Past exam of the mathematics course of the University of Cambridge / 2014 / iii / Paper 63 / 4

Use the Fourier transform convention $\hat{w}(E)={\int }_{\mathbb{R}}w(t)e^{itE}dt$. Since $w\ge 0$, it is real, so $\hat{w}(-E)=\hat{w}(E{)}^{\ast }$. The positive-frequency cutoff therefore also eliminates frequencies at or below $-\Delta$. Normalization gives $\int w=\hat{w}(0)=1$.

For spectral filtering of Hamiltonian terms, chooseIn an energy eigenbasis, its matrix elements areAt $i=j=0$ the frequency is zero, so the normalization preserves the ground-state expectation:The spectral filter is a positive weighted average of unitary conjugations; in particular the integral is bounded in operator norm by $\Vert h_Z\Vert$. We can choose it even without an extra assumed tail bound: the evenization of a nonnegative bandlimited filter proved in part (e) produces another admissible spectral filter with the same type of positive-time decay. Make that choice consistently in all the filtered terms and shell definitions below.

For an excited eigenstate, $E_i-E_0\ge \Delta$. The matrix-element identity in condition (i), together with the two-sided Fourier cutoff, givesThe reverse matrix element has frequency $E_0-E_i\le -\Delta$ and also vanishes. Equivalently, $A^{(Z)}$ is Hermitian because $h_Z$ is Hermitian and $w$ is real, so the two elements are conjugate. ThusThe spectral gap eliminates precisely the couplings needed to make the unique ground state an eigenvector of every filtered term. Couplings between excited states with smaller energy differences can remain; the spectral filter need not diagonalize the whole operator.

Sum the spectral filtering of Hamiltonian terms over the original finite decomposition. The total Hamiltonian operator commutes with its own evolution, soThe absence of all ground-to-excited matrix elements, proved in part (a), makes each filtered term block diagonal relative to $P_0=|{\varphi }_0⟩⟨{\varphi }_0|$ and its orthogonal complement. ThereforeA filtered term may act on the whole system, since unitary time evolution spreads its original support. Neither commutation with $P_0$ nor preservation of its ground-state expectation asserts that the global ground state minimizes each individual term. In particular, this construction does not generally turn a frustrated decomposition into a frustration-free Hamiltonian.

For a fixed original term, define $F_d(h_Z)=\int w(t)e^{it{H}_d}{h}_Z{e}^{-it{H}_d}dt$. The shell increment is exactly $A^{(Z,d)}=F_d(h_Z)-F_{d-1}(h_Z)$. Hence the telescoping local-shell decomposition givesUse the printed endpoint convention $H_0=h_Z$ and $H_D=H$. The first endpoint is $F_0(h_Z)=h_Z\int w=h_Z$, because $h_Z$ commutes with its own evolution, and the second is the operator chosen in part (b). ThusThere is a distance-convention issue in the endpoint assertion. The usual minimum distance between supports is zero for overlapping distinct interactions, so a literal $H_0={\sum }_{Y:d(Z,Y)=0}{h}_Y$ need not equal $h_Z$ or commute with it. To realize the stated $H_0=h_Z$, index the neighborhoods by distance between interaction terms: the central term has shell zero, and other terms begin in positive shells. For example, for distinct terms use one plus their support interaction distance. With the ordinary overlapping-support convention left unchanged, the exact formula instead begins with $F_0(h_Z)$, not necessarily with $h_Z$. The telescoping identity itself is valid in either convention.

Put $\lambda =2ks$. For $t\ge 0$, apply a valid Lieb-Robinson bound to each commutator in the supplied comparison of the two evolutions. For a disjoint shell at support distance $d$,In the usual uniform interaction-norm setting, truncation preserves the strength bound, so the same constants apply to $H_d$. The given polynomial shell-weight hypothesis and $e^{\lambda \tau }-1\le e^{\lambda t}$ now giveConsequentlyThe question explicitly supplies a bound for the full $H$; such a bound alone should not be assumed to transfer to $H_d$. A full-evolution comparison for truncated dynamics avoids that additional assumption. The Duhamel comparison formula givesNow use only the full-Hamiltonian bound, and sum the shell weights. Since ${\sum }_{q\ge d}{q}^{\alpha }{e}^{-\mu q}=O(d^{\alpha }{e}^{-\mu d})$, the errors for $j=d$ and $j=d-1$ are each bounded by $O(\Vert h_Z\Vert |t|d^{\alpha }{e}^{-\mu d+\lambda |t|})$. The triangle inequality comparing both evolutions to ${\tau }_t^H(h_Z)$ proves the same boxed estimate from the stated full-system assumption. Negative times follow by the reverse-time Duhamel formula; the commutator bounds use $|t|$.

For the interaction-term shells used to enforce $H_0=h_Z$, shell $d$ corresponds to support distance $d-1$ for distinct terms. Replace the displayed commutator estimate by the coarser bound with $e^{\lambda \tau }$, retaining the equal-time contribution for overlaps. Its distance factor is $e^{-\mu (d-1)}=e^{\mu }{e}^{-\mu d}$, and the fixed factor $e^{\mu }$ is absorbed by the big-O constant. This supplies the same claimed shell estimate without incorrectly using a zero equal-time bound for overlapping terms.

The filtered shell integral uses both signs of time, while the stated decay hypothesis controls only positive time. Here is a way to choose an even admissible spectral filter from the given one, rather than silently assume that its negative tail is controlled. Call the supplied spectral filter $f$. Reality and its Fourier cutoff imply that $\hat{f}$ is supported in $[-\Delta ,\Delta ]$. It is bounded and integrable in frequency, so Fourier inversion supplies a bounded continuous representative of $f$. That representative is real analytic, since its Fourier support is compact, and is not identically zero.

Define the evenization of a nonnegative bandlimited filter byThe normalizing integral is finite and positive: boundedness and integrability give finiteness, while a nonzero real-analytic nonnegative function cannot vanish on an interval, so the product is positive on some interval. The new spectral filter is even, nonnegative and normalized. Each factor has Fourier support in $[-\Delta /2,\Delta /2]$, and the convolution rule for the product therefore gives support in $[-\Delta ,\Delta ]$, including vanishing at the outer endpoints. Its positive tail is bounded byIt has the same required tail form, with a rescaled positive constant $\beta$. Since $w$ is even, its two-sided tail is twice its positive tail. The preceding parts use this chosen $w$ consistently.

For the almost-exponential locality of filtered Hamiltonian terms, split the integral defining $A^{(Z,d)}$ at $|t|=T_d$. Write $\lambda =2ks>0$ and choose $T_d=\mu d/(2\lambda )$. On the short-time part, part (d) and $\int w=1$ giveFor the long-time part, each conjugated operator has norm $\Vert h_Z\Vert$, so their difference has norm at most $2\Vert h_Z\Vert$. The two-sided spectral filter tail givesLet $c=\beta \mu /(2\lambda )>0$, so $\beta T_d=cd$. The logarithmic prefactor is bounded by $O(d^{\alpha +1})$, and the first, exponentially decaying contribution is asymptotically smaller than this almost-exponential contribution. ThereforeThis is an asymptotic statement for large $d$; small shells have the elementary bound $2\Vert h_Z\Vert$, avoiding the meaningless substitution $cd=1$ into the logarithmic expression. If $ks=0$, there is no dynamical spreading and the shell increments vanish. Filtering gives almost-exponentially decaying shells despite the filtered operator's potentially global support.