm{meV}$ and prepare $|\uparrow\downarrow\rangle$ for two localized spins. Find the possible exchange energies, their probabilities, and the mean energy.

Invert the two coupled combinations to obtain $|\uparrow\downarrow\rangle=(|1,0\rangle+|0,0\rangle)/\sqrt2$ . The two exchange-energy sectors therefore have equal amplitudes, not definite simultaneous total-spin values.
Insert the coefficient into the exchange eigenvalues. The triplet energy is $4/4=1\,\mathrm{meV}$ and the singlet energy is $-3(4)/4=-3\,\mathrm{meV}$ . The energy gap is $4\,\mathrm{meV}$ .
Square the amplitudes to obtain probability $1/2$ for each energy. The mean is $(1-3)/2=-1\,\mathrm{meV}$ . This mean is not itself an available energy outcome in this ideal model.

The result

Energy measurements give +1 meV or −3 meV equally often; their expectation is −1 meV.

The living notebook

Explore the model

Interactive lab

Interactive lab · quantum

Inspect the two opposite-spin amplitudes

Use the two lab basis labels only as an encoding of the opposite-spin products. Reset and set both controls to $(90^\circ,0^\circ)$ , then independently to $(90^\circ,180^\circ)$ . Predict identical half probabilities and opposite Bloch x values, representing the triplet and singlet relative signs in this encoded subspace. This single pure-state display does not simulate two independent spin measurements or exchange dynamics.

Bloch vector / P(0)P(1)Phase latitude

Probability |0⟩: 71.1%
Probability |1⟩: 28.9%
Bloch (x, y, z): 0.78, 0.45, 0.42

Mixing angle θ65°Relative phase φ30°

t = 0 s

A normalized two-level pure state, not an interacting field simulation. Playback adds one radian per second to the relative phase; basis probabilities stay fixed. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.

Look forThe same product probabilities can conceal different total-spin states.

Pause & reason

Check your understanding

Interactive checks need JavaScript. You can still read the explanation.

Read the checkpoint explanation

The total labels range from the absolute difference, one half, to the sum, three halves, in unit steps. Their dimensions are two and four, giving the six product states.

Your turn

Try it yourself

Start from the stretched state $|J=3/2,M=3/2\rangle=|m_1=1,m_2=1/2\rangle$ for $j_1=1$ , $j_2=1/2$ . Lower it once and find both product probabilities in the normalized $M=1/2$ state.

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

The individual lowering coefficients are square root two times hbar and hbar, while the total lowering coefficient is square root three times hbar.

Compare with the solution

Applying the sum of the two lowering operators gives $\hbar(\sqrt2|0,1/2\rangle+|1,-1/2\rangle)$ . Dividing by the total factor $\sqrt3\hbar$ gives amplitudes $\sqrt{2/3}$ and $1/\sqrt3$ . The product probabilities are $2/3$ and $1/3$ , which sum to one and both obey the same total magnetic label $M=1/2$ .

Keep the idea

Take it with you

Clebsch–Gordan coefficients are amplitudes connecting two bases of one tensor-product space.
A singlet and a middle triplet have the same local z probabilities but different coherent structure.
Isotropic exchange can split total-spin sectors while preserving degeneracy inside a triplet.

Read further

Original sources and deeper explanations to follow at your own pace.

Robert G. Littlejohn, Berkeley Physics 221: Coupling of Ket Spaces and the Addition of Angular Momenta, sections 7–12

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