The Hamiltonian for a particle of spin 21 in a magnetic field B is
H=−21ℏγB⋅σ where σx=(0110),σy=(0i−i0),σz=(100−1)
and γ is a constant (the motion of the particle in space can be ignored). Consider a magnetic field which is independent of time. Writing B=Bn, where n is a unit vector, calculate the time evolution operator and show that if the particle is initially in a state ∣χ⟩ the probability of measuring it to be in an orthogonal state ∣χ′⟩ after a time t is
∣⟨χ′∣n⋅σ∣χ⟩∣2sin22γBt.
Evaluate this to find the probability for a transition from a state of spin up along the z direction to one of spin down along the z direction when B=(Bx,0,Bz).
Now consider a magnetic field whose x and y components are time-dependent but small:
B=(Acosαt,Asinαt,Bz).
Show that the probability for a transition from a spin-up state at time zero to a spin-down state at time t (with spin measured along the z direction, as before) is approximately
(γBz+αγA)2sin22(γBz+α)t
where you may assume ∣A∣≪∣∣∣Bz+αγ−1∣∣∣. Comment on how this compares, when α=0, with the result for a time-independent field.
[The first-order transition amplitude due to a perturbation V(t) is
−ℏi∫0tdt′ei(E′−E)t′/ℏ⟨χ′∣V(t′)∣χ⟩
where ∣χ⟩ and ∣χ′⟩ are orthogonal eigenstates of the unperturbed Hamiltonian with eigenvalues E and E′ respectively.]