Coherent spectral minimum and a frozen Gibbs energy shift

Let $U=\exp (iH\mathrm{/}T)$. Its signed eigenphase is $\varphi =\lambda \mathrm{/}(2\pi T)$, with $\mathrm{\mid }\varphi \mathrm{\mid }<1\mathrm{/}(4\pi )$ because $\mathrm{\mid }\lambda \mathrm{\mid }\le K$ and $T=2K+1$. Thus the principal representative is injective on the promised spectrum. By [F7], choose a certified rational ${\pi }_{+}\in (\pi ,4)$ (for example $22\mathrm{/}7$) and let $M$ be the least dyadic integer at least $128{\pi }_{+}T$, so $128\pi T\le M<256{\pi }_{+}T$. Choose an odd $r\ge 8\ln (1\mathrm{/}\nu )+8$ using a certified rational logarithm upper enclosure of additive width at most one, with the least such odd choice giving $r=O(1+\ln (1\mathrm{/}\nu ))$. Doubling and integer comparisons give finite schedules as in [F3].

Prepare ${\mathrm{\Phi }}_d$ once. Apply $r$ QPE trials to its same first factor, with fresh phase registers; preserve all phase outputs and arithmetic history. In an orthonormal eigenbasis $u_j$, ${\mathrm{\Phi }}_d=d^{-1\mathrm{/}2}{\sum }_j{u}_j\otimes \overline{u_j}$. The second factor is untouched and supplies a uniform orthogonal index in analysis. Conditional on $j$, controlled evolutions preserve $u_j$ and each fresh phase register has the kernel of [F1]; their conditional joint law is its $r$-fold product. Thus the median estimates one eigenvalue repeatedly, rather than independently redrawing spectral eigenvalues. No eigenbasis knowledge is required; repeated eigenvalues use any orthonormal basis within their eigenspaces.

Set $\eta =1\mathrm{/}(32\pi T)$. Then $M\eta \ge 4$, so [F2] bounds one trial's circular error probability by $1\mathrm{/}6<1\mathrm{/}4$. On circular success its signed estimate has distance below $\eta$ from $\varphi$ and lies in $(-1\mathrm{/}2,1\mathrm{/}2)$: indeed $\mathrm{\mid }\varphi \mathrm{\mid }+\eta <1\mathrm{/}(4\pi )+1\mathrm{/}(32\pi )<1\mathrm{/}2$. Hence it is exactly the principal representative, and decoding by $2\pi T$ gives error below $1\mathrm{/}16$. This derives the trial bound for this actual $M$ directly, without substituting it into a least-$M$ theorem.

Conditional on an eigenindex, let $X_i$ indicate an accurate exact trial and $S={\sum }_i{X}_i$. By step 2.1 and step 2.2 the indicators are independent and each has success probability at least $3\mathrm{/}4$. Markov's inequality applied to $3^{-S}$ gives $\mathrm{Pr}(S\le r\mathrm{/}2)\le 3^{r\mathrm{/}2}\mathbb{E}{3}^{-S}\le (\sqrt{3}\mathrm{/}2{)}^r$. For the elementary exponential estimate, apply [F5] on $[1,8\mathrm{/}7]$: some $c\in (1,8\mathrm{/}7)$ satisfies $\ln (8\mathrm{/}7)=L(8\mathrm{/}7)-L(1)=1\mathrm{/}(7c)>1\mathrm{/}8$. Since $2\mathrm{/}\sqrt{3}>8\mathrm{/}7$, monotonicity of $\ln$ (also immediate from [F5] and the mean value theorem) gives $\ln (2\mathrm{/}\sqrt{3})>1\mathrm{/}8$, so [F6] gives $(\sqrt{3}\mathrm{/}2{)}^r\le \exp (-r\mathrm{/}8)\le \nu \exp (-1)$. Applying [F5] on $[1,2]$ gives $\ln 2=1\mathrm{/}{c}_2<1$ for some $c_2\in (1,2)$, hence [F6] gives $\exp (-1)<1\mathrm{/}2$. The conditional failure is therefore below $\nu \mathrm{/}2$. An inaccurate odd median requires at least half the trials to be inaccurate. Therefore the exact decoded median has error at most $1\mathrm{/}16$ with conditional failure below $\nu \mathrm{/}2$. This is the explicit conditional median argument also used in [F3].

For each phase output $y$, set $z=y\mathrm{/}M$ if $y\mathrm{/}M<1\mathrm{/}2$, and $z=y\mathrm{/}M-1$ otherwise. Approximate $2\pi Tz$ using a certified rational approximation to $\pi$ of error at most $1\mathrm{/}(32T)$, followed by rational rounding of error at most $1\mathrm{/}32$, and clip to $[-K,K]$. Since $\mathrm{\mid }z\mathrm{\mid }\le 1\mathrm{/}2$, the first error is at most $1\mathrm{/}32$ and the total is at most $1\mathrm{/}16$. Clipping cannot increase distance from the true eigenvalue. The median of entries each perturbed by at most $h$ changes by at most $h$: at least half the new entries remain below the old median plus $h$, and at least half remain above the old median minus $h$. Thus the rational median label has error at most $1\mathrm{/}8$ on the event of step 3.1. Retain all rounded values, comparison bits and scratch for reversible computation.

Allocate total operator-error sum at most $\nu \mathrm{/}4$ to the implemented controlled evolutions and other synthesis slots of a complete spectral sampler. With $S_0$ slots, error at most $\nu \mathrm{/}(4S_0)$ per slot suffices. Telescoping unitary products bounds conditional output-vector error by that sum, and measured total variation by at most twice the sum. Together with step 3.1, the bad-label probability is at most $\nu$ at every spectral index. Exact reference preparation is assumed; an approximate reference may instead be covered by a separately allocated whole-algorithm hybrid since its error can destroy exact index preservation. Retain every QPE target, phase register, arithmetic register and coin, so reversing the whole finite circuit, including reference preparation, provides its inverse. Controlled implementations are supplied or constructed gatewise with their actual costs.

The uniform-index cleaning proof of [F4] is basis independent: in analysis its per-index rotations can act in the orthogonal $\overline{u_j}$ reference sectors from step 2.1. No such unknown-basis rotation or spectral projector is implemented. The declared $\nu$ is smaller than $(\delta \mathrm{/}(16Q_{\mathrm{c}\mathrm{a}\mathrm{p}}){)}^2$. Apply [F4] with $N=d$. With probability at least $1-\delta$, it returns a good rational score $s$ not strictly larger than every good label at any one index. Its goodness gives $s\ge {\lambda }_{\min }(H)-1\mathrm{/}8$. At an index of minimum eigenvalue, every good label is at most ${\lambda }_{\min }(H)+1\mathrm{/}8$; a larger selected score would contradict nondomination. Thus the stated two-sided bound on $s$ holds without a probability lower bound on any individual label.

In power-evolution access, a QPE trial uses controlled $\exp (iH{2}^{\mathrm{\ell }}\mathrm{/}T)$ for $0\le \mathrm{\ell }<{\log }_{2}M$. Their total physical evolution time is ${\sum }_{\mathrm{\ell }}{2}^{\mathrm{\ell }}\mathrm{/}T=(M-1)\mathrm{/}T<256{\pi }_{+}$, and each individual time is bounded by this constant. Thus each trial has total physical time $O(1)$. With repeated base-$U$ access there are instead $M-1=O(K+1)$ base calls; the two models have distinct costs. One must not multiply the power-time sum by another factor $M$. With supplied evolution cost $C_H(t,\epsilon )$, the precise charge is ${\sum }_{\mathrm{\ell }}{C}_H(2^{\mathrm{\ell }}\mathrm{/}T,{\epsilon }_{\mathrm{\ell }})$ per trial with slot errors summed as in step 5.1, plus QFT, control and decoding work.

Measure $s$ once and put $\tilde{\lambda }=s-1\mathrm{/}8$. Step 6.1 implies ${\lambda }_{\min }(H)-1\mathrm{/}4\le \tilde{\lambda }\le {\lambda }_{\min }(H)$, so ${\lambda }_{\min }(G)={\lambda }_{\min }(H)-\tilde{\lambda }+3\mathrm{/}2\in [3\mathrm{/}2,7\mathrm{/}4]$. In particular $I⪯G$, its smallest eigenvalue is at most two, and the triangle inequality gives the finite norm bound in the Statement. Optionally $\lfloor 8s\rfloor \mathrm{/}8-1\mathrm{/}8$ yields the weaker lower estimate greater than ${\lambda }_{\min }(H)-3\mathrm{/}8$ and hence ${\lambda }_{\min }(G)<15\mathrm{/}8$, still sufficient. The formulas include $K=0$ and rational rounding ties.

Freeze this measurement, the shift, the norm certificate and all ensuing coefficient encodings, cutoffs and synthesis precisions once per MMW round, before making coherent indexed trace circuits. These later circuits use one fixed $G$; they do not rerun or remeasure the spectral minimum. Since an identity energy shift multiplies numerator and partition function by the same positive scalar, $e^{-G}\mathrm{/}\mathrm{tr}(e^{-G})=e^{-H}\mathrm{/}\mathrm{tr}(e^{-H})$. The spectral good-event failure remains part of the round's confidence budget.

The minimum theorem charges $O(\sqrt{d}(1+\ln (1\mathrm{/}\delta )))$ complete sampler, inverse and reflection accesses. Each sampler includes its reference preparation at declared cost $C_{\mathrm{\Phi }}$, $r$ QPE trials and all coherent decoding, comparisons, median arithmetic and history storage; inverses reverse all of this work and include reference unpreparation. These are charged quantum gates and workspace, not free classical preprocessing. If an independently proved sparse-simulation theorem bounds the sum in step 6.2 by $\tilde{O}(d_{H}K)$, substitute that cost to obtain $\tilde{O}(\sqrt{d}{d}_{H}K)$ evolution cost, adding preparation and every other charged coordinate; at $K=0$ retain any supplied nonzero overhead. This conditional substitution supplies neither an unproved simulation theorem nor a total-gate guarantee, and introduces no $K^2$ factor. The finite $Q_{\mathrm{c}\mathrm{a}\mathrm{p}}$ is chosen first, then $\nu$, then the confidence and synthesis budgets, so no unknown stopping time sets precision retrospectively. Together with step 7.1 this proves the claims.