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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.

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.

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.

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.

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 $\mathrm{Tr}(U_j^{†}{U}_k)/2={\delta }_{jk}$.

Charlie's Hadamard basis measurement, Alice's Bell-basis measurement and communicated correction bits give deterministic GHZ-assisted quantum teleportation. Let the unknown input be $|\chi {⟩}_Q=\alpha |0⟩+\beta |1⟩$. Charlie measures $C$ in the Hadamard basis and labels $+$ by $c=0$, $-$ by $c=1$. The identityshows that $AB$ is conditionally a Bell pair with phase label $c$. Charlie sends that one classical bit to Bob. The procedure for localizing a Bell pair from a GHZ state supplies precisely the missing conditional information; forgetting $c$ leaves the mixed reduced state of the preceding part.

Alice performs the usual Bell-basis measurement on $QA$. Operationally, she applies ${\mathrm{CNOT}}_{Q\to A}$, applies $H_Q$, and measures $Q$ and $A$ in the computational basis, obtaining bits $s,t$. She sends both bits to Bob. To track the correction without a convention ambiguity, conditional on $s,t$ the normalized state on $BC$ isProjecting Charlie's factor onto his recorded Hadamard basis state gives Bob, up to the global phase $(-1{)}^{ct}$,Consequently the inverse Pauli correction isBob applies $X^t$ first and $Z^{s\oplus c}$ second. The measurements on Charlie and on Alice commute because they act on different systems, so deriving the conditional state with Alice's measurement first gives the same protocol as Charlie measuring first. Each triple $(s,t,c)$ has probability $1/8$, independent of the unknown input, and every triple succeeds after correction.

This uses one previously shared GHZ state, two classical bits from Alice and one from Charlie, with no quantum transmission during the transfer. The standard Bell-measurement/Pauli-correction mechanism was established in the original quantum teleportation proposal.^[ref] Here Charlie's record supplies the Bell-pair phase, giving a controlled quantum teleportation protocol. Alice's input is consumed by her measurement; Bob reconstructs its state without learning $\alpha ,\beta$ and without making a second surviving copy.

Use entanglement-assisted discrimination of a Pauli basis. Define the known single-qubit unitary operator $V$ by $V|0⟩=|\psi ⟩$ and $V|1⟩=|{\psi }^{\perp }⟩$. Prepare the device input $A$ and a reference ancilla qubit $R$ inApply the unknown device once to $A$, then apply the known inverse $V^{†}$ to $A$. Sincethe possible two-qubit outputs areThese are the four mutually orthogonal Bell states. A Bell-basis measurement therefore identifies the device with certainty. Explicitly, apply a controlled-NOT gate $A\to R$, then a Hadamard gate on $A$, followed by computational-basis measurements. The readouts $00,10,01,11$ identify $U_0,U_1,U_2,U_3$, respectively. Thus one use of the device and one reference qubit suffice.

Combine the hypothetical protocol with superdense coding. Before a message is chosen, Alice and Bob share a further Bell pair $|{\varphi }_{+}{⟩}_{QR}$, with $Q$ at Alice and $R$ at Bob, in addition to any resources needed for the hypothetical quantum teleportation protocol. To encode two classical bits $u,v$, Alice applies the Pauli gates $Z_Q^u{X}_Q^v$, preparingThe four choices are the four mutually orthogonal Bell states. Alice now uses the hypothetical protocol to teleport $Q$ to Bob using only one classical bit. Its assumed preservation of the joint quantum state gives Bob both $B$ and $R$ inBob performs a Bell-basis measurement. Explicitly, a controlled-NOT gate $B\to R$, followed by a Hadamard gate on $B$, givesA computational-basis measurement therefore recovers both message bits with certainty after only one transmitted classical bit. The initial Bell pairs are shared before the message is chosen and carry no message themselves. This proves the classical communication lower bound for exact qubit teleportation: a perfect protocol preserving an external reference needs at least two transmitted classical bits.

Let the unknown input qubit be $|\psi {⟩}_S=\alpha |0⟩+\beta |1⟩$, with $|\alpha {|}^2+|\beta {|}^2=1$. Alice holds $S,A$, Bob holds $B$, and the shared Bell pair is $|{\varphi }^{+}{⟩}_{AB}$. All three-qubit strings below use the order $S,A,B$. Initially,The Bell-basis teleportation identity givesThus a perfect Bell state analyzer on Alice's two qubits identifies the Pauli operator affecting Bob's quantum state.

To display the intermediate three-qubit states explicitly, realize that Bell-basis measurement by a controlled-NOT gate $S\to A$, a Hadamard gate on $S$, and two computational-basis measurements. After the controlled-NOT gate,After the Hadamard gate,Measuring $S,A$ gives $(m,n)$ with probability $1/4$ for each outcome, independently of $\alpha ,\beta$. Conditional on that result, the normalized three-qubit post-measurement state isIf the Bell state analyzer is instead regarded as an ideal direct projective measurement preserving the measured qubits, replace $|mn{⟩}_{SA}$ here by the corresponding Bell state; Bob's factor is identical. A destructive optical analyzer records the same label but absorbs Alice's photons.

Alice sends the two classical bits $(m,n)$ to Bob. During transmission the above conditional three-qubit state is unchanged. Bob then applies the inverse Pauli operator $Z^m{X}^n$:The final conditional three-qubit state is therefore $|mn{⟩}_{SA}\otimes |\psi {⟩}_B$. The protocol transmits two classical bits per teleported qubit, consuming one shared Bell pair. Without Alice's record Bob averages over four Pauli operators and has density matrix $I/2$, so quantum teleportation respects the no-communication theorem. Alice's input has been measured and is no longer an independent copy, consistent with the no-cloning theorem.

For teleportation with an ambiguous Bell-state phase label, write $\rho =|\psi ⟩⟨\psi |$. The Bell-basis measurement outcomes ${\varphi }^{+}$ and ${\varphi }^{-}$ each occur with probability $1/4$. In either case a correct guess gives Bob $\rho$, while the wrong guess differs by the Pauli Z gate and gives $Z\rho Z$. The other two Bell states are identified correctly. Since Alice guesses each ambiguous label with probability $1/2$, the overall wrong-correction probability is $(1/2)(1/2)=1/4$. Averaging over the actual quantum measurement outcome and her guess gives the phase-flip channelConditional only on an ambiguous analyzer result, the output is $(\rho +Z\rho Z)/2$; that conditional channel has weight $1/2$ in the displayed unconditional result. There is no rejection of those rounds.

Use the teleportation fidelity for a pure state target, namely the squared quantum fidelityThis compares the input with the resulting mixed state; averaging state vectors before forming a density matrix would give a different, incorrect answer. Direct substitution givesThus the maximum is $1$, attained by $|0⟩$ and $|1⟩$ up to a global phase, because these are eigenstates of the Pauli Z gate. The minimum is $3/4$, attained by every equal-weight quantum superpositionon the equator of the Bloch sphere. The unsquared quantum fidelity is $\sqrt{F_{\psi }}$ if that alternative convention is used. No average over different input quantum states is needed for the requested state-dependent answer.

In quantum dense coding, a shared Bell pair lets Alice send two classical bits by transmitting one qubit. Use photon polarization as the computational basis, $|0⟩=|H⟩$, $|1⟩=|V⟩$, and define the Bell statesAlice and Bob initially share $|{\varphi }^{+}{⟩}_{AB}$. Alice encodes a two-bit message $(m,n)$ by applying the Pauli operator $X^n{Z}^m$ to her qubit:The last minus sign is a global phase. Alice transmits her photon to Bob, who now holds both qubits and uses a Bell-basis measurement to identify the message. The four Bell states are orthogonal, so an ideal Bell state analyzer decodes them without error. The resource cost is one previously shared Bell pair and one transmitted qubit per two classical bits; the shared quantum entanglement is consumed. Before transmission Bob's reduced density matrix is $I/2$ for every message, consistent with the no-communication theorem.

The BBO crystal supplies the polarization-entangled photon pair. An ultraviolet laser pumps this nonlinear optical crystal; spontaneous parametric down-conversion produces signal and idler photons with ${\omega }_p={\omega }_s+{\omega }_i$ and phase matching ${\mathbf{k}}_p={\mathbf{k}}_s+{\mathbf{k}}_i$. In a noncollinear type-II phase matching arrangement, collect the intersections of the two emission cones. The alternatives $|H{⟩}_A|V{⟩}_B$ and $|V{⟩}_A|H{⟩}_B$ must be indistinguishable in arrival time, spectrum and spatial mode. Compensation of birefringence and adjustment of the relative quantum phase then give $(|HV⟩+|VH⟩)/\sqrt{2}$; a local Pauli X gate converts it to $|{\varphi }^{+}⟩$. Half-wave plates and other waveplates implement Alice's Pauli gates. The BBO is the pair source; the message is encoded subsequently by the local optical elements.^[ref]

A basic passive linear optics Bell state analyzer with no extra resources identifies only two of the four polarization Bell states unambiguously. Thus the ideal two-bit capacity presupposes complete Bell-basis measurement; that basic optical implementation alone does not attain it deterministically.

In entanglement swapping, start with independent Bell pairs $|{\varphi }^{+}{⟩}_{12}|{\varphi }^{+}{⟩}_{34}$. Send qubits 2 and 3 to a central Bell state analyzer, retaining 1 and 4 at separate endpoints. Reordering the tensor product factors to group 23 and 14 gives the identityConsequently each ideal Bell-basis measurement outcome has probability $1/4$ and projects 14 onto the corresponding Bell state, even though those two photons have never interacted. Communicating the two-bit result permits an endpoint Pauli gate to convert every outcome into the same chosen Bell pair. Without that record, the endpoints have ${\rho }_{14}=I_4/4$, a product state, rather than a known entangled state.

The optical Bell state analyzer overlaps photons 2 and 3 at a balanced beam splitter, with matched spatial modes, spectra and arrival times, and resolves the output photon polarizations using polarizing beam splitters and single-photon detectors. Two-photon interference maps Bell states to different output patterns. The antisymmetric $|{\psi }^{-}⟩$ gives one photon in each output with opposite polarizations. The symmetric $|{\psi }^{+}⟩$ gives two opposite-polarization photons in the same output; the two polarization detectors there can resolve this pattern. The $|{\varphi }^{\pm }⟩$ states give same-polarization pairs and remain mutually ambiguous in this basic setup. The detected pattern therefore heralds a particular endpoint Bell state. With ideal detection this arrangement succeeds unambiguously on two of four equally likely inputs, with probability $1/2$; the successful events can be selected. A complete analyzer would accept all four outcomes. The endpoint quantum entanglement is conditional on the measurement record, so it enables no instantaneous signalling.^[ref]

In superdense coding, Alice applies a local Pauli operator to her half of the shared Bell state. Choose a phase bit $u$ and a parity bit $v$, and encode them using $X^v{Z}^u$ on Alice's qubit. With the standard definitions of the Bell states,Alice sends her qubit to Bob. He now holds both halves and performs a Bell-basis measurement, recovering $uv$ without error. Equivalently, a CNOT gate with Alice's qubit as control, followed by a Hadamard gate on that qubit, maps these states to their computational labels up to global phases. Four distinguishable alternatives give two classical bits per transmitted qubit, with one shared entangled pair consumed per use.

The initial entanglement distribution is independent of the later message and can be completed beforehand. Bob's local reduced density matrix before Alice's transmission is $I/2$ for every encoding, so quantum no-signalling prevents it from carrying her choice. This does not make the entangled pair free: if a qubit transmission is used to distribute it, counting that transmission as well gives two qubit transmissions for the two-bit message. The quoted dense-coding rate counts the subsequent message-bearing transmission with entanglement already shared.

The actual resource is a two-component Bell resource in dense coding. Applying the same local encoder $X^v{Z}^u$ togives a superposition of the intended Bell state with amplitude magnitude $2/\sqrt{5}$ and its phase-label partner with amplitude magnitude $1/\sqrt{5}$. A Bell-basis measurement therefore reports the wrong Bell state with probability $1/5$, independently of $uv$. With our labeling, only the phase bit $u$ flips:The first bit of each pair passes through a binary symmetric channel with error $1/5$; the second bit is correct. A uniformly chosen bit from the full stream consequently has

This bit error probability depends on the Bell labeling. The Bell-label Hamming distance and bit errors rule multiplies the wrong-symbol probability by the number of bit positions changed. For example, relabel ${\Phi }^{+},{\Phi }^{-},{\Psi }^{+},{\Psi }^{-}$ as $00,11,01,10$. Both wrong-state partner pairs then differ in two bits, giving bit error probability $1/5$ instead of $1/10$. The physical wrong-Bell-state probability remains $1/5$.

For the entropy-matched independent source in part 2, the transmitted bit stream is fair, and consecutive two-bit labels are uniform. Write the binary entropy as $h_2(p)=-p{\log }_{2}p-(1-p){\log }_2(1-p)$. If a bit position is chosen uniformly and its phase/parity position is forgotten, the pooled corresponding-bit pair is a binary symmetric channel with error $1/10$. Its mutual information is

The bit-position information in a binary channel distinguishes this pooled quantity from the information in the decoded blocks. Alice and Bob know which bit is first in each Bell label. Conditioned on that position $J$, the first bit gives $1-h_2(1/5)$ bits and the second gives one bit. ThusEquivalently, the joint mutual information between the two-bit input and output labels is $2-h_2(1/5)\simeq 1.278072$ bits per transmitted Bell label for the prescribed Bell measurement. Pooling without the position loses usable side information; it is not interchangeable with this block rate. The stated fair-input values use the optimal source distribution above; biased source bits require their actual joint distribution.

Use the ordered polarization basis $(|H⟩,|V⟩)$ and define the Bell statesRegrouping the four-photon quantum state by channel pairs $(1,4)$ and $(2,3)$ givesFor instance, projecting channels 2 and 3 on $⟨{\Psi }^{-}|$ retains the two original terms with inner polarizations $HV$ and $VH$, with opposite signs, leaving $(|HV⟩-|VH⟩{)}_{14}/(2\sqrt{2})$. Its squared norm is $1/4$.

By the same orthonormality argument, each Bell-basis measurement outcome prepares the identically named Bell state in channels 1 and 4, withBefore communication of the outcome, Alice and Bob's density matrix is $I_4/4$; the usable conditional entanglement is established by the announced Bell-basis measurement result. This assumes an ideal complete Bell-basis measurement; ordinary passive linear optics alone does not deterministically distinguish all four polarization Bell states.

Each source has density matrixAssume the two sources and the detector's random choices are independent. In the $p^2$ branch both resources are ideal. If either resource is $I_4/4$, the remote pair remains $I_4/4$ after any specified Bell-basis measurement outcome. An ideal detector and the correction of part (b) therefore yieldWhen the detector instead reports an independent uniformly random label, the corresponding four Pauli gates average the conditional pair to $I_4/4$. ThusHere $q$ is the probability of the reliable-detector branch, not its total chance of being correct: that chance is $q+(1-q)/4$.

For the usual two-setting CHSH inequality, the spin singlet state has correlation $-\mathbf{a}\cdot \mathbf{b}$ and the white-noise part has zero correlation, so $E(\mathbf{a},\mathbf{b})=-v\mathbf{a}\cdot \mathbf{b}$. With Alice's axes $z,x$ and Bob's axes $(z+x)/\sqrt{2},(z-x)/\sqrt{2}$, the absolute CHSH inequality combination is $2\sqrt{2}v$. These settings are optimal for an isotropic correlation tensor, givingThis is the threshold for the standard CHSH inequality with optimized projective polarization measurements. The unqualified phrase “the Bell inequality” needs this choice of test; entanglement alone is a weaker requirement, and no claim about every possible Bell inequality is needed for this calculation.

After the first Bell-basis measurement and the announced correction, the shared Werner state has visibility $v=p^{2}q$. Ideal quantum teleportation through that resource is the depolarizing channelOne way to see this is to separate the singlet branch, which teleports perfectly with the appropriate known correction convention, from $I_4/4$, which has no correlations and yields $I_2/2$.

The additional Bell-basis measurement at Alice's station has its own independent reliable branch of probability $q$. In its random-label branch the correction averages the four Pauli gates, again giving $I_2/2$. Hence the actual teleported density matrix isOn sifted BB84 rounds, Bob measures in Alice's preparation basis. The preserved component gives the correct bit with certainty; the maximally mixed component gives an error with probability $1/2$. ThereforeFor $p=q=1$ it vanishes; if either parameter is zero it is $1/2$. The result presumes successful full Bell-basis measurement outcomes and independent stated imperfections, without extra loss, polarization drift or dark counts.

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.

Under the intended assumption that one use of the quantum channel transmits one qubit without noise, Alice can communicate two classical bits with certainty if she and Bob already share a Bell pair. She uses superdense coding. The prior distribution of the Bell pair must be a resource available before the permitted channel use; distributing it through that same channel would consume another use.

Take the shared Bell state to be $|{\Phi }^{+}⟩=(|00⟩+|11⟩)/\sqrt{2}$, with Alice holding the first qubit. For a two-bit message $(a,b)\in \{0,1{\}}^2$, Alice applies the local unitary operation $U_{ab}=X^a{Z}^b$ to her qubit. The four encoded pure states are, up to irrelevant global phases,They are orthogonal: for any matrix $V$ on the first qubit, $⟨{\Phi }^{+}|(V\otimes I)|{\Phi }^{+}⟩=\mathrm{Tr}V/2$, and the four Pauli operators are orthogonal in the trace inner product. Explicitly,Alice sends her qubit through the quantum channel once. Bob now holds both qubits and performs a Bell-basis measurement. Its four orthogonal projectors identify $(a,b)$ with certainty, givingBefore the transmission, Bob's reduced density matrix is $I/2$ for every message, so shared entanglement alone does not convey the message.

Without shared entanglement, a noiseless single-qubit transmission cannot perfectly distinguish four messages: perfect discrimination requires orthogonal state supports, and its two-dimensional Hilbert space has room for at most two nonzero mutually orthogonal supports. The Holevo bound also limits its accessible classical information to one bit. For the usual protocol starting with a shared two-qubit pure state and using local unitary encodings, exact two-bit transmission requires a maximally mixed Bob marginal. Indeed all encoded joint states are pure and have the same ${\rho }_B$, while Subadditivity of Von Neumann entropy gives their average entropy at most $1+S({\rho }_B)$. Four equally likely perfectly distinguishable messages have average Von Neumann entropy two, forcing $S({\rho }_B)=1$ and hence a maximally entangled state.

The specification “quantum channel” alone does not determine its dimension or noise, so the one-qubit, noiseless interpretation is needed for this particular conclusion. A noiseless four-dimensional channel could send four orthogonal states without any shared entanglement; a completely depolarizing qubit channel cannot transmit the message even with a shared Bell pair, since its output is independent of Alice's encoding. Shared entanglement is therefore the extra resource for the intended single-qubit channel, not a guarantee for every unspecified quantum channel.

The Pauli matrices act by $X|0⟩=|1⟩$, $X|1⟩=|0⟩$, $Z|0⟩=|0⟩$, and $Z|1⟩=-|1⟩$. Consequently $X\otimes X$ swaps $00$ with $11$ and $01$ with $10$. The plus combinations are its $+1$ eigenvectors, and the minus combinations its $-1$ eigenvectors. Meanwhile $Z\otimes Z$ is $+1$ on the even-parity basis vectors and $-1$ on the odd-parity ones. Thus, with the paper's labels,This verifies all four Bell states directly. The two observables commute: $XZ=-ZX$ on each factor, so the two minus signs cancel in their tensor products. Their joint eigenvalues distinguish the four states, as used in a Bell-basis measurement.