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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.

Distributing one half of $|{\Phi }^{+}⟩$ through a quantum bit-flip channel produces $p{\Pi }_{{\Phi }^{+}}+(1-p){\Pi }_{{\Psi }^{+}}$. Ideal quantum teleportation with error-free classical correction bits induces the same bit-flip map on the input: conjugating the resource error by any Pauli gate correction changes at most its sign. For a pure target the squared quantum fidelity is $p+(1-p)|⟨X⟩{|}^2$. This describes noise in resource distribution, not a perfect resource supplied by an independent channel.

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.

Initialize two qubits in $|00⟩$. A Hadamard gate on the first followed by a controlled-NOT gate $1\to 2$ prepares the Bell state $|{\varphi }_{+}⟩$. The permitted T gates and Hadamard gates implement the needed Pauli gates exactly:Apply $T_1^4$, followed by $H_2,T_2,T_2,T_2,T_2,H_2$. These are a Pauli Z gate on the first qubit and a Pauli X gate on the second, givingEvery operation is in the allowed gate set; no quantum measurement or probabilistic branch is needed.

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.

Write the arbitrary initial pure state asThe reference vectors need not be normalized or orthogonal. By linearity, the same quantum teleportation calculation, with the identity operator on $R$, gives the post-gate quantum stateThus each outcome still has probability $1/4$. Bob's Pauli gate correction leaves the normalized joint quantum stateAll operations act trivially on $R$, so the result is independent of its holder or location. This establishes teleportation as an identity channel on a reference, preserving quantum entanglement with the untouched reference. Since every corrected branch is the same, discarding the classical outcomes also leaves the required joint density matrix.

Alice holds the input qubit $Q$ in $|\psi {⟩}_Q=a|0⟩+b|1⟩$ and her half $A$ of a shared Bell pair $|{\varphi }_{+}{⟩}_{AB}$; Bob holds $B$. Alice applies a controlled-NOT gate $Q\to A$, then a Hadamard gate on $Q$. In the system order $Q,A,B$, the initial quantum state isAfter these two quantum gates, direct expansion yields the quantum teleportation identityAlice performs computational-basis measurements on $Q,A$, and sends the two classical bits $m,n$ to Bob. Each outcome has probability $1/4$, independently of the input pure state. Bob applies the indicated Pauli gates: first $X^n$, then $Z^m$, so his total correction is $Z^m{X}^n$. In every branch,This proves exact quantum teleportation using one Bell pair and two classical bits. Alice's original qubit has been measured; the protocol transfers the quantum state without leaving her another copy.

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]

Define the Hadamard basis $|\pm ⟩=(|0⟩\pm |1⟩)/\sqrt{2}$ and write $s=e^{-i\varphi }$. After $U_{\varphi }$ the two-qubit quantum state isApplying the Hadamard gate to the first qubit gives the normalized quantum stateFor $\varphi =n\pi$, $s=(-1{)}^n$, soThe first is a product state. The second is exactly the minus Bell state in the printed phase gate convention; its reduced density matrices are $I/2$.

For three qubits, let $U_{12}$ use the ordered pair $(1,2)$ and $U_{23}$ the ordered pair $(2,3)$. The two diagonal quantum gates commute. For $\varphi =n\pi$ their action before the final Hadamard gates isIf $n$ is even this is $|+++⟩$, and $H_1{H}_3$ gives $|0⟩|+⟩|0⟩$. If $n$ is odd, group the quantum superposition by the middle qubit:Since $H_{\mathrm{Had}}|+⟩=|0⟩$ and $H_{\mathrm{Had}}|-⟩=|1⟩$, the final states areThe odd case is a GHZ state: each individual qubit has reduced density matrix $I/2$, and it is related to $(|000⟩+|111⟩)/\sqrt{2}$ by the local Pauli gates $Z_1{X}_3$. Conversely, applying those gates converts $|G⟩$ to that standard form. They have not been applied in the printed network, so the actual output used in the next part is $|G⟩=(|001⟩-|110⟩)/\sqrt{2}$.

In superdense coding, Alice and Bob initially share $|{\Psi }^{+}{⟩}_{AB}$. Alice selects the same one-qubit Pauli gate $I,Z,X,XZ$ according to her message and sends her single encoded qubit to Bob. He now has both halves and measures in the Bell basis. For the chosen assignment,which reaches two bits for the uniform distribution. A circuit implementation first converts the resource convention with $X_B$, then applies ${\mathrm{CNOT}}_{A\to B}$ and $H_A$, and measures both in the computational basis. The four outcomes are distinct; relabeling them restores the selected message order.

Before Alice's transmission, Bob's reduced density operator is $I/2$ for every choice of her local operation. Thus distributing the message-independent entangled resource has transmitted no information about the subsequently chosen message. It is nevertheless a real physical resource: the two-bit rate counts the one additional transmitted qubit after entanglement has been shared, not a free initial distribution. The deterministic conditional probabilities above establish perfect correlation rather than merely matching output and input entropies.

A resonant Rabi pi pulse givesIt swaps $|0⟩$ and $|1⟩$ with a common global phase, implementing the Pauli gate $X$, or NOT. Its Bloch sphere action is a half-turn about the $x$-axis.

The extra Pauli gate on the control is removed by $Z_{a}G=C_Z$. Conjugating the target by a Hadamard gate then givesThus, from left to right in the network, apply $H_b$, then $G$, then $Z_a$ and $H_b$; the last two gates commute.

For a direct proof, the control-zero block of $G$ is $I_b$ and its control-one block is $-Z_b$. Multiplication by $Z_a$ changes the latter into $Z_b$. Since $H_b{I}_b{H}_b=I_b$ and $H_b{Z}_b{H}_b=X_b$, the result iswhich is exactly the CNOT gate on all inputs, including quantum superpositions.

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.

Define $E|0⟩=|D_k^N⟩$ and $E|1⟩=|D_{k+1}^N⟩$. From the branch calculation in part (b), outcome $R_0=0,R_1=1$ leavesThus the effective input teleported was $X|\psi ⟩=\beta |0⟩+\alpha |1⟩$. A Pauli X gate flips every bit and hence maps a Dicke state with $j$ excitations to one with $N-j$ excitations. Since $N-k=k+1$, applying $X$ at every receiving party interchanges the two logical basis states and restores $E|\psi ⟩$. ThereforeIn particular Bob's uncorrected reduced density matrix is $X\rho X$, and his correction returns it to the matrix in part (c). The remaining branches can likewise be corrected: $Z^{\otimes N}$ implements logical $Z$ up to the common global phase $(-1{)}^k$, while $X^{\otimes N}$ implements logical $X$. Thus the outcome dependence can be removed by local Pauli gates even though the resulting copies remain imperfect.

For the known bit string $y$, define the Pauli gate product $T_y={⨂}_{j=0}^{N-1}{Z}_j^{y_j}$. Its diagonal action is $T_y|x⟩=(-1{)}^{x\cdot y}|x⟩$, where the binary inner product is taken modulo two. The Hadamard transform produces $|{\varphi }_y⟩=H^{\otimes N}|y⟩=T_y|\varphi ⟩$. Thus the required vectors for Grover search with a nonzero reflection label areThey remain orthonormal and $|{\varphi }_y⟩=\sin \theta |{\tilde{\Psi }}_g⟩+\cos \theta |{\tilde{\Psi }}_b⟩$. The new iterate is $G_y=(2|{\varphi }_y⟩⟨{\varphi }_y|-I)U_f$. Since $T_y$ commutes with the diagonal marked-state phase oracle and $T_y^2=I$,Consequently its matrix on the signed basis is exactly the rotation matrix in part (a), proving the claimed equivalence of the two evolutions.

Quantum teleportation transfers the quantum state of an unknown qubit from a sender's carrier to a receiver's carrier. It also transfers correlations with any external reference. It does not determine classical amplitudes or send the original particle.

The standard deterministic protocol consumes one shared maximally entangled Bell pair and two classical bits from Alice to Bob. Alice performs a local Bell-basis measurement on the input and her half of the pair; Bob applies the corresponding local Pauli gate. The classical message is essential: before receiving it, Bob has the outcome-averaged density operator $I/2$ and cannot recover or detect the unknown state. The original input is consumed by the measurement, so quantum teleportation is compatible with the no-cloning theorem.

Write the arbitrary joint pure state of the input $A$ and an external reference $R$ asThe reference vectors need not be normalized or orthogonal. Share $|{\Phi }^{+}{⟩}_{aB}$. Label Alice's Bell states byProjecting onto this Bell basis gives the unnormalized stateEach outcome has probability $1/4$. Alice sends $(r,s)$ and Bob applies $Z_B^r{X}_B^s$, reversing the two Pauli gates. The final state is exactly $|\Omega {⟩}_{BR}$ for every outcome. In particular all input entanglement and other correlations with $R$ now belong to $B$, while the reference marginal is unchanged. This is teleportation as an identity channel on a reference.

Conditioned on the record, Alice's two measured carriers are in $|{\beta }_{rs}⟩$ and factor from $BR$. Thus $A$ no longer retains its original correlations with $R$. There is no second copy, even when the teleported qubit was initially entangled with a larger system.