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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.

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.

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.

Label the unknown input $1$, Alice's resource photon $2$, and Bob's resource photon $3$. First distribute photon $3$ through the channel while keeping photon $2$ with Alice; at $p=1$ this leaves the perfect shared Bell state $|{\Phi }^{+}{⟩}_{23}=(|00⟩+|11⟩)/\sqrt{2}$. When the input arrives, the three-qubit state isAlice applies ${\mathrm{CNOT}}_{1\to 2}$, givingShe next applies $H_1$, producingThese two gates followed by computational measurement implement a perfect Bell state analyzer. For outcome $(m,n)$ the normalized conditional state is $|mn{⟩}_{12}{X}^n{Z}^m|\psi {⟩}_3$, with probability $1/4$. Alice sends the two classical bits $m,n$; Bob applies $Z^m{X}^n$, recovering $|\psi ⟩$ exactly. Averaging before reception of the bits gives Bob $I/2$, independent of the input, so this protocol provides no faster-than-light communication.

For $p=1$, directly sending the original qubit is already perfect. Quantum teleportation therefore has no fidelity advantage in this ideal case and consumes a shared entangled pair plus classical communication. Its operational advantage is that entanglement can be distributed beforehand: once the unknown input appears, only classical information need travel, and the original input photon need not itself propagate to Bob. The Bell measurement consumes the input; it does not create a second copy.

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.

For a memoryless quantum channel $\mathcal{N}$, successive uses are described by ${\mathcal{N}}^{\otimes n}$. Its entanglement-assisted classical capacity is the supremum of classical bits per use achievable with error tending to zero when sender and receiver may use unlimited preshared quantum entanglement, independent of the message. Only the channel is used to transmit the message.

For input state ${\rho }_A$, choose a purification $|{\psi }_{\rho }{⟩}_{RA}$ and set ${\omega }_{RB}=({\mathrm{id}}_R\otimes \mathcal{N})(|{\psi }_{\rho }⟩⟨{\psi }_{\rho }|)$. The capacity formula isPurifications differ only by an isometry on the reference, so this expression depends only on the input density matrix. The quantity optimized is quantum mutual information.

The binary quantum erasure channel sends a qubit to a qubit subspace plus an orthogonal erasure flag:On the nonerased branch it transmits the entire state, including its coherences. Put $s=S(\rho )$. The entropy of an orthogonal quantum mixture givesThe reference-output state has two orthogonal branches:The first branch is pure; the second has entropy $S({\rho }_R)=s$. Thus $S({\omega }_{RB})=h_2(p)+ps$ andA qubit has entropy at most one bit, with equality at $\rho =I_2/2$. We concludeThe endpoint capacities are two and zero. As an operational check, superdense coding encodes four classical messages in one member of a shared Bell pair. A successful channel use lets the receiver identify the message; an erased qubit yields the flag. This induces a four-symbol classical erasure channel of capacity $(1-p){\log }_{2}4=2(1-p)$, agreeing with the entropy calculation.

Yes: use superdense coding, with a shared Bell state. Let the sender and receiver initially hold the first and second qubits of $|{\Phi }^{+}⟩=(|00⟩+|11⟩)/\sqrt{2}$. To encode bits $(a,b)$, the sender applies $Z^a{X}^b$ to her qubit. These Pauli operators produce four orthogonal states:She sends this qubit through the noiseless quantum channel. The receiver then possesses both qubits. He applies a controlled-NOT gate from the received qubit to his original qubit, followed by a Hadamard gate on the received qubit. This maps the four states in the table to $|ab⟩$. A measurement in the computational basis therefore recovers the two bits with certainty. No classical message is sent.

The resource assumption deserves care: one-qubit superdense coding needs a Bell pair already shared before encoding. If none is initially available, the sender can prepare a Bell pair locally and send one half through the same noiseless quantum channel in advance. The protocol then uses two transmitted qubits in total, including distribution of the entanglement, still with no classical communication. Alternatively, sending two computational basis qubits directly also communicates two bits. With a pre-shared Bell pair, one transmitted qubit suffices; without it, the entanglement-distribution cost must be counted. An unassisted single transmitted qubit cannot perfectly encode four classical messages: its Holevo bound is at most ${\log }_{2}2=1$ bit.

Use superdense coding. For the two-bit message $(a,b)\in \{0,1{\}}^2$, Alice applies $Z^a{X}^b$ to her half of the shared Bell state, where $X,Z$ are the Pauli X gate and Pauli Z gate. The encoded states areThese four Bell states are orthonormal. Alice then transmits her one qubit to Bob, who now holds both qubits. A Bell-basis measurement identifies the message with probability one.

More explicitly, Bob can apply a CNOT gate with the received qubit as control and his original qubit as target, followed by a Hadamard gate on the control. The CNOT gate sends the encoded state toand the Hadamard gate sends this to $|a⟩\otimes |b⟩$. Computational-basis projective measurements therefore read out $a,b$ exactly. One transmitted qubit, assisted by one previously shared Bell pair, communicates two classical bits.

This is superdense coding. To encode $00,01,10,11$, Alice applies respectively $I,X,Z,XZ$ to her half of $|{\Phi }^{+}⟩$. These local Pauli operators transform the shared state into the four mutually orthogonal Bell states $|{\Phi }^{+}⟩,|{\Psi }^{+}⟩,|{\Phi }^{-}⟩,|{\Psi }^{-}⟩$, up to irrelevant global phases. Alice sends her one qubit to Bob. Bob now holds both qubits and performs a Bell-basis measurement, whose outcome identifies Alice's two-bit message with certainty. One transmitted qubit communicates two classical bits because one shared Bell pair was available beforehand.

Assume no previously shared entanglement. Alice's encoded ensemble $\{p_x,{\rho }_x\}$ lives in a Hilbert space of dimension $2^n$. Its Holevo quantity satisfiesusing nonnegativity of Von Neumann entropy and the maximum entropy of a quantum state. The data-processing inequality for quantum relative entropy implies ${\chi }_{\mathrm{out}}\le {\chi }_{\mathrm{in}}$ for the transmission quantum channel. Finally, the Holevo bound applied to Bob's quantum measurement givesThis argument also works when the output system has larger dimension than the input. Prior shared entanglement changes the communication resource: superdense coding can transmit two classical bits per sent qubit, so the absence of that resource is part of this bound's setting.

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.

A pure bipartite quantum state is entangled if it cannot be written as a tensor product $|a⟩\otimes |b⟩$. More generally a mixed quantum state is entangled if its density matrix cannot be expressed as a convex mixture of product density matrices. Two applications are quantum teleportation, which transfers an unknown qubit using a shared entangled pair plus classical communication, and superdense coding, which uses a shared entangled pair and transmission of one qubit to communicate two classical bits. Neither permits faster-than-light communication.

Superdense coding uses a shared Bell state to communicate two classical bits by transmitting one qubit. The sender selects one of four orthogonal Bell states with a local Pauli operator, and the receiver performs a Bell-basis measurement. For an ideal shared Bell pair, the transmitted qubit alone has a maximally mixed state for every message and reveals no information about the chosen encoding.