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Write $|\alpha ⟩=a|0⟩+b|1⟩$. Expanding in the orthonormal Bell basis of systems $CA$ gives the Bell-basis teleportation identityThus, uniformly for $x,z\in \{0,1\}$,The states are unique because the four Bell states are an orthonormal basis of the two-qubit space. In particular,

This is superdense coding. To send the two classical bits $(z,x)$, Alice applies $Z^z{X}^x$ to her half of the shared Bell state $|{\beta }_{00}⟩$, producing $|{\beta }_{zx}⟩$, and sends that one qubit to Bob. Bob applies $\mathrm{CX}$ followed by $H$ on the first qubit, as in part (i), and measures both qubits in the computational basis. The outcome is $zx$ with certainty, so one transmitted qubit communicates two classical bits using the previously shared quantum entanglement.

The required identity isthe first exponent in the scanned statement should be $z$. Indeed, $X^x$ changes the parity label from $0$ to $x$, while $Z^z$ multiplies the component whose first qubit is $1$ by $(-1{)}^z$, creating the phase label. Equivalently, the hint permits either one-qubit operator to be transferred to the second half of the Bell state by transposition.

The four Bell states form an orthonormal basis, so their uniform mixture isIt is a product state and therefore

Distinct Bell states are orthogonal. For orthogonal pure states, the trace distance is one and the fidelity is the absolute overlap, zero. Hence

Alice adjoins $|0{⟩}_{A^{\prime }}$ and applies a local unitary whose action on the relevant inputs isMeasuring $A^{\prime }$ and obtaining zero leaves the unnormalized stateso the normalized state is the Bell state $|{\varphi }^{+}⟩$. The success probability is $2s^2>0$.

Applying this entanglement concentration independently to $n$ pairs yields an expected $2s^{2}n$ Bell pairs. Alice and Bob measure successful pairs in the computational basis to obtain identical secret random bits; public communication identifies successful positions without revealing outcomes. These bits form a secret key for a one-time pad of expected length $2s^{2}n$.

The pure state $|\Psi {⟩}_{AB}$ is entangled when it cannot be written as a product state $|a{⟩}_A|b{⟩}_B$, equivalently when its Schmidt rank exceeds one. Its subsystem states are the reduced density matriceswhere the partial trace is characterized by $\mathrm{Tr}(M_A{\rho }_A)=⟨\Psi |M_A\otimes I_B|\Psi ⟩$ for every observable $M_A$.

Let Bob's measurement have Kraus operators $M_b$ satisfying ${\sum }_b{M}_b^{†}{M}_b=I_B$. If Alice does not learn Bob's random outcome, her state after the measurement isThis no-communication theorem means that Bob can change Alice's conditional state after she learns his outcome, but cannot change any local outcome distribution available to her alone. Standard nonrelativistic quantum mechanics is therefore operationally compatible with the prohibition of superluminal signalling in special relativity, despite its nonlocal conditional-state updates.

An exact nondisturbing state readout would destroy this protection if it reported the globally collapsed state on an absolute-time slice. For example, Alice and Bob may share Bell states. At a prearranged time Bob encodes a bit by measuring his qubit in either the computational or Hadamard basis. The usual projection postulate assigns Alice respectively one of $|0⟩,|1⟩$ or one of $|+⟩,|-⟩$. A device returning the exact pure state lets Alice identify which basis Bob chose without waiting for his outcome, producing a superluminal signal. Equivalently, Bob may choose whether to measure, and the device distinguishes the resulting proper pure state from the original improper maximally mixed local state.

A causal alternative is a quantum state readout device located at a spacetime point $x$ that reports the local quantum state under objective collapse: take reduced states on spacelike hypersurfaces through $x$ and let those hypersurfaces approach the past light cone of $x$. The result includes localized collapses in $J^{-}(x)$ and excludes spacelike-separated collapses. This assumes a fixed Minkowski spacetime, localized collapse events, ordinary local unitary dynamics between them, and outputs that may control only operations in their causal future. The postulate is logically consistent because it adds a classical record of this local state without changing the state or any standard measurement probability. It also cannot support an indirect signalling algorithm. Inductively through the algorithm's events, every readout at $x$ depends only on operations, collapses, and earlier readouts in $J^{-}(x)$; any operation selected from that output lies in the readout's future light cone. Composing readouts, unitary evolutions, and measurements therefore never carries a controllable dependence outside a future light cone, so relativistic causality is preserved.

Let Alice and Bob share the Bell stateAlice may measure in the computational basis, remotely preparing Bob's ensemble $\{|0⟩,|1⟩\}$, or in the Hadamard basis, preparing $\{|+⟩,|-⟩\}$. Without her outcome, both ensembles have density matrix $I/2$, as no-signalling requires.

If Bob could clone all four states, the resulting two-qubit ensembles would beandThey are distinguishable: a $Z\otimes Z$ measurement is perfectly correlated for the first and only equally likely correlated or anticorrelated for the second. Bob could infer Alice's basis choice instantaneously, violating no-signalling. Thus no-signalling implies no-cloning for this set.

The controlled-NOT leaves computational-basis signals unentangled:For the two Hadamard-basis choices it givesBoth outputs are maximally entangled Bell states.

A Bell measurement is the projective measurement of two qubits in the orthonormal Bell state basisIts four classical outcomes identify the corresponding rank-one projector.

Expanding either Bell state and using $⟨0|1{⟩}_B=0$ removes the cross terms:Thus the two expectation values agree. Equivalently, both states have the same reduced density matrix $I/2$ on qubit $A$; the positivity of $Q$ makes it a possible measurement effect but is not needed for the algebraic equality.

Apply the gates from left to right. The first Hadamard gate and the first controlled-NOT gate produce the Bell stateThe parallel $Z\otimes H$ gates then produceThe final controlled-NOT has the lower qubit as control; it interchanges $|01⟩$ and $|11⟩$, whose amplitudes here are equal. Thus the output isBy quantum measurement in the computational basis, every basis outcome has probability equal to the squared modulus of its amplitude:

Let the unknown qubit be $|\alpha ⟩=a|0⟩+b|1⟩$. Alice holds this qubit and her half of the shared Bell state $|{\Psi }^{+}⟩$, while Bob holds the other half.

Alice performs a measurement of her two qubits in the Bell basis and sends the two-bit outcome to Bob through the classical channel. For the shared state $|{\Psi }^{+}⟩$, the teleportation with the psi-plus Bell state correction table isup to physically irrelevant global phases. Here $X$ and $Z$ are the Pauli X gate and Pauli Z gate. After the indicated correction, Bob's qubit is $|\alpha ⟩$. This is quantum teleportation; no physical copy of the unknown qubit is sent.

Let $D$ denote Alice's unknown input qubit and $A$ her GHZ qubit. Expanding the four-qubit state in the Bell basis of $D,A$ giveswhere a global sign in the last term is immaterial.

For outcomes ${\Phi }^{+}$ and ${\Phi }^{-}$, Bob and Charlie already haveFor either $\Psi$ outcome, both apply the Pauli X gate. Since $X\otimes X$ exchanges $|00⟩$ and $|11⟩$, the ${\Psi }^{+}$ branch becomes $|{\varphi }_1⟩$ and the ${\Psi }^{-}$ branch becomes $|{\varphi }_2⟩$, up to a global phase. This proves the Bell measurement on one qubit and one leg of a GHZ state result.

Finally, tracing out Charlie givesfor either sign. This reduced state has rank two exactly when $a\ne 0$ and $b\ne 0$. By the entanglement criterion for a two-term correlated state,

Across $A|BC$ the state is the product $|0{⟩}_A\otimes |{\varphi }^{+}{⟩}_{BC}$, so its Schmidt rank is one. Across $B|AC$ it isand the two states on each side are orthonormal, so the Schmidt rank is two. The same argument with $B$ and $C$ exchanged handles $C|AB$. This is precisely the entanglement of the Bell state shared by $B$ and $C$.

Projecting the three-qubit $W$ state onto Charlie's outcome $0$ leavesAfter normalization this is $(|01⟩+|10⟩)/\sqrt{2}$, a Bell state and hence an entangled state of Alice and Bob.

Using $T=2\mathrm{SWAP}-I$, the symmetric triplet Bell state have eigenvalue $1$, while the antisymmetric singlet has eigenvalue $-3$. With the ordering shown, ${\mathcal{B}}_1,{\mathcal{B}}_2,{\mathcal{B}}_3$ are triplet combinations and ${\mathcal{B}}_4$ is the singlet.

A density operator describes both a quantum state prepared in one pure state and a statistical preparation whose members occupy different pure states. If normalized vectors $|{\psi }_i⟩$ occur with probabilities $p_i$, defineThe resulting operator is Hermitian, positive semidefinite and has trace one. Positivity follows because $⟨v|\rho |v⟩={\sum }_i{p}_i|⟨{\psi }_i|v⟩{|}^2\ge 0$. Conversely, the spectral theorem gives such a representation of any finite-dimensional operator with these properties, using its eigenvectors and eigenvalues. The matrix representing $\rho$ depends on the chosen basis; the operator and its predictions do not.

One operator gives every measurement probability. For a projective measurement with orthogonal projections ${\Pi }_a$ and outcomes $a$,For example, the first formula follows by averaging $⟨{\psi }_i|{\Pi }_a|{\psi }_i⟩$ over the preparation. The trace expression is basis independent. A general measurement replaces ${\Pi }_a$ by positive measurement effects that sum to the identity, without changing the probability formula. Thus $\rho$ specifies the statistics accessible to measurements on that system.

The distinction between pure and mixed states is especially transparent in the eigenvalues. A pure state has $\rho =|\psi ⟩⟨\psi |$, ${\rho }^2=\rho$, and $\mathrm{Tr}{\rho }^2=1$. A mixed state has at least two positive eigenvalues and $\mathrm{Tr}{\rho }^2<1$. For a qubit,pure states lie on the Bloch sphere; mixed states lie inside the Bloch ball. The maximally mixed qubit $\rho =I/2$ can be prepared by choosing equally between the two $z$ spin eigenstates, or equally between the two $x$ spin eigenstates. These ensembles are different preparation procedures but give the same density operator. No measurement on the resulting qubit alone can identify which ensemble was used.

quantum coherence must be distinguished from merely having two populated levels. In the $z$ basis, the state $|+x⟩$ hasBoth give equal probabilities for the two $z$ outcomes. A measurement along $x$ gives the positive result with probability one for the first state and one half for the second. The off-diagonal entries encode the relative phase information responsible for this difference. Whether a matrix entry is off diagonal is basis dependent; purity is not.

For an isolated system, each component evolves with the same unitary time-evolution operator. Therefore $\rho (t)=U(t)\rho (0)U^{†}(t)$ andUnitary evolution preserves the eigenvalues and purity of $\rho$. A density operator commuting with a time-independent Hamiltonian operator is stationary. A thermal equilibrium example is the canonical ensemble ${\rho }_{\beta }=e^{-\beta H}/Z$, where $Z=\mathrm{Tr}(e^{-\beta H})$ and $\beta =1/(k_{\mathrm{B}}T)$. expectation values then reproduce quantum statistical mechanics without assigning the system a definite energy eigenstate on every run.

A mixed state can also arise without classical randomness in the joint preparation. For a bipartite quantum system, the reduced density operator ${\rho }_A={\mathrm{Tr}}_B{\rho }_{AB}$ gives every measurement probability accessible to subsystem $A$. For the Bell state $(|00⟩+|11⟩)/\sqrt{2}$, the joint state is pure, whereas ${\rho }_A=I/2$. Its local mixedness reflects entanglement with the inaccessible system; the partial trace discards correlations that matter for joint measurements. More generally, interaction with an environment can suppress off-diagonal entries in a selected basis, producing decoherence while the combined system still evolves unitarily.

Finally, state evolution conditional on a measurement outcome differs from isolated evolution. An ideal projective measurement yielding $a$ updates the state by the Lüders rule $\rho \mapsto {\Pi }_a\rho {\Pi }_a/P(a)$. Discarding the outcome gives ${\sum }_a{\Pi }_a\rho {\Pi }_a$, which generally removes coherence between different eigenspaces. The Von Neumann entropy $-\mathrm{Tr}(\rho \ln \rho )$ quantifies mixedness: it vanishes for pure states and reaches $\ln d$ for $I/d$. Multiplying by $k_{\mathrm{B}}$ gives thermodynamic entropy units. These uses make the density operator a common language for preparation uncertainty, measurement, thermal states and entangled subsystems, while retaining their distinct physical origins.

An eigenvalue of an observable in a known corresponding eigenstate is a simple example satisfying the Einstein–Podolsky–Rosen criterion of reality: its value is predictable with certainty. For example, a spin angular momentum component in a prepared spin-up eigenstate has a definite predicted value. For a distant system, Albert Einstein, Boris Podolsky and Nathan Rosen invoked locality: a measurement performed on a sufficiently separated, noninteracting partner should not disturb that system. In the relativistic version here, the measurement events have spacelike separation.

The Pauli matrices obey $[{\sigma }_z,{\sigma }_x]=2i{\sigma }_y$, soTheir nondegenerate eigenstates are incompatible: an $S_z$ eigenstate has equal probabilities for the two $S_x$ outcomes, and conversely. Thus no single quantum state has both components sharply determined. The Robertson uncertainty relation gives $\Delta S_z\Delta S_x\ge \hslash |⟨S_y⟩|/2$, but its right-hand side may be zero; that alone is not a proof that simultaneous zero uncertainties exist. The absence of a common eigenstate supplies the stronger conclusion for this spin-half pair.

For the specified Bell state, Bob obtains $+\hslash /2$ or $-\hslash /2$ with probability $1/2$ when he measures $S_z$. Conditional on his result, Alice obtains the same sign with certainty if she measures $S_z$. Under the EPR locality premise, this certainty supplies an element of reality for her $S_z$; it is a conditional prediction after Bob learns his own random result, not a premeasurement prediction of its sign.

The normalized $S_x$ eigenstates are $|+⟩=(|0⟩+|1⟩)/\sqrt{2}$ and $|-⟩=(|0⟩-|1⟩)/\sqrt{2}$. Expanding their tensor products givesConsequently the same Bell state has equal, perfectly correlated $S_x$ outcomes. Bob's two outcomes again each have probability $1/2$, and the same locality-and-certainty argument applies separately to Alice's $S_x$.

The disputed step is to combine predictions conditional on two incompatible choices into two simultaneous, measurement-independent pre-existing values. Quantum theory provides the two conditional ensembles, not a joint sharp assignment of $S_x$ and $S_z$. This is quantum steering. Bob cannot measure both incompatible observables sharply on the same member of the pair. Nor does conditional steering allow signalling: ignoring his result leaves Alice's reduced density matrix equal to $I/2$ for either measurement choice, by the no-communication theorem.

To test the stronger claim of a local hidden-variable theory, Alice needs several correlation settings, not just perfect agreement along the two common axes. For the CHSH inequality, choose observables with outcomes $\pm 1$:In the given Bell state, $⟨{\sigma }_z\otimes {\sigma }_z⟩=⟨{\sigma }_x\otimes {\sigma }_x⟩=1$ and the crossed correlations vanish. Henceexceeding the local bound $|\mathcal{S}|\le 2$. The bound assumes Bell local causality and measurement independence, with the appropriate treatment of sampling and settings.

Experimental Bell tests support violations rather than that local assignment. The Aspect switching experiment changed analyzer settings during separated photon flights, while leaving sampling and switching limitations. By the time of this examination, the Delft electron-spin Bell test had used electron spins at stations $1.3\mathrm{km}$ apart, efficient spin readout and rapid random basis choices. The combination addressed both detection and locality loopholes, providing evidence against the tested local model class.^[ref]

This does not prove that every conceivable notion of reality is false, or that noncommutation alone rules out every hidden-variable theory. A nonlocal theory can evade the local premise. What Alice can challenge is Bob's claim that locality, measurement-independent simultaneous values and all observed correlations can be retained together. An ordinary measurement disturbance at Bob's lab is not by itself a local explanation of Alice's conditional ensemble.

For classical events, $A\setminus C\subset (A\setminus B)\cup (B\setminus C)$: an outcome in $A$ but not in $C$ either fails $B$ or belongs to $B\setminus C$. Monotonicity and the union bound prove the Fréchet-Bell inequalityMalus' law gives transmission ${\cos }^2\gamma$ for a photon polarized by the first filter, so its conditional rejection probability at the second is $\boxed{{\sin }^2\gamma }$. Conditional and joint probabilities must be distinguished: $P(A\setminus B)=P(A)P(B^c\mid A)$.

For equal unbiased marginals $P(A)=P(B)=1/2$, the quantum two-setting rejection probabilities are $\frac{1}{2}{\sin }^2({\theta }_A-{\theta }_B)$. A classical common assignment of all three outcomes would therefore require ${\sin }^2({\theta }_A-{\theta }_B)+{\sin }^2({\theta }_B-{\theta }_C)\ge {\sin }^2({\theta }_A-{\theta }_C)$. Choose directions $0,\gamma ,2\gamma$. ThenFor example $\gamma =\pi /6$ gives $1/2<3/4$. More generally ${\sin }^{2}a+{\sin }^{2}b-{\sin }^2(a+b)=-2\sin a\sin b\cos (a+b)$. In a triangle whose side directions are these filters, the internal angle supplementary to $a+b$ is obtuse exactly in this violation range. This specifies the angle convention behind the triangle description: with polarization axes regarded as unoriented lines, a claim that every sufficiently large physical axis separation violates the inequality would not be correct. The contradiction concerns a single classical joint model for all settings, not the validity of ordinary probability or a comparison of arbitrarily renormalized sequential experiments.

Use $|0⟩=|{\alpha }_0⟩$ and $|1⟩=|{\alpha }_1⟩$. Projecting the first photon with $|0⟩⟨0|$ gives the unnormalized product vector $c_0|00⟩$, with success probability $|c_0{|}^2$. Projection of the second onto $|\theta ⟩=\cos \theta |0⟩+\sin \theta |1⟩$ givesConditioned on success, normalize these vectors; a zero amplitude means that outcome never occurs. For the maximally entangled equal-amplitude pair, rotation of both real polarization bases gives the same correlations for arbitrary settings. Comparing the appropriately normalized coincidence/rejection frequencies at three setting pairs tests the inequality above. A general unequal-amplitude state need not reproduce the unbiased rotationally invariant probabilities, so the maximally entangled choice supplies the clean explicit example.

For swapping, label the four systems in order ${\xi }_1,{\eta }_1,{\xi }_2,{\eta }_2$ and let $|\Phi {⟩}_{\eta }=2^{-1/2}{\sum }_{j=0}^1|{\eta }_j{\eta }_j⟩$. Its overlap with $|{\eta }_r{\eta }_s⟩$ is ${\delta }_{rs}/\sqrt{2}$. Applying its rank-one projector to Alice's two systems givesAfter regrouping the remote and local systems, this factorizes into $[{\sum }_r{b}_r^2|{\xi }_r{\xi }_r⟩/\sqrt{2}]\otimes |\Phi {⟩}_{\eta }$. The selected outcome has probability $p=\frac{1}{2}{\sum }_r|b_r{|}^4$. Therefore the normalized conditional remote state isIt has Schmidt rank at least two whenever the original pair was entangled, since squaring a nonzero coefficient does not make it zero. Thus the two photons that never interacted become entangled conditional on Alice's joint measurement outcome. This is entanglement swapping. Alice is projecting onto a specified Bell state, not performing a universal test of whether an arbitrary unknown state is entangled. Averaging all her outcomes leaves the remote reduced density operator unchanged, so no faster-than-light signal is available without her classical outcome label.