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Suppose one qubit is prepared in either of the known pure states $|0⟩$ and $|+⟩=(|0⟩+|1⟩)/\sqrt{2}$, with equal prior probabilities. Measuring in the computational basis seems natural, but a tilted projective measurement gives better quantum state discrimination. The two density operators and their difference areThe Pauli matrices give ${\Delta }^2=I/2$, and the eigenvalues of $\Delta$ are ${\lambda }_{\pm }=\pm 1/\sqrt{2}$. Their absolute sum gives trace distance $D=1/\sqrt{2}$, agreeing with trace distance between qubit states.

The positive spectral projector points along the unit Bloch vector $\mathbf{n}=(-1,0,1)/\sqrt{2}$:The outcome labels $+$ and $-$ here refer to the sign of the eigenvalue of $\Delta$; they do not denote a Hadamard basis measurement. Guess $|0⟩$ after $P_{+}$ and $|+⟩$ after $P_{-}$. This sign choice is the equal-prior Helstrom measurement for two pure states, a case of the Holevo–Helstrom theorem. ^[ref]

An explicit orthonormal basis for this projective measurement iswith $P_{\pm }=|h_{\pm }⟩⟨h_{\pm }|$. Thus the real coefficient rotation angle is $\beta =-\pi /8$ relative to the computational basis, while the Bloch vector of $P_{+}$ is tilted by $\pi /4$ from the positive $z$ axis toward negative $x$. State-vector and Bloch sphere angles differ by a factor of two.

Write $a=(1+1/\sqrt{2})/2$ and $b=(1-1/\sqrt{2})/2$, so $a+b=1$. The Born rule gives the conditional outcome probabilitiesThe event $P_{+}$ has probability gap $a-b=1/\sqrt{2}=D$, attaining the variational characterization of trace distance. Both preparations are identified correctly with conditional probability $a$, so

For comparison, a $Z$ projective measurement in the computational basis has conditional outcome probabilities $(1,0)$ on ${\rho }_0$ and $(1/2,1/2)$ on ${\rho }_{+}$. Guess zero on outcome zero and plus on outcome one. The average success isand the measured total variation distance is only $1/2$. The optimal tilted projective measurement gains about $0.103553$ in absolute success probability by balancing the two conditional error rates instead of identifying $|0⟩$ perfectly at the expense of $|+⟩$.

These are statistics for alternative measurements on independently prepared copies. Applying $Z$ first and then the tilted projective measurement is a different experiment, because the first measurement in quantum mechanics changes the density operator.

The Lüders rule specifies the ideal quantum instrument for a projective measurement. Conditional on an outcome with nonzero probability, it givesfor a density operator and a pure state, respectively. The second formula applies only to a pure state input. If the result is discarded, the output is the nonselective projective measurement ${\sum }_i{P}_i\rho P_i$.

The crucial feature for a degenerate eigenvalue is preservation of quantum coherence within its eigenspace. For a qutrit with orthonormal basis $|0⟩,|1⟩,|2⟩$, considerOn $|\chi ⟩=(|0⟩+|1⟩)/\sqrt{2}$, outcome zero has probability one and the Lüders rule leaves $|\chi ⟩$ unchanged. Measuring all three rank-one projectors and then forgetting whether the label was zero or one instead givesBoth procedures always report the coarse value zero, but a subsequent test with projector $|\chi ⟩⟨\chi |$ succeeds with probability one for the first and one half for the second. Resolving extra information destroyed coherence. This explains why a quantum nondemolition measurement of a degenerate quantity must specify what information the apparatus actually records.

Gerhart Lüders proposed this state-change rule in 1951, explicitly discussing degenerate quantities and distinguishing selected subensembles from the aggregate measured ensemble. Kirkpatrick's translation was first posted in 2004; equations (7) and (8) give the unnormalized selective update and the nonselective update. ^[ref].

Repeatability follows briefly from $P_j{P}_i={\delta }_{ij}{P}_i$: after outcome $i$, $\mathrm{Tr}(P_j{\rho }_i)={\delta }_{ij}$. This assumes a second application of the same ideal instrument without intervening dynamics. An instrument with the same effects can rotate or reset the output and need not satisfy this conclusion; see POVM does not determine the post-measurement state.

Measurement backaction is the effect of a measurement in quantum mechanics on later predictions for the measured system. A quantum instrument with one Kraus operator $K_m$ per recorded outcome hasThe conditional density operator ${\rho }_m$ exists only when $p_m>0$. Its nonlinear normalization expresses conditioning; the unnormalized map $\rho \mapsto K_m\rho K_m^{†}$ is linear. More generally, an unresolved outcome can contain several Kraus operators: replace the numerator by ${\sum }_a{K}_{ma}\rho K_{ma}^{†}$ and take its trace for $p_m$. ^[ref].

Keeping the outcome gives a selective quantum measurement. Discarding it gives a nonselective quantum measurement with quantum channel $\rho \mapsto {\sum }_m{p}_m{\rho }_m$. The latter describes the whole measured population without its labels, whereas each ${\rho }_m$ describes a selected subpopulation. Neither description requires the measured quantity to have possessed its eventual value before the interaction.

For a concrete test, prepare the Pauli X eigenstate $|+⟩=(|0⟩+|1⟩)/\sqrt{2}$. Its $X=+1$ probability is one. An ideal $Z$ projective measurement gives zero or one with equal probability, leaving $|0⟩$ or $|1⟩$ under the Lüders rule. Discarding the result leaves the maximally mixed state $I/2$. A subsequent $X$ measurement now gives either sign with probability one half, even if the $Z$ result was kept. This observable change in statistics is backaction. Measuring $X$ on the original $|+⟩$ instead leaves it unchanged: disturbance depends on both the preparation and the instrument. Measurement-based quantum feedback uses the recorded outcome to choose a subsequent control operation, so both parts of the instrument matter.

A mixed state is a density operator that is not a rank-one orthogonal projection. In finite dimension it is characterized by $\mathrm{Tr}({\rho }^2)<1$, so its purity of a density operator distinguishes it from a pure state.

Compare a pure state quantum superposition with an equal classical choice between two computational-basis states of a qubit:A $Z$ projective measurement in the computational basis gives $P(0)=P(1)=1/2$ for both density operators. An $X$ projective measurement in the Hadamard basis $|\pm ⟩=(|0⟩\pm |1⟩)/\sqrt{2}$ distinguishes them: the Born rule gives $P(+)=1$ for ${\rho }_{+}$ and $P(+)=1/2$ for ${\rho }_{\mathrm{mix}}$. These comparisons use separate copies of each preparation; measuring $Z$ first changes the state tested by a subsequent $X$ measurement.

The off-diagonal entries of ${\rho }_{+}$ encode quantum coherence in the computational basis. Such quantum coherence depends on the basis: ${\rho }_{+}$ is diagonal in the Hadamard basis but remains a pure state. Mixedness is independent of the basis, because a unitary change of basis preserves the eigenvalues and the purity of a density operator.

Even the quantum state ensemble is not unique:The off-diagonal entries of the two Hadamard basis projectors cancel. Thus the density operator retains precisely what measurements on the system can reveal, while an accessible preparation label carries additional information. A mixed state can also be the reduced density matrix of an entangled state, even when the whole system is in a pure state; classical ignorance about a selected component is not its only physical origin. ^[ref]

Write the Steane code basis as uniform quantum superpositions over a binary linear code $D$ and a disjoint coset $t+D$, where $D$ is the even-weight part of the length-seven Hamming code. For $Z(w)={⨂}_{j=1}^7{Z}_j^{w_j}$, diagonal Pauli Z gates cannot connect the two cosets. Within coset $t_a+D$, the overlap isIf $w\notin D^{\perp }$, translation by a word $x_0\in D$ with $w\cdot x_0=1$ negates this sum, so it vanishes. Since the dual code $D^{\perp }$ is the length-seven Hamming code of minimum Hamming distance three, every nonzero $w$ of Hamming weight one or two has vanishing overlap. Taking $u,v$ from $\{0,e_1,\dots ,e_7\}$ proves the displayed identity. The eight two-dimensional error images are orthogonal, so their projective measurement determines the error syndrome without revealing the logical state; applying the identified phase flip again recovers every code state. This proves nondegenerate quantum error-correcting code behaviour for the identity and all single-site phase flips, including their coherent linear span.

Choose a real unitary matrix and undo it at the second stage:After the first rotation the photon qubit is $\sqrt{q}|0⟩+\sqrt{p}|1⟩$. A dud leaves this quantum superposition unchanged, so the inverse rotation restores $|0⟩$. Its final projective measurement can never return $1$.

For a live bomb the absorbing component explodes with probability $p$. If the photon survives, the normalized state is $|0⟩$, and the second rotation makes it $\sqrt{q}|0⟩-\sqrt{p}|1⟩$. The unconditional probabilities of the three possible live-bomb outcomes are consequentlyThey sum to one. The nonexplosive result $1$ certifies a live bomb because its dud probability is exactly zero. The inconclusive result $0$ resets the qubit to its original state, allowing another identical trial.

The repeated weak-rotation bomb-test efficiency follows by summing the geometric series over the first successful trial:This is conditional on testing a live bomb, not an average over an unspecified prior fraction of duds. To make the finite-run success probability exceed $1/2-\epsilon$, choose positive $p$ so that $p/[2(2-p)]<\epsilon /2$, then choose $N$ so that $q^{2N}<\epsilon /2$. The first condition controls the limiting deficit and the second controls the remaining tail. Exactly $p=0$ yields no explosion but also no information, so the limit must be approached with nonzero weak rotations.

The half-efficiency limit is consistent with the structure of this Elitzur-Vaidman bomb tester. Suppose generally $U|0⟩=a|0⟩+b|1⟩$. A final outcome that certifies a live bomb must be orthogonal to the dud output $U^{\prime }U|0⟩$. Pulling this outcome backwards through $U^{\prime }$ gives the normalized vector orthogonal to $(a,b)$. Its squared overlap with $|0⟩$ is $|b{|}^2$. Thus the safe-certification probability is $|a{|}^2|b{|}^2$, whereas the explosion probability is $|b{|}^2$. The former never exceeds the latter. Summing these mutually exclusive terminal events over repeated, possibly adaptive trials gives $P_{\mathrm{safe}}\le P_{\mathrm{explosion}}$ and hence $P_{\mathrm{safe}}\le 1/2$. This concerns the specified rotate–interact–rotate–measure protocol.

A guaranteed exact overlap device would also distinguish two remotely prepared ensembles and permit a forbidden signal. Feed it a known reference $|{\psi }_1⟩=|0⟩$ and Bob's unknown state $|{\psi }_2⟩$, obtained from Alice's projective measurement on a shared Bell state. A device promised to work for arbitrary unknown inputs must work on this fixed-reference subset.

Alice chooses either the computational basis or the diagonal basis $|+⟩,|-⟩$. By the same remote preparation in conjugate bases, Bob receives a uniformly random member of the chosen basis. The possible exact squared overlap outputs areThese output sets are disjoint. Bob would infer Alice's basis choice from one run, without her communicating the random measurement result. Spacelike operation would contradict special relativity. Thus exact pure-state overlap readout would allow superluminal signalling, and no guaranteed exact overlap device is compatible with the stated assumptions.

This argument rules out even the phase-invariant magnitude squared. The raw complex inner product has an additional issue: replacing either pure state representative by a global phase changes that complex number without changing the physical input. The no-signalling contradiction therefore remains valid even if some phase convention is supplied. Statistical estimation from many identically prepared copies is not the promised exact single-pair readout.

Share the Bell state $|{\Phi }_{+}⟩$ between Alice and Bob, and allow their actions to be spacelike separated. The projective measurement postulate gives the remote preparation in conjugate bases identityIf Alice measures in the computational basis, Bob is prepared in $|0⟩$ or $|1⟩$, each with probability one half. If she instead measures in the conjugate basis $U^{\ast }|0⟩,U^{\ast }|1⟩$, Bob is prepared in $U|0⟩$ or $U|1⟩$, again equally likely. The conjugation is essential for a general complex unitary operator.

Interpreting distinct states as distinct physical rays, the two candidate pairs are disjoint. A hypothetical perfect identifier would tell Bob which member he received and therefore which basis Alice chose. Alice could transmit a bit by that basis choice, without sending her outcome or any other message. Bob could read it while spacelike separated from Alice, violating the no-superluminal-signalling requirement of special relativity. This establishes that perfect two-basis discrimination would allow superluminal signalling and rules out the device using the requested physical assumptions.

Ordinary measurement in quantum measurements do not signal: the two equally weighted ensembles have identical reduced density matrixThe contradiction arises from the hypothesized ability to label the ensemble's individual pure members. Global-phase variants of a vector do not count as distinct physical pure states; treating them as distinguishable labels would already introduce an unphysical extra readout.

Let $P$ be the orthogonal projector onto the quantum code $\mathcal{X}$, and choose an orthonormal basis $|i_L⟩$ for it. The necessary and sufficient Knill--Laflamme condition is that one matrix $C$, independent of the logical labels, satisfiesEquivalently, $⟨i_L|E_a^{†}{E}_b|j_L⟩=C_{ab}{\delta }_{ij}$. This makes the information in the error label independent of the encoded quantum state. It applies to the whole linear span of the errors, and hence to every noise quantum channel with Kraus operators in that span.

Here is a proof of both directions. For necessity of the Knill-Laflamme condition, let $R_{\mu }$ be Kraus operators of a common recovery, so ${\sum }_{\mu }{R}_{\mu }^{†}{R}_{\mu }=I$. Since recovery after $E_a$ must return the pure input state, every vector $R_{\mu }{E}_a|\psi ⟩$ is parallel to $|\psi ⟩$. A linear operator on the code which sends every vector to a multiple of itself is scalar: apply it to two basis vectors and their sum to force the two multipliers to agree. Therefore $R_{\mu }{E}_{a}P=c_{\mu a}P$ with no dependence on $|\psi ⟩$. It follows thatThis proves necessity without assuming in advance that the physical errors have orthogonal syndromes.

For sufficiency, $C$ is positive semidefinite: for any normalized code state and coefficients $v_a$, $v^{†}Cv=\Vert {\sum }_a{v}_a{E}_a|\psi ⟩{\Vert }^2\ge 0$. Diagonalize $C$ by a unitary change of the error basis to get operators $F_j$ with $P{F}_j^{†}{F}_{l}P=c_j{\delta }_{jl}P$. If $c_j>0$, $V_j=F_{j}P/\sqrt{c_j}$ is a linear isometry from the code to an error-image subspace, and distinct such images are orthogonal. If $c_j=0$, then $F_{j}P=0$. A projective measurement of these image subspaces followed by $V_j^{†}$ recovers every logical state. Complete the recovery on the unused orthogonal complement in any trace-preserving manner. For a coherent error ${\sum }_j{v}_j{F}_j$, the measurement branch $j$ contains $v_j\sqrt{c_j}|\psi ⟩$; the branch probability is independent of the logical state, and discarding the syndrome leaves the original state. This is constructive recovery from the Knill-Laflamme condition.

For the usual family of distinct unitary Pauli errors modulo scalar phase, nondegenerate quantum code correction means that different physical errors have orthogonal image subspaces. The precise condition isFor nonunitary errors, replace the unit diagonal by positive norm factors. In an arbitrary error basis the Gram matrix need not initially be diagonal; its positive-eigenvalue modes give the independent syndrome spaces just constructed. A kernel of $C$ exhibits a linear combination of errors acting as zero on the code. In particular two distinct physical errors with the same action on the code cannot be distinguished and give degenerate correction, as in part (c).

Here the number of errors means the number of affected physical qubits: assume that all errors supported on at most $t$ qubits are jointly correctable. It does not mean the cardinality of an arbitrary selected error list. Let $F$ be any Pauli operator of weight at most $2t$. Split its support into disjoint sets $A,B$, each of size at most $t$, and write $F=F_A{F}_B$. Each Pauli factor is Hermitian, so take $E_a=F_A$, $E_b=F_B$ to get $F=E_a^{†}{E}_b$, up to an irrelevant scalar phase if different Pauli representatives are used. The Knill--Laflamme condition givesThis is precisely quantum error detection: a projective measurement with outcomes $P,I-P$ either detects departure from the code or leaves the accepted logical state unchanged up to a scalar. Some errors can be harmless scalars on the code, particularly for degenerate correction; detection is not required to flag such an error with certainty.

Any linear operator supported on at most $2t$ qubits is a linear combination of Pauli operators on those qubits, so its code compression is scalar by linearity. ThusThis is correction implies detection at twice the weight. The premise about all locations and all error types is essential: merely correcting one selected operator gives no universal statement about every operator of twice its weight.

Use the normalized three-qubit states $|G_{\pm }⟩=(|000⟩\pm |111⟩)/\sqrt{2}$. The Shor code has logical basisIt encodes $\alpha |0⟩+\beta |1⟩$ as $\alpha |0_L⟩+\beta |1_L⟩$, preserving the unknown amplitudes by a linear isometry. It concatenates a three-block phase-flip repetition code with the three-qubit bit-flip repetition code inside each block. Six stabilizer generators, $Z_1{Z}_2,Z_2{Z}_3,Z_4{Z}_5,Z_5{Z}_6,Z_7{Z}_8,Z_8{Z}_9$, test bit errors inside the blocks. The other two stabilizer generators areAll eight commute and have eigenvalue $+1$ on both logical basis states. They define the $[[9,1,3]]$ stabilizer code. The inner checks identify single bit flips, and the outer checks identify a block phase flip; combined they also correct a Pauli Y gate error, since $Y=iXZ$.

For the requested phase flip, a $Z$ on any one qubit interchanges $|G_{+}⟩$ and $|G_{-}⟩$ in its block. It commutes with the six $Z$-pair checks. It anticommutes with $A$ or $B$ exactly when its block is contained in that check. The Shor-code phase-flip syndrome is thereforeA projective measurement of the two commuting checks yields the displayed error syndrome without learning $\alpha$ or $\beta$: both logical components have the same check eigenvalues after the error. Applying the indicated Pauli Z gate restores every encoded quantum state.

The correction is genuinely degenerate. For $i,j$ in the same block, $Z_i{Z}_j$ fixes both $|000⟩$ and $|111⟩$, so $Z_i{Z}_{j}P=P$ and consequently $Z_{i}P=Z_{j}P$. In particular $Z_1$ and $Z_2$ are distinct physical errors but have identical action on every code state; their image subspaces coincide, not merely their observed syndrome labels. In the Knill--Laflamme condition for $I,Z_1,\dots ,Z_9$, the error Gram matrix has $C_{00}=1$, $C_{0i}=C_{i0}=0$, andThe zero overlaps between different blocks follow because a phase flip changes a different pattern of orthogonal $G_{+},G_{-}$ factors; cross-logical overlaps also vanish. Thus $C$ consists of one $1\times 1$ block and three all-ones $3\times 3$ blocks and has rank four, not ten. This both proves correctability of the entire single-phase-error span and exhibits its degeneracy. A coherent sum of phase errors in one block has the same logical action with its coefficients summed, so inability to locate the individual qubit loses no logical information. The two block checks suffice to correct every single phase flip, even though the within-block errors are indistinguishable.

A measurement in quantum mechanics with outcomes $m$ is specified, in the single-operator-per-outcome description, by linear operators $M_m$ satisfying ${\sum }_m{M}_m^{†}{M}_m=I$. Its effects $F_m=M_m^{†}{M}_m$ form a positive operator-valued measure. The two measurement postulates are the Born rule for outcomes and the conditional state update:For a density operator the corresponding formulas are $p_m=\mathrm{Tr}(M_m\rho M_m^{†})$ and ${\rho }_m=M_m\rho M_m^{†}/p_m$. More general quantum instruments have multiple operators for each outcome and sum these terms. For a projective measurement of an observable $A={\sum }_m{a}_m{P}_m$, take $M_m=P_m$; the outcome value is $a_m$ and the postmeasurement state lies in its eigenspace. A POVM alone specifies probabilities, whereas the measurement operators or instrument also specify the backaction.

For quantum teleportation, Alice and Bob initially share the Bell pair $|{\Phi }^{+}{⟩}_{AB}=(|00⟩+|11⟩)/\sqrt{2}$. Alice also holds the input $|\psi {⟩}_C=\alpha |0⟩+\beta |1⟩$, with $|\alpha {|}^2+|\beta {|}^2=1$. The shared entanglement is a resource; the two transmitted classical bits alone could not transfer an arbitrary unknown quantum state.

Alice applies a controlled-NOT gate with $C$ as control and $A$ as target, then a Hadamard gate to $C$. Starting from $|\psi {⟩}_C|{\Phi }^{+}{⟩}_{AB}$, direct expansion givesFor example, the four receiver vectors are $\alpha |0⟩+\beta |1⟩$, $\beta |0⟩+\alpha |1⟩$, $\alpha |0⟩-\beta |1⟩$, and $-\beta |0⟩+\alpha |1⟩$. Thus the gate calculation fixes the ordering of the Pauli corrections, including the last branch's sign.

Alice measures $C,A$ in the computational basis, obtaining $(a,b)$. The Born rule gives each outcome probability $1/4$, independent of the input amplitudes. Bob's normalized conditional state is $X^b{Z}^a|\psi ⟩$. Alice sends the two outcome bits. Bob applies the inverse, $Z^a{X}^b$, and henceThe same circuit is a Bell-basis measurement on Alice's two qubits, expressed using ordinary gates followed by computational-basis measurements. It restores the state without Alice or Bob knowing $\alpha ,\beta$.

By linearity, the calculation also holds when the input is entangled with a reference: each outcome has branch map $|\psi ⟩\mapsto X^b{Z}^a|\psi ⟩/2$, and after correction every branch is one-half of the identity map. Summing the four corrected density operators gives exactly the original input density operator, with all reference correlations preserved. This is teleportation as an identity channel on a reference. Without the two classical bits Bob must average the four uncorrected states, giving $\frac{1}{4}{\sum }_{a,b}{X}^b{Z}^a\rho Z^a{X}^b=I/2$, so the process cannot signal before the message arrives. The Bell pair is consumed and Alice retains no copy of the unknown state, in accordance with the no-cloning theorem.

The generalized measurement postulate specifies operators $M_i$ with ${\sum }_i{M}_i^{†}{M}_i=I$. On a density matrix $\rho$, the outcome probability and conditional state areThe probabilities are nonnegative and sum to one. More generally an outcome may contain several Kraus operators $M_{i\alpha }$, in which case one sums over $\alpha$ in both formulas. A projective measurement with the usual Lüders update is the special case $M_i=P_i$, where $P_i=P_i^{†}$, $P_i{P}_j={\delta }_{ij}{P}_i$ and ${\sum }_i{P}_i=I$. Projection-valued effects alone fix the projective outcome probabilities, but an additional outcome-dependent unitary can change the conditional state; for the standard projective update the measurement operators themselves are the projectors.

A POVM is a family of positive effects $E_i$ summing to $I$. Its probabilities are $p_i=\mathrm{Tr}(\rho E_i)$. A generalized measurement gives effects $E_i=M_i^{†}{M}_i$, or $E_i={\sum }_{\alpha }{M}_{i\alpha }^{†}{M}_{i\alpha }$. Conversely $M_i=\sqrt{E_i}$ realizes any POVM. The POVM does not determine the post-measurement state: different operators with these same effects have the same probabilities and can have different state updates.

A pure POVM has rank-one nonzero effects, $E_i=w_i|{\psi }_i⟩⟨{\psi }_i|$, with normalized pure states $|{\psi }_i⟩$ and positive $w_i$. Here pure means rank-one effects, not the different property of extremality in the convex set of POVMs.

The POVM–ensemble duality for the maximally mixed state is a bijection between POVMs in dimension $d$ and ensemble decompositions of $I/d$. Given effects $E_i$, setomitting zero effects. Each ${\sigma }_i$ is a density matrix, ${\sum }_i{p}_i=\mathrm{Tr}I/d=1$, andConversely, if an ensemble satisfies ${\sum }_i{p}_i{\sigma }_i=I/d$, then $E_i=d{p}_i{\sigma }_i$ are positive and sum to $I$, hence form a POVM. Taking traces recovers the original probabilities, so the transformations are inverse. Pure POVMs correspond exactly to pure-state ensemble decompositions of the maximally mixed state. For such a POVM, $p_i=w_i/d$ and ${\sum }_i{w}_i=d$; it need not be an orthogonal measurement.

By the Schmidt decomposition theorem, suitable local unitary operations put the normalized pure state into the formIt is entangled precisely when both Schmidt coefficients are nonzero. Put $s=2{\lambda }_0{\lambda }_1$, so $0<s\le 1$ for the case of interest. The Pauli operatorssatisfy $⟨Z\otimes Z⟩=1$ and $⟨X\otimes X⟩=s$ in this pure state: $Z\otimes Z$ fixes both basis terms, and $X\otimes X$ exchanges them.

Choose Alice's two observables as $A_0=Z$, $A_1=X$, and Bob's asThese are valid dichotomic projective measurements. Indeed, $XZ+ZX=0$ and $X^2=Z^2=I$, so $B_0^2=B_1^2=I$; each observable has outcomes $\pm 1$. For the CHSH inequality convention $|⟨A_0\otimes (B_0+B_1)+A_1\otimes (B_0-B_1)⟩|\le 2$, the quantum value isFor the original pure state, conjugate these four observables by the corresponding local unitary operators. Their outcomes and Pauli correlators are unchanged, and each party still measures only their own qubit. Thus every entangled pure two-qubit state violates CHSH for suitable local measurements, proving Gisin's theorem directly. The CHSH axes for an entangled pure two-qubit state approach a nonviolating configuration continuously as one Schmidt coefficient tends to zero; the strict violation therefore need not be large for weak entanglement.

There is a necessary distinction between a failure bound averaged over the unknown state, a worst-case bound, and a separate bound for each candidate state. The last interpretation is false: a forger can discard the inputs and always prepare $|{\psi }_0{⟩}^{\otimes N}$. If the actual state is $|{\psi }_0⟩$, every authenticity test passes with probability one. The meaningful uniform statement concerns worst-case failure, or average failure under any fixed prior giving both states positive probability. The PDF does not specify that prior; the argument below gives an explicit bound for either interpretation.

Put $c=|⟨{\psi }_0|{\psi }_1⟩|\in (0,1)$. Any unconditional strategy producing $N$ output qubits, including ancillas, adaptive measurement in quantum measurements and correlated outputs, is a quantum channel $\mathcal{E}$. Its inputs and ideal outputs are the density operatorsThe projective measurements on the different banknotes commute, and their all-pass effect is exactly $P_i$. Therefore the probability of at least one rejection, conditional on input $i$, isThis remains true for entangled outputs; no independence of their individual test outcomes is assumed.

Use trace distance $D(\rho ,\sigma )=\frac{1}{2}\Vert \rho -\sigma {\Vert }_1$. The pure-state trace distance and fidelity identity givesFor completeness, two pure-state projectors differ by a matrix with nonzero eigenvalues $\pm \sqrt{1-|⟨u|v⟩{|}^2}$, which proves this identity. Also the pure-target upper bound on trace distance gives $D({\rho }_i,P_i)\le \sqrt{e_i}$. To see the mixed-state step, write ${\rho }_i={\sum }_j{q}_j|u_j⟩⟨u_j|$ and use convexity of the trace norm followed by the Cauchy-Schwarz inequality:Finally, trace-distance contraction under quantum channels gives $D({\rho }_0,{\rho }_1)\le D({\sigma }_0,{\sigma }_1)$. This follows from $D(\rho ,\sigma )={\max }_{0\le E\le I}\mathrm{Tr}[E(\rho -\sigma )]$: the adjoint channel is positive and unital, so its image of any effect $E$ is still an effect.

The triangle inequality now yieldsThus the cloning failure bound from trace-distance contraction isand in particular $\max (e_0,e_1)\ge {\Delta }^2/4$. For positive priors ${\pi }_0+{\pi }_1=1$, another Cauchy-Schwarz inequality givesso the average rejection probability satisfiesFor equal priors, one may take $p_0={\Delta }^2/8$; the same choice is a strict lower bound on worst-case rejection. This proves a quantitative version of the no-cloning theorem for the bank's all-pass test. It concerns an unconditional attempt: heralded success cannot be postselected while omitting failed attempts from the accounting.

Compute the honest one-qubit density operators in the computational basis. Writing $|+⟩=(|0⟩+|1⟩)/\sqrt{2}$ and $|u⟩=(5|0⟩+|1⟩)/\sqrt{26}$, the two quantum state ensembles giveThe equality includes the off-diagonal coherence; comparing only the basis outcome probabilities would not suffice.

The protocol is perfectly hiding against Bob when Alice is honest. Independence of the honest preparations makes his full received density operator ${\rho }_0^{\otimes N}={\rho }_1^{\otimes N}$. For any POVM $\{E_z\}$, including collective measurements and measurements using ancillas, the Born rule gives the same outcome probabilities $\mathrm{Tr}(E_z{\rho }_0^{\otimes N})=\mathrm{Tr}(E_z{\rho }_1^{\otimes N})$. Thus no measurement supplies information about the committed bit.

It is not binding against Alice. Here is an explicit ensemble-steering attack on quantum bit commitment. For each position, Alice prepares the normalized two-qubit pure stateretains $A$ and sends $B$. Its partial trace on $A$ is the honest common density operator. She sends the $B$ halves of $N$ independent copies while retaining all the $A$ halves, and chooses her bit only at unveiling.

To open one, she makes a projective measurement of each retained qubit in the computational basis. Outcome zero has probability $13/20$ and prepares Bob's state $|u⟩$; outcome one has probability $7/20$ and prepares $|1⟩$. She announces that list. It has exactly the honest bit-one distribution, and every declared-state test passes.

To open zero, she instead measures each retained qubit using a three-outcome POVM with effects $E_j=|v_j⟩⟨v_j|$, whereAll effects are positive. Their diagonal sums are $4/5+1/5=1$ and $4/70+16/70+5/7=1$, and the off-diagonal sum is zero, so $E_0+E_{+}+E_1=I$. This establishes that the proposed POVM is physically valid.

To verify the conditional states directly, the coefficient matrix of $|\Omega ⟩$, with Bob's index as the row and Alice's as the column, isA real rank-one effect $|v_j⟩⟨v_j|$ on Alice's half gives the unnormalized Bob state vector $C{v}_j$. For the three effects above,Their squared norms are therefore $1/2,1/4,1/4$, exactly the honest bit-zero probabilities, and Bob's conditional pure states are the corresponding declared states. Independent measurements on the retained halves produce the honest distribution of the entire list, so even checks on the declared frequencies do not defeat the attack. Every declared-state projector test again succeeds.

Alice can consequently unveil either bit with acceptance probability one, without having chosen it when sending the qubits. This is the Hughston–Jozsa–Wootters theorem mechanism implemented by an explicit measurement, rather than merely an appeal to the general impossibility of perfectly hiding and binding quantum bit commitment.

Let $D={\mathcal{C}}_{\mathrm{ev}}$, and write ${\mathcal{C}}_{\mathrm{odd}}=t+D$ for any odd-weight word $t\in {\mathcal{C}}_H$. The parity functional on the four-dimensional binary linear code ${\mathcal{C}}_H$ is nonzero, so its kernel $D$ has dimension three and its other fibre is the coset $t+D$. For example $t=(1,\dots ,1)$ belongs to the standard length-seven Hamming code and has odd Hamming weight. Denote the two logical basis states by $|{\psi }_0⟩$ and $|{\psi }_1⟩$, corresponding to these two cosets. The eight computational orthonormal basis vectors in each sum are distinct, so both logical states have norm one; their disjoint supports make them orthogonal.

For a binary vector $w\in {\mathbb{F}}_2^7$, define the product of Pauli Z gateswhere the dot product is computed modulo two. The key diagonal overlap, with $t_0=0$ and $t_1=t$, isIf $w\in D^{\perp }$, every summand is one. If $w\notin D^{\perp }$, choose $x_0\in D$ with $w\cdot x_0=1$. Translation $x\mapsto x+x_0$ permutes $D$ and negates every summand. The sum must then be zero. ThusOff-diagonal logical overlaps vanish for every $w$:because a diagonal Pauli operator cannot move a computational basis word from one coset to the disjoint other coset.

The possible errors are $E_0=I$ and $E_j=Z(e_j)$ for $j=1,\dots ,7$. Set $e_0=0$. Each phase flip is its own adjoint and inverse, soIf $i\ne j$, the vector $e_i+e_j$ is nonzero and has Hamming weight one or two. By the given dual code identity $D^{\perp }={\mathcal{C}}_H$ and the minimum Hamming distance three of the Hamming code, it cannot belong to $D^{\perp }$. Hence all its diagonal logical overlaps vanish. If $i=j$, the product is the identity and its logical overlaps are ${\delta }_{ab}$. Combining both cases givesThis proves the Knill--Laflamme condition with an identity error-overlap matrix, not merely a possibly degenerate scalar matrix. In particular the eight two-dimensional error subspaces $E_i{\mathcal{X}}_{\mathrm{Steane}}$ are mutually orthogonal. This is precisely nondegenerate phase-flip correction in the Steane code.

To exhibit recovery rather than only the criterion, let $P$ be the orthogonal projector onto the code and $P_i=E_{i}P{E}_i^{†}$. The projective measurement consisting of these eight mutually orthogonal projectors, completed by $I-{\sum }_i{P}_i$, determines which error occurred. For an arbitrary encoded pure state $|\chi ⟩=\alpha |{\psi }_0⟩+\beta |{\psi }_1⟩$ affected by $E_j$, outcome $j$ occurs with certainty and leaves $E_j|\chi ⟩$ unchanged. Applying $E_j$ again recovers $|\chi ⟩$. The outcome gives the error syndrome and does not reveal $\alpha$ or $\beta$, so recovery preserves logical quantum superpositions.

An equivalent explicit error syndrome uses three commuting X-type stabilizer generators. Choose a parity-check matrix for ${\mathcal{C}}_H$ with the seven different nonzero binary columns:Its rows span ${\mathcal{C}}_H^{\perp }=D$. For a row $h_l$, the operator $X(h_l)={⨂}_j{X}_j^{(h_l{)}_j}$ permutes each of $D$ and $t+D$, and therefore has eigenvalue $+1$ on both logical states. A phase flip at position $j$ changes its eigenvalue to $(-1{)}^{H_{lj}}$, since $X(h_l)Z_j=(-1{)}^{H_{lj}}{Z}_{j}X(h_l)$. Measuring the three commuting stabilizer generators gives the $j$th column of $H$ as a binary syndrome; no error gives the zero column. All eight outcomes are different, and applying the identified $Z_j$ restores the code state. Measuring these stabilizer generators never distinguishes the two logical states.