One unknown input and a separate held receiver

For an oscillator of mass $m$ and frequency $\omega$, let $\phi$ be its real normalized ground state, so $\sigma^2=\hbar/(2m\omega)$. For a prescribed $C^3$ center $d(t)$, set

Substitution into the Schrödinger equation proves the formula: the $\phi'$ terms match, the oscillator acts on $\phi$ with energy $\hbar\omega/2$, and the acceleration term matches the time derivative of $m\dot d x$. Its current is $j=\dot d|\psi_d|^2$. The same construction applies to the receiver with its own mass, frequency and width $a$.

Choose the flat smooth step (3.4) on each finite stage interval. The displayed Schrödinger calculation requires only a $C^3$ center, so the polynomial schedule given earlier is also sufficient for the endpoint formulas and all estimates. In qubit sector $s$ use the block-diagonal sum of (7.9) for the corresponding writer and receiver centers. First move only the writer from $0$ to $(-1)^{s+1}D$; then hold it and move only the receiver from $0$ to $(-1)^{s+1}A$; then hold the receiver and return the writer to zero.

The two internal sectors eliminate every interference cross term in the complete positional density and current, even when the reference vectors overlap. Each moving-coordinate velocity is a convex combination of the two prescribed center velocities. Active densities are strictly positive at every finite coordinate, including $p=0,1$, and the velocities are smooth locally and bounded on each finite time interval. Thus the active ODE has a unique global solution for every finite entrance coordinate. The old coordinates remain under their admitted held parent. During copy the writer has zero current; during return the receiver has zero current.

For a one-dimensional conservative flow, its density CDF is constant along a trajectory: differentiating the CDF in time gives $-j$, and the advective term is $\rho\dot q=j$. Applying this identity first to the writer mixture, then to the receiver at the frozen writer coordinate, and finally to the returning writer at the frozen receiver coordinate gives (7.5)–(7.7). All the CDFs involved are strictly increasing. The final joint wave is

with known spatially constant branch phases. Its receiver marginal is $M_p$ and hence its receiver reference quantile is (7.8).

For a direct Jacobian check, write

It factors both as $\rho_p(q)$ times the conditional receiver density and as $M_p(z)$ times the conditional split-writer density. Consequently the Jacobian of $(u,v)$ with respect to $(q,z)$ is $L$, and that of $(r,b)$ with respect to $(q,z)$ is also $L$. Positivity gives determinant one for $T_p$ and the successive inverse CDFs give its inverse. This proves reference-area preservation independently of any actual-law assertion.

Finally let $U(t)$ be a spatially constant internal unitary that stores the measured qubit in a fresh internal register. For an explicit SWAP, take $K=\pi(I-\mathrm{SWAP})/2$ and $U(t)=\exp[-i s(t)K]$, where $s$ increases smoothly from zero to one. The transformed parent is

The receiver holding projectors must be conjugated in this expression. Then $U\Psi$ is the exact evolved wave. Spatial constancy makes its complete density and canonical current identical to those of $\Psi$. In particular, $j_z=0$ is preserved through the SWAP. The statement for a later finite hold requires subsequent controls to preserve the stored sector and receiver hold; it does not cover arbitrary future couplings to that receiver. No unknown state is cloned or erased.

The split writer has density $\rho_p$. Its band $|q|<c_w\sigma$ has mass at most $\Phi(-(d_*-c_w))$, since the two translated Gaussians give the same band integral. On $q<-c_w\sigma$, the integral of the wrong posterior is

On the right the corresponding integral is $p\Phi(-(d_*+c_w))$. Markov's inequality bounds the two bad-posterior sets by at most the second term in (7.12); the factor two is a convenient conservative bound. There is no division by $p$ or $1-p$.

On the remaining left region, $w_p\geq1-\gamma$. The copy equation gives

Since $\Phi(-c_z)<\eta/2$, this implies $z<(c_z-a_*)a<0$. The reflected argument gives $z>0$ on the right good region. Independence of $u$ from the full $\nu$ means that the discarded writer mass is multiplied by $m$, regardless of correlations between $N$ and $v$. This proves the failure bound. Moreover

Coupling $Y$ with this earlier $S$ proves (7.13). Under the cap, the omitted receiver-end intervals have mass at most $2C\eta$, which proves the last statement.

For a nonzero ideal branch weight let $\eta_s=|s\chi_s\rangle/\|\chi_s\|$. For a zero-weight branch choose any unit vector in its qubit sector. On the physical region $Y(z)=s$, the pure-state trace distance from (7.14) to $\eta_s$ is $\sqrt{W(z)}$, where

This follows from the squared overlap with the orthogonal correct qubit sector. It also holds for a zero ideal branch weight: the conditional vector in that region is orthogonal to the chosen comparison sector. Direct integration gives the exact identity

The returned writer integrates to one. Area preservation of $T_p$ thus gives

In the capped case, the comparison density is dominated by $C\,du\,dv\,\kappa(dN)$. Equation (7.23) implies $\mathbb E W\leq C\Phi(-a_*)$, and Cauchy gives a conditional-state replacement cost at most $\sqrt{C\Phi(-a_*)}$. After replacement, only the classical branch weights differ from (7.16); their trace distance is the outcome TV from Lemma 7.2. Finally (7.17) adds $\delta$ once for replacement of the actual entrance law by $du\,\nu$. The copy-history bound follows from the same entrance-law replacement applied to the path event $\mathcal F$.

For the boxed assertion fix a spectator value $N$ and set $m_N=\int f(N,v)dv$ and $V_N=\operatorname{Var}_v f(N,\cdot)$. The one-dimensional BV inequality $f(N,v)\leq m_N+V_N$ almost everywhere gives

This is equally the active-square $L^2(du\,dv)$ bound, since the comparison density is independent of $u$. Cauchy and (7.23), followed by integration in $N$, bound the entire subprobability state-replacement cost by

There is no need to normalize rare spectator conditionals or to assign them a common cap. The outcome cost for this comparison law is $m(B+\Phi(-d_*))$. The boxed actual entrance law differs by at most $\Delta_B$, so (7.17) bounds its instrument distance from $m\Lambda(\chi)$ by the last three terms in (7.21). The outside-box output and $\tau\Lambda(\chi)$ are positive operators of the same trace $\tau$; their trace distance is at most $\tau$. Adding them charges the original tail once. The same decomposition gives (7.20), with outside-box paths charged at most their mass. This proves both claims.

The original complete box has outside mass $\tau\leq10^{15}/10^{20}=10^{-5}$. The preparation estimates give

The quantity $\Delta_B$ compares the actual boxed warmup endpoint to a uniform writer times its own boxed nuisance marginal. It does not already include the outside-box tail.

The unused receiver coordinate $v$ does not participate in warmup. Each warmup map is independent of $v$ and preserves the reference volume of the other coordinates. Therefore its pushforward preserves the integrated $v$-directional BV norm of the initial boxed density; this can be seen by commuting the distributional $v$ derivative with the pushforward and using volume preservation. Marginalizing the final writer contracts that norm. The physical score-to-BV estimate therefore gives

where $J_z\leq100$. The source remains supported on $v\in[\Phi(-10),1-\Phi(-10)]$ under every warmup map and its marginalization. Thus it already satisfies the receiver guard for $\eta=10^{-110}$; no new source cut or second moment-tail charge is used.

Apply Theorem 7.3 with $d_*=a_*=50$, $c_w=3$, $c_z=24$ and $\gamma=10^{-130}$. The inequalities $\gamma\leq\eta/2$ and $\Phi(-24)<\eta/2$ hold, and the left good receiver finishes below $-26a$. Gaussian tail bounds give

For reproducibility, the elementary inequality $\Phi(-x)<e^{-x^2/2}/(x\sqrt{2\pi})$, obtained by bounding $1\leq t/x$ in the Gaussian tail integral, proves these comparisons. The receiver guard follows also from $\Phi(-10)\geq0.01\,\phi(10.01)>10^{-110}$, by integrating just over $[10,10.01]$. Together with (7.26), substitution in (7.21) and (7.20) yields the strict ceiling (7.25).