Reading a retained archive into different receivers

5 Reading a retained archive into different receivers

The objective in this section is a joint path event. At a specified entrance an archive has an actual position. Its first $n$ binary digits are to be written into $n$ different receivers, and every written receiver must remain in its assigned separated region until a common final time. The construction retains the scratch, archive, receivers, reset modes and all earlier records. Its reset restores a quantum factor; it does not assume a new actual population.

We use dimensionless positions in units $\sigma$ and time $\tau=\omega t$, with $\sigma^2=\hbar/(2m\omega)$. Put

Thus $[q,P]=2i$, the oscillator is $H_{\rm o}=-\partial_q^2+q^2/4$, and its scalar canonical current is $j=2\operatorname{Im}(\Psi^\dagger\partial_q\Psi)$. During a compact baker control the additional current of Theorem 3.1 is retained; a real stationary wave does not then imply zero current.

The scratch $x$ and archive $y$ are two coordinates of one planar carrier. Each $z_j$ is the longitudinal coordinate of a different physical receiver, and each $r_j$ is a distinct retained reset mode. The reference entrance contains the factors

tensored with the specified remaining wave. All initially unused modes remain oscillator factors until used. Every genuine positional copy or reference belongs to the full configuration; a finite internal fibre is included in the wave norm. The actual entrance law satisfies

on that complete state, or conditionally on a retained external parameter with the same uniform $C$. Equivariance propagates this global inequality. It is never asserted after conditioning on a selected recorded bit. Proposition 10.2 distinguishes this premise from an earlier preparation statement conditional on a bank. For the full-flow assertion, additional positional quantum factors must have propagated regularity sufficient for a $C^1$ normalized, conserved complete spacetime current and for the finite current integrals specified in the proof of Proposition 6.3. Propagation of all finite weighted Sobolev orders under their specified decoupled dynamics is one sufficient choice. A passive finite internal reference is unrestricted. The constructed Gaussian bank meets these conditions directly. A retained external conditioning parameter is fixed during the protocol; it is not assigned a wave amplitude merely by being retained.

Proposition 5.1 (Actual digits, independent of the scratch population)

Let $u=\Phi(x)$ and $v=\Phi(y)$. Off dyadic boundaries the inverse baker map is

$$s=\lfloor2v\rfloor,\qquad v^+=2v-s,\qquad u^+=(u+s)/2.$$(5.3)

It is invertible and area preserving, and $\lfloor2u^+\rfloor=s$ for every $0<u<1$. Repeating it exposes the successive digits of the actual entrance archive, even if every subsequent scratch position is correlated with all previous records. For the smooth inverse of Proposition 2.3, the endpoint is exactly (5.3) on

$$u\in[2h,1-2h],\qquad v\in[h/2,(1-h)/2]\cup[(1+h)/2,1-h/2].$$(5.4)

Its reference exceptional area is

$$\beta_h=1-(1-4h)(1-2h)=6h-8h^2 .$$(5.5)

Proof

On each half-strip the determinant is $(1/2)2=1$. Recover $s$ from $\lfloor2u^+\rfloor$, then $u=2u^+-s$, $v=(v^++s)/2$. The archive update contains no $u$. It follows by induction that the exposed bits are the binary digits of $v_0$ regardless of the scratch values. The set (5.4) is the image of the forward baker's two retained rectangles. Reversing its complete smooth flow proves the endpoint assertion; its area gives (5.5).

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For comparison with an earlier warmup event, suppose the forward archive recursion was $v_k=(v_{k-1}+a_k)/2$, with actual declarations $a_k\in\{0,1\}$. Then

The inverse reader emits $a_n,a_{n-1},\ldots,a_1$. Accordingly receiver names must be reversed for a chronological decoder. An already priced event on which a smooth warmup differs from this history relation is added once; the reader itself does not manufacture a missing relation to an earlier writer event.

5.1 An exact sign-preserving scalar loader

For $d\geq0$ define

The square root is taken after adding densities. The sum of the two Gaussian amplitudes is a different wave and will be used only as a controlled comparator below.

Proposition 5.2 (Loader, domain and stationary final hold)

Choose

$$d(\tau)=A\,s(\tau/T_L),\qquad s(a)=35a^4-84a^5+70a^6-20a^7,\quad 0\leq a\leq1,$$(5.6)

with static extensions. This center is $C^3$ across its joins; the resulting potential is $C^1$ in time and smooth in position. If $35A/(16T_L)<1/\sqrt2$, the scalar Hamiltonian $H_L=-\partial_x^2+V_L$ below has a common oscillator strong domain and an exact normalized solution

$$\begin{align}\psi_L(x,\tau)&=R_{d(\tau)}(x)e^{iS(x,\tau)},& S&=\frac{d'}{2d}\log\cosh(dx),\tag{5.7}\\ v_L&=d'\tanh(dx),& V_L&=\frac{R_d''}{R_d}-S_\tau-\frac{v_L^2}{4}. \tag{5.8}\end{align}$$

The expressions at $d=0$ have their continuous limits. Every actual trajectory preserves $F_d(x)$ and its initial sign. At the endpoint the exact wave is the positive $R_A$, and its holding Hamiltonian is

$$H_A=-\partial_x^2+U_A(x)=Q_A^\dagger Q_A,\qquad Q_A=\partial_x-R_A'/R_A,\qquad U_A=R_A''/R_A .$$(5.9)

Its complete scalar current vanishes pointwise.

Proof

Writing $L=\log R_d$ gives

$$L=\text{constant}-(x^2+d^2)/4+\tfrac12\log\cosh(dx),\qquad U_d=-\tfrac12+\tfrac{d^2}{2}\operatorname{sech}^2(dx) +\tfrac14(x-d\tanh(dx))^2 .$$

The mixture satisfies the exact conservation identity

$$\partial_\tau\rho_d+ \partial_x\!\left\{\tfrac{d'}2[\gamma(x-d)-\gamma(x+d)]\right\}=0.$$

Its positive-density current divided by $\rho_d$ is $v_L$. Since $2S_x=v_L$, substitution in the imaginary and real parts of $i\partial_\tau\psi_L=H_L\psi_L$ gives precisely (5.8). This verifies both equations, not only a prescribed real potential.

Here are sufficient global domain estimates. Set

$$P_1=\frac{35A}{16T_L},\quad P_2=\frac{8A}{T_L^2},\quad P_3=\frac{53A}{T_L^3},\quad P_{13}=140^3(5/14)^5(9/14)^9 A^2/T_L^3 .$$

They bound $|d'|,|d''|,|d'''|,(d')^3/d$, respectively. Indeed $s'=140a^3(1-a)^3$, its maximum is $35/16$, the acceleration maximum is $84/(5\sqrt5)<8$, and the jerk maximum is $105/2<53$. The polynomial $s(a)/a^4$ decreases from $35$ to $1$; hence $(s')^3/s\leq140^3a^5(1-a)^9$, with maximum at $a=5/14$. The last ratio tends to zero at the initial join.

The inequalities

$$U_d\geq x^2/8-d^2/4-1/2,\qquad |S_\tau|\leq P_2|x|/2+P_1^2x^2/4$$

give, with $a_0=1/8-P_1^2/4>0$,

$$V_L\geq (a_0/2)x^2- \left(A^2/4+1/2+P_1^2/4+P_2^2/(8a_0)\right).$$

Conversely

$$|V_L|\leq(1/4+P_1^2/4)x^2+ (A/2+P_2/2)|x|+1/2+3A^2/4+P_1^2/4.$$

The elementary bound $|a\operatorname{sech}^2(a)\tanh(a)|\leq1/2$ gives

$$|V_{L,xx}|\leq(AP_2+2P_1^2)/2+P_1^2A^2 +1/2+3A^2/2+A^4.$$

Finally, twice differentiating the integral expression $S=\frac12\int_0^x d'\tanh(dy)\,dy$ shows

$$|S_{\tau\tau}|\leq P_3|x|/2+ (3P_1P_2+P_{13})x^2/4.$$

Differentiating the remaining terms gives a uniform $|V_{L,\tau}|\leq C(1+x^2)$, continuously through both joins.

Choose a common shift so that $W=V_L+c\geq a x^2+1$. For a compact smooth test,

$$\|(-\partial_x^2+W)f\|^2 =\|f''\|^2+\|Wf\|^2+ 2\int W|f'|^2-\int W''|f|^2 .$$

Together with the preceding upper bound this makes the Hamiltonian graph equivalent to $\|f''\|+\|x^2f\|+\|f\|$. The real resolvent cutoff identity for an $L^2$ distributional solution of $(H_L+\lambda)f=0$ is

$$\int |(\chi_Rf)'|^2+(V_L+\lambda)|\chi_Rf|^2 =\int|\chi_R'|^2|f|^2\longrightarrow0.$$

Local elliptic regularity justifies the test. Positivity excludes a nonzero such solution; the semibounded closure is self-adjoint. Cutoff and mollifier approximation and the graph equivalence give the common domain $H^2(\mathbb R)\cap\{x^2f\in L^2\}$. The time estimate makes $H_L(\tau)$ continuously differentiable on this domain, supplying common-domain unitary evolution by the common-domain theorem in [5]. The explicit wave already constructed is its unique solution.

The velocity is globally bounded by $|d'|$, with derivative bounded by $|dd'|$, so its ordinary trajectories are complete and unique. Continuity and the conservation equation give $dF_d(x(\tau))/d\tau=0$. Symmetry fixes $x=0$. At the endpoint $d'=d''=0$, $S=0$, and direct multiplication gives the factorization (5.9). Thus $R_A$ is its normalized zero-energy state and has identically zero canonical velocity.

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For an inverse bit $s$, the endpoint obeys $F_A(x_L)=(u+s)/2$. In particular its sign is the actual extracted bit. With $\xi_A=\Phi(-A/2)-\Phi(-3A/2)$, symmetry gives $2F_A(-A/2)=1-\xi_A$ and $2F_A(A/2)=1+\xi_A$. If $\xi_A\leq2h$, every good scratch in (5.4) is also loaded beyond $|x_L|=A/2$. The receiver theorem below needs only the sign, and prices its own separated endpoint without assigning the scratch a Born population.