Past exam of the mathematics course of the University of Cambridge / 2015 / iii / Paper 67 / 4

Here is a finite-dimensional form of the Kitaev geometrical lemma. Let $A,B$ be positive operators, with nullspaces $S_A,S_B$ and all positive eigenvalues at least $\gamma >0$. First assume $S_A\cap S_B=\{0\}$. Let $P,Q$ be the orthogonal projections onto those nullspaces, and define their smallest angle between two subspaces byIf either nullspace is zero, set $\cos \vartheta =0$. Then

To prove it, spectral decomposition gives $A\ge \gamma (I-P)$ and $B\ge \gamma (I-Q)$. Write $C=P+Q$ and $c=\cos \vartheta$. For every vector $z$,Thus $C^2\le (1+c)C$. Since $C$ is positive semidefinite, each of its eigenvalues lies in $[0,1+c]$, so $2I-C\ge (1-c)I$. Consequentlyas claimed.

If there is a common nullspace $S=S_A\cap S_B$, restrict to $S^{\perp }$ and define the angle between $S_A\cap S^{\perp }$ and $S_B\cap S^{\perp }$ there. The same proof bounds the smallest positive eigenvalue of $A+B$ by $2\gamma {\sin }^2(\vartheta /2)$, while $S$ is its zero-energy subspace. This version gives a spectral gap above a possibly degenerate ground state space.

Use the Fourier transform convention $\hat{w}(E)={\int }_{\mathbb{R}}w(t)e^{itE}dt$ and define the spectral filtering of Hamiltonian terms byThe normalization $\hat{w}(0)=1$ means $\int w=1$. Since $w$ is real and nonnegative, $\hat{w}(-E)=\overline{\hat{w}(E)}$. Thus the stated positive-frequency cutoff also implies $\hat{w}(E)=0$ for $E\le -\Delta$. The integral exists in operator norm, because its integrand has norm at most $w(t)\Vert h_Z\Vert$.

Let $H|{\varphi }_i⟩=E_i|{\varphi }_i⟩$, with $E_i-E_0\ge \Delta$ for $i>0$. In this eigenbasis,Every matrix element coupling the unique ground state to an excited state vanishes, in either direction. HenceMoreover, linearity and conservation of $H$ under its own Heisenberg picture evolution giveEach filtered term is Hermitian, but it need not remain supported on its original set $Z$. The tail assumption supplies stronger long-time control than the integrability used here; no additional tail estimate is needed for the requested identities.

For each interaction neighbourhood, setThen the shell term is $g^{(Z,s)}=G_s^{(Z)}-G_{s-1}^{(Z)}$. Since $H_0=h_Z$, its conjugation leaves $h_Z$ fixed, so $G_0^{(Z)}=h_Z\int w=h_Z$. Since $H_D=H$, $G_D^{(Z)}$ is precisely the choice of $g^{(Z)}$ from (b). Therefore the telescoping local-shell decomposition givesNo limiting interchange is needed: the sum has finitely many shells. Each $G_s^{(Z)}$ is supported within the union of supports appearing in $H_s$ together with $Z$, because its unitary time evolution acts trivially outside that neighbourhood. The individual shell terms need not commute with the full ground state projector; the full filtered sum does.

The filtered decomposition is not necessarily frustration-free. Commuting with $P_0$ makes $|{\varphi }_0⟩$ an eigenvector of each term; it does not make that eigenvector a lowest-energy state of each term. This is the distinction expressed by commuting with a ground projector does not imply frustration freeness.

For an explicit positive, two-local counterexample on three qubits, let $P_0^{(i)}=|0⟩⟨0{|}_i$ and $P_1^{(i)}=|1⟩⟨1{|}_i$, and use the two distinct interaction sets $\{1,2\}$ and $\{2,3\}$:Both are positive semidefinite and commute with their sum. The total energy on a computational basis vector isThus $H=h_{12}+h_{23}$ has unique ground state $|000⟩$, energy $1$, and spectral gap $1$. Since $[H,h_{12}]=[H,h_{23}]=0$, spectral filtering of Hamiltonian terms leaves both terms unchanged for every normalized filter:But $h_{12}|000⟩=|000⟩$, whereas $h_{12}|010⟩=0$. The global ground state therefore does not minimize $g^{(12)}$, proving that the resulting decomposition is not a frustration-free Hamiltonian.

There is a useful positive result if the original decomposition already is a frustration-free Hamiltonian. If $a_Z={\lambda }_{\min }(h_Z)$ and $h_Z|{\varphi }_0⟩=a_Z|{\varphi }_0⟩$, then $h_Z-a_{Z}I\ge 0$. Averaging its unitary conjugates with nonnegative $w$ gives $g^{(Z)}-a_{Z}I\ge 0$, while $g^{(Z)}|{\varphi }_0⟩=a_Z|{\varphi }_0⟩$. Thus filtering preserves existing frustration freeness; it does not create it for an arbitrary gapped Hamiltonian.