Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 324 / 3 / c

Solution

Each summand of

J_{X}

acts on one qubit, while each summand

Z_{i} Z_{j}

J_{Z}

acts on two. Therefore

J = J_{X} + J_{Z}

∥ J_{X} ∥ = O (n), ∥ J_{Z} ∥ \leq (2 n) = O (n^{2}) .

Split

t

into

r

steps of length

δ = t / r

S_{2} (δ) = e^{- i δ J_{X} /2} e^{- i δ J_{Z}} e^{- i δ J_{X} /2} .

For one step,

Λ = O (δ n^{2})

in the stated estimate, so the spectral-norm error is

O (δ^{3} n^{6})

. The error bound for a product of unitary operators makes the total error

O (r δ^{3} n^{6}) = O (\frac{n ^{6} t ^{3}}{r ^{2}}) .

It is therefore enough to choose

r = O (\frac{n ^{3} t ^{3/2}}{ϵ}),

with

r

also large enough that the small-step estimate applies.

All

Z_{i} Z_{j}

terms commute. A factor

e^{- i δ Z_{i} Z_{j}}

O (n^{2})

gates. The complete Hamiltonian simulation consequently has size

O (\frac{n ^{5} t ^{3/2}}{ϵ}) .

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