A spin- particle sits in a magnetic field whose Hamiltonian switches once: for

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\frac12 particle sits in a magnetic field whose Hamiltonian switches once: H^(t)=ℏω2σ^x\hat H(t) = \frac{\hbar\omega}{2}\hat\sigma_x for 0≤t<T/20\le t < T/2, then H^(t)=ℏω2σ^z\hat H(t) = \frac{\hbar\omega}{2}\hat\sigma_z for T/2≤t<TT/2 \le t < T, with ωT=π\omega T = \pi. The spin starts in ∣−y⟩=12(1−i)|{-y}\rangle = \tfrac{1}{\sqrt2}\begin{pmatrix}1\\-i\end{pmatrix}, the eigenstate of σ^y\hat\sigma_y with eigenvalue −1-1. ((a)) Someone argues: "H^(t)\hat H(t) is Hermitian at every instant, so U^(T,0)=exp⁡ ⁣(−iℏ∫0TH^(t′) dt′)\hat U(T,0) = \exp\!\left(-\frac{i}{\hbar}\int_0^T \hat H(t')\,dt'\right) is always valid." Using this formula, find the probability of measuring spin up along zz at t=Tt=T. ((b)) Construct the correct U^(T,0)\hat U(T,0) as the ordered product of the two constant-Hamiltonian pieces, and find the true probability of spin up along zz at t=Tt=T. ((c)) The same person then says: "Fine, the naive exponential is wrong in the Schrodinger picture, but if I instead worked entirely in the Heisenberg picture the ordering problem would disappear and I could exponentiate directly." Is this correct? Justify your answer in one or two sentences.

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