Superdense Coding

Superdense coding is an entanglement-assisted communication protocol in which Alice sends one of four classical messages by applying a local Pauli operator to her half of a shared Bell pair and then physically transmitting that qubit to Bob. Once Bob holds both qubits, a Bell-basis decoder identifies the message without error.

Bennett and Wiesner (1992) introduced the protocol. Its ideal resource statement is

$$[qq] +[q\to q] \succeq 2[c\to c].$$

One shared ebit and one noiseless qubit transmission produce two classical bits of forward communication. The ebit is consumed. The statement is not that an isolated qubit always carries two accessible classical bits, nor that entanglement sends a message by itself.

Required background. Bell states supplies the entangled basis. Single-qubit gates and multi-qubit gates define the Pauli, Hadamard and controlled-NOT operations used below.

Encoding four messages in a Bell pair

Alice and Bob initially share $|\Phi^+\rangle_{AB}$. After this pair is distributed, Alice receives the two-bit message $(a,b)$. She encodes it on $A$, physically sends $A$ through a noiseless qubit channel, and Bob measures both qubits jointly. The ideal protocol recovers every message with probability one. Uniform messages carry two bits of Shannon entropy; the same decoder also works for nonuniform messages.

The shared pair is independent of the later message. If distributing it uses the same counted channel, that earlier transmission must also be counted.

The phase bit and the parity bit

Alice encodes the phase bit $a$ and parity bit $b$ with

$$U_{ab} = Z^aX^b.$$

The rightmost operator acts first, so the circuit applies $X^b$ and then $Z^a$. Acting on Alice’s half of $|\Phi^+\rangle$ gives four codewords:

$a$	$b$	$U_{ab}$	encoded state
$0$	$0$	$I$	$\lvert\Phi^+\rangle$
$1$	$0$	$Z$	$\lvert\Phi^-\rangle$
$0$	$1$	$X$	$\lvert\Psi^+\rangle$
$1$	$1$	$ZX$	$\lvert\Psi^-\rangle$

The possible global phase of the last row under a different Pauli ordering has no observable effect. What matters is that the four messages map to four orthogonal joint states.

A compact expression is

$$\begin{aligned} |\beta_{ab}\rangle&=(Z^aX^b\otimes I)|\Phi^+\rangle,\\ &=\frac1{\sqrt2}\sum_{j=0}^1(-1)^{a(j\oplus b)} |j\oplus b,j\rangle. \end{aligned}$$

Here $\oplus$ denotes addition modulo two, and the ket entries are ordered $A,B$. The bit $b$ selects equal versus different computational-basis values, while $a$ selects the relative phase.

Orthogonality proof

For a maximally entangled qubit pair,

$$\langle\Phi^+| (M\otimes I) |\Phi^+\rangle = \frac12\operatorname{Tr}M.$$

Therefore

$$\begin{aligned} \langle\beta_{ab}|\beta_{a'b'}\rangle &= \frac12 \operatorname{Tr} \left( U_{ab}^\dagger U_{a'b'} \right) \\ &= \delta_{aa'}\delta_{bb'}, \end{aligned}$$

because the four Pauli operators are orthogonal under the Hilbert–Schmidt inner product. Bob can distinguish the codewords perfectly once both qubits are in his laboratory.

Circuit and decoder

Qubit superdense coding. Alice encodes $(a,b)$ with $X^b$ followed by $Z^a$ and sends qubit $A$. Bob applies $\operatorname{CNOT}_{A\to B}$ and $H_A$, the inverse Bell transform, then measures both qubits to recover $a$ and $b$.

The decoding unitary is

$$D = (H_A\otimes I_B) \operatorname{CNOT}_{A\to B}.$$

Apply the controlled-NOT to the explicit codeword:

$$\begin{aligned} &\operatorname{CNOT}_{A\to B}|\beta_{ab}\rangle\\ &= \frac{1}{\sqrt2} \sum_{j=0}^{1} (-1)^{a(j\oplus b)} |j\oplus b\rangle_A |b\rangle_B \\ &= H|a\rangle_A \otimes |b\rangle_B. \end{aligned}$$

The final Hadamard gives

$$D|\beta_{ab}\rangle = |a\rangle_A|b\rangle_B.$$

A computational-basis measurement therefore reads the original two-bit message. This derivation also fixes the bit convention: the first measured wire is the phase bit $a$, and the second is the parity bit $b$. Software frameworks may print bitstrings in the opposite order, so an implementation must state its wire and readout conventions.

Where the two bits reside

The message is not locally visible in either half of the encoded Bell pair. For every $(a,b)$,

$$\begin{aligned} \operatorname{Tr}_B|\beta_{ab}\rangle\langle\beta_{ab}|&=\frac I2,\\ \operatorname{Tr}_A|\beta_{ab}\rangle\langle\beta_{ab}|&=\frac I2. \end{aligned}$$

Before Alice sends qubit $A$, Bob holds only $B$ and sees the same state $I/2$ for all four messages. This is required by No-Cloning and No-Signaling. After the transmission, Bob holds the joint system $AB$, and the message is available in correlations between its two parts.

This is an example of a recurring principle: local states can be identical while joint states are perfectly distinguishable. Entanglement changes which global codewords Alice can reach by acting locally before she sends her subsystem.

Holevo information and the dimension bound

Without preshared entanglement, Alice can encode a classical variable $M$ into qubit states $\rho_m$. The Holevo quantity obeys

$$\begin{aligned} \chi &= S\left(\sum_m p_m\rho_m\right) - \sum_m p_m S(\rho_m) \\ &\leq \log_2\dim\mathcal H_A = 1. \end{aligned}$$

Thus the unassisted classical capacity of a noiseless qubit channel is one bit per channel use, even though the Bloch sphere contains continuously many pure states.

In dense coding, Bob decodes an ensemble on the joint four-dimensional system $AB$. For four equiprobable Bell states,

$$\overline\rho_{AB} = \frac14 \sum_{a,b} |\beta_{ab}\rangle\langle\beta_{ab}| = \frac{I_{AB}}4.$$

Each codeword is pure, so

$$\chi_{\mathrm{DC}} = S\left(\frac{I_{AB}}4\right) -0 = 2.$$

There is no violation of the Holevo bound. The relevant receiver system has dimension four because Bob combines the transmitted qubit with his preshared half. The extra dimension was not free: it came with the entangled resource.

The cost of distributing the pair

The two-bit protocol needs a Bell pair before the message is chosen, one Pauli encoding, one forward qubit transmission and a joint decoder. The compact resource inequality in the lead assumes this prior assistance. It does not charge the earlier distribution to the same use of the forward channel.

If Bob creates $|\Phi^+\rangle$ locally and sends half to Alice through the same channel being counted, then one use distributes the ebit and a second use returns Alice’s encoded half. Sending two classical bits required two qubit transmissions across that link. Dense coding becomes an advantage when entanglement is available as prior assistance, perhaps created during idle time, supplied by a separate link, or amortized within a larger resource protocol.

The protocol also consumes the ebit. After ideal decoding, the state is the product codeword $|ab\rangle$, and the Bell entanglement is gone. Reusing the protocol requires another shared entangled pair.

Teleportation has the complementary resource inequality

$$[qq] +2[c\to c] \succeq [q\to q].$$

The two protocols exchange the roles of classical and quantum communication, but neither makes communication free. Dense coding still sends a qubit; teleportation still sends two classical bits.

Qudit generalization

Let Alice and Bob share the maximally entangled state

$$|\Phi_d\rangle = \frac1{\sqrt d} \sum_{j=0}^{d-1} |j\rangle_A|j\rangle_B.$$

Define generalized shift and phase operators

$$\begin{aligned} X|j\rangle&=|j+1\bmod d\rangle,\\ Z|j\rangle&=\omega^j|j\rangle,\qquad\omega=e^{2\pi i/d}. \end{aligned}$$

Alice applies one of the $d^2$ Weyl operators

$$U_{mn} = Z^mX^n, \qquad m,n\in\{0,\ldots,d-1\}.$$

The resulting states

$$|\beta_{mn}^{(d)}\rangle = (U_{mn}\otimes I)|\Phi_d\rangle$$

are orthonormal because

$$\begin{aligned} \langle\beta_{mn}^{(d)}|\beta_{m'n'}^{(d)}\rangle &=\frac1d\operatorname{Tr}(U_{mn}^\dagger U_{m'n'})\\ &=\delta_{mm'}\delta_{nn'}. \end{aligned}$$

One transmitted $d$-level system can therefore select among $d^2$ messages, or carry

$$\log_2 d^2 = 2\log_2 d$$

classical bits, given one preshared maximally entangled qudit pair and a complete generalized Bell measurement (Werner, 2001). For $d$ not a power of two, $2\log_2d$ describes the uniform message entropy and asymptotic bit rate. A single $d^2$-symbol alphabet can carry $\lfloor2\log_2d\rfloor$ arbitrary fixed-length binary message bits with zero error. For example, a qutrit offers nine distinct messages but cannot encode all sixteen four-bit strings.

Imperfect shared entanglement

Suppose Alice and Bob share an arbitrary state $\rho_{AB}$, Alice’s subsystem has dimension $d_A$, the forward channel is ideal, and Alice uses a complete orthogonal unitary ensemble with equal probabilities. The encoded states are

$$\rho_{mn} = (U_{mn}\otimes I) \rho_{AB} (U_{mn}^\dagger\otimes I).$$

Unitary invariance gives

$$S(\rho_{mn}) = S(\rho_{AB}).$$

The complete unitary average depolarizes Alice’s subsystem:

$$\overline\rho_{AB} = \frac{I_A}{d_A} \otimes \rho_B.$$

The Holevo quantity of this standard dense-coding ensemble is therefore

$$\begin{aligned} \chi_{\mathrm{DC}} &= S(\overline\rho_{AB}) - S(\rho_{AB}) \\ &= \log_2d_A +S(\rho_B) -S(\rho_{AB}) \\ &= \log_2d_A -S(A|B)_\rho. \end{aligned}$$

The equal-probability orthogonal-unitary encoding is optimal within the local-unitary encoding class (Bowen, 2001; Hiroshima, 2001). This formula assumes that class and ideal transmission; it is an asymptotic Holevo rate when collective code construction and decoding are allowed. It should not be read as a guarantee of that many perfectly distinguishable messages in one shot.

This unitary scheme gives a dense-coding advantage over the ordinary $\log_2d_A$-bit baseline precisely when

$$S(A|B)_\rho = S(\rho_{AB})-S(\rho_B) < 0.$$

Negative quantum conditional entropy implies entanglement, but not every entangled mixed state has negative conditional entropy. Entanglement is therefore not automatically useful for this particular coding task.

If the formula falls below the unassisted baseline, Alice can ignore the poor resource and use an ordinary orthogonal encoding. With that option included, the available rate is at least

$$\max \left\{ \log_2d_A,\, \log_2d_A-S(A|B)_\rho \right\}.$$

Partially entangled pure pair

For

$$|\psi_\lambda\rangle = \sqrt\lambda\,|00\rangle +\sqrt{1-\lambda}\,|11\rangle,$$

the joint entropy vanishes and Bob’s entropy is the binary entropy

$$h_2(\lambda)={}-\lambda\log_2\lambda -(1-\lambda)\log_2(1-\lambda).$$

The asymptotic dense-coding rate is

$$\chi_{\mathrm{DC}} = 1+h_2(\lambda).$$

It ranges from one bit for a product state to two bits for a Bell pair. Except at maximal entanglement, the four Pauli codewords are not all orthogonal, so this fractional asymptotic rate is not a deterministic four-message one-shot protocol.

A Pauli-noise example

Let the transmitted qubit pass through a Pauli channel

$$\begin{aligned} \mathcal P(\rho)&=\sum_{P\in\{I,X,Y,Z\}}p_P P\rho P,\\ \sum_Pp_P&=1. \end{aligned}$$

A Pauli error on Alice’s transmitted half permutes the Bell basis. After the ideal decoder, Bob sees the intended two-bit label plus an additive two-bit error determined by $P$. With uniform messages, the induced classical channel has rate

$$I(M;\widehat M) = 2 -H_4(p_I,p_X,p_Y,p_Z),$$

where

$$H_4(\mathbf p) = -\sum_P p_P\log_2p_P.$$

The formula has useful checks:

• if one known Pauli occurs with probability one, it can be calibrated away and the rate is two bits;
• if all four Paulis are equally likely, the Bell label is fully randomized and the rate is zero;
• intermediate noise turns the perfect decoder into a four-symbol confusion channel.

For general noisy channels and unrestricted asymptotic coding, the relevant quantity is the entanglement-assisted classical capacity

$$\begin{aligned} \sigma_{RB}&=(\mathcal I_R\otimes\mathcal N)(\phi_{RA}),\\ C_E(\mathcal N)&=\max_{\rho_A} I(R;B)_\sigma. \end{aligned}$$

Here $\phi_{RA}$ purifies the input $\rho_A$, and the capacity permits unlimited prior entanglement. Bennett et al. (2002) prove this capacity theorem; Communication with Quantum Systems explains its assumptions. The Bell-pair protocol here is its most elementary noiseless qubit example.

Bell measurement and lost trials

An experiment needs both coherent encoding and a joint measurement capable of distinguishing the chosen alphabet. Mattle et al. (1996) demonstrated a three-message photonic alphabet. Three equiprobable symbols have entropy $\log_2 3\approx1.585$ bits; the achieved mutual information also depends on decoding errors and losses.

The measurement model matters. For two photonic qubits encoded in two optical modes each, a passive linear-optical analyzer with vacuum auxiliary modes and photon counting cannot unambiguously discriminate four equiprobable Bell states with average success above $1/2$ (Calsamiglia and Lütkenhaus, 2001). This is a bound on that analyzer class and on unambiguous discrimination. Additional photons, other degrees of freedom or nonlinear interactions change the available resources; the bound is not a prohibition on Bell measurements in general.

The full confusion matrix describes which message Bob reports for every message Alice sends. Include a separate failure symbol when an attempt produces no usable label. If an independent erasure occurs with probability $1-\eta$, the failure flag itself carries no message information, and

$$I_{\mathrm{attempt}}=\eta I_{\mathrm{accepted}}.$$

Multiplying by the attempt frequency converts this quantity to bits per second for the specified coding model. If acceptance depends on the message, the accepted message distribution changes and the failure flag can itself carry information. Then use the complete channel, including failures, rather than multiplying a postselected rate by a single efficiency. Exercise 7 works through the independent-erasure case.

Exercises

1. Verify the decoder algebra

Show explicitly that $(H\otimes I)\operatorname{CNOT}|\Psi^-\rangle=|11\rangle$.

Solution

First,

$$\begin{aligned} \operatorname{CNOT}|\Psi^-\rangle &=\frac{|01\rangle-|11\rangle}{\sqrt2}\\ &=|-\rangle|1\rangle. \end{aligned}$$

Since $H|-\rangle=|1\rangle$,

$$(H\otimes I) \operatorname{CNOT} |\Psi^-\rangle = |11\rangle.$$

The top measurement returns the phase bit and the bottom measurement returns the parity bit.

2. Why a partially entangled pair cannot carry four perfect labels

For $|\psi_\lambda\rangle$ defined above, compare the codewords obtained by $I_A$ and $Z_A$. Compute their inner product at $\lambda=0.9$ and explain the limitation for deterministic discrimination.

Solution

The vectors are

$$\begin{aligned} |u\rangle&=\sqrt\lambda|00\rangle+\sqrt{1-\lambda}|11\rangle,\\ |v\rangle&=\sqrt\lambda|00\rangle-\sqrt{1-\lambda}|11\rangle. \end{aligned}$$

Their overlap is $\langle u|v\rangle=2\lambda-1=0.8$. Distinct nonorthogonal pure states cannot be distinguished with certainty in every trial. This pair alone rules out perfect discrimination of all four Pauli labels. It does not rule out an asymptotic code with a rate between one and two bits per use.

3. Locate the message before transmission

Prove that Bob’s original qubit contains no information about $(a,b)$ before Alice’s qubit arrives.

Solution

Every encoded state is a Bell state, and every Bell state has maximally mixed one-qubit marginals:

$$\rho_B^{ab} = \operatorname{Tr}_A |\beta_{ab}\rangle\langle\beta_{ab}| = \frac I2.$$

Thus the ensemble available to Bob before transmission is independent of the message. Any local measurement has the same outcome distribution for all $(a,b)$. The message becomes accessible only after Bob receives $A$ and can measure $AB$ jointly.

4. Partially entangled resource

For $\lambda=0.9$, compute the asymptotic dense-coding rate of $|\psi_\lambda\rangle$ and explain why it is not a four-message one-shot protocol.

Solution

The binary entropy is

$$\begin{aligned} h_2(0.9)&=-0.9\log_2 0.9-0.1\log_2 0.1\\ &\approx0.469. \end{aligned}$$

Hence

$$\chi_{\mathrm{DC}} = 1+h_2(0.9) \approx 1.469$$

bits per use asymptotically. Because the state is not maximally entangled, the four Pauli-encoded states are not all orthogonal. A single joint measurement cannot identify four equiprobable labels perfectly.

5. Pauli-channel rate

Let $p_I=0.85$ and $p_X=p_Y=p_Z=0.05$. Find the mutual information of the uniform Bell-label scheme.

Solution

The four-symbol error entropy is

$$\begin{aligned} H_4 &= -0.85\log_2 0.85 -3(0.05\log_2 0.05) \\ &\approx 0.848. \end{aligned}$$

Therefore

$$I(M;\widehat M) = 2-H_4 \approx 1.152$$

bits per channel use for this encoding and decoder.

6. Count the distribution cost

Bob prepares a Bell pair, sends qubit $A$ to Alice, and later Alice uses dense coding to send $A$ back. If both transmissions use the same noiseless link, what is the net number of classical bits per forward-or-back channel use for this one-shot cycle?

Solution

The cycle uses the link once to distribute Alice’s half and once to return the encoded qubit. It communicates two classical bits in two quantum-channel uses, so the net rate is one classical bit per counted use. The ideal two-bits-per- use statement treats the Bell pair as prior assistance and counts only the message-bearing forward use.

7. Symmetric decoder errors

Suppose the correct two-bit label is reported with probability $1-\epsilon$, and each of the other three labels is reported with probability $\epsilon/3$. For uniform inputs, find the mutual information. Then include an independent erasure with success probability $\eta$ and evaluate $\eta=0.6$, $\epsilon=0.1$.

Solution

The output is uniform by symmetry, so $H(\widehat M)=2$. The conditional entropy is the entropy of one row of the confusion matrix:

$$H(\widehat M|M) = h_2(\epsilon) +\epsilon\log_2 3.$$

Thus

$$I(M;\widehat M) = 2 -h_2(\epsilon) -\epsilon\log_2 3.$$

Now add an independent erasure with probability $1-\eta$. The receiver’s alphabet contains the four labels and a failure symbol. Since the failure probability is independent of $M$, the chain rule gives

$$\begin{aligned} I_{\mathrm{attempt}}&=\eta I_{\mathrm{accepted}},\\ I_{\mathrm{accepted}}&=2-h_2(\epsilon)-\epsilon\log_2 3. \end{aligned}$$

For $\eta=0.6$ and $\epsilon=0.1$, the accepted-trial information is about $1.373$ bits and the attempted-trial information is about $0.824$ bits. There is no extra information in the independent failure flag.

References

• C. H. Bennett and S. J. Wiesner, “Communication via one- and two-particle operators on Einstein–Podolsky–Rosen states,” Physical Review Letters 69, 2881–2884 (1992), doi:10.1103/PhysRevLett.69.2881.
• C. H. Bennett, P. W. Shor, J. A. Smolin, and A. V. Thapliyal, “Entanglement-assisted capacity of a quantum channel and the reverse Shannon theorem,” IEEE Transactions on Information Theory 48, 2637–2655 (2002), doi:10.1109/TIT.2002.802612.
• G. Bowen, “Classical information capacity of superdense coding,” Physical Review A 63, 022302 (2001), doi:10.1103/PhysRevA.63.022302.
• J. Calsamiglia and N. Lütkenhaus, “Maximum efficiency of a linear-optical Bell-state analyzer,” Applied Physics B 72, 67–71 (2001), doi:10.1007/s003400000484.
• T. Hiroshima, “Optimal dense coding with mixed state entanglement,” Journal of Physics A: Mathematical and General 34, 6907–6912 (2001), doi:10.1088/0305-4470/34/35/316.
• K. Mattle, H. Weinfurter, P. G. Kwiat, and A. Zeilinger, “Dense coding in experimental quantum communication,” Physical Review Letters 76, 4656–4659 (1996), doi:10.1103/PhysRevLett.76.4656.
• R. F. Werner, “All teleportation and dense coding schemes,” Journal of Physics A: Mathematical and General 34, 7081–7094 (2001), doi:10.1088/0305-4470/34/35/332.