Past exam of the mathematics course of the University of Cambridge / 2007 / iii / Paper 58 / 2

The Hamiltonian operator annihilates every vector orthogonal to both of its defining states:Every positive power also annihilates that vector. The exponential series therefore givesIt is a stationary zero-energy state under this unitary time evolution. The displayed evolution uses $\hslash =1$, as in the question. For $N=1$ the two independent defining vectors already span the whole qubit space, so no nonzero vector satisfies both orthogonality conditions; for larger $N$ their orthogonal complement has dimension $2^N-2$.

Write $a=⟨\psi |w⟩=2^{-N/2}$, which is real and satisfies $0<a<1$ for $N\ge 1$. The invariant two-dimensional span has orthonormal eigenvectorswith eigenvalues $\Delta (1+a)$ and $\Delta (1-a)$. In this basis,Transfer therefore requires the two evolved eigenphases to differ by an odd multiple of $\pi$:The first positive time isEquivalently, exponentiating on the invariant span gives the full evolutionAt $t_0$ this is $-i{e}^{-i\Delta t_0}|w⟩$, so the requested global phase can be chosen as $\alpha =-\Delta t_0-\pi /2$ modulo $2\pi$. Since the two vectors are linearly independent, their relative coefficient cannot produce exact transfer at an earlier positive time. This is continuous-time quantum search, with time proportional to the square root of the search-space dimension for fixed $\Delta$. Restoring units multiplies the stated time by $\hslash$. The degenerate zero-qubit case would already have $\psi =w$, so there would be no smallest positive transfer time; the formula concerns the nontrivial $N\ge 1$ search.

For the normalized vector defining the orthogonal projection $P=|\varphi ⟩⟨\varphi |$, $P^k=P$ for every $k\ge 1$. Thuswhich proves the projector phase rotation identity in this case:It changes the sign of the component along $\varphi$ and fixes the orthogonal complement, so it is a reflection operator.

For a marked computational basis vector $w$, $U_{P_w}=I-2|w⟩⟨w|$ is the marked-state phase oracle. With $|\psi ⟩=H^{\otimes N}|0^N⟩$, the other reflection isThe Hadamard gates thus implement its change of basis from the known all-zero state. The usual Grover diffusion operator is $D=2|\psi ⟩⟨\psi |-I=-U_{P_{\psi }}$. Consequently a Grover search algorithm iteration isThe minus sign is an irrelevant global phase. These two reflections rotate the search state in the span of the marked vector and the unmarked uniform vector, with the Grover rotation angle controlled by the initial marked amplitude. Iterating them approximately $\pi \sqrt{2^N}/4$ times gives high-probability search. The same pair of projectors appears additively in the continuous-time Hamiltonian, but the product of full reflections and evolution under their sum are distinct dynamics.

An oracle can recognize a solution without disclosing it in advance. If an efficiently evaluable or black-box predicate has the promised unique marked input, write $f(x)=1$ exactly for that input. A Boolean quantum oracle computes this predicate reversibly asIts diagonal projector on the search register is ${\sum }_{x}f(x)|x⟩⟨x|=|w⟩⟨w|$. To implement a selectable phase, compute $f(x)$ into a zero quantum ancilla, apply a known phase to its $1$ component, and uncompute. The search register then undergoesThis requires evaluation access to the predicate, not classical knowledge of which input passes it. The $P_{\psi }$ term is known from the Hadamard transform. With Hamiltonian-oracle access, or by controlled short-time simulations of the two terms, their sum can realize continuous-time quantum search.

The unknown projector is supplied through an oracle or a physically encoded recognition rule. It cannot be manufactured for an arbitrary hidden $w$ with no such input-dependent resource. The oracle assumption and the fixed energy scale $\Delta$ are essential when interpreting the square-root search time as a computational advantage.