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First explain the stated local conversion. The four equally weighted correlated local Pauli operations $I\otimes I$, $X\otimes X$, $Z\otimes Z$, and $Y\otimes Y$ remove coherences in the Bell states basis, since those states have distinct joint parity eigenvalues. This leaves four Bell probabilities. One-party Pauli operations permute the Bell projectors, so the largest probability can be placed on the singlet ${\Phi }_{-}$. Call it $p$; necessarily $p\ge 1/4$.

The two common unitaries found above generate permutations of the three triplet projectors: $S\otimes S$ exchanges ${\Psi }_{+}$ and ${\Psi }_{-}$, and $HSH\otimes HSH$ exchanges ${\Psi }_{+}$ and ${\Phi }_{+}$. Averaging the six triplet permutations replaces their weights by their common mean, while collective-unitary covariance of the two-qubit singlet fixes the singlet projector. This is Bell twirling followed by triplet symmetrization, producing the specified two-qubit Werner state. Shared randomness coordinates the local choices, so the procedure is allowed by local operations and classical communication. Common unitaries alone would preserve the original singlet weight, which need not initially be at least $1/4$; the preceding one-party permutation is what ensures that range.

For the eigenvalue calculation, put $P_s=|{\Phi }_{-}⟩⟨{\Phi }_{-}|$, $a=(1-p)/3$, and $b=p-a=(4p-1)/3$. Then $\rho =aI+b{P}_s$. In the ordered computational basis $00,01,10,11$, transposing the second subsystem givesThe $00,11$ block has eigenvalues $a-b/2$ and $a+b/2$, while the other two basis vectors also have eigenvalue $a+b/2$. Thus the partial transpose spectrum isThe last three are positive throughout the given range, so the partial transpose is not positive exactly when $p>1/2$. A separable quantum state is a convex sum of product density operators, each still positive after transposing one factor. The negative eigenvalue therefore proves entanglement.

For completeness, separability on the other side can be proved explicitly rather than inferred solely from a theorem name. Let $P_{j,\pm }=(I\pm {\sigma }_j)/2$ be the one-qubit Pauli eigenprojectors. At $p=1/2$, the mixture of six antiparallel product states isusing $P_s=(I-{\sum }_j{\sigma }_j\otimes {\sigma }_j)/4$. It is manifestly separable. For $1/4\le p\le 1/2$,with nonnegative coefficients summing to one; $I/4$ is also separable. ConsequentlyAt the boundary $p=1/2$ the partial transpose has a zero eigenvalue and the explicit decomposition shows separability.

Write $|0⟩=|\uparrow ⟩$, $|1⟩=|\downarrow ⟩$, and use $X_j,Y_j$ for the Pauli matrices on particle $j$. Their actions on the computational basis areSet $P_1=X_1{Y}_2{Y}_3$, $P_2=Y_1{X}_2{Y}_3$, $P_3=Y_1{Y}_2{X}_3$, and $P_4=X_1{X}_2{X}_3$. Any two distinct operators in this list differ by an $X$ versus $Y$ at exactly two sites. At each such site $XY=-YX$, while factors on different particles commute. Swapping the two full products therefore introduces two minus signs, givingThis is the Pauli-string commutation parity rule. Each product is Hermitian and squares to the identity, so its only eigenvalues are $\pm 1$.

For the negative-phase Greenberger–Horne–Zeilinger state, put $|G_{-}⟩=(|000⟩-|111⟩)/\sqrt{2}$. The two $Y$ factors in each of $P_1,P_2,P_3$ contribute $i^2=-1$ on $|000⟩$ and $(-i{)}^2=-1$ on $|111⟩$. Thus these operators take $|000⟩$ to $-|111⟩$ and $|111⟩$ to $-|000⟩$. By contrast, $P_4$ flips these two basis states without a phase. ConsequentlyThe GHZ state is therefore a simultaneous eigenstate of the four commuting products. As a sign check, direct multiplication givesThis is compatible with the displayed quantum eigenvalues.

Now apply the EPR criterion of reality together with locality for the spacelike separated particles. In the context with local settings $X_1,Y_2,Y_3$, the product of the three outcomes is certainly $+1$. Therefore the two remote $Y$ outcomes predict the first particle's $X$ outcome with certainty, without disturbing it under the locality assumption. The other perfect-correlation contexts similarly predict each particle's $X$ or $Y$ outcome from measurements performed only on the other particles. The EPR criterion of reality then assigns predetermined local values $x_j,y_j\in \{+1,-1\}$ to these six observables. Locality makes the value assigned at one site independent of which measurements were chosen at the other sites.

To reproduce the three certain mixed-setting quantum predictions, every such assignment must satisfyMultiplying these scalar equations cancels each squared $y_j$ and gives $x_1{x}_2{x}_3=+1$. But the all-$X$ quantum prediction, from the $P_4$ eigenvalue, is $x_1{x}_2{x}_3=-1$ with certainty. HenceThe perfect GHZ correlations contradict the simultaneous local elements of reality inferred by the EPR argument. No statistical Bell inequality is needed. The locality assumption is essential to that inference; the certainty criterion without the premise that distant choices cannot disturb the local element does not by itself imply a local hidden-variable model. Nor does the argument require simultaneous measurement of $X$ and $Y$ on a single particle: different experimental contexts establish the perfect predictions, and it is the EPR/locality interpretation that attempts to assign all their values at once.

Write $N=2k+1$ and abbreviate the Dicke states by $|D_j^n⟩=|{\Psi }_j^n⟩$. Alice's shared resource has decompositionFollow the actual circuit order: a controlled-NOT gate from the input to $q_0$, a Hadamard gate on the input and a Pauli X gate on $q_0$, followed by computational basis measurements. Just before measurement, the joint vector isThe Dicke states in each branch are orthogonal, so each branch has squared norm $1/4$. Conditioning on outcome $00$ and normalizing therefore gives the state of the other $N$ parties:This is the encoding used in Dicke-resource telecloning. The extra Pauli X gate on Alice's resource qubit is essential to the displayed ordering of the amplitudes.

Put $L=2^N$ and define $\theta =\arcsin \sqrt{M/L}$. The normalized good and bad vectors have disjoint computational basis supports, so they are orthogonal. Counting the terms in the uniform quantum superposition givesSince $V_0=I-2|0^N⟩⟨0^N|$, the Hadamard transform gives $-H^{\otimes N}{V}_0{H}^{\otimes N}=2|\varphi ⟩⟨\varphi |-I$, the Grover diffusion operator. On the ordered orthonormal basis $(|{\Psi }_g⟩,|{\Psi }_b⟩)$, the two reflection operators have matricesTheir product isThis is a plane rotation through angle $-2\theta$ in the displayed coordinate convention, with the invariant good-bad plane preserved by both operators. In particular its action sends $(\sin a,\cos a)$ to $(\sin (a+2\theta ),\cos (a+2\theta ))$. Induction therefore gives the Grover rotation angle formula

Measuring the state from part (a) in the computational basis succeeds with probability $p_n={\sin }^2((2n+1)\theta )$. In the sparse-solution regime $0<M\le 2^{N-1}$, one has $0<\theta \le \pi /4$. The first success interval with $p_n\ge 1/2$ is $\pi /4\le (2n+1)\theta \le 3\pi /4$. Entering it requires $n\ge \pi /(8\theta )-1/2$, so the first half-probability Grover iterate isFor this integer, the angle is at least $\pi /4$ and less than $\pi /4+2\theta \le 3\pi /4$, so the rounding cannot skip the required interval. Every smaller nonnegative integer has angle below $\pi /4$ and hence success below $1/2$, proving minimality. For $M\ll 2^N$, the leading size is $n_{\min }\sim (\pi /8)\sqrt{2^N/M}$. If $M\ge 2^{N-1}$, the initial measurement already succeeds with probability at least one half and the smallest count is zero; if $M=0$ no search can succeed.

Initialize the register inThis can be prepared by Pauli X gates on the one-bits of $y$, starting from $|0^N⟩$, followed by Hadamard gates on all qubits. Alternatively, prepare the original uniform state and apply Pauli Z gates at those same bit positions. The conjugation calculation givesThe final diagonal sign operator changes no computational basis outcome probabilities. Thus the success probability and the iteration count in part (b) are unchanged. No final correction is needed when the objective is to measure a marked string.

With access to the uncontrolled physical gate, use single-qubit process tomography. Prepare the positive eigenstate of each of $X,Y,Z$, apply $V$, and on independent copies measure each of $X,Y,Z$. Arbitrary single-qubit rotation gates let one prepare these inputs and rotate each measurement into the computational basis. The nine measured means areThey determine the real rotation matrix $R$ of the Bloch vector, and hence the action of $V$ on every qubit state. To see precisely what this identifies, suppose another unitary $U$ has the same action. Then $V^{†}U$ commutes with every Pauli matrix. Commuting with $Z$ makes it diagonal; commuting with $X$ makes its diagonal entries equal. Therefore $U=e^{i\gamma }V$.

Consequently tomography determines $V$, and its eigenvalues, up to a common phase. More explicitly, write a representative as $V=e^{i\gamma }(\cos aI-i\sin a\mathbf{n}\cdot \mathbf{\sigma })$. Its eigenvalues are $e^{i(\gamma -a)}$ and $e^{i(\gamma +a)}$, and its Bloch vector action is rotation through $2a$ about $\mathbf{n}$. The measured rotation identifies the unordered relative eigenphase information; for example $\mathrm{Tr}R=1+2\cos (2a)$. Once a representative with a specified overall phase is supplied, diagonalizing that representative gives the corresponding two eigenvalues.

The common phase itself cannot be recovered with the stated uncontrolled access. Indeed, $(e^{i\gamma }V)\rho (e^{i\gamma }V{)}^{†}=V\rho V^{†}$ for every input, including inputs entangled with ancillas. Replacing $V$ by $e^{i\gamma }V$ therefore leaves every such preparation-and-measurement experiment unchanged, whereas it multiplies both eigenvalues by $e^{i\gamma }$. For example, $I$ and $e^{i\gamma }I$ are indistinguishable physical gates under these operations. This proves that the request for absolute eigenvalue phases needs an additional phase reference or controlled-gate assumption: the global phase of an uncontrolled quantum gate is unobservable. The controlled powers in the earlier parts supply that extra resource, since $\mathrm{diag}(I,e^{i\gamma }V)$ changes a relative phase between the two control branches.

The practical advantages are that no unknown eigenstate preparation, controlled powers, entangling gates or inverse quantum Fourier transform are needed: a fixed set of single-qubit preparations and measurements suffices. The disadvantages are statistical estimation and the phase ambiguity just proved. Estimating the bounded Pauli measurement means to additive accuracy $ϵ$ at fixed confidence requires order ${ϵ}^{-2}$ independent repetitions, giving the usual sampling cost for this direct procedure. The earlier quantum phase estimation circuit uses $N$ phase qubits and high controlled powers to resolve scale $2^{-N}$ in one coherent experiment with a constant success bound; constructing those powers from controlled uses of $V$ costs $2^N-1$ gate calls, so its exponential-looking precision is not obtained with only $N$ uses of $V$. Repetitions can raise confidence. Finally, full process tomography grows exponentially with the number of system qubits, whereas the eigenstate-based phase procedure can estimate a selected eigenphase without reconstructing the entire operator.

The EPR criterion of reality proposes that a quantity has an element of physical reality if its value can be predicted with certainty without disturbing the system. To use remote predictions this way, one additionally assumes locality: the choice and execution of a measurement in quantum mechanics on separated particles cannot disturb the local physical properties of the remaining particle. The contradiction below is with this combination of the criterion and locality, rather than with a definition of reality alone.

Use the computational basis $|0⟩=|\uparrow ⟩$, $|1⟩=|\downarrow ⟩$ and the conventional Pauli matricesFor each site $j$, let $O_j$ be the tensor product of $X$ at that site and $Y$ at all other sites. For $N=4M+3$, the number of $Y$ factors is $N-1=4M+2$, soThe minus sign in the given GHZ state then impliesHence every one-$X$, remaining-$Y$ product has measurement outcome $+1$ with certainty, while the all-$X$ product has outcome $-1$ with certainty. These are experimentally separate local Pauli measurements on identically prepared copies, not joint measurements of $X$ and $Y$ on the same particle.

For a chosen site, the other $N-1$ sites can determine its $X$ outcome with certainty by measuring $Y$ and using the corresponding $O_j$ relation. They can determine its $Y$ outcome with certainty by selecting an $O_{\ell }$ with $\ell \ne j$, measuring $X$ at $\ell$ and $Y$ at all other remote sites. Since $N\ge 3$, both choices are available. Locality and the EPR criterion of reality therefore assign each site definite values $x_j,y_j\in \{+1,-1\}$, independent of which remote choice is made, even though quantum theory does not jointly measure the local $X$ and $Y$.

The certain one-$X$ correlations require those assigned values to obeyMultiplying these scalar equations givessince $N-1$ is even and $y_k^2=1$. Thus the all-$X$ product would necessarily be $+1$. But its quantum value is $-1$, givingThe equations cannot hold even for one assignment of values, so allowing a probability distribution over assignments cannot remove the contradiction. This is the GHZ contradiction for N congruent to 3 modulo 4 for every $M\ge 0$. Multiplying classical assigned values is crucial: one must not commute the incompatible local Pauli operators as though they were ordinary numbers. Their noncommutativity is exactly what permits the quantum correlations to evade the classical product argument.

The physical space of $n$ qubits is ${\mathcal{H}}_n=({\mathbb{C}}^2{)}^{\otimes n}$. An error operator is a linear operator on this space, for example one Kraus operator of a noise quantum channel. The usual channel description assumes initially uncorrelated system and environment, with the environment subsequently ignored; more general initial correlations need not define a channel on arbitrary system inputs. Finite dimension and the tensor decomposition into identifiable qubits allow an exact Pauli expansion of a quantum error. No independence between errors on different qubits is assumed by this expansion.

The $4^n$ tensor products of $I,X,Y,Z$ form an orthogonal basis of the operator space under the Hilbert-Schmidt inner product. Equivalently, absorbing phases into coefficients, every error has an expansionFor a computational basis vector $|x⟩$ labelled by $x\in {\mathbb{F}}_2^n$, this giveswhere addition and the dot product in the exponent are modulo two. Thus $X$ changes a bit, $Z$ changes its phase, and $Y$ performs both up to a scalar phase. A general error is a coherent linear combination of such actions; it need not be a classical random choice of Pauli operators. A restriction to at most $t$ faulty qubits is an additional locality assumption, expressed by the span of Pauli operators of weight at most $t$.

The weight of a Pauli operator is the number of tensor factors different from $I$, ignoring global phase. Let $P$ be the orthogonal projector onto a $2^k$-dimensional code $\mathcal{X}$. Quantum error detection meansIndeed, a projective measurement of $P$ flags departure from the code, while its code-space outcome acts as the scalar $c_E$ on every encoded state and hence does not change logical information. Harmless scalar actions are included in this criterion. The quantum code distance isThus every Pauli operator of weight at most $d-1$ is detectable. The condition is linear in $E$, so it holds for any error in their span. This proves detection of arbitrary errors affecting at most $d-1$ qubits, including coherent errors.

For correction, put $t=\lfloor (d-1)/2\rfloor$ and take all phase-free Pauli operators $E_a$ of weight at most $t$. Their pairwise products satisfyThese are the Knill--Laflamme conditions. Here is their recovery construction, rather than merely using the criterion's name. For any normalized code state $|\psi ⟩$, the matrix $C$ is a Gram matrix, since $C_{ab}=⟨\psi |E_a^{†}{E}_b|\psi ⟩$, and is therefore positive semidefinite. Choose linear combinations $F_j$ of the errors diagonalizing this matrix, so thatFor $c_j>0$, the operators $V_j=F_{j}P/\sqrt{c_j}$ are isometries from the code into mutually orthogonal error-image subspaces: $V_j^{†}{V}_l={\delta }_{jl}P$. Measure the projectors $Q_j=V_j{V}_j^{†}$ and, on outcome $j$, apply $V_j^{†}$. Complete this recovery arbitrarily on the remaining orthogonal subspace to make a trace-preserving quantum channel.

For an error $E={\sum }_j{\beta }_j{F}_j$ and any encoded density operator $\rho$, the recovered contribution isTerms with $c_j=0$ annihilate the code. For a physical noise channel whose Kraus operators belong to this span, summing the displayed scalar over those operators gives one by trace preservation. The resulting recovered state is exactly $\rho$. This proves the constructive recovery from the Knill-Laflamme condition and the requested guaranteeThis argument allows degenerate codes; distinct physical errors may have identical actions on the code.

For the Shor code, define $|G_{\pm }⟩=(|000⟩\pm |111⟩)/\sqrt{2}$. Its two normalized basis codewords areSix stabilizer generators are $Z_1{Z}_2,Z_2{Z}_3,Z_4{Z}_5,Z_5{Z}_6,Z_7{Z}_8,Z_8{Z}_9$; the two phase checks areAll eight have eigenvalue $+1$ on both basis codewords. A phase error $Z_4$ commutes with the six $Z$ checks and anticommutes with both $g_7$ and $g_8$. Their projective measurements therefore returnThe two negative phase checks identify the middle three-qubit block. Apply the Pauli Z gate $Z_4$ to restore the state, since $Z_4^2=I$. This procedure acts identically on every logical superposition, and the checks reveal no logical amplitudes. The syndrome cannot distinguish $Z_4$ from $Z_5$ or $Z_6$, but all three have the same action on the code: their pairwise products lie in the stabilizer group. Correcting with any one of them is sufficient. Thus the degeneracy in the syndrome is harmless.

Yes: use superdense coding, with a shared Bell state. Let the sender and receiver initially hold the first and second qubits of $|{\Phi }^{+}⟩=(|00⟩+|11⟩)/\sqrt{2}$. To encode bits $(a,b)$, the sender applies $Z^a{X}^b$ to her qubit. These Pauli operators produce four orthogonal states:She sends this qubit through the noiseless quantum channel. The receiver then possesses both qubits. He applies a controlled-NOT gate from the received qubit to his original qubit, followed by a Hadamard gate on the received qubit. This maps the four states in the table to $|ab⟩$. A measurement in the computational basis therefore recovers the two bits with certainty. No classical message is sent.

The resource assumption deserves care: one-qubit superdense coding needs a Bell pair already shared before encoding. If none is initially available, the sender can prepare a Bell pair locally and send one half through the same noiseless quantum channel in advance. The protocol then uses two transmitted qubits in total, including distribution of the entanglement, still with no classical communication. Alternatively, sending two computational basis qubits directly also communicates two bits. With a pre-shared Bell pair, one transmitted qubit suffices; without it, the entanglement-distribution cost must be counted. An unassisted single transmitted qubit cannot perfectly encode four classical messages: its Holevo bound is at most ${\log }_{2}2=1$ bit.

In the ordered computational basis $00,01,10,11$, the first two qubits of $|{\psi }_{0000}⟩$ have amplitudes $(1,1,1,1)/2$, while those of $|{\psi }_{1111}⟩$ have amplitudes $(-1,1,1,1)/2$. Both have the same normalized answer qubit $|-⟩$. Their inner product is thereforeA common unitary operator $U\otimes I$ preserves this nonzero inner product. To distinguish the alternatives with certainty using the prescribed quantum measurement in the computational basis, their first-register outcome supports would have to be disjoint: every outcome possible under one alternative must be impossible under the other. Vectors with disjoint computational basis supports have zero inner product, contradicting its preservation. Thus no such $U$ exists. This is also an instance of perfect discrimination of pure states requires orthogonality; allowing a more general measurement in quantum mechanics would not rescue this pair of states. The obstruction concerns these prepared oracle-output states, rather than every conceivable oracle-query procedure.

Write $N=|A|$. For $N>1$, the unmarked computational basis vectors in $A$ give the normalized vectorIts norm is one and $⟨a|\omega ⟩=0$, so these two vectors form an orthonormal basis for $\mathcal{V}$. Splitting the uniform superposition state into its marked and unmarked terms givesThe nonnegative coefficients select the specified angle in $[0,\pi /2]$.

There is one degenerate boundary case: if $N=1$, then $|{\psi }_A⟩=|a⟩$ and $\mathcal{V}$ is one-dimensional. No vector in $\mathcal{V}$ can complete $|a⟩$ to a two-element orthonormal basis. Thus the printed two-dimensional description needs $N>1$. For $N=1$ the search itself is already solved, with $\theta =\pi /2$ and no query.

Label the chain cluster state from top to bottom by $1,2,3,4$, and let $k,l,m$ be the phase-measurement outcomes on the first three vertices, with raw computational basis outcome $s$ on vertex four. Repeated one-bit teleportation leaves the last qubit, up to global phase, inThis sequential description applies even though the whole graph state is prepared first: a later Controlled-Z gate acts only on unmeasured qubits and commutes with earlier single-qubit measurement operators. It can therefore be moved past those measurements for the purpose of calculating each conditional branch.

Take $\alpha =\theta$ and $\beta =0$, so $U(0)=H$. After the second measurement the logical state isFor binary $p,q$, the identities proved above imply the Pauli-frame propagation along a measurement wire rulewhere $\simeq$ means equality up to global phase. Choose the third angle adaptively as $\gamma =(-1{)}^{l\oplus 1}\varphi$. Then $(-1{)}^l\gamma =-\varphi$, andReading the target circuit in temporal order gives its premeasurement stateConsequentlyThe Pauli Z gate changes only phases in the computational basis, while the Pauli X gate exchanges its two measurement outcomes. Thus the explicit choices areIndeed, for every earlier outcome branch,Hence averaging over the earlier random outcomes leaves exactly the target probability distribution. This implements the desired measurement-based quantum computation without physically removing the final Pauli frame.

Let $\varphi =j\pi /2$ for an integer $j$, and put $q=j\mathrm{mod}2$. Use the fixed phase-measurement angles $(\theta ,0,-\varphi )$ and a fixed final computational basis measurement. Denote their outcomes by $k,l,m,s$. The adaptive construction would instead use $-\varphi$ when $l=0$ and $+\varphi$ when $l=1$.

For the printed measurement vectors,Therefore $+\varphi$ and $-\varphi$ define the same unordered pair of rank-one linear projections; the labels are exchanged if and only if $q=1$. In the branch $l=1$, the outcome that would be called $m_{\mathrm{ad}}$ in the adaptive experiment is thus $m\oplus q$. In both branches,The output correction from the preceding part becomesIf $j$ is even this reduces to $s\oplus k\oplus m$; if $j$ is odd an additional $l$ is included. This equatorial measurement relabeling at Clifford angles changes only the classical interpretation, not any physical measurement basis.

All four fixed single-qubit projective measurements act on different vertices, so their operators commute. They can be performed simultaneously, with the displayed classical correction applied afterward. The first angle $\theta$ may be arbitrary: it was never adaptive. Only the sign choice for the third angle needed to be replaced. Thus the simulation is nonadaptive for every $\varphi \in (\pi /2)\mathbb{Z}$.

For nonnegative integers $n$, assume the storage channel over successive intervals of length $\tau$ composes as $D_{ϵ}^n$. This is the memoryless interpretation of the specified storage law. In the computational basis, write the density operator asConjugation by the Pauli Z gate reverses the off-diagonal entries and leaves the diagonal entries unchanged, soAfter $n$ applications of the phase-flip channel, the multiplier is $(1-2ϵ{)}^n$. Comparing with the multiplier $1-2{ϵ}_n$ of a single phase-flip channel givesFor $n=0$ this gives the identity channel. Equivalently, composing two consecutive phase errors cancels them because $Z^2=I$, so their effective probability obeys ${ϵ}_{n+1}=ϵ+(1-2ϵ){ϵ}_n$. The same expression solves this recurrence. The iterated phase-flip channel formula presumes this composition law; specifying a channel at one time alone would not constrain a memory-bearing environment at later times.

Since $0<1-2ϵ<1$, the off-diagonal multiplier of the iterated phase-flip channel tends to zero. ThusLet $P_0=|0⟩⟨0|$ and $P_1=|1⟩⟨1|$ be the computational basis measurement projectors. The nonselective projective measurement channel iswhich is exactly the displayed limit. Its diagonal entries retain the original outcome probabilities; the measurement outcome is ignored and the coherences are erased. The limiting evolution is therefore the completely dephasing channel.

Let $R=|\psi ⟩⟨\psi |$ and choose an orthonormal basis $\{|e_{\mu }⟩\}$ of ${\mathcal{H}}_2$. The reduced density matrix on $S_1$ is the partial traceHere $I_1$ is the identity on ${\mathcal{H}}_1$, and the bras and kets in the sum contract only the second tensor factor. Equivalently, if $|\psi ⟩={\sum }_{a,\mu }{c}_{a\mu }|f_a⟩\otimes |e_{\mu }⟩$, then $({\rho }_1{)}_{ab}={\sum }_{\mu }{c}_{a\mu }{c}_{b\mu }^{\ast }$. The definition is independent of the chosen basis: it is characterized by $\mathrm{Tr}(M{\rho }_1)=\mathrm{Tr}[(M\otimes I_2)R]$ for every operator $M$ on ${\mathcal{H}}_1$.

The generalized measurement postulate takes $A_i$ to be operators on ${\mathcal{H}}_2$ satisfying ${\sum }_i{A}_i^{†}{A}_i=I_2$. The Born rule and the conditional state update giveIf the outcome is not supplied to $S_1$, the appropriate post-measurement density matrix is $R^{\prime }={\sum }_i(I_1\otimes A_i)R(I_1\otimes A_i^{†})$. For any local operator $M$, cyclicity of the full trace givesSince this holds for every $M$, $\boxed{{\mathrm{Tr}}_2{R}^{\prime }={\rho }_1}$. This is the no-communication theorem.

The invariance concerns the nonselective measurement in quantum mechanics. A conditional reduced density matrix ${\rho }_{1|i}={\mathrm{Tr}}_2|{\psi }_i⟩⟨{\psi }_i|$ can change, although ${\sum }_i{p}_i{\rho }_{1|i}={\rho }_1$. For example, measuring one half of a Bell pair in the computational basis changes the other half conditionally to $|0⟩$ or $|1⟩$; before learning the outcome, its density matrix remains $I/2$. Thus distant measurement in quantum measurements cannot transmit a controllable signal without classical communication. This quantum no-signalling is compatible with special relativity, even though some entangled states have correlations incompatible with a local hidden-variable theory.