Copying and retention for the exact loaded wave

E Copying and retention for the exact loaded wave

All estimates in this appendix concern the same normalized enlarged wave $\Psi$ and its original actual law. Write $C=270000/67499$, $m=2^{52}$, $h=1/128$, $M=10\,\mathrm{kg}$, $v=1\,\mathrm{m/s}$, $T_0=0.03\,\mathrm s$, $\ell_0=10^{-4}\,\mathrm m$, and $\mu=\mu_Z=0.01$. Radial and pointer coordinates are dimensionless; $S$ and the elapsed time $t$ below have physical units. The origin $t=0$ is the handoff at laboratory time $-0.002\,\mathrm s$. The transferred entrance estimates are those proved in Section B.5; no law or wave is assigned anew there.

E.1 Gaussian derivatives and the complete clock residual

Put $b(s)=24B_9(s/w)$, $w=0.003$, with the constant extensions specified in the model. Direct differentiation gives the following rational ceilings $b_j\geq\|b^{(j)}\|_\infty$:

Indeed $B_9'=630x^4(1-x)^4$; the first three derivative bounds follow from $x(1-x)\leq1/4$ and $|1-2x|\leq1$, and the coefficient sum of $B_9^{(4)}$ is $1360800$. These bounds apply on both constant extensions because $B_9$ has four matching endpoint derivatives.

For the exact sector-one pointer put $y=Z-b$ and

Then $g_b'=(cy+d)g_b$ and

In particular the scalar phase is retained. Define

The Gaussian moments $\|y^j g_b\|^2=(2j-1)!!/2^j$ prove $\|g_b^{(j)}\|\leq g_j$ for $j=1,2$. The sector-zero pointer is constant, so the same ceilings apply to both sectors, uniformly in the normalized qubit input.

The characteristic comparison through readout is

Here $u_i$ is the exact smooth-activation radial family, continued backwards by the quiet Hamiltonian at negative ages. On the entire compact support of $\phi$, the preparation phase potential has ended. Substitution into the boosted Schrödinger equation leaves precisely the residual $\pm\hbar^2G_{SS}/(2M)$. The sign is immaterial for the norm estimates, but both pointer and activation derivatives in $G_{SS}$ must be included.

Let $N_0=1$, $N_1=V+1$, $N_2=1001(V+1)^2/1000$ and $N_3=1001(V+1)^3/1000$, where $V=2^{35}$; these are the energy-graph bounds proved in Section C.3. Set

The form inequality $H_\gamma\geq p_r^2/(2m)-F_0$, applied to $H_\gamma^ju_i$, proves $\|\partial_rH_\gamma^ju_i\|\leq K_j$. Only the third Hamiltonian graph is needed. Since

the product-rule bounds, including the gate derivative, are

For the normalized mollified clock packet, convolution contraction and its normalization factor at most $1.000001$ give

as ceilings on $\|\phi'\|$ and $\|\phi''\|$. Therefore

bound $\|G_{SS}\|$ and $\|\partial_rG_{SS}\|$.

Write $E=\Psi-G$. Its entrance norm $e_h$ is $D_h$ from Section B.5 plus the backward-quiet discrepancy. For that discrepancy, if $e=(e^{-\ii\sigma H_F}-1)\chi$ and $|\sigma|\leq1/8$, then

The quiet tail estimates in Section C.4 make its physical momentum correction less than $10^{-100}\,\mathrm{kg,m/s}$. Let $P_{r,h}$ be the old ordinary radial error momentum, the loaded-minus-old momentum from Section B.5, and this quiet correction added together.

For completeness the global radial force used in propagating $E$ is explicit. Put $m_{\rm phys}=m\hbar T_0/\ell_0^2$, $\omega=1/(mT_0)$, $R=0.201\,\mathrm m$, $a=140\,\mathrm{s}^{-1}$, $a'=43600\,\mathrm{s}^{-2}$ and $F_{\max}=2R^2$. Define

These follow by differentiating the displayed preparation, weak-trap and capped detector potentials; all remote preparation terms remain. The writer has zero radial derivative. Unitary Duhamel, followed by the ordinary radial momentum commutator, consequently gives

The odd extension at the Dirichlet origin has no boundary term. Approximation on the common smooth core and these uniform bounds justify the differentiated evolution without a fourth radial graph. At $T_c=0.0404\,\mathrm s$, denote the right sides by $e_*$ and $P_{r,*}$ and put $p_*={\ell_0P_{r,*}}/({\hbar mh})$. Evaluation gives

E.2 A closed clock and annihilator estimate through the hold

The exact boosted generator contains $p_s+\epsilon p_s^2/2$, $\epsilon=\hbar/(Mv^2T_0)$. Define $\mathcal A=\partial_Z+Z-\Pi_1(b+\ii\mu b')$. Since the radial potentials commute with this operator, its complete commutator is

For example $[\mathcal A,H_{\rm ptr}]=(\mathcal A+\Pi_1c)/\mu -\Pi_1F$ and $[\mathcal A,p_s]=-\ii\Pi_1c'$, where $c=b+\ii\mu b'$; $c/\mu-F-\ii c'=0$ cancels the prescribed driving terms. The clock Laplacian supplies both terms on the right of (115).

Let $P(t)=\|-\ii\hbar\partial_S\Psi(t)\|$ be the boosted clock momentum, $n(t)$ the moving left-collar norm of width $g=0.0005\,\mathrm m$, and $R_Z(t)=(\|Z\Psi\|^2+\|p_Z\Psi\|^2)^{1/2}$. The entrance estimates allow the conservative values

in SI units where applicable; the much sharper actual entrance $A_h=\|\mathcal A\Psi(0)\|$ is retained. Set

The Heisenberg equations for $(Z,p_Z)$ are a rotation with bounded forcing $\Pi_1F$; its variation-of-constants formula gives $R_Z(t)\leq2+F_*t/T_0$, without exponentiating the oscillator frequency. The translating collar and clock commutator give

Here $F_p$ and $F_a$ have units of force. Explicit ceilings from the original potentials, with $a''=70141750\,\mathrm{s}^{-3}$, are

The $C_*'$ term retains the clock force of the scalar compensation. All inequalities apply to the full wave, including remote clock tails. The oscillator estimates extend from finite-energy approximants by the form-domain bounds just obtained.

For $q=0.0002\,\mathrm{kg,m/s}$ and $k=0.04\,\mathrm{s}^{-1}$, $Y=P+qn$ satisfies $Y'\leq kY+f_0+f_1t$, where

Indeed $q/(Mg)=F_p/q=k$. With $T=0.092\,\mathrm s$ and $E_T=(1-kT)^{-1}\geq e^{kT}$, positive integration proves

The last line is unitary Duhamel applied to (115). The resulting ceilings are

Equivariance and Cauchy–Schwarz imply $\E_{|\Psi|^2}\int_0^T|\dot S-v|\dd t\leq\mathcal P/M$. Thus actual-law domination and Markov's inequality bound the failure of the entire-path deviation $\int|\dot S-v|\leq10^{-5}\,\mathrm m$ by $C\mathcal P/(10^{-5}M)<1.067\times10^{-13}$. The handoff core $|S_h-0.102|\leq0.000839\,\mathrm m$ has failure at most

To see this, normalize the unsmoothed $\cos^4$ packet on its $0.001\,\mathrm m$ half-width. Its squared density has coefficient $64/35$ after scaling to unit half-width; the two edges give the factor $128/35$. On either edge use $\sin x\leq x$, integrate the eighth power, and enlarge the edge width by the smoothing radius $10^{-10}\,\mathrm m$. Positive convolution and Jensen preserve this two-sided bound, with the displayed normalization factor. The $D_h$ triangle then uses the exact handoff wave, rather than replacing its marginal.

On the intersection of these two good events, the first possible crossing of $S_{\rm on}$ is after $0.036601\,\mathrm s$, and the last possible completion of the pulse is before $0.038389\,\mathrm s$. More directly, the pathwise bound $|S_t-(0.104+vt_{\rm lab})|\leq0.000849\,\mathrm m$ implies $S_t\geq0.14154\,\mathrm m$ for every $t_{\rm lab}\in[0.0384,0.09]$. This proves completion throughout the whole hold, including the exclusion of later returns. The same two exceptional events are charged only once below.

E.3 The actual earlier label and a positive capture event

Let $H_r(r)\in[0,1]$ be the soft radial classifier, agreeing with the sharp label $\ell(r)$ outside its bands $B$ and having $|H_r'|\leq10/h$. Put $I_0=[-8,8]$, $I_1=[16,32]$ and

For each absolutely continuous exact trajectory, with witness and readout at laboratory times $t_a=0.0366$, $t_c=0.0384$,

Outside $B$, $H_r(r_a)$ is the binary earlier label; replacing this coefficient by $H_r(r_c)$ changes the mismatch by at most its total variation. This proves the inequality without assigning a sampled internal sector.

Throughout this copy interval the entire comparison packet has conditional age in $[71/60,4/3]$. Use the $B_9$-taper lens expressions $D_9,Q_9$ of (86), evaluated at $4/3$, and denote these by $D,Q$; set

They give $D<0.007640094$, $Q<262597634.037$ and $q_r<0.003820673$. The contraction scale is at most $\cos(\sigma-1/300)$. Since $\sigma-1/300\geq59/50$ and $\cos(59/50)<2/5$, the compact comparison and its radial derivative vanish on the bands $[h/5,3h/10]$ and their translates. Hence $\|\one_BG\|\leq D$ and $\|\one_B\partial_rG\|\leq mhq_r$. Expanding the exact current of $G+E$, taking its modulus before integrating any hidden variable, gives

For example the four terms before division by $m$ are $D(mhq_r)$, $D\|\partial_rE\|$, $e_*(mhq_r)$ and $e_*\|\partial_rE\|$.

On the completed-clock region, let $P_g$ project in each internal sector onto the normalized real Gaussian centered at $0$ or $24$. The oscillator ladder spectrum proves the fibre inequality

Both operators commute with completed-clock localization. Thus the localized excited component has norm at most $A_*/\sqrt2$. The ground-state wrong-inner probability is $q_I=\operatorname{erfc}(8)$, and contraction and the norm triangle give a pointer cost at most $C(\sqrt{q_I}+A_*/\sqrt2)^2$. To combine it with the radial mismatch, use the pointwise inequality

and sum against the exact orthogonal component densities. The square-root multiplier of the first term kills the compact sector-$i$ lens comparison, so its total contribution is at most $C(D+e_*)^2$. Together with the initial band and (121), this proves the copy cost, apart from the already identified exceptional events,

Here $10(t_c-t_a)/T_0=3/5$. Each endpoint and variation refers to the same exact trajectory and its own earlier radial label.

E.4 Absolute current over the entire holding interval

At each full configuration, including $S,r,Z$, the exact identity is

The second term has not been discarded: it vanishes on every visited point of a good clock path because the pulse is completed there. For $R_0=[-10,10]$, $R_1=[14,34]$, a path starting in $I_i$ and leaving $R_i$ crosses one of the four width-two bands $[-10,-8]$, $[8,10]$, $[14,16]$, $[32,34]$. Linear cutoffs of slope $1/2$ on these disjoint bands require unit variation for such an exit. Restricted equivariance, actual-law domination and then Cauchy–Schwarz therefore give

All hidden-coordinate moduli are taken before integration. The variation counts recrossings and every time of the interval; an endpoint estimate is not being substituted for a path estimate. The global annihilator norm bounds the restricted current, so no unfinished part of the comparison wave has been removed.

E.5 The loaded radial exit and the evaluated sum

For the radial barrier increasing from zero at $0.099\,\mathrm m$ to one at $0.100\,\mathrm m$, the slope is $1000\,\mathrm m^{-1}$. Let

Here $d_j$ are the half-line Gaussian derivative norms, equivalently the norms of its unitary odd extension. For an integrable absolutely continuous $f$ the grid sum obeys

This follows by comparing the value at the chosen point of each cell with its integral and then summing the fundamental theorem of calculus bounds. With $f=|\chi_{100}^{(j)}|^2$, $\int|f'|\leq2d_jd_{j+1}$ proves the folded bound $B_j^2$. The periodic reference has mass $h/2$ per cell and circle norm one. Its kinetic energy starts at zero and increases by at most $V$ under the nondecreasing gate, proving $\|\partial_ru_{{\rm ref},i}^{\gamma}\|_{L^2(0,2)}\leq mh/2$. Consequently the norms of $\sqrt2\chi_{100}^{(j)}u_{{\rm ref},i}^{\gamma}$ and $\sqrt2\chi_{100}'\partial_ru_{{\rm ref},i}^{\gamma}$ are bounded by $B_j$ and $B_1mh/2$, respectively. For the auxiliary $\sqrt2\chi_{100}u_{{\rm ref},i}^{\gamma}$, the product-rule residual consists respectively of the second Gaussian derivative, the cross derivative, the outer cutoff tail and the weak trap, and their sum is exactly $\beta$ above. For the decreasing Gaussian density beyond $990$, periodic cell masses give a folded tail at most $\operatorname{Tail}(990)+h\rho(990)$. Writing $z=9.9$, integration by parts gives

The radial density is $4y^2e^{-y^2}/\sqrt\pi$ in this scaled coordinate. Thus $\operatorname{Tail}(990)<21/7^{49}$ and $\rho(990)<4/7^{49}$ in detector coordinates, by $e^2>7$ and $\sqrt\pi>1$. Since $h<1/4$, their folded sum is below $T_G^2$. Unitary integration for age at most $4/3$ yields the full-wave bound

Negative quiet ages obey the same larger bound by the quiet estimate used in (114); pointer tensor factors have norm one.

The same loaded clock estimate gives a uniform left-collar bound $n_*=n_h+\mathcal P/(Mg)$. Thus the exact radial momentum starts below $P_{r,0}=\hbar\sqrt{3/2}/0.01+P_{r,h}$ and grows at most at rate $F_r=W_r+Q_r+V_{p,r}n_*$. There is no writer radial force. The initial outer tail plus the absolute soft-barrier current prove

This bounds initial outside configurations as well as every later exit through readout under the loaded wave.

For a directly evaluable decomposition, define

The inner-tail replacement follows from $\operatorname{erfc}(8)<e^{-64}/(8\sqrt\pi) <1/(14\,7^{32})$. All constants in these expressions have now been specified by rational operations, square roots, $\pi$, and the lens functions $D_9,Q_9$ in (86). Outward evaluation gives

Their sum is strictly below $0.000552421956$. The event proof is (119), the positive endpoint bound, (125), and the union with the three common exceptional events (clock core, clock deviation, radial exit). When combined with the earlier-label theorem, those same events are counted once. None of the estimates supplies an independent internal-sector variable, a new actual-law marginal, or an additional physical source beyond the stated effective Hamiltonian.