Question and first prediction
How can an unchanging Hamiltonian generate a moving state?
A constant magnetic field defines a fixed axis in spin space. A state prepared along that axis is an energy eigenstate, and its spin expectation stays fixed. A transverse preparation is a superposition of the two energy eigenstates; the resulting relative dynamical phase drives a periodic change in the Bloch vector. This simple evolution already contains the central ideas of qubit rotations, the Heisenberg picture and magnetic resonance.
Recommended background: Pauli matrices, unitary time evolution U(t)=exp(−iHt/ℏ), and the Bloch-vector parameterization.
Results that must agree with the equations
Computed observables
Three quantum observables
The Bloch components evolve together
The longitudinal projection r·n remains constant even as laboratory-frame components oscillate.
Exact Rodrigues rotation of r(t) by angle Ωt about the field axis n. All components are expectations of Pauli operators, dimensionless between −1 and +1.
A conserved quantity
Why the energy expectation does not oscillate
The Hamiltonian itself generates rotations about its own Bloch vector; this projection is preserved.
The longitudinal spin 〈σn〉=r·n remains fixed. The magnitude |r×n| of the transverse component is conserved under pure-state unitary evolution.
Mathematical explanation
A Pauli Hamiltonian generates an SO(3) rotation
Choose H=(ℏΩ/2)n·σ with unit vector n fixed in the laboratory x–z plane. Since (n·σ)²=I, exponentiating the Hamiltonian requires only two terms:
The state ρ=(I+r·σ)/2 transforms as ρ(t)=Uρ(0)U†. The Bloch equation from [H,ρ] is dr/dt=Ω n×r. Therefore the exact solution is a rigid rotation:
Multiplying by n shows n·r(t)=n·r₀. Also |r(t)|=|r₀|=1 for the pure preparation used here, so its purity stays one. The Hamiltonian expectation 〈H〉=(ℏΩ/2)n·r is constant; only other spin components need oscillate.
In the Heisenberg picture it is the Pauli observable σ(t)=U†σU that evolves, while the initial ket is fixed. Both pictures give identical expectations. This can be checked by comparing either calculation against the plotted Rodrigues formula.
Test an exactly solvable limit
If n and r₀ are parallel or antiparallel, n×r₀=0 and the state is stationary up to a global phase. If n is perpendicular to r₀, its laboratory projection can change sign after Ωt=π. At Ωt=2π the Bloch vector returns to its original direction, while the underlying spinor acquires a minus sign.
Specify the model's validity
Ideal isolated spin-½, constant effective magnetic field, no relaxation, dephasing or spatial motion. Ω has angular-frequency units and all plotted Pauli expectations are dimensionless. A 2π spinor sign is observable in a relative-phase interference experiment, not by a single spin expectation.
Follow a self-directed learning path
Three quantitative investigations
- Follow a full revolution. Fix the field along x, prepare z-up and increase time. Record the first zero and minimum of 〈σz〉. Compare them with Ωt=π/2 and π.
- Find a stationary state. Align the prepared vector with the field. Explain why every spin component remains time independent, despite the nonzero Hamiltonian eigenenergy.
- Change the Larmor frequency. Keep the same initial spin and field, then double Ω. Verify that the precession period halves while the energy-axis projection remains unchanged.
Check your understanding: For initial +z and a field along +x, what is 〈σz〉 at Ωt=π?
Extension: derive the spinor unitary under a 2π rotation and design an interferometer that measures the resulting relative minus sign.
Connect back to quantum theory
What you can conclude
The Bloch vector obeys a classical-looking rotation equation, but this motion comes from coherent relative phase evolution of a quantum spinor. The model provides the dynamical foundation for qubit gates, Rabi oscillations, Berry cycles and the Schrödinger–Heisenberg equivalence.
Lecture notes: Unitary time evolution · Schrödinger and Heisenberg pictures.