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ID: past-exam-of-the-atomic-and-laser-physics-course-of-the-university-of-oxford/2021/1/b/solution

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Past exam of the atomic and laser physics course of the University of Oxford / 2021 / 1 / b / Solution by

Codex  0 2026-10-25

Within a fixed

L, S, J

angular momentum multiplet, the component of

L

along

J

follows from

2 L \cdot J = J^{2} + L^{2} - S^{2}

. The corresponding result for

S

follows by interchanging

L

and

S

. For the quantum state

∣ L S J M_{J} ⟩

,

⟨ L_{z} ⟩ ⟨ S_{z} ⟩ = M_{J} \frac{J ( J + 1 ) + L ( L + 1 ) - S ( S + 1 )}{2 J ( J + 1 )}, = M_{J} \frac{J ( J + 1 ) + S ( S + 1 ) - L ( L + 1 )}{2 J ( J + 1 )} .

These are dimensionless expectation values in the convention of part (a). Substituting into the first-order time-independent perturbation theory formula

Δ E = ⟨ H_{Z} ⟩

gives

Δ E = g_{J} μ_{B} B M_{J}, g_{J} = 1 + (g_{s} - 1) \frac{J ( J + 1 ) + S ( S + 1 ) - L ( L + 1 )}{2 J ( J + 1 )} .

For

g_{s} = 2

, the Landé g-factor becomes

g_{J} = 1 + \frac{J ( J + 1 ) + S ( S + 1 ) - L ( L + 1 )}{2 J ( J + 1 )} .

The magnetic sublevels must first be diagonalized within a degenerate multiplet; choosing

M_{J}

along the magnetic field already does this. For

J = 0

the displayed fraction is undefined, but all vector expectation values vanish and there is no first-order Zeeman splitting. A higher-order shift may remain.

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