The physical question
When does total angular momentum remain a good quantum number, and how does a perturbation mix its multiplets?
Two spin-½ systems admit a four-dimensional tensor product, which can equally be reorganized into a spin-one triplet and a spin-zero singlet. Exchange symmetry and total spin become transparent in the coupled basis, while a magnetic-field gradient mixes the total-spin sectors. An exactly solvable two-state block makes this transition visible.
Prerequisites: the corresponding lecture-note treatment of eigenstates, phases, normalization and elementary differential equations. Work through the calculation on paper before exploring parameter sweeps.
Quantitative predictions
Live computed observables
All four coupled-state energy levels
The m=±1 triplets remain at J/4, while the two m=0 states anticross as the field gradient varies.
Four exchange-coupled spin energy levels versus applied spin-gradient Δ
The m=0 block eigenvalues are −J/4±√[(J/2)²+Δ²]; the m=±1 triplet energy is J/4.
Singlet versus triplet character
Inspect the probability of projecting onto the coupled basis and compare with the original product basis.
Singlet and triplet weights for the two m=0 energy eigenstates versus gradient
At Δ=0 the eigenstates have exact total spin S=0 or S=1 (except degeneracy). A nonzero gradient mixes them.
Which spin-resolved outcome appears?
The product measurement probabilities contain interference of singlet and triplet amplitudes and are distinct from their total-spin weights.
Joint outcomes up-down and down-up versus applied field gradient
The coupling convention is |T₀〉=(|↑↓〉+|↓↑〉)/√2 and |S〉=(|↑↓〉−|↓↑〉)/√2.
Recover the mathematics
Derivation and physical meaning
The two-spin Hilbert space decomposes as ½⊗½=1⊕0. In the product basis, the coupled triplet is |T+〉=|↑↑〉, |T₀〉=(|↑↓〉+|↓↑〉)/√2, and |T−〉=|↓↓〉; the singlet is |S〉=(|↑↓〉−|↓↑〉)/√2. The states are exactly orthonormal and resolve the identity.
Consider H=J S₁·S₂+Δ(S₁z−S₂z) in units ℏ=1. The gradient vanishes on |T±〉 but couples |T₀〉 and |S〉. In that ordered basis the relevant block is H₀=[[J/4,Δ],[Δ,−3J/4]], with exact energies E±=−J/4±√[(J/2)²+Δ²]. The gap √(J²+4Δ²) controls the mixing.
Within the m=0 subspace define σz=|T₀〉〈T₀|−|S〉〈S| and σx=|T₀〉〈S|+h.c. The eigenstate Bloch vector aligns or anti-aligns with (Δ,0,J/2). Its triplet probability P(T₀)=[1±(J/2)/√((J/2)²+Δ²)]/2. Total-spin expectation is 〈S²〉=2P(T₀), while spin-correlation expectation is 〈σ₁·σ₂〉=4P(T₀)−3. An external gradient breaks total-spin conservation while preserving total Sz=0.
Check an exact limiting case
At Δ=0 and J>0 the ground state is the singlet and the excited m=0 state is triplet; for J<0 their ordering reverses. At |Δ|≫|J| the eigenstates approach product spin configurations |↑↓〉 or |↓↑〉. At J=Δ=0 the m=0 block is exactly degenerate and its individual eigenvectors are not uniquely determined.
Scope and validity of the model
Two ideal spin-½ particles with isotropic Heisenberg exchange plus a static opposite local z field, using ℏ=1 and no spin-orbit anisotropy or decoherence. The model includes the full triplet spectrum but computes wavefunction weights in the m=0 sector; the degenerate origin is explicitly undefined.
Independent investigations
Explore and explain
- Verify Clebsch–Gordan structure. At Δ=0, identify the triplet and singlet projectors, confirm the exchange energy difference J and determine 〈S²〉 of each.
- Break total-spin symmetry. Increase |Δ| at fixed positive J. Explain why the energy levels avoid crossing and why the corresponding eigenstates approach product states.
- Recover a product measurement. Expand |T₀〉 and |S〉 back into |↑↓〉 and |↓↑〉. Compare the two product probabilities with the coupled-basis probabilities.
Self-check: For J>0 and Δ=0, which state has the lower exchange energy?
For a deeper challenge, reproduce the displayed model independently with a symbolic or numerical notebook and ask an AI tutor for a critical review of each approximation, boundary condition and sign.
Synthesis and references
What the experiment establishes
The coupled basis is a physical classification by total spin. Its usefulness is decided by Hamiltonian symmetry: isotropic exchange respects S², while opposing fields mix S=0 and S=1. Students can compute both descriptions and confirm that probabilities and energies remain consistent.
Lecture notes: Corresponding lecture-note chapter ↗.