Past exam of the mathematics course of the University of Cambridge / 2015 / iii / Paper 67 / 1

Let $R=X(l)$ and $C$ be its complement, with Hilbert-space dimension $d_C$. Use Haar twirling conditional expectation to defineThis is an operator supported on $R$. The normalization $1/d_C$ is essential: the partial trace alone would not fix an operator already supported on $R$.

Subtract the integrand from $A_X(t)$ and use the operator norm triangle inequality. Since $A-UA{U}^{†}=[A,U]U^{†}$ and every unitary operator has norm one, the assumed Lieb-Robinson bound, applied to a unitary supported on the entire complement, givesThe factor $\min (|X|,|C|)$ was bounded by $|X|$; there is no sum over sites and hence no volume-dependent prefactor.

ChooseFor $t,l\ge 0$, $e^{2kst}-1\le 2kst{e}^{2kst}$, and thereforeThus $A_{X(l)}(t)=A_R(t)$ is the desired Lieb-Robinson localization by Haar twirling. The displayed choice assumes the usual positive constants $\mu ,s$; if the interaction bound has $s=0$, the error is identically zero and any positive $v$ works. For negative times the same argument uses $|t|$.

Write $P_0=|{\psi }_0⟩⟨{\psi }_0|$ and $Q_0=I-P_0$. Frustration freeness implies that every local orthogonal projection fixes $|{\psi }_0⟩$ on both the left and right. HenceThe spectral gap makes $\mathrm{supp}H=Q_0\mathcal{H}$. The supplied detectability lemma givesConsequentlyNo assumption that $K$ is Hermitian or normal is needed.

A stronger estimate will keep this same $\alpha$ when we count the two projection layers in (ii). On $Q_0\mathcal{H}$, put $P=P_{\mathrm{odd}}-P_0$ and $Q=P_{\mathrm{even}}-P_0$. They are orthogonal projections, with $\Vert PQ\Vert \le q$. For $m\ge 1$,so the powers of a product of two orthogonal projections satisfyThis extra estimate uses the projection structure, rather than generic submultiplicativity alone.

Fix an integer radius $m\ge 0$ and put $r=\lfloor m/2\rfloor$. Starting with $A_x$, scan the $2r$ layers of $K^r$ from right to left. In each layer retain precisely those bond orthogonal projections whose supports meet the current support; multiply them onto the current operator and enlarge that support to include their bonds. Discard every other projection: it commutes through the current operator and acts as the identity on $|{\psi }_0⟩$, by frustration freeness.

If $L_r$ denotes the ordered product of the retained projections, defineEach layer expands the support by at most one lattice spacing. Therefore the local projector cone in a frustration-free chain lies within radius $2r\le m$ of $x$, andThe last equality holds because every retained projection also fixes the ground-state bra.

For $r\ge 1$, the stronger estimate from (i) givesFor $r=0$, choose $A_{x(m)}=A_x$; the error is at most $\Vert A_x\Vert$, so the same bound holds. HenceHere $\alpha$ is exactly the value in (i), and the prefactor is independent of $m$ and chain length. Counting $K^r$ as $r$ rather than $2r$ layers would give an incorrect support claim; the powers of a product of two orthogonal projections are what avoid a lost factor of two in the decay rate.

Let $d=|x-y|\ge 1$ and take $m=d-1$, so the local projector cone in a frustration-free chain for $A_{x(m)}$ does not contain $y$. It follows that $[A_{x(m)},B_y]=0$. The exact bra identity from (ii) and this commutation giveSince $P_0{A}_x|{\psi }_0⟩=⟨A_x⟩|{\psi }_0⟩$, subtracting the product of expectations yieldsThus the exponential clustering from layered projectors bound isThe connected correlation function therefore has correlation length at most $1/\alpha$ in lattice-spacing units. In particular $\alpha =\Delta /6+O({\Delta }^2)$ for a small spectral gap. The same-site case is covered by the elementary bound $2\Vert A_x\Vert \Vert B_y\Vert$.