Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 324 / 3

The spectral norm is the operator norm induced by the Euclidean norm:The Pauli operators $X_1$ and $Z_2$ act on different qubits, and each is a unitary operator. Their product is therefore unitary and preserves the norm of every vector. Hence

For finite-dimensional matrices $A_1,\dots ,A_m$, the Lie-Trotter product formula stateswhere convergence is in operator norm. In particular, for Hermitian matrices $H_j$, set $A_j=-i{H}_j$ to obtain a product of unitary operators approximating $e^{-it{\sum }_j{H}_j}$.

The printed sum contains $Z_n$ although the stated labels end at $n-1$. We use the natural periodic convention $Z_n=Z_0$, with $n\ge 3$. If an open chain was intended, omitting the final term gives the same asymptotic bound.

Set $H_j=X_{j-1}{Z}_j$, with indices modulo $n$, and implement the product-formula Hamiltonian simulationEach factor acts on two qubits, so it is a two-qubit unitary operator. More explicitly, if $a=j-1$, $b=j$, and $C_{a\to b}$ is a controlled-NOT gate,where ${\mathsf{H}}_a$ is a Hadamard gate. This gives a constant number of one-qubit and two-qubit gates per factor; one-qubit gates can also be viewed as two-qubit gates tensored with the identity.

The spectral norm obeys the triangle inequality, submultiplicativity of the operator norm, and invariance under multiplication by unitary operators. In particular, the telescoping bound for products of operators gives $\Vert V^k-W^k\Vert \le k\Vert V-W\Vert$ for unitary $V,W$. The first-order unitary product-formula error bound consequently yieldsOnly neighboring terms can have a nonzero commutator, because all other supports are disjoint. There are $n$ neighboring unordered pairs, and $\Vert [H_j,H_l]\Vert \le 2\Vert H_j\Vert \Vert H_l\Vert =2$. Thus $\Vert U-\tilde{U}\Vert \le n/k$. Choosing $k>n/\epsilon$ givesThe coarser bound that counts all $\left(\genfrac{}{}{0ex}{}{n}{2}\right)$ pairs also proves the often-used $O(n^3/\epsilon )$ construction. Locality improves that cubic estimate to the quadratic bound above; neither assertion is a lower bound on the best possible circuit.

Use a clean quantum ancilla and apply the quantum circuit $C$, then the Pauli Z gate on the ancilla, then the inverse quantum circuit $C^{†}$:This compute-phase-uncompute construction implements $A\otimes I$ on every input with the ancilla initially $|0⟩$, and by linearity on any superposition of those inputs. The action on ancilla-$|1⟩$ inputs need not coincide with $A\otimes I$; the question only requires the clean-ancilla subspace.

Since $A|x⟩=(-1{)}^{f(x)}|x⟩$, the spectral decomposition givesReplace the middle gate of the previous compute-phase-uncompute construction bya one-qubit rotation about the z-axis. ThenThe circuit uses the gates of $C$, their inverses, and one fixed rotation; it returns the quantum ancilla to $|0⟩$. The same construction implements $e^{itA}$ with middle gate $e^{itZ}$, an instance of simulation of a computable diagonal Hamiltonian.