 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2021/ii/paper-1/35b/b/iv/solution

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

Since

- V_{0} < V (x) \leq 0

except at the single point

x = 0

, and the kinetic-energy expectation is nonnegative, every normalized wavefunction satisfies

⟨ H ⟩ > - V_{0} .

Therefore

E_{0} > - V_{0} .

For the upper bound, use the admissible choice

a = V_{0}^{2} .

Then

E (V_{0}^{2}) = \frac{1}{2} V_{0}^{2} - \frac{V _{0}^{2}}{1 + V _{0}^{2}} = - \frac{1}{2} V_{0}^{2} + \frac{1}{2} V_{0}^{4} + O (V_{0}^{6}) .

- V_{0} < E_{0} \leq - \frac{1}{2} V_{0}^{2} + O (V_{0}^{3}) .

Indeed, the displayed calculation provides the sharper remainder

O (V_{0}^{4})

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