 Solution

ID: past-exam-of-the-quantum-physics-course-of-the-university-of-oxford/2016/10/a/solution

With

B = B_{0} \hat{z}

and

S_{z} = (ℏ/2) σ_{z}

H_{0} = μ_{B} B_{0} (10 0 - 1) .

Thus

E_{+} = μ_{B} B_{0}, ∣ + ⟩ = (10); E_{-} = - μ_{B} B_{0}, ∣ - ⟩ = (01) .

For

B_{0} > 0

, spin down is the lower energy eigenstate. The negative electron magnetic moment is already accounted for in the positive coefficient multiplying

S \cdot B

in the given Hamiltonian.

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