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In an ontological model of a quantum system, a preparation of $|\psi ⟩$ gives a probability distribution ${\mu }_{\psi }$ over a physical state $\lambda$. A fixed quantum measurement has response probabilities ${\xi }_j(\lambda )\ge 0$ with ${\sum }_j{\xi }_j=1$, reproducing the Born rule after averaging over ${\mu }_{\psi }$. The PBR theorem says that, assuming preparation independence and the quantum predictions, distributions for distinct pure states cannot overlap with positive probability. Thus the quantum state is determined by the physical state in the psi-ontic model sense.

Preparation independence says that separately prepared systems have independent physical states: a product preparation is represented by the product of their individual ontic distributions. This is an assumption about the underlying physical states, not merely a statement that experimental preparation choices are independent. The excluded psi-epistemic model hypothesis is that the same physical state can occur with positive probability for two different pure-state preparations. The theorem neither excludes additional hidden variables nor claims that every interpretation which speaks of information is ruled out without these assumptions.

For the two given qubit preparations, consider the following PBR exclusion measurement for zero and plus, written in the ordered basis $00,01,10,11$:Direct inner products show that these four vectors are an orthonormal basis. Each labelled vector is orthogonal to the correspondingly labelled product preparation, so the associated projective measurement satisfiesThis is an example of antidistinguishable quantum states: every outcome excludes one possible preparation, although the preparations cannot be perfectly distinguished.

To prove the contradiction without requiring deterministic quantum measurement responses, choose a common dominating measure for ${\mu }_0,{\mu }_{+}$ and write their densities $m_0,m_{+}$. If they overlap, $\nu (\lambda )=\min (m_0,m_{+})$ has mass $\epsilon >0$. By preparation independence, every one of the four product densities dominates $\nu ({\lambda }_1)\nu ({\lambda }_2)$. Its total mass is ${\epsilon }^2$. The zero Born rule probability for outcome $xy$ implies that its nonnegative response ${\xi }_{xy}({\lambda }_1,{\lambda }_2)$ vanishes almost everywhere for its matching preparation, hence also under this common product measure. All four responses would then vanish on a set of positive measure, contradicting their sum being one. Therefore ${\mu }_0$ and ${\mu }_{+}$ are mutually singular.

For the second pair, group the $2n$ independent preparations into two blocks of $n$. Define $|{\Psi }_i⟩=|{\psi }_i{⟩}^{\otimes n}$. The tensor-power reduction of PBR overlap givesEach block therefore has an effective two-dimensional Hilbert space. Explicitly,are orthonormal and $|{\Psi }_2⟩=(|e_0⟩+|e_1⟩)/\sqrt{2}$. Embed the four-vector exclusion basis above into the tensor product of these two block spaces. To obtain a complete quantum measurement on all $2n$ qubits, add the orthogonal complement of that four-dimensional subspace to one of its four projectors. All four allowed preparations lie in the subspace, so the four forbidden probabilities remain zero.

If the single-copy preparation distributions overlap with common mass $\epsilon >0$, their $2n$-fold products all dominate the common measure ${\nu }^{\otimes 2n}$ of mass ${\epsilon }^{2n}>0$. The same zero-response contradiction now applies to the four block preparations. ThusThe argument is exact for the ideal devices specified in the question; no finite experimental resolution or noisy overlap bound is assumed.

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.

For any linear operator, the trace norm is $\Vert X{\Vert }_1=\mathrm{Tr}\sqrt{X^{†}X}$, the sum of its singular values. For a Hermitian operator,The Holevo–Helstrom theorem concerns minimum-error discrimination of two density operators $\rho ,\sigma$. If their prior probabilities are $p$ and $1-p$, the maximum success probability over all quantum measurements isAn optimal binary POVM declares $\rho$ on the positive spectral subspace of $p\rho -(1-p)\sigma$ and declares $\sigma$ on its negative spectral subspace; zero-eigenvalue vectors can be assigned either way. For equal priors the formula becomes $P_{\mathrm{succ}}^{\ast }=\frac{1}{2}+\frac{1}{4}\Vert \rho -\sigma {\Vert }_1$.

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.

The path graph state is $|{\psi }_3⟩=E_{23}{E}_{12}|+{⟩}_1|+{⟩}_2|+{⟩}_3$. The factor $Z_3^r$ commutes with both the Controlled-Z gates and the quantum measurement of vertex $1$. Treat $Z_1^r|+{⟩}_1$ as the input of one-bit teleportation. The equatorial qubit measurement result $s$ therefore gives the normalized stateHere the subscript on $Z_2^r$ expresses the teleported input factor; it is not an extra operation on the already measured vertex. Use $J(\alpha )Z^r=X^{r}J(\alpha )$ to obtainThe exclusive or exponent is equivalent to the printed sum $r+s$ because $X^2=I$. The conditional probability of $s$ is one half for either $r$, by the one-bit teleportation branch norm. This calculation concerns an equatorial measurement of a graph-state leaf; no third remaining qubit or extra $|-⟩$ factor is present.

The basic one-bit teleportation primitive uses the input as qubit 1 and a fresh $|+⟩$ quantum ancilla as qubit 2. Apply the Controlled-Z gate $E_{12}$ and measure qubit 1 in the equatorial qubit measurement basis with angle $\alpha$. If $s\in \{0,1\}$ is the result, the normalized post-measurement state on qubit 2 isBoth results have probability one half. For example, writing $|\psi ⟩=a|0⟩+b|1⟩$, the unnormalized branch is $(a|+⟩+(-1{)}^s{e}^{i\alpha }b|-⟩)/\sqrt{2}=X^{s}J(\alpha )|\psi ⟩/\sqrt{2}$. Thus $J(\alpha )$ is applied deterministically as a logical gate with known Pauli frame $X^s$; a permitted conditional Pauli X gate would remove this byproduct physically.

The literal resource restriction does not need an unstated direct Pauli X gate. A Pauli Z gate can be enacted by applying $E$ to the data and a fixed $|1⟩$ quantum ancilla. Two consecutive angle-zero one-bit teleportations, with results $p,q$, map an arbitrary current state $|\chi ⟩$ to $X^{q}H{X}^{p}H|\chi ⟩=X^q{Z}^p|\chi ⟩$. Apply the available $Z^p$ to remove the latter factor, up to an irrelevant global phase. This implements a heralded $X^q$: if $q=0$, the input is unchanged; if $q=1$, the required Pauli X gate has been applied.

If the original result is $s=1$ and a physically corrected output is required, repeat this heralded Pauli X correction using controlled-Z and measurements until $q=1$. Each attempt succeeds with probability one half independent of the input, so it terminates with probability one, using two attempts on average. Only $E$, fixed ancillary quantum states and single-qubit quantum measurements are used. This exact physical correction has no finite worst-case measurement bound; the standard finite deterministic realization is the logical Pauli frame version, which suffices for the later output-simulation parts.

Let the ideal two-qubit output before final quantum measurements beThe PDF places $J(\alpha )$ only on the upper wire, followed by the Controlled-Z gate and a quantum measurement in the computational basis on each output wire. Measure vertex $0$ of the square in the computational basis, obtaining $r$, then vertex $1$ of the surviving path in the fixed equatorial qubit measurement basis at angle $\alpha$, obtaining $s$. Put $t=r\oplus s$. The previous part gives $E_{23}{X}_2^{t}J(\alpha {)}_2{Z}_3^r|++⟩$.

Commute its Pauli frame through the Controlled-Z gate. Since $E_{23}{X}_2^t=X_2^t{Z}_3^t{E}_{23}$ and $E_{23}$ commutes with $Z_3^r$, the actual output isFinally measure vertices $2,3$ in the computational basis, with raw results $d_2,d_3$. A Pauli X gate flips a computational basis result, whereas a Pauli Z gate changes only its phase. Thus the purely classical correction isThis four-cycle graph-state simulation of an entangle-and-measure circuit reproduces the full joint output distribution, not just each marginal. All measurement bases are fixed beforehand, and no physical byproduct correction is necessary. As a check, the ideal Controlled-Z gate is diagonal in the output basis, so $P(k,l)=\frac{1}{2}{\cos }^2(\alpha /2)$ for $k=0$ and $\frac{1}{2}{\sin }^2(\alpha /2)$ for $k=1$, independently of $l$. The measurement procedure gives exactly these probabilities after its classical relabelling.

The logical depth of a measurement pattern counts sequential layers of quantum measurements forced by dependence of measurement bases on earlier outcomes. Depth one means every basis can be fixed before measurements start, allowing all quantum measurements on distinct qubits to be performed in parallel. This does not require depth-one resource-state preparation or constant-depth classical parity processing.

Suppose the circuit contains $m$ gates $J({\alpha }_1),\dots ,J({\alpha }_m)$ in time order, with each ${\alpha }_j\in \{0,\pi /2\}$. Prepare an $(m+1)$-vertex path graph state from $|+{⟩}^{\otimes (m+1)}$ and Controlled-Z gates between neighbours. Its first qubit supplies the specified input $|+⟩$. Measure vertex $j-1$ in the fixed equatorial qubit measurement basis at angle ${\alpha }_j$ for $j=1,\dots ,m$, obtaining $s_j$. Measure the final vertex in the computational basis, obtaining $d$.

Track the Pauli frame as $X^{a_j}{Z}^{b_j}$, starting with $a_0=b_0=0$. A hypothetical sequential reading of the same quantum measurements gives the step $X^{s_j}J({\alpha }_j)X^{a_{j-1}}{Z}^{b_{j-1}}$. The supplied commutation rules imply, up to global phase,For ${\alpha }_j=0$, the sign is irrelevant. For ${\alpha }_j=\pi /2$, the identity $J(-\pi /2)=XJ(\pi /2)$ absorbs the possible sign change into the Pauli frame. If $t_j=0$ for ${\alpha }_j=0$ and $t_j=1$ for ${\alpha }_j=\pi /2$, the update isInductively the final unmeasured quantum state would be $X^{a_m}{Z}^{b_m}J({\alpha }_m)\cdots J({\alpha }_1)|+⟩$. The final computational basis outcome is corrected by $a_m$ and is unaffected by $b_m$. The induction on a sequential interpretation proves the joint statistics; projectors on distinct vertices commute, so the identical fixed-basis pattern can actually be measured simultaneously, including its final vertex.

The first printed commutation formula has an incorrect exact scalar phase: with the printed matrix definition, $J(\alpha )X=e^{i\alpha }ZJ(-\alpha )$, rather than the negative-exponent prefactor. For example, at $\alpha =\pi /4$ the two versions differ by a factor $i$. They agree up to global phase, so this error does not affect the Pauli frame recurrence or any quantum measurement probability.

Consequently a fixed-basis path-state pattern of logical depth one simulates every such circuit. Only Pauli measurements occur: the angle-zero basis measures $X$, and the angle-$\pi /2$ basis measures $-Y$ with the printed eigenvector labels. Classical exclusive or processing suffices for all output corrections. These gates are Clifford gates, which explains why angle adaptation can be eliminated. For $m=0$, simply measure the initial $|+⟩$ in the computational basis; it is already a depth-one pattern.

Assume no previously shared entanglement. Alice's encoded ensemble $\{p_x,{\rho }_x\}$ lives in a Hilbert space of dimension $2^n$. Its Holevo quantity satisfiesusing nonnegativity of Von Neumann entropy and the maximum entropy of a quantum state. The data-processing inequality for quantum relative entropy implies ${\chi }_{\mathrm{out}}\le {\chi }_{\mathrm{in}}$ for the transmission quantum channel. Finally, the Holevo bound applied to Bob's quantum measurement givesThis argument also works when the output system has larger dimension than the input. Prior shared entanglement changes the communication resource: superdense coding can transmit two classical bits per sent qubit, so the absence of that resource is part of this bound's setting.

Let $P_{+}$ be the maximizing projection in the variational characterization of trace distance. Use the binary POVM $\{P_{+},I-P_{+}\}$ and its measurement channel. The difference between its two output probability vectors iswhere $D=D(\rho ,\sigma )$ and the second equality uses $\mathrm{Tr}(\rho -\sigma )=0$. Since trace distance between diagonal density operators is half the L1 norm of their probability-vector difference,Thus a binary quantum measurement can preserve the distinguishability of this particular pair exactly. This is the trace-distance-preserving binary measurement, which may depend on the two states.

With equal priors, the probability of a correct result from this quantum measurement isThe two required amplitudes areThereforeIts maximum occurs at $\beta =-\pi /8$ and is

Here $|⟨a_0|a_1⟩|=1/\sqrt{2}$, so $\theta =\pi /4$ and $\sin \theta =1/\sqrt{2}$. The stated Helstrom-Holevo bound is consequently exactly the same value, and the basis above is the Helstrom measurement for the zero and plus states.

In quantum channel discrimination, prepare a density operator ${\rho }_{HR}$ on the channel input $H$ and an optional quantum ancilla $R$. Under hypothesis $j\in \{1,2\}$ the output isUse a two-outcome quantum measurement $\{Q,I-Q\}$ and decide for $T_1$ on outcome $Q$. The conditional Type I error and Type II error areWith prior probabilities $p$ and $1-p$, symmetric Bayesian discrimination minimizes the average error $p\alpha +(1-p)\beta$ over the input and measurement.

For input pairs other than $11$, either local output is the equiprobable ensemble $\{|+⟩,|-⟩\}$; for input $11$, it is $\{|0⟩,|1⟩\}$. Both ensembles have the same density operator,Every local quantum measurement therefore has the same statistics for every remote input. The joint outputs have unusual correlations, but observing them requires the parties to compare results through an ordinary causal channel. Hence this device also does not necessarily allow superluminal signalling.

One obtains a logically consistent extension by specifying three rules: ordinary quantum evolution and localized objective collapses determine the global state; the past-light-cone prescription determines a unique local density operator at every readout event; and a readout copies a classical description of that operator into a classical register without changing the quantum state. These rules assign an output to every finite-dimensional input and every spacetime history.

The device is outside the class of ordinary quantum measurements and permits tasks such as learning and then preparing a copy of an unknown pure state, but violating the operational no-cloning theorem for two pure states does not make the enlarged axioms contradictory. Feed-forward operations can be defined from earlier readouts in the same causal region. Because every output depends only on collapse events in its past light cone, no closed causal dependence or spacelike signal is introduced. The extension is therefore mathematically coherent, although it is hypothetical and goes beyond standard quantum theory.

Postselection conditions a quantum measurement on a specified outcome. If its branch acts by a linear operator $K$ on a normalized input $|b⟩$, its probability is $\Vert K|b⟩{\Vert }^2$, and its normalized output is $K|b⟩/\Vert K|b⟩\Vert$. This conditional state is defined only when the probability is nonzero. Physically unsuccessful trials are discarded; postselection is not a deterministic implementation of the nonlinear normalization map.