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The Ghirardi--Rimini--Weber theory modifies unitary time evolution by random Gaussian position localizations. Each constituent collapses at a small fixed rate, while the rate for an entangled macroscopic body grows with its number of constituents.

The operator support is the set of subsystems on which an operator may act nontrivially; it acts as the identity outside that set. Unitary time evolution can enlarge the support, while a Lieb-Robinson bound controls the error of replacing the evolved operator by one with smaller support. The support diameter and the number of sites in the support are different quantities.

Let $B={\sum }_k{g}_k{Z}_k$. In the Zurek spin-bath model, $H=-Z_D\otimes B$, so unitary time evolution with $\hslash =1$ is $U(t)=e^{it{Z}_D\otimes B}$. The bath Hamiltonian terms commute, and device states $|0⟩,|1⟩$ have $Z_D$ eigenvalues $+1,-1$. Hencewhere the two normalized conditional bath states areTake each bath factor normalized, $|{\alpha }_k{|}^2+|{\beta }_k{|}^2=1$, and $|a{|}^2+|b{|}^2=1$. This entails no restriction: if only the product is initially normalized, divide each nonzero factor by its norm; the product of these norms is one.

Taking the partial trace over the bath gives the reduced density matrixThe orientation of this conditional environment overlap fixes the sign of the phase in the upper-right entry. Factorizing the overlap yields the decoherence factorThe populations are conserved because $[H,Z_D]=0$. Only phase coherence can be reduced. In particular,Quantum decoherence here results from distinguishable conditional bath states, although the complete system remains in a pure state under unitary time evolution.

Write $g_{\ast }>0$ for the upper endpoint called $a$ in the question, to distinguish it from the device amplitude. For a single realization of a finite bath,The literal claim of convergence to zero as time tends to infinity is false for every fixed finite $N$. Random-coupling spin-bath decoherence must distinguish individual realizations, ensemble averages and the large-bath limit.

First establish finite spin-bath coherence recurrence without assuming commensurate couplings. Fix a reference time $t_0>0$ and consider $Q^N+1$ points with coordinates $j{g}_k{t}_0/\pi$ modulo $1$, for $0\le j\le Q^N$. Partition the unit cube into $Q^N$ smaller cubes of side $1/Q$. Two points share a cube, so their difference provides an integer $1\le q\le Q^N$ withAt time $q{t}_0$ every cosine is arbitrarily close to $1$. As $Q$ increases, either these $q$ have an unbounded subsequence, or a bounded subsequence supplies a fixed $q$ with all distances exactly zero; its arbitrarily large multiples are then exact recurrences. In both cases there are unbounded times $t_j$ such that $z_N(t_j)\to 1$. This is the simultaneous Dirichlet approximation theorem argument; it applies equally to typical irrational random couplings. Part (i) supplies a particularly simple exact periodic counterexample.

The intended suppression is valid after ensemble averaging. Independence and the uniform density give the ensemble spin-bath coherenceThe continuous value at $t=0$ is $1$. For $t\ne 0$ its magnitude is strictly below $1$ to the power $N$, and for fixed $N$ it has an envelope of order $(2g_{\ast }t{)}^{-N}$ as $t\to \infty$. Thus the ensemble mean tends to zero with time.

The mean square distinguishes actual loss of coherence from cancellation of signs in that mean:At long times this tends to $2^{-N}$, exponentially small for $N\gg 1$, rather than exactly zero for a finite bath. At any fixed $t\ne 0$, the bracket $q(t)$ is strictly less than $1$, since ${\cos }^2(2gt)<1$ on a set of positive measure. The Markov inequality then givesThis proves small coherence for typical large baths at a fixed nonzero time. An even stronger fixed-time formulation uses the strong law of large numbers:The logarithmic singularities at isolated cosine zeros are integrable, and an exact zero has probability zero, so the law applies. Typical coherence consequently decreases exponentially with bath size.

The initial time scale is also explicit. When $g_{\ast }|t|\ll 1$, $\log \cos (2g_{k}t)=-2g_k^2{t}^2+O(g_k^4{t}^4)$, so the short-time Gaussian spin-bath decoherence approximation isHere $N^{-1}\sum g_k^2\to g_{\ast }^2/3$; on the scale $t\sim (g_{\ast }\sqrt{N}{)}^{-1}$ the displayed remainder tends to zero. The coherence is therefore rapidly suppressed for a large bath and is usually tiny at later fixed times, while rare recurrences still prevent a finite-realization long-time zero limit. Ensemble decay or a specified large-$N$ limit is the correct qualification of the printed assertion, consistent with reversible global unitary time evolution.

Let $U_1=e^{-iH(t-t_1)/\hslash }$ and $U_2=e^{-iH(t_2-t)/\hslash }$ describe unitary time evolution. For an ideal projective measurement with the Lüders rule, its unconditioned outcome probability is $p_j=\Vert P_j{U}_1|\psi ⟩{\Vert }^2$. After that outcome, the normalized state is $P_j{U}_1|\psi ⟩/\sqrt{p_j}$. The Born rule probability of successful postselection is then $|⟨{\psi }^{\prime }|U_2{P}_j{U}_1|\psi ⟩{|}^2/p_j$. Multiplying gives the joint probabilityConditional probability therefore gives the Aharonov-Bergmann-Lebowitz rule:The denominator must be positive; otherwise the selected subensemble does not occur. Define the forward-evolved ket $|a⟩=U_1|\psi ⟩$ and backward-evolved ket $|b⟩=U_2^{†}|{\psi }^{\prime }⟩$. The numerator becomes $|⟨b|P_j|a⟩{|}^2$, which is unchanged by interchanging $a,b$. Equivalently, with ${\rho }_a=|a⟩⟨a|$ and ${\rho }_b=|b⟩⟨b|$,This expresses the boundary-state symmetry explicitly. It follows from the ordinary time-asymmetric preparation, Born rule, and state update; it does not posit an additional backward dynamical collapse.

Restore the post-selected vector omitted entirely from the TeX aid by reading the original PDF. Write $D=N^2-N+1$ andIts norm is one because $N+(N-1{)}^2=D$. For the uniform prestate and $H=0$, put $c=1/\sqrt{(N+1)D}$. The individual transition amplitudes areIn experiment $E_i$, the complement amplitude is $c-c=0$. Thus for every $i\le N$,The successful postselection rate in this experiment is $c^2=1/((N+1)D)$.

In the fully resolved experiment $E_0$, the ABL rule squares the individual amplitudes before summing. Their squared sum is $D{c}^2=1/(N+1)$, givingThese probabilities sum to one. For $N=1$ they reduce to a single certain outcome in either quantum measurement. For $N\ge 2$, the N-box pre- and post-selection paradox is that each separate binary question can be answered affirmatively with certainty, although the fully resolved quantum measurement cannot give all those outcomes at once.

There is no inconsistency. In $E_i$, the unresolved complement preserves coherent cancellation between its basis contributions under the Lüders rule. In $E_0$, those alternatives are resolved, so their squared amplitudes add instead. The different projective measurements disturb the state differently and have different postselection success rates. Merely merging the recorded $E_0$ outcomes afterwards does not reproduce $E_i$. This is the context dependence of pre- and post-selected measurements; certainties in mutually alternative experiments do not describe simultaneous measurement-independent properties.

Let $U(t)=e^{-itH}$. Define the second initial state by $|{\tilde{\psi }}_0⟩=U(t{)}^{†}|{\psi }_2⟩$, while $|{\psi }_0⟩=U(t{)}^{†}|{\psi }_1⟩$. The unitary time evolution preserves their inner product:Both initial states are normalized and evolve to the chosen final pair. This supplies the orthogonal partner needed for topological quantum order. The remaining requirement is local indistinguishability, which is proved in the next condition's solution.

Sum the spectral filtering of Hamiltonian terms over the original finite decomposition. The total Hamiltonian operator commutes with its own evolution, soThe absence of all ground-to-excited matrix elements, proved in part (a), makes each filtered term block diagonal relative to $P_0=|{\varphi }_0⟩⟨{\varphi }_0|$ and its orthogonal complement. ThereforeA filtered term may act on the whole system, since unitary time evolution spreads its original support. Neither commutation with $P_0$ nor preservation of its ground-state expectation asserts that the global ground state minimizes each individual term. In particular, this construction does not generally turn a frustrated decomposition into a frustration-free Hamiltonian.

Choose boundary states with nonzero overlap with the lowest-energy state in the vacuum sector $Q=0$ and in the one-kink topological sector $Q=1$. In a finite spatial box, a fixed scalar field configuration eigenstate may be used formally, with temporal endpoints equal to the chosen configuration. More generally, smear the endpoints with wavefunctionals ${\Psi }_Q$. Their unitary time evolution kernels areThe inner scalar field path integral remains in the chosen topological sector, with $\varphi (+\infty )-\varphi (-\infty )=2\pi Q$. A zero-total-momentum projection can be included to select the rest state; alternatively, the translational prefactor does not change the large-time exponential. After a Wick rotation to physical Euclidean time $\tau$, the energy eigenstate expansion gives $K_Q^E(\tau )\sim C_Q{e}^{-E_Q\tau }$. Thus the exact vacuum-subtracted soliton mass isEquivalently it is $i{T}^{-1}\log (K_1/K_0)$ at large time with a damping prescription. The ratio subtracts the vacuum energy; the Hamiltonian operator and action here are the fully regulated and renormalized ones, not merely their classical approximations.

With the dimensionless coordinates of this paper, the correctly normalized classical action isFor the static Sine-Gordon kink, ${\varphi }_K^{\prime }=2\mathrm{sech}x$ and $1-\cos {\varphi }_K=2{\mathrm{sech}}^{2}x$, so $M_{\mathrm{cl}}=8m/{\beta }^2$. Write $\eta =\delta \varphi$. Expanding and integrating by parts givesThe first variation is ${\beta }^{-2}\int ({\varphi }_K^{\prime \prime }-\sin {\varphi }_K)\eta$, plus boundary terms. It vanishes because the kink satisfies the Euler-Lagrange field equation and the fluctuations have the prescribed temporal endpoints and admissible spatial boundary behavior. This is the principle of stationary action, not a symmetry assumption about $\eta$.

The printed expansion omits ${\beta }^{-2}$ despite the stated Lagrangian density. Its displayed form is the expansion of ${\beta }^{2}S$; it is not the physical $S$ at arbitrary coupling. Alternatively, writing $\eta =\beta \chi$ puts the quadratic term in canonical normalization, while leaving the classical term $-8mT/{\beta }^2$ unchanged. This normalization repair does not change ${\Delta }_x$ or the physical fluctuation frequencies.

The Sine-Gordon kink fluctuation operator has the useful factorizationIt is nonnegative, and $A{\psi }_0=0$ gives the normalized translational zero mode of a sine-Gordon kink:A displacement $a$ changes ${\varphi }_K(x-a)$ by $-a{\varphi }_K^{\prime }(x)$. Thus the zero mode in field theory is the position collective coordinate of the kink, reflecting translation invariance. It has no restoring force and no oscillator zero-point energy. Integrate that collective coordinate separately rather than inserting a zero factor into the Gaussian functional determinant.

In the Gaussian fluctuation approximation, the formal oscillator contribution to the one-loop soliton mass correction, before adding any counterterm, isUse a common regulator for the two sums, include all discrete modes, and treat the translation mode as above. The vacuum sum is essential: subtracting only classical vacuum energy would leave an extensive oscillator energy. A periodic fluctuation and its derivative are matched at the two ends of the large box; the one-kink background lies in the twisted topological sector, and its infinite-line profile is accurate up to exponentially small boundary corrections.

For a continuum scattering wavefunction, equality of its two asymptotic values gives the periodic-box phase-shift quantizationAway from the threshold, expand at a matched mode number:Since consecutive free wave numbers are separated by $2\pi /L$, replacing the continuum-mode sum by an integral proves the displayed continuum contribution:The ultraviolet cutoff is retained until the counterterm is added.

There is a finite threshold issue if this expression is identified with the complete oscillator correction. It can be settled directly using the factorization: a continuum eigenfunction is $f_k\propto (\tanh x-ik)e^{ikx}$. Its transmission phase obeyswith the odd phase branch that tends to zero at large $|k|$. For $k>0$, the continuum roots have labels $n=1,2,\dots$; there is no periodic continuum root at $k=0$, since the limiting eigenfunction $\tanh x$ has opposite signs at the two ends. The bound state at ${\omega }_0=0$ replaces the free $k=0$ oscillator with ${\omega }_0^{(0)}=1$. Consequently, in mode-number regularization of soliton masses,The PDF's continuum-only formula misses this finite $-m/2$ under this standard phase and mode-counting convention. It has the correct logarithmic ultraviolet divergence, but the missing term is not suppressed by large $L$. Changing the phase branch requires changing the mode labels and endpoint terms consistently; it cannot erase a physical mode from the formal spectrum sum.

At high momentum, $\delta (k)=2/k+O(k^{-3})$, so both expressions have divergent part $-(m/\pi )\log \Lambda$. The canonical field $\phi =\varphi /\beta$ has a quartic interaction with coupling $-m^2{\beta }^2$. Its vacuum tadpole diagram shifts the squared mass by $-m^2{\beta }^{2}J(\Lambda )/4$, whereThe Sine-Gordon vacuum tadpole counterterm has $\delta m^2=+m^2{\beta }^{2}J/4$ and adds potential energy density $\delta m^2(1-\cos \varphi )/{\beta }^2$. Since $\int (1-\cos {\varphi }_K)dx=4$, its vacuum-subtracted kink energy iswhich cancels the logarithmic ultraviolet divergence. Finite parts require a specified renormalization condition. As a consistency check, with this tadpole subtraction and matched mode-number cutoff, integration by parts yieldsThe complete semiclassical soliton mass is then $8m/{\beta }^2-m/\pi +O(m{\beta }^2)$ in that convention. This last finite result uses the bound-mode term and is additional to the requested ultraviolet cancellation.

For each interaction neighbourhood, setThen the shell term is $g^{(Z,s)}=G_s^{(Z)}-G_{s-1}^{(Z)}$. Since $H_0=h_Z$, its conjugation leaves $h_Z$ fixed, so $G_0^{(Z)}=h_Z\int w=h_Z$. Since $H_D=H$, $G_D^{(Z)}$ is precisely the choice of $g^{(Z)}$ from (b). Therefore the telescoping local-shell decomposition givesNo limiting interchange is needed: the sum has finitely many shells. Each $G_s^{(Z)}$ is supported within the union of supports appearing in $H_s$ together with $Z$, because its unitary time evolution acts trivially outside that neighbourhood. The individual shell terms need not commute with the full ground state projector; the full filtered sum does.

For spin $1/2$, $\mathbf{S}=\hslash \mathbf{\sigma }/2$. The Pauli matrices satisfy ${\sigma }_i{\sigma }_j={\delta }_{ij}I+i{ϵ}_{ijk}{\sigma }_k$. The antisymmetric term vanishes on contraction with $n_i{n}_j$, and $|\mathbf{n}|=1$, so $(\mathbf{n}\cdot \mathbf{\sigma }{)}^2=I$. Separating the even and odd terms of the exponential series givesFor the single-particle Hamiltonian, unitary time evolution consequently acts asFor two spins the joint eigenstates of total angular momentum are the spin-one-half triplet statewith $J^2=2{\hslash }^2$ and $J_3=m\hslash$, and the spin-one-half singlet stateThese follow by adding the two spin-$1/2$ angular momenta, or by direct action of the Pauli matrices.

The two terms of the specified Hamiltonian commute, so its evolution factorizes into rotations with opposite angles on the two tensor factors. The unique initial $J_3=\hslash$ state is $|\uparrow \uparrow ⟩$. With $c=\cos (\lambda t/2)$, $s=\sin (\lambda t/2)$, its evolved state isThe singlet amplitude is $i\sqrt{2}cs$, giving the requested probability

Each energy eigenstate acquires the phase $e^{-i{E}_{n}t/\hslash }$ under unitary time evolution. Consequently at the requested time $t=1$,and the wavefunction is zero outside the well. Equivalently, in elementary functions,The distinct phases retain their relative sign from the initial expansion; an overall global phase would not change measurement probabilities.