An electron ( , Bohr magneton ) sits in a magnetic field of fixed magnitude whose…
PHYSICS · Multi-part · Fermi University Physics · Quantum Mechanics · Quantum dynamics and its formulations · Quantum dynamics and its formulations
The Structure of Quantum Mechanics, Ch Slow change, sudden change and the Berry phase
Problem
An electron (g=2, Bohr magneton μB=9.274×10−24JT−1) sits in a magnetic field of fixed magnitude B0=50mT whose direction is swept once around a cone of half-angle β=45∘ about a fixed axis, returning to its starting direction after a time T. The instantaneous spin eigenstates ∣n±(t)⟩ of H^(t)=−μ⋅B(t) have energies E±(t)=±21ℏωL at every instant, with Larmor frequency ωL=gμBB0/ℏ fixed throughout (only the direction of B moves). The spin starts in ∣n+⟩. The electron's moment points opposite to its spin, so this higher-energy state is the one with spin along B, the state written ∣+⟩ in the text. Use the spin-one-half geometric phase result γ±=∓21Ω(C), where Ω(C)=2π(1−cosβ) is the solid angle enclosed by the cone.
((a)) Compute ωL, and check that a sweep taking T=1.0μs satisfies the adiabatic condition, taking the rate of turning of B to be 2π/T.
((b)) Find the dynamical phase θ+(T) and the geometric phase γ+ accumulated by ∣n+⟩ over one full circuit, for T=1.0μs and again for T′=3.0μs.
((c)) State, with a one-line reason, whether θ+ and whether γ+ change between the two sweep times, and whether γ+ would still be a well-defined, convention-independent number if the field direction stopped a quarter of the way round the cone instead of completing the circuit.