A spin- particle sits in a magnetic field that switches direction once

PHYSICS · Multi-part · Fermi University Physics · Quantum Mechanics · Quantum dynamics and its formulations · Symmetry, angular momentum and spin

The Structure of Quantum Mechanics, Ch Time evolution as a unitary group, and the three pictures

Problem

A spin-12\tfrac12 particle sits in a magnetic field that switches direction once. For 0≤t<τ0 \le t < \tau the Hamiltonian is H^(t)=12ℏωσ^z\hat H(t) = \tfrac12\hbar\omega\hat\sigma_z (field along zz); for τ≤t<2τ\tau \le t < 2\tau it is H^(t)=12ℏωσ^x\hat H(t) = \tfrac12\hbar\omega\hat\sigma_x (field along xx), with ωτ=π/2\omega\tau = \pi/2. The particle starts in the state ∣+x⟩|{+x}\rangle, the eigenstate of σ^x\hat\sigma_x with eigenvalue +1+1.Someone argues: "The three pictures of quantum mechanics never disagree on a measurable prediction. Time-ordering, naive exponentiation, these are just different ways of writing the same evolution operator, so they must be different 'pictures' of the same physics too. Therefore U^(2τ,0)=exp⁡ ⁣[−iℏ∫02τH^(t) dt]=exp⁡ ⁣[−iπ4(σ^x+σ^z)]\hat U(2\tau,0) = \exp\!\left[-\dfrac{i}{\hbar}\int_0^{2\tau}\hat H(t)\,dt\right] = \exp\!\left[-\dfrac{i\pi}{4}(\hat\sigma_x+\hat\sigma_z)\right] must give the same measurable probabilities as the correct evolution." ((a)) Explain what is wrong with this argument, distinguishing the claim it makes about time-ordering from the genuine equivalence of the three pictures. ((b)) Compute the correct (time-ordered) probability of finding σ^y=+1\hat\sigma_y = +1 at t=2τt=2\tau, using the fact that H^(t)\hat H(t) is piecewise constant so the exact U^\hat U is a product of two ordinary exponentials. ((c)) Compute the probability of finding σ^y=+1\hat\sigma_y=+1 predicted by the naive exponential above, and confirm that it differs from part (b).

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