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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.

Superdense coding uses a shared Bell state to communicate two classical bits by transmitting one qubit. The sender selects one of the four Bell states with a local Pauli operation, and the receiver performs a Bell-basis measurement.