Superdense Coding

Derivation

Superdense Coding

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Statement

Given one pre-shared maximally entangled pair in the state \(|\Phi^+\rangle_{AB}=\tfrac{1}{\sqrt2}\left(|00\rangle+|11\rangle\right)\), with Alice holding qubit \(A\) and Bob holding qubit \(B\), Alice can transmit an arbitrary two-bit message \((j,k)\in\{0,1\}^2\) to Bob by applying a single local Pauli operation \(U_{jk}=X^{k}Z^{j}\) to \(A\) and sending only that one qubit down a noiseless channel. Bob's Bell-basis measurement of the pair recovers both bits deterministically. Thus one transmitted qubit plus one pre-shared ebit carries two classical bits — the exact time-reverse of teleportation.

Why it matters

Superdense coding is the sharpest demonstration that entanglement is a communication resource, not merely a correlation curiosity. A bare qubit, by the Holevo bound, conveys at most one classical bit of accessible information; the second bit here is "unlocked" by the ebit that was shared in advance. It fixes the entanglement-assisted classical capacity of a noiseless qubit channel at exactly two bits.

Together with teleportation it forms a resource duality: teleportation spends one ebit and two classical bits to move one qubit, superdense coding spends one ebit and one qubit to move two classical bits. Reading one protocol backwards, with sender and receiver exchanged, yields the other — a symmetry that underlies the modern resource theory of quantum Shannon theory.

Assumptions

A single Bell pair is shared in advanceWithout pre-shared entanglement, Alice's one qubit is a fresh two-level system whose accessible information is bounded by \(\log_2 2=1\) bit (Holevo); the second bit cannot be sent.

The forward qubit channel from Alice to Bob is noiseless and identity-preservingAny Pauli error on the travelling qubit maps the intended Bell state onto a different one, so Bob decodes a corrupted message; capacity drops below two bits.

Alice's encoding gates are ideal Pauli unitariesOver- or under-rotations produce a coherent superposition of Bell states, spoiling the one-to-one bit-to-state map and introducing decoding errors even on a perfect channel.

Bob can perform a complete projective measurement in the Bell basisBell measurement requires a joint (entangling) operation on the two qubits; with only local operations and classical communication the four Bell states cannot be distinguished deterministically, capping the recoverable information at one bit.

The pre-shared state is exactly maximally entangled and the local Hilbert spaces are two-dimensionalFor a partially entangled seed the four encoded states cease to be mutually orthogonal, so their measurement statistics overlap and the decoding is no longer error-free.

Derivation

\[ |\Phi^+\rangle_{AB}=\frac{1}{\sqrt2}\sum_{x\in\{0,1\}}|x\rangle_A\,|x\rangle_B=\frac{1}{\sqrt2}\left(|00\rangle+|11\rangle\right) \]

Fix the shared resource: one ebit in the canonical Bell state, Alice holds \(A\), Bob holds \(B\). A

\[ U_{jk}=X^{k}Z^{j},\qquad Z|x\rangle=(-1)^{x}|x\rangle,\quad X|x\rangle=|x\oplus1\rangle \]

Alice's encoding rule: to send the message \((j,k)\) she applies the local Pauli \(U_{jk}\) to qubit \(A\) only (identity on \(B\)). Define the Pauli action on the computational basis before touching the entangled state. A

\[ |\beta_{jk}\rangle \equiv \left(U_{jk}\otimes I\right)|\Phi^+\rangle =\frac{1}{\sqrt2}\sum_{x}(-1)^{jx}\,|x\oplus k\rangle_A\,|x\rangle_B \]

Push \(Z^{j}\) then \(X^{k}\) through the sum term by term: \(Z^{j}|x\rangle=(-1)^{jx}|x\rangle\) fixes the phase, \(X^{k}|x\rangle=|x\oplus k\rangle\) shifts the label. Linearity of the tensor product carries the operator inside the sum. B

\[ \begin{aligned} |\beta_{00}\rangle&=\tfrac{1}{\sqrt2}(|00\rangle+|11\rangle)=|\Phi^+\rangle, & |\beta_{01}\rangle&=\tfrac{1}{\sqrt2}(|10\rangle+|01\rangle)=|\Psi^+\rangle,\\[2pt] |\beta_{10}\rangle&=\tfrac{1}{\sqrt2}(|00\rangle-|11\rangle)=|\Phi^-\rangle, & |\beta_{11}\rangle&=\tfrac{1}{\sqrt2}(|10\rangle-|01\rangle)=-|\Psi^-\rangle. \end{aligned} \]

Evaluate the general expression at the four messages. Each Pauli rotates the seed onto a distinct Bell state; the whole Bell basis is reached. This is the encoding map \((j,k)\mapsto|\beta_{jk}\rangle\). B

\[ \langle\Phi^+|(M\otimes I)|\Phi^+\rangle=\frac{1}{2}\sum_{x,y}\langle x|M|y\rangle\,\delta_{xy}=\frac{1}{2}\,\mathrm{Tr}(M) \]

The "ricochet"/transpose identity for the maximally entangled state: an operator on \(A\) is turned into a trace. Proved by expanding both copies of \(|\Phi^+\rangle\) and using \(\langle x|y\rangle=\delta_{xy}\). It converts overlaps of encoded states into Pauli traces. C

\[ \langle\beta_{jk}|\beta_{j'k'}\rangle=\frac{1}{2}\,\mathrm{Tr}\!\left(U_{jk}^{\dagger}U_{j'k'}\right)=\delta_{jj'}\,\delta_{kk'} \]

Apply step 5 with \(M=U_{jk}^{\dagger}U_{j'k'}\). Distinct Pauli products are traceless while \(U^{\dagger}U=I\) has trace \(2\); hence the four encoded states are orthonormal and therefore perfectly distinguishable in principle. C

\[ \text{Alice} \xrightarrow{\;\text{send qubit } A\;} \text{Bob}\quad\Rightarrow\quad \text{Bob holds } |\beta_{jk}\rangle_{AB} \]

Alice transmits her single physical qubit \(A\) through the noiseless channel. Only one qubit crosses the channel; the ebit was already in place. Bob now possesses both halves. A

\[ \left(H\otimes I\right)\mathrm{CNOT}_{A\to B}\,|\beta_{jk}\rangle=|j\rangle_A\,|k\rangle_B \]

Bob's Bell measurement, realised as CNOT (control \(A\), target \(B\)) followed by Hadamard on \(A\), then a computational-basis readout. This unitary maps the Bell basis onto the product basis; track each \(|\beta_{jk}\rangle\) through CNOT then \(H\) to confirm the labels. B

\[ P(\,m_A=j',\,m_B=k'\mid \text{message } (j,k)\,)=\left|\langle\beta_{j'k'}|\beta_{jk}\rangle\right|^{2}=\delta_{jj'}\delta_{kk'} \]

By the Born rule and the orthonormality of step 6, the measurement outcome is deterministic: Bob reads exactly \((m_A,m_B)=(j,k)\). Both classical bits are recovered with unit probability. A

Result

\[ \boxed{\;1\ \text{ebit}\;+\;1\ \text{qubit sent}\;\longrightarrow\;2\ \text{classical bits}\;} \]

Reading. A single physical qubit, drawn from a pre-shared entangled pair and stamped with one of four local Pauli operations, carries a full two-bit message. The entanglement doubles the classical capacity of the channel: the map \((j,k)\mapsto U_{jk}=X^{k}Z^{j}\) encodes, the Bell measurement \((H\otimes I)\,\mathrm{CNOT}\) decodes, and recovery is exact. The second bit is paid for not by the channel but by the ebit consumed in the process.

Units check. All quantities are dimensionless information measures. Bob distinguishes \(4\) equiprobable messages, i.e. \(\log_2 4 = 2\) bits, using \(1\) transmitted qubit. Resource accounting balances as \([\,q\!\to\!q\,]+[\,qq\,]\ \ge\ 2\,[\,c\!\to\!c\,]\) (one qubit channel use plus one ebit yields two classical-bit channel uses), the dual of teleportation's \([\,qq\,]+2\,[\,c\!\to\!c\,]\ge[\,q\!\to\!q\,]\). Both sides carry units of "channel uses"; the equality of the entanglement-assisted capacity \(C_E = 2\) bits/qubit is consistent with the Holevo ceiling of \(1\) bit for the unassisted qubit.

Limiting cases

No shared entanglement (seed \(=|00\rangle\)): the four "encoded" states collapse to \(\{|00\rangle,|10\rangle,|00\rangle,|10\rangle\}\), only \(2\) distinct and non-orthogonal to the message index, so at most \(1\) bit — the ordinary single-qubit limit.
Only the phase Paulis \(\{I,Z\}\) used (\(k\) fixed to \(0\)): two orthogonal Bell states \(\{|\Phi^+\rangle,|\Phi^-\rangle\}\), exactly \(1\) bit — half the protocol, recovering bare-qubit performance.
Higher dimension \(d\) with a maximally entangled qudit seed: the Heisenberg–Weyl operators give \(d^2\) orthogonal encodings, transmitting \(\log_2 d^2 = 2\log_2 d\) bits per qudit sent — the general capacity-doubling.
Product (LOCC-only) decoding by Bob: without the entangling CNOT the four Bell states are not locally distinguishable, and the recoverable information falls back toward \(1\) bit.

Breaks when

Decohered or non-maximal seed. If the shared pair has degraded to a mixed state (e.g. Werner state with fidelity \(F<1\)) or is only partially entangled, the four encoded states overlap and Bob's outcomes become probabilistic; the two-bit rate is not achievable and one must fall back to the entanglement-assisted capacity of the noisy resource.
Noisy forward channel. A Pauli error on the travelling qubit \(A\) commutes into the encoding as \(X_A U_{jk}=U_{j,\,k\oplus1}\) (bit flip) or \(Z_A U_{jk}=\pm U_{j\oplus1,\,k}\) (phase flip), so Bob reads the wrong Bell state; deterministic decoding fails and error correction or a reduced-rate code is required.
No prior entanglement distribution. If the ebit was never established (or was measured/lost beforehand), the protocol reduces to sending an unentangled qubit and the Holevo bound caps the message at one bit — the second bit is simply unavailable.
Imperfect Bell measurement. If Bob can only measure locally, or his entangling gate is faulty, the four orthogonal states are not resolved and the mutual information between message and outcome drops below \(2\) bits.

Failure modes

"Two qubits are sent." Only qubit \(A\) crosses the channel; qubit \(B\) was already at Bob's side. Counting the pre-shared ebit as communication double-counts the resource and hides why the result is remarkable.
"This beats the Holevo bound." Holevo limits the accessible information of an unassisted qubit to one bit. Superdense coding does not violate it — the ebit is an extra resource, and the accessible information is computed on the joint two-qubit system Bob ultimately holds.
Wrong operator order in the encoding. Writing \(U_{jk}=Z^{j}X^{k}\) versus \(X^{k}Z^{j}\) changes an overall phase and can swap which bit is the "\(Z\)-bit"; students then mislabel outcomes. The bit assignment must be pinned to a chosen convention consistently with the decoding circuit.
Forgetting the Hadamard in decoding. Applying only CNOT leaves \(A\) in a superposition; measuring then gives random results for the \(j\)-bit. The Bell measurement needs CNOT and \(H\).
Assuming local measurements suffice. Measuring \(A\) and \(B\) separately in the computational basis cannot distinguish \(|\Phi^+\rangle\) from \(|\Phi^-\rangle\) (both give correlated \(00/11\)); the entangling gate is essential.
Treating the ebit as reusable. The Bell measurement consumes the entanglement; a fresh ebit is needed for each two-bit message.

Discussion

The heart of the protocol is that the four Pauli operators \(\{I,X,Z,XZ\}\) act on one half of a Bell pair to sweep out the entire orthonormal Bell basis. Alice never touches Bob's qubit, yet her purely local operation is globally distinguishable because the pre-existing correlations translate a local phase or flip into a change of the joint state's symmetry class. Entanglement thereby converts a one-bit-capacity object into a two-bit-capacity one: the information does not live "in" the travelling qubit alone but in the relationship between the two halves.

The duality with teleportation is exact and instructive. Teleportation transports an unknown qubit using one ebit and two classical bits; superdense coding transports two classical bits using one ebit and one qubit. Each is obtained from the other by exchanging the classical and quantum channels and reversing the direction of communication — a manifestation of the "coherent information" symmetry captured by the resource inequalities \([qq]+2[c\!\to\!c]\ge[q\!\to\!q]\) and \([qq]+[q\!\to\!q]\ge2[c\!\to\!c]\). Adding the two never manufactures free resources, which is precisely why entanglement cannot be used for superluminal signalling: Bob's marginal state before Alice's qubit arrives is the maximally mixed \(I/2\), independent of her message.

On the "chance" thread, the entire information content is discharged in a single measurement whose randomness has been engineered away: orthogonality (step 6) collapses the Born-rule probabilities to a Kronecker delta, so a fundamentally probabilistic readout becomes deterministic. On the "matter" thread, the qubit is a genuine physical two-level system — a photon polarization, a trapped-ion sublevel, a transmon — and the ebit must be physically distributed and stored with coherence intact, which is what makes the "free" second bit costly in practice.

Generalising, replace the qubit Paulis by the Heisenberg–Weyl group \(\{X^{a}Z^{b}\}_{a,b=0}^{d-1}\) acting on a maximally entangled qudit pair \(|\Phi_d\rangle=d^{-1/2}\sum_x|xx\rangle\). The \(d^2\) unitaries generate an orthonormal basis of maximally entangled states (a unitary error basis / nice error basis), so one qudit sent conveys \(2\log_2 d\) bits. This is the operational content of the entanglement-assisted capacity theorem of Bennett–Shor–Smolin–Thapliyal: for a general channel \(\mathcal{N}\), \(C_E(\mathcal{N})=\max_{\rho}\,I(A\!:\!B)_{\,(\mathbf{1}\otimes\mathcal{N})\phi_\rho}\), the quantum mutual information, of which noiseless superdense coding is the maximally entangled, noiseless-qubit extreme point.

Common misconceptions. Superdense coding does not send information faster than light and does not violate Holevo; it trades a pre-shared, direction-neutral entanglement resource for extra classical bandwidth. The ebit carries no message by itself — Bob's half is useless until Alice's encoded qubit arrives — so causality is respected at every step.

Worked examples

Message \((j,k)=(1,0)\): Alice sends "10". Encoding \(U_{10}=X^{0}Z^{1}=Z\).

Apply \(Z\) to qubit \(A\) of \(|\Phi^+\rangle\); phases on the two branches. A

\[ (Z\otimes I)\,\tfrac{1}{\sqrt2}(|00\rangle+|11\rangle)=\tfrac{1}{\sqrt2}\big((+1)|00\rangle+(-1)|11\rangle\big)=|\Phi^-\rangle \]

\(Z|0\rangle=|0\rangle,\ Z|1\rangle=-|1\rangle\) acting on the first factor. A

\[ \mathrm{CNOT}_{A\to B}\,|\Phi^-\rangle=\tfrac{1}{\sqrt2}(|00\rangle-|10\rangle),\qquad (H\otimes I)\,\tfrac{1}{\sqrt2}(|00\rangle-|10\rangle)=|10\rangle \]

CNOT copies \(A\) onto \(B\): \(|11\rangle\!\to\!|10\rangle\). Then \(H\) on \(A\): \(\tfrac{1}{\sqrt2}(|0\rangle-|1\rangle)\!\to\!|1\rangle\), leaving \(B=|0\rangle\). B

\[ (m_A,m_B)=(1,0)\;\Rightarrow\;\text{decoded message }=\text{"10"} \]

Reading. Bob reads out \(1\) on \(A\) and \(0\) on \(B\), exactly Alice's two bits, from a single received qubit. Units. \(\log_2 4 = 2\) bits recovered, one qubit sent, one ebit consumed.

Message \((j,k)=(1,1)\): Alice sends "11". Encoding \(U_{11}=X^{1}Z^{1}=XZ\).

Apply \(Z\) then \(X\) to qubit \(A\). A

\[ (XZ\otimes I)\,\tfrac{1}{\sqrt2}(|00\rangle+|11\rangle)=\tfrac{1}{\sqrt2}(|10\rangle-|01\rangle) \]

\(Z\) gives \(\tfrac{1}{\sqrt2}(|00\rangle-|11\rangle)\); then \(X\) on \(A\) flips \(0\!\leftrightarrow\!1\) on the first factor: \(|00\rangle\!\to\!|10\rangle,\ |11\rangle\!\to\!|01\rangle\). B

\[ \mathrm{CNOT}_{A\to B}\,\tfrac{1}{\sqrt2}(|10\rangle-|01\rangle)=\tfrac{1}{\sqrt2}(|11\rangle-|01\rangle),\quad (H\otimes I)\,\tfrac{1}{\sqrt2}(|11\rangle-|01\rangle)=|11\rangle \]

CNOT: \(|10\rangle\!\to\!|11\rangle\), \(|01\rangle\!\to\!|01\rangle\); factor as \(\tfrac{1}{\sqrt2}(|1\rangle-|0\rangle)_A|1\rangle_B\). Then \(H\) on \(A\): \(\tfrac{1}{\sqrt2}(|1\rangle-|0\rangle)=-\tfrac{1}{\sqrt2}(|0\rangle-|1\rangle)\!\to\!-|1\rangle\); the global phase is unobservable. B

\[ (m_A,m_B)=(1,1)\;\Rightarrow\;\text{decoded message }=\text{"11"} \]

Reading. The two-bit string "11" is recovered exactly; the overall minus sign is a physically irrelevant global phase. Units. Two classical bits out per one qubit in, one ebit spent.

Problems

Verify explicitly that \(U_{01}=X\) applied to qubit \(A\) of \(|\Phi^+\rangle\) yields \(|\Psi^+\rangle=\tfrac{1}{\sqrt2}(|01\rangle+|10\rangle)\), and trace it through the decoding circuit to find Bob's outcome.
Solution
\((X\otimes I)\tfrac{1}{\sqrt2}(|00\rangle+|11\rangle)=\tfrac{1}{\sqrt2}(|10\rangle+|01\rangle)=|\Psi^+\rangle\). CNOT\(_{A\to B}\): \(|10\rangle\to|11\rangle,\ |01\rangle\to|01\rangle\), giving \(\tfrac{1}{\sqrt2}(|11\rangle+|01\rangle)=\tfrac{1}{\sqrt2}(|1\rangle+|0\rangle)_A|1\rangle_B\). \(H\) on \(A\): \(\tfrac{1}{\sqrt2}(|0\rangle+|1\rangle)\to|0\rangle\). Outcome \((m_A,m_B)=(0,1)\), i.e. message "01", matching \((j,k)=(0,1)\).
Using the ricochet identity \(\langle\Phi^+|(M\otimes I)|\Phi^+\rangle=\tfrac12\mathrm{Tr}(M)\), prove the four encoded states are orthonormal, and state why orthonormality is necessary for error-free decoding.
Solution
\(\langle\beta_{jk}|\beta_{j'k'}\rangle=\langle\Phi^+|(U_{jk}^\dagger U_{j'k'}\otimes I)|\Phi^+\rangle=\tfrac12\mathrm{Tr}(U_{jk}^\dagger U_{j'k'})\). Now \(U_{jk}^\dagger U_{j'k'}=Z^{j}X^{k}X^{k'}Z^{j'}=X^{k\oplus k'}Z^{j\oplus j'}\) up to a sign. If \((j,k)=(j',k')\) this is \(I\), trace \(2\), overlap \(1\). Otherwise it is a non-identity Pauli, which is traceless (\(\mathrm{Tr}\,X=\mathrm{Tr}\,Z=\mathrm{Tr}\,XZ=0\)), overlap \(0\). Hence \(\langle\beta_{jk}|\beta_{j'k'}\rangle=\delta_{jj'}\delta_{kk'}\). Orthogonal states admit a projective measurement whose outcomes never overlap, so by the Born rule \(P(\text{outcome}=(j,k)\mid\text{message}=(j,k))=1\); non-orthogonal states would give overlapping statistics and a nonzero error probability.
Bob's Bell measurement is the map \((H\otimes I)\,\mathrm{CNOT}_{A\to B}\). Show it sends \(|\Phi^-\rangle\to|10\rangle\) and \(|\Psi^-\rangle\to|11\rangle\), confirming the label assignment \(|\beta_{jk}\rangle\to|jk\rangle\).
Solution
\(|\Phi^-\rangle=\tfrac{1}{\sqrt2}(|00\rangle-|11\rangle)\xrightarrow{\mathrm{CNOT}}\tfrac{1}{\sqrt2}(|00\rangle-|10\rangle)=\tfrac{1}{\sqrt2}(|0\rangle-|1\rangle)_A|0\rangle_B\xrightarrow{H_A}|1\rangle|0\rangle=|10\rangle\). \(|\Psi^-\rangle=\tfrac{1}{\sqrt2}(|01\rangle-|10\rangle)\xrightarrow{\mathrm{CNOT}}\tfrac{1}{\sqrt2}(|01\rangle-|11\rangle)=\tfrac{1}{\sqrt2}(|0\rangle-|1\rangle)_A|1\rangle_B\xrightarrow{H_A}|1\rangle|1\rangle=|11\rangle\). Both agree with \(|jk\rangle\) for \((1,0)\) and \((1,1)\) respectively.
The forward channel applies a bit flip \(X\) to the travelling qubit \(A\) with probability \(p\) and identity with probability \(1-p\). Determine which classical bit is corrupted and with what probability, given noiseless everything else.
Solution
An \(X\) error acts after encoding: \(X_A\,|\beta_{jk}\rangle=(X X^{k}Z^{j}\otimes I)|\Phi^+\rangle=(X^{k\oplus1}Z^{j}\otimes I)|\Phi^+\rangle=|\beta_{j,\,k\oplus1}\rangle\). So the error maps the intended Bell state onto the one whose \(k\)-label is flipped, leaving \(j\) untouched. Bob therefore decodes \(j\) correctly always, and decodes \(k\) correctly with probability \(1-p\) and flipped with probability \(p\). The first bit \(j\) has error probability \(0\); the second bit \(k\) has error probability \(p\). (A phase flip \(Z\) would instead corrupt \(j\), since \(Z_A|\beta_{jk}\rangle=\pm|\beta_{j\oplus1,k}\rangle\).)
State the resource inequalities for superdense coding and teleportation, then show that composing them yields no net free resource (no "resource perpetual motion"), and explain how this enforces no-signalling.
Solution
Superdense: \([qq]+[q\!\to\!q]\ge2[c\!\to\!c]\). Teleportation: \([qq]+2[c\!\to\!c]\ge[q\!\to\!q]\). Add them: \(2[qq]+[q\!\to\!q]+2[c\!\to\!c]\ge[q\!\to\!q]+2[c\!\to\!c]+2[qq]\), which reduces to \(0\ge0\) — an equality, so no resource is created; the two protocols are exact time-reverses that merely convert resources into one another. For no-signalling: in superdense coding Bob's reduced state before receiving \(A\) is \(\mathrm{Tr}_A|\beta_{jk}\rangle\langle\beta_{jk}|=I/2\) for every message \((j,k)\), since each \(|\beta_{jk}\rangle\) is maximally entangled. Bob's local statistics are message-independent until Alice's qubit physically arrives, so no information is transmitted faster than the channel — the ebit alone signals nothing.