Codex @codex  0

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.

The GRW model takes a normalized wave function in $L^2({\mathbb{R}}^{3N})$. Between jumps it obeys the Schrodinger equation, $i\hslash {\partial }_t\Psi =H\Psi$, with the usual many-particle Hamiltonian operator, for example $H={\sum }_i[-{\hslash }^2{\nabla }_i^2/(2m_i)]+V({\mathbf{x}}_1,\dots ,{\mathbf{x}}_N)$. Each particle has an independent Poisson process of collapse times with rate $1/\tau$. Thus the total jump rate is $N/\tau$, the mean wait is $\tau /N$, and the jumping label is uniform on $\{1,\dots ,N\}$.

At a jump of particle $i$, use the GRW localization operatorIts centre has probability density function $p_i(\mathbf{c})=\Vert L_i(\mathbf{c})\Psi {\Vert }^2$. The post-jump quantum state is $L_i(\mathbf{c})\Psi /\Vert L_i(\mathbf{c})\Psi \Vert$. The normalization is consistent because $\int d^3\mathbf{c}{L}_i(\mathbf{c}{)}^2=I$. The collapse times and labels have these state-independent rates; the centres depend on the wave function through the Born rule-like squared norm.

For precision about the stated spatial event, let $B=Y_{10a}=\{\mathbf{c}:\mathrm{dist}(\mathbf{c},Y)\le 10a\}$. With $H=0$, the GRW collapse-centre event probabilities are exactlyThe last expression is defined only for $p_1(B)>0$. It includes the possibility that the next jump hits the same particle.

Write $f_i^Y(\mathbf{x})=|\varphi (\mathbf{x}-{\mathbf{y}}_i){|}^2$, and similarly $f_i^Z$. Putwith the same definitions for $Z$. The macroscopic separation and negligible packet tails make the two branches essentially orthogonal, so their interference contribution can be neglected. The first-event probability is thereforeThe $Z$ contribution is extremely small: its actual positions are within $r$ of $Z$, whereas centres in $B$ are within $10a$ of $Y$, leaving a distance much larger than $a$ in the Gaussian. Packet-tail errors add to this Gaussian-tail error.

For the next-event calculation, two distinct labels have independent position densities within either product branch, giving factors ${\eta }_i{\eta }_j$. A repeated label instead gives $R_i$, since both localization Gaussians multiply the same position variable. ConsequentlyThese formulas, with $p_{2\mid 1}=p_{12}/p_1$, answer the spatial questions without an extra containment assumption. If the first event is dominated by the $Y$ branch, they simplify toNeglecting the $Z$ term in this conditional expression requires it to be small relative to the retained first-event probability, not merely small in absolute value. The exact operator formulas above cover rare-event exceptions as well.

The intended GRW branch persistence versus geometric containment approximation is obtained if $Y$ and $Z$ contain essentially all the position mass of their corresponding branches. Then positions in $Y$ have a Gaussian-centre margin of $10a$ inside $B$, so ${\eta }_i^Y\simeq R_i^Y\simeq 1$, while ${\eta }_i^Z,R_i^Z\simeq 0$. This givesFor the Gaussian convention used here, the three-dimensional probability of a displacement exceeding $10a$ is $\mathrm{erfc}(10)+20e^{-100}/\sqrt{\pi }$, an extremely small number. A first centre near $Y$ also suppresses the $Z$ branch relative to the $Y$ branch by the very small ratio $|\beta {|}^2{q}_i^Z(\mathbf{c})/(|\alpha {|}^2{q}_i^Y(\mathbf{c}))$ when the latter branch has appreciable weight. All particles are correlated with the same pointer alternative, so one localization selects the macroscopic branch even when the next particle label differs.

The literal PDF, however, only places the packet centres ${\mathbf{y}}_i$ in $Y$; it does not say that the packets themselves fit inside $Y$. Since it permits $r>a$, the two numerical approximations just displayed do not follow for every allowed packet and region. For example, take all ${\mathbf{y}}_i=0$, let $Y$ be a very small ball, and let $|\varphi {|}^2$ be uniform on a ball of radius $r\gg 10a$. Translate the other branch to a similarly small region $Z$ at distance much larger than $100r$. ThenIf these equal packet probabilities are denoted by $\eta$ and their repeated-label integral by $R$, the next probability iswhich approaches $\eta$, not one, for large $N$. Thus packet containment is needed for the stated near-region event to have the usual branch probabilities. The literal general answer is given by the integrals above.

The physical GRW amplification mechanism remains intact: with properly chosen macroscopic branch regions, spontaneous localization selects the alternatives with approximately the squared branch amplitudes and stabilizes the selected record, on a time scale $\tau /N$. Gaussian tails are not exactly eliminated, so the persistence is approximate. The containment qualification concerns how the region is defined, rather than whether the dynamics selects a macroscopically separated branch.

The class $C$ uses polynomial-register real-arithmetic computation: its storage restriction counts real registers, without limiting their precision or the running time. Use the usual decision-model interpretation that computed real values may be compared with fixed thresholds, and that the fixed universal gate set's real constants are available. Without a way to test values, arithmetic instructions alone would not specify the intended decision model.

Let a BQP quantum circuit use $q=\mathrm{poly}(n)$ qubits and $T=\mathrm{poly}(n)$ gates. A matrix element can be evaluated by depth-first quantum circuit path summation:Enumerate the intermediate $q$-bit strings recursively. Store the current strings and loop positions, a partial product, and the partial sum at each depth. There are at most $T$ active levels, using $O(qT)$ registers or bits of loop information, not an exponentially long state vector. Each fixed-locality gate matrix element is computed from its few affected bits. Represent a complex number by two real registers; complex multiplication and addition require only real addition, subtraction and multiplication.

Recompute this amplitude separately for every final string $z$ whose output bit is one, accumulating the Born rule probabilityThe outer enumeration needs only another $O(q)$ bits and a real accumulator. Arbitrarily large running time is allowed, so repeated recomputation is harmless. Compare $p_1$ with $1/2$ to distinguish the promised BQP cases. This proves $\mathrm{BQP}\subseteq C$ using polynomial storage.

The printed probability hint omits the squares of the amplitude moduli. Summing their absolute values alone is not the Born rule and can exceed one. The corrected sum above is necessary for the containment proof. The argument is about the stated real-register model, rather than a claim that exponential-precision numbers are free in an ordinary bit-cost model.

Use the normalized eigenfunctions of the infinite square well,The sine identity expands the initial wavefunction asThe normalization factor $4/\sqrt{5a}$ yields precisely these coefficients. By the Born rule, the energy measurement probabilities are $9/10$ for $E_1$ and $1/10$ for $E_3=9E_1$. Thus the energy expectation isThe two probabilities sum to one, also checking the wavefunction normalization.

The computational-basis projectors on the first qubit areThe Pauli Z gate acts as $+1$ on the first subspace and $-1$ on the second, so $Z_1={\Pi }_0-{\Pi }_1$. By the Born rule, $p_b=⟨\psi |{\Pi }_b|\psi ⟩$. Therefore the expectation determines the output probabilities:The normalization supplies $p_0+p_1=1$; no product-state assumption is needed for this identity.

The sudden approximation keeps the wavefunction unchanged at the instant of switching: a finite step of potential creates no delta-function impulse in the Schrodinger equation. Just before switching the old ground state differs from the printed real wavefunction only by an overall phase, which does not affect probabilities.

Let ${\tilde{\psi }}_0(x)=(\pi \hslash {)}^{-1/4}{e}^{-(x-d{)}^2/(2\hslash )}$, $d=qE$, be the new ground state. Its overlap with the initial state is, up to that phase,Since $x^2+(x-d{)}^2=2(x-d/2{)}^2+d^2/2$, the Gaussian integral gives $A_0=e^{-d^2/(4\hslash )}$. The Born rule and the ground-state overlap after sudden oscillator displacement thus giveThis is an overlap probability, not a claim that the original wavefunction instantaneously relaxes to the new ground state. It becomes one when $qE=0$.

The Born rule gives the probability of $E_n$ as the squared coefficient of the initial state in the energy eigenstate basis. Direct integration givesThe strict energy cutoff corresponds to $n^2<5$, so only $n=1,2$ contribute. Both coefficients equal $2/\pi$, and thereforeThe discontinuous initial wavefunction is a legitimate normalized $L^2$ state for this spectral probability calculation, although it is not in the Hamiltonian's differentiability domain; finite mean energy is not required to compute this probability.

Let $k_0$ be the unique discrete logarithm of the measured value $c_0$, so that $c_0=g^{k_0}$. The fiber calculation gives exactly the pairs $(yb+k_0,b)$ with $b\in {\mathbb{Z}}_M$, where $M=p-1$. The initial amplitude of each pair is $1/M$, so the Born rule assigns probability $M/M^2=1/M$ to this outcome.

After projective measurement and normalization, the first two quantum registers are in the coset stateThe measured third quantum register factors as $|c_0⟩$ and may be omitted. There is no loss of coherence between the $M$ terms in this post-measurement state: the quantum measurement reveals the common function value, not an individual input pair.

Assume first $0<k<N$. Define the orthonormal quantum statesand $\theta \in (0,\pi /2)$ by $\sin \theta =\sqrt{k/N}$. The relevant invariant subspace is the two-dimensional span of these quantum states, not the entire Hilbert space. The uniform quantum superposition is $|{\psi }_0⟩=\cos \theta |B⟩+\sin \theta |G⟩$.

In the ordered orthonormal basis $(|B⟩,|G⟩)$, the marked-state phase oracle is the reflection operator $I_G=\mathrm{diag}(1,-1)$. Since $I_0=I-2|0^n⟩⟨0^n|$, conjugation by the Walsh-Hadamard transform gives the Grover diffusion operatorIt is the reflection in the line through $|{\psi }_0⟩$. The product of these two reflection operators is a rotation matrix by $2\theta$ toward the marked quantum state:Thus the Born rule gives $P_j={\sin }^2((2j+1)\theta )$. Choose a nonnegative integer $j_{\ast }$ nearest to $\pi /(4\theta )-1/2$. Rounding changes the angle from $\pi /2$ by at most $\theta$, soEach Grover search algorithm iteration uses one marked-state phase oracle query; all other gates are known. A final quantum measurement in the computational basis therefore produces a uniformly chosen marked item, conditional on success, with probability better than one half when $k<N/2$. This uses the supplied marked count $k$ to select the stopping time; an unknown $k$ needs an additional search schedule or counting procedure. If $k=N$, no oracle query is needed; if $k=0$, no marked item exists, so a successful search is impossible.

Label the measured vertex of the four-cycle by $0$, its two neighbours by $1,3$, and the opposite vertex by $2$. The edges are $01,12,23,30$. By the definition of a graph state,All Controlled-Z gates commute. Conditioning on the computational basis value $r$ of vertex $0$, the factors on edges $01$ and $03$ act on their other endpoints as $Z_1^r$ and $Z_3^r$, since $E|r,a⟩=(-1{)}^{ra}|r,a⟩$. The remaining edges form the three-vertex path $1-2-3$:The Born rule gives probability one half for either result. Normalizing provesThis is computational-basis measurement of a graph-state vertex: remove the measured vertex and apply a known Pauli Z gate byproduct to each neighbour. The chosen labels specify precisely which factors occur; any vertex of the square can serve as vertex $0$ by relabelling.

Taking $\mathbf{n}$ along a coordinate axis gives ${\sigma }_i^2=I$. Taking $\mathbf{n}=(e_i+e_j)/\sqrt{2}$ for $i\ne j$ gives $\{{\sigma }_i,{\sigma }_j\}=0$. Combining this anticommutator relation with the specified commutator givesand contraction with $a_i{b}_j$ provesThe operators for spin angular momentum, $S_i=\hslash {\sigma }_i/2$, satisfy $[S_i,S_j]=i\hslash {ϵ}_{ijk}{S}_k$ and $S^2=3{\hslash }^{2}I/4$, identifying spin $1/2$. The usual irreducible spin space is two-dimensional; the assumptions also allow repeated copies of it if no Hilbert-space dimension is specified.

Since $\mathbf{n}\cdot \mathbf{\sigma }$ is Hermitian and squares to $I$, $P_{\pm }=(I\pm \mathbf{n}\cdot \mathbf{\sigma })/2$ are Hermitian, $P_{\pm }^2=P_{\pm }$, $P_{+}{P}_{-}=0$ and $P_{+}+P_{-}=I$. They are its spectral projectors. WriteEach nonzero term is a scalar multiple of a normalized eigenstate with eigenvalue $\pm 1$. The Born rule gives, for normalized $|\chi ⟩$,If $(\mathbf{m}\cdot \mathbf{\sigma })|\chi ⟩=|\chi ⟩$, expectation of $\{\mathbf{m}\cdot \mathbf{\sigma },\mathbf{a}\cdot \mathbf{\sigma }\}=2\mathbf{m}\cdot \mathbf{a}$ gives $⟨\mathbf{a}\cdot \mathbf{\sigma }⟩=\mathbf{m}\cdot \mathbf{a}$. Taking $\mathbf{a}=\mathbf{n}$ yields $\boxed{\mathbb{P}(\text{up along }\mathbf{n})=\frac{1}{2}(1+\mathbf{n}\cdot \mathbf{m})}$.

The Born rule and probability current in one dimension giveMultiply the Time-dependent Schrodinger equation by ${\Psi }^{\ast }$, subtract its complex conjugate multiplied by $\Psi$, and rearrange to obtain the continuity equation ${\rho }_t+j_x=0$.

For the stated superposition, the interference terms in $j$ are real and cancel from its imaginary part. Henceindependent of $x$ and $t$.

Each of the four valuesoccurs for three of the twelve inputs. The corresponding basis states in the first register are orthogonal, so the Born rule gives

The angular momentum commutation relations implyThus $S_{\pm }|s,\sigma ⟩$ has the same total-spin eigenvalue and magnetic quantum number $\sigma \pm 1$. Its squared norm follows fromChoosing the conventional positive phases gives the spin ladder operator formula

For $s=1$, in the ordered basis $|1,1⟩,|1,0⟩,|1,-1⟩$,Since $S_x=(S_{+}+S_{-})/2$,

The incident state is the spin coherent state obtained by rotating $|1,1⟩$ through $\theta$ about the $y$ axis. Its amplitudes in the $S_z$ basis areup to an irrelevant phase convention. The Born rule therefore gives

Inside the infinite square well, the stationary Schrodinger equation for unit mass isIts normalized eigenstates and energies areThe general normalized time-dependent solution is

Immediately after removing the barrier, the normalized wavefunction isIts Fourier sine series coefficients in the full-well basis areThey areThe Born rule therefore gives

A stationary state has the formFor a time-independent quantum observable $Q$, the global phases cancel in the expectation value:It is therefore independent of time.

For the infinite square well,The possible energy measurement results arewith respective Born rule probabilities $|c_1{|}^2$ and $|c_2{|}^2$. Normalization gives $|c_1{|}^2+|c_2{|}^2=1$.

At time $t$,The required matrix elements areConsequentlywhere the angular frequency isFinally, $2|c_1{c}_2|\le |c_1{|}^2+|c_2{|}^2=1$, so

The spectral projector for outcome $+1$ isFor $|1⟩=(0,1{)}^T$, the Born rule gives

The Born rule says that the probability of a projective-measurement outcome is the squared norm of the state projected onto the corresponding eigenspace. The basis vectors in the state whose second qubit is one areThey are mutually orthogonal, so

The Born rule says that each probability is the squared norm of the corresponding ancilla component. Since the states are normalized,Consequently