Preparation estimates and transfer to the enlarged Hamiltonian

B Preparation estimates and transfer to the enlarged Hamiltonian

All momenta in this appendix have the constant Galilean carrier removed. A superscript $g$ additionally removes the bounded preparation phase $\theta$. Thus an ordinary clock momentum here means $p_S-Mv$ in the laboratory frame. We give the finite coefficient construction as well as its numerical enclosures; the latter are consequences of the construction, not extra hypotheses.

B.1 Coefficient conventions and the Gaussian residual

Write $T=0.104$, $T_0=0.03$, $L_i=10^{-4}$, $L_f=10^{-2}$, $M=10$, $v=1$, $K=100$, and $\lambda=2^{52}T_0/L_i^2$, so that $m=\hbar\lambda$. Lengths and times in this appendix are in metres and seconds. Let $b_p$ denote the preparation dilation, to distinguish it from the writer displacement. Set $a=b_p'/b_p$, $f_*=0.201$, $F_*=0.080802$, $g_1=0.0005$, $g_2=0.00025$ and $\gamma_*=141$. Here $F(r)=2\int_0^r f$ is the bounded preparation profile; the subscript on $F_*$ in this subsection is unrelated to the writer force. Here $b_0(s)=1+99B_9(s/0.1)$ has its constant extensions, and $b_p$ is its convolution with the fixed $10^{-5}$-scale mollifier. The explicit clamped ninth-degree polynomial and its fixed positive mollifier give the derivative bounds

The sharper relative bounds $b_p'\le140b_p$ and $|b_p''|\le24000b_p$ follow from polynomial positivity before convolution. One finite verification is to express each of $140b_0-b_0'$, $24000b_0-b_0''$, $24000b_0+b_0''$ in Bernstein form on dyadic subintervals of $[0,1]$; bisect a subinterval if any coefficient is negative. All leaves are nonnegative by depth five. For reproducibility, if $P(x)=\sum_{j=0}^n p_jx^j$, its Bernstein coefficient of index $k$ on $[u,u+w]$ is

Positive convolution preserves these inequalities. Put

These bound $|a^{(j)}|$. Let $B_5$ be the largest absolute Bernstein coefficient on $[0,1]$ of $d^5b_0/ds^5$ and set

The normalized clock envelope has derivative norms bounded by

where $\sigma=0.001$, $q=1.000001$. The convolution scale is $10^{-10}$ and $q(1-10^{-10}p_1/q)>1$, which proves the normalization allowance. These are weak-derivative estimates for the compact envelope, not a band-limit assertion.

Let $A=2r\exp[-r^2/(2L^2)]/(\pi^{1/4}L^{3/2})$, $L=L_ib_p(S)$, and $G=\phi(S-S_c-vt)A$. Set $z=(r/L)^2$, $M_j=(2j+1)!!/2^j$ and $\nu_j=\sqrt{M_{2j}}$. Thus $\|z^jA\|=\nu_j$. For a polynomial $P=\sum c_jz^j$ define $\mathcal N(P)=\sum|c_j|\nu_j$. This coefficient norm is deliberately subadditive; signs of independently bounded derivatives cannot cancel. Use

Put $k=B_1$, $k_1=B_2+B_1^2$, $k_2=B_3+3B_1B_2+2B_1^3$, and

The three nonnegative coefficient bounds

give the residual norm coefficient $R_0=\sum_{j=0}^2r_j\nu_j$. For its radial derivative set

where the first summand is zero at $j=0$. This follows equally by the unitary radial reduction of the three-dimensional Gaussian gradient; its integration-by-parts identity includes the reduced amplitude's factor $r$. For the clock derivative set

These are obtained by differentiating the complete gauged residual

In particular, no derivative of a restored phase is added to this gauged residual.

Here are all coefficients needed to include its tails and the weak trap. Write $\omega_i=(\lambda L_i^2)^{-1}$, $q_2=m\omega_i^2/2$, $q_0=3\hbar\omega_i/2$, $q_{2S}=4a_0q_2$, $q_{0S}=2a_0q_0$, $r_c=0.201$, $c_1=5000$, and $\epsilon_{20}=10^{12}/7^{100}$. Set

The complete residual rates are

The Gaussian tail recurrence $\int_{20}^\infty y^n e^{-y^2}dy =20^{n-1}e^{-400}/2+(n-1)\int_{20}^\infty y^{n-2}e^{-y^2}dy/2$ proves the common tail norm used here; $e^2>7$ makes its stated bound rational.

B.2 First moments and preparation current

In the gauge the velocity coefficients are $u_r=af$ and $u_S=v+\delta$, $\delta=ma'F/(2M)$. Put

If $W_0=\hbar2^{35}/T_0$ and $W_1=\hbar(2^{52}/256+5\cdot2^{35})/(T_0L_i)$, the remote activation forces are bounded by $F_r=W_1+Q_r$, $F_S=25000(W_0+Q_0)+Q_S$. Let $P_j=\|p_j(\Psi^g-G)\|$ and $D=\|\Psi^g-G\|$. Differentiating the symmetrized drift gives

The collar defining $n$ translates at $v+\|\delta\|_\infty$ on the right and $v-\|\delta\|_\infty$ on the left; its convective contribution is nonpositive. Both collars vanish on the initial stock. Their final full-one edges are respectively below $0.1035000001262$ and above $0.1004999998738$, so that they cover the remote activation and the left phase support respectively. Thus the force term $F_jn$ is valid throughout preparation, including the exact wave's tails.

For the sharper clock estimate take $\alpha=10^{-12}$ and

The three scaled row sums $a_0+Kd_r+F_r/\alpha$, $u_{rS}/K+d_S+F_S/(K\alpha)$ and $\alpha K/(Mg_1)$ are less than $141$. A positive supersolution gives

Using $e^{141T}<E_*$ is legitimate since $141T<15$ and $e<11/4$. These terminal majorants are increasing and apply at every earlier time.

For the smaller radial estimate and the preparation current retain also the mass-weighted system. Put $\ell=10^{-6}$,

Indeed the weighted momentum plus $\ell n$ has growth at most $141$: the drift row bound is $a_0+\sqrt{m/M}a_1f_*+d_S$, increased by $\ell/(\sqrt M g_1)$, and $(F_r/\sqrt m+F_S/\sqrt M)/\ell<141$. Consequently the ordinary radial error at the handoff is bounded by

The phase terms here are localized to the reflected collar; no phase is set to zero on an exact wave tail.

For the Gaussian rank let $k_r=6/(7L_i)$, $k_S=2a_0$, $c_r=k_r\sqrt{3/2}/L_i$, $c_S=k_S(p_1+a_0\sqrt{3/2})$. The exact material-rank identity and the bilinear current expansion give

The first term of $\beta$ bounds $K_S\delta$ pointwise; the second bounds the nonquadratic radial tail. The remaining current has absolute value before integration over any live coordinate. This proves $I_o<1.974548968\cdot10^{-19}$ and $\beta<2.638660\cdot10^{-13}$.

B.3 Second spatial moments and the second collar

Here is an explicit construction of every second-order residual coefficient. Continue $H_j$ by $H_0=1$, $H_{j+1}=(z-3/2)H_j-2zH_j'$, and take absolute coefficients before adding independently bounded terms. The positive Gaussian clock polynomials are

Here $|P|$ means coefficientwise absolute value. To evaluate radial norms, form $P_{j,r}(y)=(\partial_y-y)^r y^{2j+1}$ and set

Only indices at least $-1$ occur. For a factor $F/L^2$ and $r\le2$ replace this by

This is the product rule using $F\le r^2$, $|F'|\le2r$, $|F''|\le2$. Define

and $J_{n,r}^F$ by replacing $\mathcal R_r$ with $\mathcal R_r^F$. With $C_j^*=\lambda L_f^2a_j/2$ for $1\le j\le4$, the sources are

For example the last formula differentiates $G_{SS}+2i\theta_SG_S+i\theta_{SS}G$ twice and retains coefficients $2,5,4,1$. The nonquadratic and trap tails after two derivatives have Gaussian degree at most seven and total frequency coefficient below $10^{30}$; their momentum contribution is less than $\hbar^2 10^{30}\epsilon_{20}<10^{-100}$ in each row. This can also be checked by the general product-rule construction below.

For clarity we spell out the second-force coefficients. Set $c_2=1.28\cdot10^8$, $q_{t1}=q_{2S}r_c^2+q_{0S}$, $q_{t2}=(4a_1+16a_0^2)q_2r_c^2+(2a_1+4a_0^2)q_0$. Then

Put $\gamma_1=25000$, $\gamma_2=(2520/64)/(0.0001)^2$, and

Let $v_{jr}$ and $v_{jS}$ denote the component derivatives of $(af,ma'F/(2M))$. The second-derivative ceilings used below are

Write $d^{\rm div}_r=a_0\,4000+ma_2f_*/M$ and $d^{\rm div}_S=a_1+ma_3F_*/(2M)$ for the first divergence ceilings. If $U_{jk},V_{jk},D_{jk}$ denote the three rows of this table, all scalar coefficients of the Hessian comparison are

These follow directly by commuting two momenta through $\{u_j,p_j\}/2+Q_c+V_A$. For example the mixed commutator contributes $\hbar\sum_l\|u_{l,jk}\|P_l$ and $\hbar(\|\partial_j\operatorname{div}u\|P_k+ \|\partial_k\operatorname{div}u\|P_j)/2$; both occur above. The three homogeneous scaled row sums are below $282$.

With $Z_2=\max(\|p_r^2E^g\|,\|p_rp_SE^g\|/K, \|p_S^2E^g\|/K^2)$, nested clock cutoffs give

Indeed a cutoff supported where the first collar is one satisfies $\|\chi p_rE^g\|\le\sqrt{n_1\|p_r^2E^g\|}$ and $\|\chi p_SE^g\|\le\sqrt{n_1\|p_S^2E^g\|}+2\hbar n_1/g_2$. These inequalities are integration by parts, not finite propagation. Let $b_n=K/(Mg_1)$, $\gamma=141$, and define

For the last display, extend positive integrals to infinity after using $n_1(t)\le g_0t+b_ne^{\gamma t}(st^2/2+cdt^3/6)$. The comparison follows by setting $y=e^{-\gamma t}\sqrt{Z_2}$: the sum of the separate positive supersolutions for $y'\le A+B/(2y)$ is a supersolution, with zero handled by regularization. The factor $e^{\gamma T}$ is enclosed by its 90-term Taylor sum and the remaining positive geometric tail with ratio $\gamma T/91$; its upper enclosure is squared when evaluating $e^{2\gamma T}$.

The second collar amplitude and the physical phase restoration are

Both reflected second collars fit before the relevant remote support: their final edges are below $0.103750000127$ and above $0.100249999873$. These same increasing majorants hold for all preceding times. In particular the ordinary derivatives on the writer support are bounded by $L_S,L_r$, because both the helper and the phase vanish there.

B.4 Fourth spatial closure and its complete coefficient ledger

Only ordinary spatial derivatives are used here. There is no assumption that the unmollified radial reference lies in the domain of a fourth power of its Hamiltonian. Use the commuting scaled momenta $D_r=p_r$, $D_S=p_S/K$, and $Z_j=\max_{a+b=j}\|D_r^aD_S^bE^g\|$.

The higher coefficient construction is finite. Define $q_j=b_p^{(j)}/b_p$. Starting with $P_0(q)=q_1$, form

Evaluate the absolute coefficients at

to obtain $\bar a_j\ge|a^{(j)}|$. For $j\le4$ these agree with the bounds above. The last two entries use convolution of $b_0^{(5)}$ with the first and second derivative of the mollifier. Similarly set $p_5=4p_4/10^{-10}$, $p_6=300p_4/10^{-20}$. The normalized unit bump has derivative polynomials generated by

Its normalization is greater than $1/4$; hence its $j$th derivative supremum is bounded by $4\sum_{l,k}|[x^lu^k]U_j|\,k^k/(8/3)^k$, with the $k=0$ factor defined as one. This proves the quoted mollifier bounds (the first derivative also uses unimodality), as well as all tent jets below.

Let the positive partial Bell coefficients be

All unspecified entries are zero. For $n=0$ put $\mathcal A_0=1$; otherwise $\mathcal A_n=\sum_{k=1}^n\mathcal B_{n,k}|H_k|$. For $\varepsilon=0,1$, define the computable Gaussian norm majorant

For the higher-core calculation one may use $\pi\le22/7$ in $p_j$ and $\hbar\le\bar h=1.055\cdot10^{-34}$. The fourth physical source coefficient is bounded by

where a binomial coefficient outside its range is zero. Direct rational arithmetic and outward square roots give $s_4<7.141\cdot10^{-127}$. The physical phase-square cancellation in the gauge is essential to this small coefficient.

We next give a complete way to check the nonquadratic tail bound, including higher derivatives of $F-r^2$. Let $b_j$ denote the bump derivative suprema just constructed and use

The disjoint jumps of the unsmoothed tent give these bounds. The scaled velocity derivative sum at order $j$ is explicitly

It gives $V_2<560001$, $V_3<1.400001\cdot10^9$, $V_4<3.669751\cdot10^{13}$, $V_5<2.786850\cdot10^{18}$.

For an entirely algebraic tail check replace every Gaussian norm in $\mathcal J_{n,r}^{(0)}$ by the absolute sum of its polynomial coefficients; call the result $\mathcal C_{n,r}$. Keep powers of $y$ as formal factors. The derivatives of $F-r^2$ have coefficient ceilings $\Delta F=(2,2,4,2f_2,2f_3)$, and those of $f-r$ have $\Delta f=(2,2,f_2,f_3,f_4,f_5)$. The unit logistic cutoff has order-$j$ derivative bounded by

To prove it on $x\le1/2$, put $u=1/x$: the exponent is at most $-u+2$, its $j$th derivative at most $2j!u^{j+1}$, and apply the Bell product rule and $\max u^ne^{-u}\le n^n/(8/3)^n$. Symmetry handles the other half. Set $t_0=1$ and $\tau_j=t_j/0.001^j$. For $p=2,4$ define $J_p(0)=1$, $J_p(n)=\sum_{k=1}^n\mathcal B_{n,k}p^k$. With $A_2=4\cdot10^{-41}$ and $A_0=1.2\cdot10^{-48}$, put

The three-entry sequence is zero outside $r=0,1,2$. A bound for the sum of all fourth-derivative tail frequency coefficients is the following finite positive sum, with $r=4-n$:

This is simply the product rule applied respectively to the two phase terms, transport tail, and $Q-Q_c$ in (20). Taking absolute coefficients before forming $\mathcal C$ prevents cancellation between independently bounded clock jets. As a rough independent size check, $p_j\le10^{9j}$, $\bar a_j<10^3 10^{9j}$, and Gaussian mixed coefficients through order six are below $10^{12}10^{9j}$. The transport term has at most five derivative slots and fewer than $10^5$ product summands, giving $10^{3+3+12+45+5}=10^{68}$. In the phase term $\hbar\theta$ replaces the large phase coefficient by the physical mass: six derivative slots give at most $10^{-12+3+3+12+54+5}=10^{65}$. The pure $G_{SS}$ term is smaller. The explicit positive sum above also checks the weak potential term. No nonzero tail is discarded.

The degree is at most twelve. Integration by parts gives

Thus the additional fourth physical source is $s_{\rm tail}=\bar h^4 10^{90}/7^{100}<10^{-130}$. The same finite product rule for the quiet potential gives, for total order $j\le4$,

For an explicit verification use the preceding $J_p(n)$ and cutoffs, replace $(L_f^2,2L_f,2)$ by $(1,1,2)$ on the cutoff support, and evaluate $\sum_{a=0}^n\sum_{b=0}^r\binom na\binom rb \tau_a\tau_b[A_2J_4(n-a)(1,1,2)_{r-b} +\one_{r=b}A_0J_2(n-a)]$ for $n+r=j$.

For the remote potential define $W_j=\bar h\,w_j/(T_0L_i^j)$, where

Set

These bounds follow by differentiating the displayed capped radial potential and activation polynomial; the last two deliberately loose bounds require only its bounded fourth weak derivative. All remote supports lie at $S\ge0.104$.

Add two smooth analytic cutoffs of width $g=10^{-4}$, with final transition intervals $(0.103751,0.103851)$ and $(0.103851,0.103951)$. They fit between the second old collar and activation. Their first two derivative bounds are $5/g$ and $128/g^2$. Put $a_c=\bar h/(Kg)$. With $n$ the second-collar amplitude, integration by parts and interpolation give local derivative bounds

For example the squared second-derivative norm is at most $nZ_4+20a_cn\sqrt{Z_2Z_4}+306a_c^2nZ_2$; its square root is bounded as displayed since $306<18^2$. The global interpolation $Z_3\le\sqrt{Z_2Z_4}$ moves one commuting derivative between the two factors in its squared norm. Odd extension removes the radial boundary term.

The useful time envelopes, with $Z_{1T}=B_T$, are

Here $D_T$ is used only in the fourth-order majorant, not to replace the sharper handoff norm in the final probability accounting. Writing $Z=Z_4$, four commutations yield

The factors $8,7,3,1/2$ in the drift terms are the sum of the first-order drift and its divergence product-rule coefficients. Putting $y=e^{-\gamma t}Z^{1/4}$ and adding the separate positive supersolutions gives

This includes the nonlinear third-derivative coupling; it is not an exponential multiplying a terminal forcing value.

For completeness all integrals in this expression can be evaluated by one finite rule. For $0<p\le1$ use subadditivity on the three terms of $n(t)$; combine with $Z_2(t)^q$ and the integrating factor, and extend each positive integral to infinity. If its time power is $u\ge0$ and its decay is $\rho>0$, with $j=\lfloor u\rfloor$, use

This is Hölder interpolation of adjacent integer moments. All decay rates in (54) are positive. Integer roots are enclosed rationally; no numerical gamma function is needed. For a final far cutoff of width $0.01$ before $S=0.14145$, put

The unshortened positive constructions define the values used in later interfaces; their shortened displays are $P_4<1.075567\cdot10^{-76}$ and $L_{SS}<2.534075\cdot10^{-57}$ uniformly through preparation.

B.5 Original-stock transfer to the writer

Let $\Psi_o$ be the exact preparation without the writer, and $\Psi_n$ the enlarged exact wave, with their common initial state $\Psi_n(0)=\Psi_o(0)g_0$. Here $g_0(Z)=\pi^{-1/4}e^{-Z^2/2}$. Write $\Delta=\Psi_n-\Psi_og_0$. The ground oscillator energy is subtracted and the additional potential is $J=(\hbar/T_0)\Pi_1[C_w(s)-F_w(s)Z]$. Thus

No comparison flow or initial law is substituted in this identity. The writer is supported on $S\ge0.14145$. Throughout preparation, the old helper and its bounded phase both vanish on this support. The source amplitudes and momentum norms are therefore bounded by $n_{2T},L_S,L_r,L_{SS}$ established above.

The following constants are exact rational majorants. Put $\mu=0.01$, $w=0.003$ and

In this subsection $b_j$ denotes a writer derivative bound, not the preparation dilation or bump derivative. Define

These bound $F_w,C_w$ and their dimensionless derivatives. Hereafter $F_{\rm old}=8\cdot10^{-6}+10^{-17}$ and $F_{{\rm old},r}=10^{-6}$. These force ceilings follow by differentiating the displayed potential: the preparation clock force is bounded by

and its radial force by $ma_1(0.201)+ma_0^2(0.201)+m^2a_1^2(0.080802)(0.201)/(2M)$. Adding the quiet and activation derivatives gives the stated ceilings; their clock contribution is below $10^{-17}$. A second differentiation gives the conservative bound $F_{{\rm old},SS}\le1$: its preparation part is

with $Q_{SS}+\gamma_2(W_0+Q_0)+2\gamma_1q_{t1}$ added.

For concise exact formulas set $n_o=n_{2T}$ and

The exact oscillator identities $\|Zg_0\|=\|p_Zg_0\|=1/\sqrt2$ and $\|Z^2g_0\|^2=\|p_ZZg_0\|^2=3/4$ prove these coefficients. Unitary Duhamel and oscillator rotation in the direct sum of its two quadratures then give increasing bounds

Here $Q_\Delta$ bounds $(\|Z\Delta\|^2+\|p_Z\Delta\|^2)^{1/2}$, not an unbounded coordinate times a bare norm estimate. In the clock inequality the term $F_1Q_\Delta$ is retained explicitly.

Let $B=\partial_Z+Z$. Its exact equations have sources $-\Pi_1F_w\Psi_n$ and $-2\Pi_1F_wB\Psi_n$ for $B\Psi_n$ and $B^2\Psi_n$, respectively. Because $B$ annihilates the old product wave, integration gives

The last quantity bounds the norm of the squared moving annihilator; it is not the square of its norm. These are operator identities on the full oscillator Gaussian domain, with no truncation of $Z$. For $f=B\Psi_n$, oscillator algebra gives $\|Zf\|^2+\|p_Zf\|^2=\|Bf\|^2+\|f\|^2$. Consequently, with integrals over $[0,T]$,

This proves $\|\partial_S(A\Psi_n)(T)\|\le U_h$. All displayed integrals are polynomials; for example $\int B_1=F_0(n_oT^2/2+d_\Delta T^3/6)/T_0$ and $\int B_2=F_0^2(n_oT^3/3+d_\Delta T^4/12)/T_0^2$.

The exact symbolic entrance values used subsequently are

Here $\epsilon_q$ is the quiet-reference comparison error in the radial momentum, as bounded in the activation appendix. In particular the larger norm $3.006684\cdot10^{-11}$ used in the fourth-moment estimate is not substituted for $D_h$.

B.6 Localized repair of the transfer Hessian

A coarse second-momentum bound is useful only to generate source-free left collars. We give it explicitly to avoid circular reuse of the improved result. Let $P=P_{\Delta,S}(T)$, $D=D_\Delta(T)$, $Q=Q_\Delta(T)$, and

The quadrature estimate $\|Z^2\Delta\|$ and its companion are controlled by $B_2+2B_1+2D_\Delta$, so that $Q_S$ bounds the mixed weighted momentum without replacing $Z$ by a bounded operator. Differentiating the inhomogeneous equation twice proves $\|p_S^2\Delta\|\le H_c$; the coarse value is below $\hbar^2(2.486519\cdot10^{27})$.

Choose decreasing left cutoffs with transitions $[0.126,0.136]$ and $[0.116,0.126]$, width $g=0.01$. The writer forcing vanishes on both supports and every old clock force is contained in their full-one regions, because it vanishes for $S\ge0.106$. Positive drift favours exit from each left region. Continuity and one integration by parts give

For the new Hessian $H(t)=\|p_S^2\Delta(t)\|$ the old-force momentum is instead localized by $\|\one_{\rm oldforce}p_S\Delta\| \le\sqrt{n_{\Delta2}H(t)}+2\hbar n_{\Delta1}/g$. Use the proved far old-wave Hessian $L_{SS}$ in the source. The differential inequality has the form $H\prime\le2\alpha\sqrt H+f$, where $\alpha=F_{\rm old}\sqrt{n_{\Delta2}}$. The function $[\alpha t+\sqrt{\int_0^t f}]^2$ is a supersolution, because its derivative is at least $2\alpha\sqrt H+f$ when evaluated at that function. Bounding the nonnegative source integral gives

It proves $\|p_S^2\Delta(T)\|\le H_\Delta$ and $H_\Delta/\hbar^2<1.828510\cdot10^{19}$. The coarse Hessian only generated the amplitude collar; it was not relabelled as the improved source or inserted into this nonlinear old-force term.

B.7 Handoff enclosures and the preparation event

The constructions above give the following outward decimal enclosures. Every endpoint in the table is a rational number. The unrounded expressions, not shortened displays, define the quantities when used in the final event sum.

B.8 Preparation history under the enlarged exact flow

The preparation rank $K(S,r)$ is independent of $Z$; hence its material derivative under the enlarged flow has no pointer-current term. Its explicit drift $\beta$ in (34) is unchanged. The error in its current must nevertheless be recomputed for that flow. In the bounded gauge the additional momentum ceilings are

The exact normalized wave stays in the derivative-error factor of the bilinear expansion, retaining the quadratic error terms. Consequently

Under the single original joint domination constant $C=270000/67499$, $N$ fixed radial cuts therefore have preparation failure probability bounded by

This follows by taking initial rank collars of width $\beta+\epsilon$, using Markov only on the unknown absolute current, and minimizing in $\epsilon$. The last term covers the terminal clock suffix where the helper vanishes. It is an estimate on the enlarged flow with its original initial law; closeness of two waves alone has not been used to assert history agreement. The initial conditional-CDF interface costs, separately,

For the final cut family, including the exterior ray, use $N=256001$ in both formulas. Evaluating the finite expressions gives

The pointer remains part of the one jointly dominated auxiliary stock; no independent pointer cap, sector sample, equilibrium reset, or new law at handoff enters these estimates.