Entanglement-Assisted Channel Capacity

The entanglement-assisted capacity has moved from a theorem to a working tool, in three arenas. First, as the one channel capacity that can actually be computed: because $I(A{:}B)$ is concave, $C_E$ is a convex optimization and the default benchmark against which the intractable classical and quantum capacities are bracketed — for many channels the only capacity known exactly. Second, as the backbone of the quantum reverse Shannon theorem, where with free entanglement any channel can simulate any other at a rate set by the ratio of their $C_E$ values, making $C_E$ the natural unit of "channel value" in resource theory. Third, in real links: entanglement-assisted communication is exactly the regime of a quantum internet, where pre-distributed Bell pairs are the free resource and $C_E$ sets the ultimate rate a repeater-linked channel can deliver. The open problems sit just past the classical version — the assisted private capacity, assisted capacities of thermal-noise bosonic channels (where $C_E$ has closed forms while $C$ and $Q$ do not), and the finite-blocklength corrections that say how long a code must be to approach $C_E$.

The noiseless prototype is superdense coding. Alice and Bob share one Bell pair; Alice applies one of the four Pauli operators to her qubit and sends it down a perfect channel; Bob, holding both halves, distinguishes the four Bell states and reads out two classical bits. One transmitted qubit plus one shared ebit conveys two cbits — entanglement has doubled the noiseless classical rate. The entanglement-assisted capacity is what this becomes when the channel is noisy: entanglement is still a free catalyst, and the mutual information of the channel's output is what survives. The device that makes the theorem work is the reference system $A$. Instead of feeding a classical ensemble into the channel, imagine feeding in one half of a pure entangled state $\phi_\rho$ whose marginal on the input is $\rho$. The channel acts on the other half $A'$, turning it into Bob's output $B$, while $A$ is never touched. Whatever correlation between $A$ and $B$ survives — measured by $I(A{:}B)=S(A)+S(B)-S(AB)$ — is exactly the classical information rate entanglement can extract. The first figure sets up this geometry and pins the noiseless number.

Why is there no limit over many channel uses? Because mutual information of independent systems adds: for a product input through $\mathcal{N}\otimes\mathcal{N}$, $I(A_1A_2{:}B_1B_2)=I(A_1{:}B_1)+I(A_2{:}B_2)$, and strong subadditivity guarantees no correlated input can beat the product optimum. This single fact — additivity — is what fails for the Holevo quantity, and why $C_E$ alone among the classical capacities is clean. The second figure shows the payoff: $C_E$ versus channel noise for a qubit, in closed form for two standard channels.

The central statement is the BSST theorem: the capacity is the maximum of one quantity — the quantum mutual information of the bipartite output obtained by sending half of a purified input through the channel. The maximization is over input density operators $\rho$; each fixes its purification $\phi_\rho$ up to an isometry that leaves the entropies unchanged.

where: $\mathcal{N}$ — the CPTP channel from input $A'$ to output $B$; $\rho_A$ — input density operator on $A'$, over which one optimizes; $\phi_\rho^{AA'}$ — a purification of $\rho$, a pure bipartite state on reference $A$ and input $A'$ with $\mathrm{Tr}_A\,\phi_\rho=\rho$; $\sigma_{AB}$ — the output state, reference $A$ untouched and $A'$ sent through $\mathcal{N}$; $I(A{:}B)$ — quantum mutual information of $\sigma_{AB}$. Reading: the capacity is the most correlation, in bits, any input can push through the channel while keeping a perfect copy in the reference. The figure of merit is the quantum mutual information, the entropic measure of total correlation. In von Neumann entropies it is the deficit between the marginals and the joint entropy — non-negative for every bipartite state, zero exactly when the systems are uncorrelated.

where: $S(\cdot)$ — von Neumann entropy $S(\rho)=-\mathrm{Tr}\,\rho\log_2\rho$, in bits; $\sigma_A,\sigma_B$ — reduced states on reference and output; $\sigma_{AB}$ — the joint output. Because the reference is never touched, $S(\sigma_A)=S(\rho)$. Reading: $I(A{:}B)$ ranges from $0$ (no surviving correlation) up to $2\log_2 d$ (a maximally entangled output on a $d$-dimensional system), the latter being the dense-coding ceiling. Mutual information splits cleanly into an output-entropy piece and the coherent information, the quantity that governs the quantum capacity — the bridge between the classical assisted capacity and the harder quantum one.

where: $I_c(A\rangle B)$ — coherent information, the central quantity of the quantum capacity; $S(\sigma_B)$ — output entropy, equal to $S(\mathcal{N}(\rho))$. Reading: wherever the coherent information is negative (a very noisy channel), $I(A{:}B)$ can still be positive. Purifying the channel by its Stinespring isometry gives a third, often most convenient, form. Because the global state on reference, output, and environment is pure, the joint entropy $S(AB)$ equals the environment entropy $S(E)$ — the output entropy of the complementary channel.

where: $V$ — Stinespring isometry $V:A'\to BE$ with $V^\dagger V=\mathbb{1}$; $E$ — the environment; $S(E)$ — entropy of the complementary-channel output $\mathcal{N}^c(\rho)$; the identity $S(AB)=S(E)$ holds because $|\Psi\rangle_{ABE}$ is pure. Reading: for a self-complementary channel $E$ mirrors $B$ and this form collapses the calculation. The keystone property — the one that removes all regularization — is additivity. The maximal mutual information of a tensor product of channels is the sum of the individual maxima, so optimizing a single channel use suffices for the asymptotic rate.

where: $\mathcal{N}_1,\mathcal{N}_2$ — arbitrary independent channels. Reading: the single-letter formula (1) already equals the many-use capacity — the property that $\chi$ and $Q$ both lack. Teleportation and dense coding tie the classical assisted capacity to its quantum counterpart by a clean factor of two: two cbits are interconvertible with one qubit plus one ebit.

where: $Q_E$ — maximal rate of reliable qubit transmission with free entanglement; the factor $\tfrac12$ — the teleportation/dense-coding conversion, $2\text{ cbits}\leftrightarrow 1\text{ qubit}+1\text{ ebit}$. Reading: unlike the unassisted case, where $C$ and $Q$ are unrelated, assistance locks them together. For the qubit depolarizing channel — the input replaced by the maximally mixed state with probability $p$ — Pauli covariance makes the maximally entangled input optimal and the output Bell-diagonal, giving a closed form.

where: $p$ — depolarizing probability, $\mathcal{N}_{\mathrm{dep}}(\rho)=(1-p)\rho+p\,\tfrac{\mathbb{1}}{2}$; the leading $2$ — the noiseless ($p=0$) dense-coding value. Reading: the four Bell-diagonal eigenvalues $\{1-\tfrac{3p}{4},\tfrac{p}{4},\tfrac{p}{4},\tfrac{p}{4}\}$ enter $S(AB)$, and $C_E=2-S(AB)$; it runs from $2$ at $p=0$ down to $0$ at $p=1$. Finally, the assisted capacity dominates the unassisted classical capacity and the Holevo quantity — entanglement can only help — and by at most a factor of two.

where: $C$ — unassisted (regularized) classical capacity; $\chi$ — single-letter Holevo capacity. Reading: computing the tractable $C_E$ brackets the intractable $C$ to within a factor of two, often the best estimate available.

1 · The noiseless qubit and superdense coding (confidence builder)

Confirm that the identity channel on a qubit has $C_E=2$ bits and identify the protocol that achieves it.

Step 1. Choose the maximally entangled input $\phi=\tfrac{1}{\sqrt2}(|00\rangle+|11\rangle)$ on reference $A$ and input $A'$. The reference marginal is $\rho_A=\mathbb{1}/2$, so $S(A)=\log_2 2=1$ bit. The reference half of a Bell pair is maximally mixed.

Step 2. The identity channel leaves the input untouched, so the output $\sigma_{AB}=\phi$ is still pure and maximally entangled. Its output marginal is again $\mathbb{1}/2$, giving $S(B)=1$ bit, while the joint state is pure, so $S(AB)=0$.

Step 3. Assemble the mutual information: $I(A{:}B)=S(A)+S(B)-S(AB)=1+1-0=\mathbf{2\ \text{bits}}$. Since the identity is the best possible channel, this is the maximum, so $C_E(\mathrm{id}_2)=\mathbf{2}$ bits per use.

Step 4. The achieving protocol is superdense coding: Alice applies one of $\{\mathbb{1},X,Y,Z\}$ to her half of a Bell pair, sends that qubit, and Bob distinguishes the four Bell states — 1 transmitted qubit + 1 shared ebit → 2 classical bits. Consistently $Q_E=\tfrac12 C_E=1$ qubit per use.

2 · The depolarizing channel at 25% noise (fully worked)

Compute $C_E$ for the qubit depolarizing channel $\mathcal{N}(\rho)=(1-p)\rho+p\,\mathbb{1}/2$ at $p=0.25$, and bracket its unassisted classical capacity.

Step 1. By Pauli covariance the maximally entangled input is optimal, so $S(A)=1$ and $S(B)=1$ (a maximally mixed input gives a maximally mixed output). The whole problem reduces to the joint entropy $S(AB)$.

Step 2. On this input the output is Bell-diagonal with eigenvalues $\{1-\tfrac{3p}{4},\tfrac p4,\tfrac p4,\tfrac p4\}$. At $p=0.25$: $1-\tfrac{3p}{4}=\mathbf{0.8125}$ and $\tfrac p4=\mathbf{0.0625}$ (three times).

Step 3. Evaluate $S(AB)=-\sum_i\lambda_i\log_2\lambda_i$. Using $\log_2 0.8125=-0.29956$ and $\log_2 0.0625=-4$ exactly:

Step 4. Assemble: $C_E=S(A)+S(B)-S(AB)=2-0.99340=\mathbf{1.0066\ \text{bits per use}}$ — the point marked on the second figure, and the closed form (7) agrees.

Step 5. Bracket the unassisted capacity with (8): $\tfrac12 C_E\le C\le C_E$ gives $\mathbf{0.503\le C\le 1.007}$ bits. Without ever computing the hard, non-additive Holevo quantity, assistance has pinned the classical capacity to within a factor of two.

3 · The erasure channel and the saturated bound

A qubit erasure channel loses its input with probability $p=0.1$. Find $C_E$ and check the sandwich.

Step 1. From the derivation, $C_E=2(1-p)\log_2 d$ with $d=2$: $C_E=2(0.9)(1)=\mathbf{1.8\ \text{bits}}$.

Step 2. The classical capacity of the quantum erasure channel is known exactly, $C=(1-p)=0.9$ bits — equal to $\tfrac12 C_E$. The erasure channel saturates the lower sandwich bound, the case where entanglement buys the full factor of two, as in noiseless dense coding. Consistently $Q_E=\tfrac12 C_E=0.9$ qubit per use, and both capacities vanish at the $p=\tfrac12$ threshold.

The entanglement-assisted capacity is the capstone of a program that turned entanglement from a paradox into a quantifiable resource — with one author at its center more than any other.

Texts. Wilde, Quantum Information Theory, 2nd ed., Ch. 21 (the definitive treatment) · Nielsen & Chuang, Quantum Computation and Quantum Information, Ch. 12 · Watrous, The Theory of Quantum Information, Ch. 8 · Preskill, Lecture Notes on Quantum Computation, Ch. 10 · Bennett, Shor, Smolin, Thapliyal, Phys. Rev. Lett. 83, 3081 (1999); and IEEE Trans. Inf. Theory 48, 2637 (2002).

Provenance. The capacity theorem (1) and additivity (5) follow BSST (PRL 1999; IEEE-IT 2002) and Wilde Ch. 21; the converse uses only data processing and strong subadditivity, each inequality checkable independently. The depolarizing form (7) is written in the convention $\mathcal{N}(\rho)=(1-p)\rho+p\,\mathbb{1}/2$, whose Bell-diagonal eigenvalues $\{1-3p/4,p/4,p/4,p/4\}$ reproduce (7) and give $C_E=0$ at $p=1$; the erasure result $2(1-p)\log_2 d$ is rederived above. All entropies are in bits; the worked point $C_E(p{=}0.25)=1.0066$ bits was recomputed from the eigenvalues.

Builds on: Von Neumann Entropy · Quantum Mutual Information · Strong Subadditivity · Quantum Channels and CPTP Maps · Data-Processing Inequality · Holevo Bound. Leads to: Complementary Channels and Degradability · Choi–Jamiołkowski Isomorphism · Stinespring Dilation.

Ask the tutor: ask me to walk through the depolarizing closed form eigenvalue by eigenvalue, or to show why superdense coding is the exact noiseless corner of the BSST theorem.