Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 324 / 3 / a

Write the local Hamiltonian as $H={\sum }_{j=1}^m{h}_j$. Since its terms commute, their matrix exponentials factor exactly:Each factor acts on at most two qubits and can be compiled over a fixed universal quantum gate set to operator norm error at most $\epsilon /m$. The telescoping bound for products of operators then bounds the total error by the sum of the factor errors, at most $\epsilon$. Because $m$ is polynomial in $n$ and the Solovay--Kitaev theorem gives gate count polynomial in $\log (m/\epsilon )$ for each fixed-dimensional factor, this is an efficient commuting local Hamiltonian simulation. Finally, the eigenvalue equation $H|\Psi ⟩=\lambda |\Psi ⟩$ impliesso $|\Psi ⟩$ remains an eigenstate and its eigenvalue is $e^{-i\lambda t}$.

Apply exact quantum phase estimation to $U(t)$ with the supplied eigenstate $|\Psi ⟩$. Sincethe promise that the phase has an $N$-bit representation makes an $N$-qubit control register recover $\varphi$ exactly. Multiplying by $-2\pi$ modulo $2\pi$ gives $\lambda t\mathrm{mod}2\pi$. Equivalently, phase estimation may be run on $U(t{)}^{†}$, whose eigenphase is $\lambda t/(2\pi )$ modulo one.

Let $A=X\otimes X\otimes I$ and $B=Z\otimes I\otimes Z$. The two Pauli operators anticommute because their local factors anticommute at exactly one qubit. Since $A^2=B^2=I$, the mixed terms cancel andThis is a scalar, or $0$-local, Hamiltonian, so the smallest value is $\boxed{k=0}$.