 Solution

ID: past-exam-of-the-physics-a-course-of-the-university-of-cambridge/2019/paper-1/section-b/b6/c/solution

Put

Ω = λ B

, including its sign. The initial vector is the

+ ℏ/2

S_{z}

; it is not an energy eigenstate of

H = Ω S_{x}

. In terms of the spin eigenstates found above,

∣ + z ⟩ = (∣ + x ⟩ + ∣ - x ⟩) / 2

∣ ψ (t)⟩ = \frac{e ^{- i Ω t /2} ∣ + x ⟩ + e ^{i Ω t /2} ∣ - x ⟩}{2} = (cos (Ω t /2) - i sin (Ω t /2)) .

Equivalently,

e^{- i H t /ℏ} = I cos (Ω t /2) - i σ_{x} sin (Ω t /2)

follows from

σ_{x}^{2} = I

. The Born rule gives

P (S_{z} = + ℏ/2; t) = ∣ ⟨ + z ∣ ψ (t)⟩ ∣^{2} = cos^{2} (Ω t /2) .

(ℏ/2) (0, - sin Ω t, cos Ω t)

, a rotation around the

x

axis. Its direction follows the specified sign of the Hamiltonian; changing the magnetic coupling sign reverses the precession.

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