 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2022/ii/paper-3/33a/d/solution

Codex  0 Created 2026-09-24 Updated 2026-09-25

Choose the

z

-axis along

E

, so the perturbation is

Δ H = e ∣ E ∣ z

E_{100}^{(2)} = e^{2} ∣ E ∣^{2} \sum_{α \neq = 100} \frac{∣ ⟨ α ∣ z ∣1 , 0 , 0 ⟩ ∣ ^{2}}{E _{1} - E _{α}} .

Parts (a) and (b) restrict the contributing bound states to

∣ n, 1, 0 ⟩, n \geq 2.

E_{n} = - \frac{R}{n ^{2}},

and hence

\frac{1}{E _{1} - E _{n}} = \frac{1}{- R + R / n ^{2}} = \frac{1}{R} \frac{n ^{2}}{1 - n ^{2}} .

Therefore

E_{100}^{(2)} = \frac{e ^{2} ∣ E ∣ ^{2}}{R} n = 2 \sum \infty \frac{n ^{2}}{1 - n ^{2}} ∣ ⟨ n, 1, 0∣ z ∣1, 0, 0 ⟩ ∣^{2} .

Every denominator is negative, so the discrete-state contribution lowers the ground-state energy.

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