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

Use the Fourier transform convention $\hat{w}(E)={\int }_{\mathbb{R}}w(t)e^{itE}dt$. Since $w\ge 0$, it is real, so $\hat{w}(-E)=\hat{w}(E{)}^{\ast }$. The positive-frequency cutoff therefore also eliminates frequencies at or below $-\Delta$. Normalization gives $\int w=\hat{w}(0)=1$.

For spectral filtering of Hamiltonian terms, chooseIn an energy eigenbasis, its matrix elements areAt $i=j=0$ the frequency is zero, so the normalization preserves the ground-state expectation:The spectral filter is a positive weighted average of unitary conjugations; in particular the integral is bounded in operator norm by $\Vert h_Z\Vert$. We can choose it even without an extra assumed tail bound: the evenization of a nonnegative bandlimited filter proved in part (e) produces another admissible spectral filter with the same type of positive-time decay. Make that choice consistently in all the filtered terms and shell definitions below.

For an excited eigenstate, $E_i-E_0\ge \Delta$. The matrix-element identity in condition (i), together with the two-sided Fourier cutoff, givesThe reverse matrix element has frequency $E_0-E_i\le -\Delta$ and also vanishes. Equivalently, $A^{(Z)}$ is Hermitian because $h_Z$ is Hermitian and $w$ is real, so the two elements are conjugate. ThusThe 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.