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ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2014/iii/paper-63/4/a/ii/solution

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Past exam of the mathematics course of the University of Cambridge / 2014 / iii / Paper 63 / 4 / a / ii / Solution by

Codex  0 Created 2026-10-03 Updated 2026-10-06

For an excited eigenstate,

E_{i} - E_{0} \geq Δ

. The matrix-element identity in condition (i), together with the two-sided Fourier cutoff, gives

⟨ ϕ_{i} ∣ A^{(Z)} ∣ ϕ_{0} ⟩ = w (E_{i} - E_{0}) ⟨ ϕ_{i} ∣ h_{Z} ∣ ϕ_{0} ⟩ = 0.

The reverse matrix element has frequency

E_{0} - E_{i} \leq - Δ

and also vanishes. Equivalently,

A^{(Z)}

is Hermitian because

h_{Z}

is Hermitian and

w

is real, so the two elements are conjugate. Thus

⟨ ϕ_{0} ∣ A^{(Z)} ∣ ϕ_{i} ⟩ = ⟨ ϕ_{i} ∣ A^{(Z)} ∣ ϕ_{0} ⟩ = 0 (i > 0) .

The spectral gap eliminates precisely the couplings needed to make the unique ground state an eigenvector of every filtered term. Couplings between excited states with smaller energy differences can remain; the spectral filter need not diagonalize the whole operator.

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