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The Bell basis is the two-qubit case of the generalized Bell basis. Its four vectors are the Bell states, and a Bell-basis measurement projects onto them.

Apply a Hadamard gate to the first qubit, then a controlled-NOT gate from the first to the second. This maps a computational-basis state $|ij⟩$ to the Bell state with phase bit $i$ and parity bit $j$. Its inverse uses the same gates in reverse order because each gate is self-inverse but they do not commute. Applying the forward sequence twice is not generally a decoding operation.

A Bell-basis measurement is the four-outcome projective measurement whose projectors are onto the four Bell states. It jointly measures two qubits and is the sender's measurement in quantum teleportation.

Give Alice the first qubit $A$ and Bob the second $B$. Compare the Bell state $|{\Phi }^{+}⟩$ with a preparation that tosses a fair classical bit and gives both parties its value. Their density operators, written in the computational basis ordered $00,01,10,11$, areThey have identical diagonal quantum state populations. The off-diagonal entries of the Bell state retain quantum coherence between $00$ and $11$; the classical mixture has none in this basis.

If both parties perform a Pauli measurement of $Z$, the Born rule gives $P(00)=P(11)=1/2$ and $P(01)=P(10)=0$ for both preparations. Perfect agreement here therefore cannot establish entanglement: an ordinary shared random bit already explains it.

Now both measure $X$, whose normalized eigenstates are $|+⟩=(|0⟩+|1⟩)/\sqrt{2}$ and $|-⟩=(|0⟩-|1⟩)/\sqrt{2}$. Changing basis givesThe Bell state again has matching outcomes, each with probability $1/2$. For either product state $|00⟩$ or $|11⟩$, each of the four $X$ outcome pairs has probability $1/4$, so their classical mixture does too. Equivalently, the Pauli correlator $⟨X\otimes X⟩$ is one for the Bell state and zero for this mixture, while $⟨Z\otimes Z⟩$ is one for both.

Taking a partial trace gives ${\rho }_A={\rho }_B=I_2/2$ for both preparations. Thus every measurement confined to Alice's system has the same statistics in the two cases, and so does every measurement confined to Bob's. They reveal the difference by comparing records through classical communication. Conditional on Alice's result, Bob can predict a matching $X$ result for the Bell state; without that result, his outcomes remain evenly split. An uncommunicated local quantum channel, including a measurement with its outcome discarded, leaves his reduced density matrix unchanged. This is quantum no-signalling, even when conditional states change. ^[ref]

The Von Neumann entropy makes the joint-versus-local distinction numerical. With base-two logarithms, the Bell state has eigenvalues $1,0,0,0$, giving $S(AB)=0$, whereas the mixture has $1/2,1/2,0,0$, giving $S(AB)=1$. Both have $S(A)=S(B)=1$. Their quantum mutual information $I(A:B)=S(A)+S(B)-S(AB)$ is consequently two and one bits, respectively. The value two is not two readable classical bits in either local qubit: the matching $Z$ records have ordinary mutual information one bit in both cases. Random records can come from a globally pure state. ^[ref]

These two matching-axis measurements distinguish the specified preparations; they alone are not a test of a Bell inequality. A local hidden-variable theory could assign shared random answers separately to these two axes. Bell theorem concerns incompatible correlations across a suitable larger collection of measurement settings. Bell's 1964 paper established the obstruction for local hidden-variable descriptions of the predicted correlations. ^[ref]

If a wrong Bell-label partner occurs with probability $ϵ$ and differs from the intended two-bit label in $w$ positions, its contribution to the error probability of a uniformly selected bit is $ϵw/2$. Relabeling the Bell states can change $w$ while preserving the physical wrong-state probability. Thus symbol error probability and bit error probability are different quantities.

A Bell pair consists of two qubits prepared jointly in a Bell state. It is a maximally entangled two-qubit resource with Schmidt rank two. Two independent Bell pairs have resource Schmidt rank four across the same party partition.

A device implementing a Bell-basis measurement on two qubits. A complete analyzer distinguishes all four Bell states. A basic passive linear optics analyzer for two polarization qubits uses a balanced beam splitter, polarizing beam splitters and single-photon detectors. Its ideal unambiguous success probability is $1/2$: the two opposite-polarization Bell states have distinct click patterns, while the two same-polarization Bell states remain mutually ambiguous. Additional physical resources can change this restriction. A destructive analyzer absorbs the measured photons; it need not preserve the detected Bell state.

The Bell states are joint eigenstates of commuting $Z_A{Z}_B$ and $X_A{X}_B$. Two shared Bell state meter pairs realize their two entanglement-assisted nondemolition parity measurements using spacelike local circuits. Their projector products give the four rank-one Bell state outcomes and preserve each input Bell state. The globally identified outcome requires later local operations and classical communication.

Suppose independent resource pairs have density matrices $v_j|{\Phi }^{+}⟩⟨{\Phi }^{+}|+(1-v_j)I_4/4$. An ideal complete Bell-basis measurement, followed by outcome-dependent Pauli gates, produces a chosen Bell state with white-noise visibility $v_1{v}_2$: either white-noise input makes the remote pair maximally mixed. If the reported label is reliable with probability $q$ and otherwise uniformly random, averaging the erroneous corrections makes that branch maximally mixed too, so $v_{\mathrm{out}}=v_1{v}_{2}q$. A second independent detector of the same quality during quantum teleportation adds another factor $q$ to the teleported Bloch vector.

Average the four correlated local Pauli operations to remove coherences in the Bell states basis. A one-party Pauli operation can permute the largest of the four resulting weights into the singlet sector; that weight is at least $1/4$. Correlated common local unitaries then permute the three triplet projectors, while collective-unitary covariance of the two-qubit singlet keeps the singlet projector unchanged. Averaging all six permutations makes the triplet weights equal, producing a two-qubit Werner state. Shared randomness or local operations and classical communication coordinates these local operations. Common unitaries alone cannot increase an initially small singlet weight, because that weight is invariant under them.

A quantum system with two specified subsystems has a tensor product Hilbert space. A joint density operator determines subsystem $A$'s measurement statistics through its partial trace ${\rho }_A={\mathrm{Tr}}_B{\rho }_{AB}$. A joint pure state need not give a pure reduced state: the Bell state $(|00⟩+|11⟩)/\sqrt{2}$ gives ${\rho }_A=I/2$.

Dik Bouwmeester and colleagues demonstrated selected-outcome photon-polarization quantum teleportation. An ultraviolet pump traversed a nonlinear crystal twice: one passage supplied an entangled pair, the other the input photon and its trigger partner. The input and one resource photon interfered at a beam splitter; output-detector coincidences selected their antisymmetric Bell state. Narrow filters made their coherence time exceed the pump duration, preserving temporal indistinguishability.

The reported pump pulses were $200\mathrm{fs}$ long at $76\mathrm{MHz}$; $788\mathrm{nm}$ photons filtered to $4\mathrm{nm}$ bandwidth had a quoted $520\mathrm{fs}$ coherence time. A polarization analyzer measured the remote output. Trigger-conditioned fourfold coincidences excluded a double-pair background without subtraction. The selected Bell result had ideal probability $1/4$; this demonstration did not distinguish and actively correct all four outcomes.^[ref]

For dichotomic observables with $A_i^2=B_j^2=I$, let $\mathcal{B}=A_0\otimes (B_0+B_1)+A_1\otimes (B_0-B_1)$. Expanding gives ${\mathcal{B}}^2=4I-[A_0,A_1]\otimes [B_0,B_1]$. Each commutator has operator norm at most two, so $\Vert \mathcal{B}{\Vert }^2\le 8$ and $|⟨\mathcal{B}⟩|\le 2\sqrt{2}$. Appropriate spin measurements of a Bell state attain equality. The local hidden-variable bound of the corresponding CHSH inequality is two.

Exact quantum teleportation that preserves an arbitrary reference requires at least two classical bits from sender to receiver, even with preshared quantum entanglement. For the essential reduction, encode two classical bits in the four Bell states by Pauli gates on the sender's half of a Bell pair. Teleportation as an identity channel on a reference transfers this half while preserving the encoding. The receiver then has both halves and recovers both bits by a Bell-basis measurement, as in superdense coding.

The classical bit bound itself has a short no-communication theorem proof. For $N$ possible messages and $K$ possible classical transmissions, let ${\sigma }_{m,c}$ be the receiver's subnormalized conditional density matrix before decoding. The no-communication theorem gives ${\sum }_c{\sigma }_{m,c}=\tau$, independent of $m$, so $0\le {\sigma }_{m,c}\le \tau$. For each transmission $c$, let $\{E_{m|c}{\}}_m$ be the decoding POVM. Averaging over equiprobable messages,Perfect decoding of four messages requires $K\ge 4$, hence at least two classical bits. This applies to the preshared resources of the proposed quantum teleportation protocol as well as the Bell pair used for the encoding.

Measuring the transmitted qubit in the computational basis and forwarding its measured state produces a dephasing channel for users who do not know the outcome. For a Bell state $|{\Psi }^{+}⟩$, the reduced state becomes $(|01⟩⟨01|+|10⟩⟨10|)/2$. It retains perfect $Z$ anticorrelation while destroying $X$ correlation. Thus a single perfectly correlated key basis need not reveal an intercept-resend attack; incompatible-basis correlation tests can.

On $|{\Phi }^{+}⟩$, measure ${\sigma }_z\cos \theta +{\sigma }_x\sin \theta$, assigning bit $a$ to eigenvalue $(-1{)}^a$. Choose Alice's angles $2(i-1)\delta$ and Bob's angles $(2i-1)\delta$, with $\delta =\pi /(2N)$. The Bell state correlation is $\cos (\theta -\varphi )$. Each adjacent disagreement term and the last agreement term therefore has probability ${\sin }^2(\pi /(4N))$. For $N\ge 2$, their total violates the chained modular Bell inequality bound one, and it tends to zero as $N\to \infty$. At $N=1$ the value is exactly one.

// Target: probability-and-statistics.bigb

For nonnegative $u,v$ with $u^2+v^2=1$, Alice applies Kraus operators $M_0=\mathrm{diag}(u,v)$ and $M_1=\mathrm{diag}(v,u)$ to one half of a Bell state. Their squared effects sum to the identity, and both outcomes have probability one half. The first branch is the target; the second becomes the target after both parties apply Pauli X gates. Sending the outcome by classical communication makes this exact deterministic entanglement dilution, rather than a postselected conversion.

The four unitary operators $I,Z,X,iY$ can be identified with one use when an input qubit is entangled with a reference ancilla qubit. Applying the unknown operation to one half of $|{\varphi }_{+}⟩$ produces the four orthogonal Bell states, which a Bell-basis measurement distinguishes exactly. For operators $V{U}_j{V}^{†}$ in any known orthonormal basis, prepare $(V\otimes I)|{\varphi }_{+}⟩$ and apply $V^{†}$ after the unknown call. This reduces to the same quantum channel discrimination protocol. Direct expansion proves the four output identities; equivalently, their overlaps are $\mathrm{Tr}(U_j^{†}{U}_k)/2={\delta }_{jk}$.

Share a Bell state meter pair, apply local system-to-meter CNOT gates, and measure each meter in the computational basis. The parity of the local meter records measures system $Z_A{Z}_B$ while preserving all coherence within either parity sector. The complete tuple of local records has Kraus operator $P_{u\oplus v}/\sqrt{2}$. Each local record is uniform, and the nonlocal result requires local operations and classical communication to compare records. A local Hadamard gate conjugation supplies the corresponding $X_A{X}_B$ measurement.

A unitary gate on two quantum systems is entangling if it maps at least one pure state that is a product state to an entangled state. The controlled-NOT gate qualifies because it maps $|+⟩|0⟩$ to a Bell state. A gate that merely exchanges the two systems need not be a local unitary operation, but cannot create quantum entanglement from a product state.