Paper VII: Quantum Seed Calculus

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1 
2 \section{Earlier compact derivations retained as alternatives}\label{legacy:section:001}\label{legacy:alternatives}
3 The following derivations retain valid mathematical arguments from the original manuscript. Their hypotheses and notation agree with the corrected finite core. The earlier abbreviated argument for error correction is replaced by the complete normalized and scaled recovery proofs above, and its substantive conclusions are included there.
4 \subsection{Duhamel proof of leakage from a fixed packet}\label{legacy:duhamel}
5 Let $H=H^*$ on a finite-dimensional complex Hilbert space, $P=P^*=P^2$, $Q=I-P$, and $t\in\mathbb R$.
6 \begin{proof}[Alternative proof of the commutator leakage bound]\phantomsection\label{legacy:proof:001}
7 Because $QP=0$,
8 \[
9 Qe^{-itH}P=Q[e^{-itH},P]P.
10 \]
11 The Duhamel identity gives
12 \[
13 [e^{-itH},P]=-i\int_0^t e^{-i(t-s)H}[H,P]e^{-isH}\,ds.
14 \]
15 Both unitary factors have norm one, so taking norms and using the integration interval's length $|t|$ yields
16 $\|Qe^{-itH}P\|\le |t|\,\|[H,P]\|$.
17 \phantomsection\label{legacy:proof:001:end}\end{proof}
18 \subsection{Proof of projected curvature with exterior forms}\label{legacy:exterior}
19 Let $P$ be a smooth idempotent field and extend the projected connection to differential forms with values in the range bundle by $\nabla\omega=P\,d\omega$. Thus $\nabla^2$ denotes the exterior covariant square, including the Lie bracket term in the exterior derivative when evaluated on vector fields.
20 \begin{proof}[Alternative proof of the curvature formula]\phantomsection\label{legacy:proof:002}
21 For a section $s$ with $Ps=s$,
22 \[
23 \nabla^2s=P\,d(P\,ds)=P(dP)\wedge ds.
24 \]
25 Since $ds=d(Ps)=(dP)s+Pds$ and $P(dP)P=0$, the surviving term is
26 \[
27 P(dP\wedge dP)Ps.
28 \]
29 \phantomsection\label{legacy:proof:002:end}\end{proof}
30 \par\Needspace{24\baselineskip}
31 \subsection{Restricted low-mode proof}\label{legacy:lowmode}
32 Let $h\ne0$, $x=kh/2$, and $0\le x\le1$.
33 \begin{proof}[Alternative proof in the original low-mode regime]\phantomsection\label{legacy:proof:003}
34 Since $0\le\sin x\le x$, the error is nonnegative. On $[0,1]$,
35 \[
36 \sin x\ge x-\frac{x^3}{6}\ge0.
37 \]
38 Therefore
39 \[
40 \sin^2x\ge\left(x-\frac{x^3}{6}\right)^2\ge x^2-\frac{x^4}{3}.
41 \]
42 Hence $x^2-\sin^2x\le x^4/3$. Multiplication by $4/h^2$ gives
43 \[
44 0\le k^2-\frac4{h^2}\sin^2(kh/2)\le\frac{k^4h^2}{12}.
45 \]
46 \phantomsection\label{legacy:proof:003:end}\end{proof}

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