 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2025/ii/paper-2/36b/b/solution

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

Use the normalized scaled trial state

ψ_{λ} (x) = λ^{1/2} ψ_{0} (λ x), λ > 0.

T_{0} = ⟨ ψ_{0}, T ψ_{0} ⟩ > 0, V_{0} = ⟨ ψ_{0}, V ψ_{0} ⟩ < 0,

then scaling derivatives and changing variables give

E (λ) = ⟨ ψ_{λ}, H ψ_{λ} ⟩ = λ^{2} T_{0} + λ^{n} V_{0} .

If a finite ground state existed,

λ = 1

would minimize this expression, and hence

0 = E^{'} (1) = 2 T_{0} + n V_{0} .

It would follow that

E_{0} = T_{0} + V_{0} = (1 - \frac{2}{n}) T_{0} > 0

for

n > 2

. But a bound state of an attractive potential tending to zero at infinity must have negative energy. This is a contradiction.

Indeed, the scaling gives the stronger collapse statement

E (λ) = λ^{2} T_{0} + λ^{n} V_{0} ⟶ - \infty (λ \to \infty)

when

n > 2

. Thus the ideal singular Hamiltonian is not bounded below and has no finite ground bound state.

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