Past exam of the mathematics course of the University of Cambridge / 2015 / iii / Paper 67 / 2 / b

The matrix $D=I-|e_0⟩⟨e_0|$ is an orthogonal projection. Its eigenvalues areIts unique ground state in $\mathcal{L}$ is $|e_0⟩=|{\psi }_0⟩|0{⟩}^{\otimes T}$, and its spectral gap is one.

The history-subspace propagation Hamiltonian has diagonal entries $1/2$ at the endpoints and $1$ in the interior, and nearest-neighbour entries $-1/2$. The spectrum of a path propagation Hamiltonian isTo see the indexing directly, set $q_k=\pi k/(T+1)$ and $f_t=\cos [q_k(t+1/2)]$. The interior difference equation gives $Ef=(1-\cos q_k)f$. The endpoint equations correspond to $f_{-1}=f_0$ and $f_{T+1}=f_T$, both satisfied by these $q_k$. The $T+1$ distinct eigenvalues and their nonzero eigenvectors exhaust the matrix.

The zero eigenvalue is simple, soThe inequality uses $\sin x\ge 2x/\pi$ on $[0,\pi /2]$ and $1-\cos q=2{\sin }^2(q/2)$.

The supplied path-matrix hint sums through $i=N$ while listing a basis ending at $N$. That last summand would introduce an extra vertex. The matrix derived in (a) has exactly $T$ edges and $T+1$ vertices, as used here; this also proves the needed spectrum without relying on the misindexed summation.

Because each term in the history-subspace propagation Hamiltonian is positive semidefinite, the uniform history vector has zero energy and is a ground state:Indeed every edge term annihilates the uniform coefficients. Conversely, a zero-energy vector must have equal coefficients across every edge of the connected path, so this computational history state is the unique ground state in $\mathcal{L}$. It coherently records every intermediate state of the quantum circuit.

In the history basis, $M(s)=(1-s)D+sE$ is a Hermitian tridiagonal matrix. Its first diagonal entry is $s/2$, its interior entries are $1$, its last entry is $1-s/2$, and each off-diagonal entry is $-s/2$.

The Gershgorin disc theorem confines the first row's eigenvalue to $[0,s]$, the interior rows to $[1-s,1+s]$, and the last row to $[1-s,1]$. For $0\le s\le 1/3$, the first disc is disjoint from all the others and therefore contains exactly one eigenvalue. Since $M(s)$ is positive semidefinite, that eigenvalue is the ground state energy. All other eigenvalues are at least $1-s$, while the lowest is at most $s$. Hence the Gershgorin gap bound for an adiabatic history path givesThis argument also covers $s=0$, where the first disc is the isolated point zero.