Coherent Information and the LSD Quantum Capacity

Most capacities in information theory are plug-in formulas: write down a single-letter optimization, evaluate it, done. Quantum capacity — the rate at which a noisy channel carries qubits faithfully enough to preserve entanglement — is the one place that programme breaks, and it breaks in a way that reveals the deep structure of quantum information. The figure of merit is the coherent information, $I_c = S(B) - S(E)$ , introduced by Schumacher and Nielsen: the entropy the receiver holds minus the entropy that leaks to the channel's environment. It is not symmetric, it can be negative, and it does not add up when you use a channel twice. Those three facts are the whole story. The central theorem, proven in three complementary stages by Lloyd, Shor, and Devetak (LSD), says the quantum capacity equals the regularized maximum of the coherent information: the per-use limit of $\frac{1}{n}\max_\rho I_c(\rho,\mathcal{N}^{\otimes n})$ as $n\to\infty$ . The regularization is not a technicality one might hope to remove. Coherent information is genuinely superadditive — two uses can carry strictly more than twice one use — and the vivid demonstration is superactivation, where two channels each of exactly zero capacity combine to a positive one. No classical channel behaves this way. This article develops the entropic identity behind $I_c$ , the LSD converse, and the one structural condition — degradability — under which the regularization collapses and the capacity becomes computable. The dephasing and erasure channels are worked out exactly; the depolarizing channel, tellingly, is not.

Key takeaways

The coherent information $I_c(\rho,\mathcal{N}) = S(\mathcal{N}(\rho)) - S(\mathcal{N}^c(\rho)) = S(B) - S(E)$ measures how much entanglement with a reference a state keeps after passing through a channel; it equals the negative conditional entropy $-S(R|B)$ of reference given output.
The Lloyd–Shor–Devetak theorem: the quantum capacity is the regularized coherent information, $Q(\mathcal{N}) = \lim_{n\to\infty}\frac{1}{n}\max_\rho I_c(\rho,\mathcal{N}^{\otimes n})$ — an asymptotic, not a single-letter, formula.
Regularization is unavoidable because $I_c$ is superadditive; the extreme case is superactivation, where two zero-capacity channels combine to give $Q > 0$ , a phenomenon with no classical analogue.
For degradable channels $I_c$ is concave and additive, so the single-letter value $Q^{(1)} = \max_\rho I_c$ is exact and computable — this gives $Q = 1 - H_2(p)$ for dephasing and $Q = 1 - 2p$ for erasure.
Coherent information can be negative, and antidegradable channels (environment can simulate the receiver) have $Q = 0$ by no-cloning; capacity obeys the quantum data-processing inequality, a consequence of strong subadditivity.
Key idea: quantum capacity is the rate at which a channel preserves entanglement with a reference, measured by the coherent information $I_c = S(B) - S(E)$ ; faithful qubit transmission is possible exactly when the environment learns less than the receiver, and the asymptotic, regularized maximum of that gap is the capacity.

The core insight

Quantum capacity is the rate at which a channel preserves entanglement with a reference, measured by the coherent information $I_c = S(B) - S(E)$ ; faithful qubit transmission is possible exactly when the environment learns less than the receiver, and the asymptotic, regularized maximum of this gap is the capacity.

Where this lives today

Coherent information is the yardstick against which real quantum communication is measured. For the pure-loss bosonic channel — photons down an optical fibre — the quantum capacity is a degradable-channel coherent information, $\max\{0,\log_2[\eta/(1-\eta)]\}$ for transmissivity $\eta$ , and this pins the repeaterless PLOB bound that quantum repeaters and satellite links must beat to distribute entanglement over continental distances. For fault tolerance, the hashing bound $1 - H_2(p) - p\log_2 3$ of the depolarizing channel is a coherent-information rate that sets the ultimate code rate a physical qubit's noise permits, calibrating how far real surface-code thresholds sit from the information-theoretic limit. The open frontier is the depolarizing channel itself: neither degradable nor antidegradable, its coherent information superadditive, its exact quantum capacity is after two decades still unknown — not known to be computable by any finite procedure at all. And the machinery of $I_c$ powers entanglement distillation directly: the one-way hashing rate from many noisy pairs $\rho_{AB}$ is exactly $I_c(A\rangle B) = -S(A|B)$ , the same negentropy that certifies a channel can carry qubits.

The physical picture

To send a qubit reliably is to send its entanglement. Prepare a maximally entangled pair, keep one half as an inaccessible reference $R$ , and push the other half through the channel. If, at the far end, the receiver $B$ still shares the full entanglement with $R$ , then any unknown state could have ridden along undamaged — that is what "faithful qubit transmission" means. Every quantum channel is secretly a three-way split: an isometry $V$ takes the input and fans it out into the receiver $B$ and an environment $E$ that carries away whatever the receiver did not get. The coherent information $I_c = S(B) - S(E)$ is exactly the receiver's share of the input's quantum content minus the environment's share. Positive means the receiver is winning the tug-of-war; negative means the environment ran off with the entanglement.

Figure

The Stinespring picture of a channel: a purification of the input ρ on reference R passes through the isometry V, splitting into the receiver output B and the environment E, so the coherent information I_c = S(B) − S(E) is the receiver's information minus the environment's — for the dephasing channel at p = 0.1 it equals 1 − 0.469 = 0.531 qubits per use

The asymmetry with Shannon's world is total. Classical mutual information $I(X{:}Y)$ is symmetric and its capacity is a clean single-letter maximum, computable once and for all. For qubits, no-cloning forbids receiver and environment from both holding the same quantum information, so the natural figure of merit is directional and refuses to add. When the environment holds more than the receiver, $I_c < 0$ : no positive-rate transmission is possible for that input — but a smarter input, or many parallel uses coded jointly, can rescue it. That is why the capacity is a maximization over inputs and a limit over uses. The second figure shows what happens once you can compute the answer: for the two channels that happen to be degradable, the whole regularized limit collapses to a single clean curve.

Figure

Single-letter quantum capacities for two degradable qubit channels: the dephasing channel Q = 1 − H₂(p) and the erasure channel Q = 1 − 2p plotted against noise parameter p — both are exactly single-letter because degradability kills the regularization, and both vanish at p = 1/2 where the channel turns antidegradable

The core equations

The definition is entropic. Purify the input $\rho$ on system $A$ with a reference $R$ , dilate the channel to an isometry so it produces both a receiver output $B$ and an environment $E$ , and compare the two output entropies. The coherent information is what the receiver gained minus what the environment gained.

I_c(\rho,\mathcal{N}) \;=\; S\big(\mathcal{N}(\rho)\big) \;-\; S\big(\mathcal{N}^c(\rho)\big) \;=\; S(B)_\omega - S(E)_\omega \qquad (1)

where: $\rho$ — input density operator on $A$ ; $\mathcal{N}$ — the channel (CPTP map) $A\to B$ ; $\mathcal{N}^c$ — the complementary channel $A\to E$ from the same Stinespring isometry $V$ ; $S(\cdot) = -\mathrm{Tr}(\sigma\log_2\sigma)$ — von Neumann entropy in bits; $\omega^{RBE}$ — the pure global state after applying $V$ to a purification of $\rho$ . Reading: $I_c$ is receiver-entropy minus environment-entropy, so it is large when the output is disordered but the environment is not. On the pure tripartite state the same quantity is a conditional entropy — and a negative one whenever $R$ and $B$ are entangled. This is the signature that a classical figure of merit can never reproduce.

I_c(\rho,\mathcal{N}) \;=\; -\,S(R|B)_\omega, \qquad S(R|B) = S(RB) - S(B) \qquad (2)

where: $S(R|B)$ — quantum conditional entropy of the reference given the receiver; it can be negative, and $I_c > 0$ is exactly the statement $S(R|B) < 0$ . Reading: coherent information is the negentropy of reference-given-output; positive coherent information is residual entanglement. The Lloyd–Shor–Devetak theorem names the operational capacity. It is a maximization over inputs and a regularization over channel uses, because the single-letter quantity is not additive.

Q(\mathcal{N}) \;=\; \lim_{n\to\infty}\frac{1}{n}\,Q^{(1)}(\mathcal{N}^{\otimes n}), \qquad Q^{(1)}(\mathcal{N}) = \max_{\rho}\,I_c(\rho,\mathcal{N}) \qquad (3)

where: $Q(\mathcal{N})$ — quantum capacity in qubits per channel use; $Q^{(1)}$ — single-letter (one-shot optimized) coherent information; $\mathcal{N}^{\otimes n}$ — $n$ parallel uses. Reading: the true rate is set by the best block code on many uses, not by the best single-use input. Composing a second channel onto the output can only destroy coherent information. This quantum data-processing inequality is the engine of the converse, and it is nothing but strong subadditivity in disguise.

I_c(\rho,\mathcal{N}_2\circ\mathcal{N}_1) \;\le\; I_c(\rho,\mathcal{N}_1) \qquad (4)

where: $\mathcal{N}_1,\mathcal{N}_2$ — composable channels; $\circ$ — composition, $\mathcal{N}_2$ processing the output of $\mathcal{N}_1$ . Reading: post-processing is never free — every extra channel is a potential entanglement leak. For a degradable channel — one whose receiver can further process its output into a copy of the environment — the coherent information is concave and additive, and the regularization in (3) collapses to a single letter.

\mathcal{N}\ \text{degradable} \;\Longrightarrow\; Q(\mathcal{N}) = Q^{(1)}(\mathcal{N}) = \max_\rho I_c(\rho,\mathcal{N}) \qquad (5)

where: degradable means there is a channel $\mathcal{D}$ with $\mathcal{D}\circ\mathcal{N} = \mathcal{N}^c$ ; the receiver can simulate the environment. Reading: degradability is the structural condition that makes quantum capacity computable. The mirror image is antidegradability: if the environment can simulate the receiver, then a positive capacity would clone an unknown state, so the capacity must be zero.

\mathcal{N}\ \text{antidegradable} \;\Longrightarrow\; I_c(\rho,\mathcal{N}) \le 0 \ \ \forall\rho \;\Longrightarrow\; Q(\mathcal{N}) = 0 \qquad (6)

where: antidegradable means $\mathcal{N} = \mathcal{D}\circ\mathcal{N}^c$ for some $\mathcal{D}$ ; the environment reconstructs the output. Reading: no-cloning sets an exact zero, not merely a small number.

Derivation ACoherent information is minus the conditional entropy of reference given output

The goal is to prove the identity $I_c(\rho,\mathcal{N}) = S(B) - S(E) = -S(R|B)$ that lets us slide freely between three pictures — output-versus-environment, negative conditional entropy, and residual entanglement — using only the purity of the global state and the Schmidt symmetry of complementary marginals. This is the load-bearing algebraic fact of the whole subject: every appearance of data processing, degradability, and the converse ultimately rewrites some inequality about $S(R|B)$ as an inequality about $I_c$ , so establishing it once cleanly pays for everything downstream.

Step 1. Dilate the channel to an isometry $V:A\to BE$ and apply it to a purification $|\phi_\rho\rangle^{RA}$ of the input, producing a pure global state $|\omega\rangle^{RBE}$ . Because the total state is pure, any subsystem and its complement share the same spectrum, so $S(RB) = S(E)$ and $S(B) = S(RE)$ . Why: purity is the one extra structure a quantum channel carries that a classical one does not — it ties the environment's entropy to the reference-plus-receiver marginal, which is what makes the environment a genuine competitor for the input's information.

Step 2. Expand the conditional entropy directly from its definition, $S(R|B) = S(RB) - S(B)$ , and substitute the purity relation $S(RB) = S(E)$ from Step 1 to obtain $S(R|B) = S(E) - S(B)$ . Why: this rewrites a two-body correlation quantity entirely in terms of the two single-body output entropies we can actually compute channel-by-channel.

Step 3. Read off $I_c = S(B) - S(E) = -\,S(R|B)$ . Since $S(R|B)$ is negative exactly when $R$ and $B$ are entangled, positive coherent information is literally a measure of surviving entanglement, and it is bounded above by $2\log_2 d_A$ (the maximal violation of subadditivity). Why: this is the bridge between an information rate and a physical resource — a positive number here certifies that entanglement with the reference got through the channel intact.

Go deeper · Derivation B · 1 minThe converse — capacity cannot beat the regularized coherent information

Derivation BThe converse — capacity cannot beat the regularized coherent information

The goal is to show that any block code transmitting $nR$ qubits through $\mathcal{N}^{\otimes n}$ with vanishing error must obey $R \le \frac{1}{n}Q^{(1)}(\mathcal{N}^{\otimes n}) + o(1)$ , so the achievable rate can never exceed the regularized coherent information. The argument uses only three ingredients already in hand — the identity of the previous derivation, the data-processing inequality (4), and Alicki–Fannes continuity — and it is the half of LSD that needs no coding cleverness, only the monotonicity strong subadditivity guarantees.

Step 1. Set up the ideal test. Feed one half of a maximally entangled state $\Phi$ of $nR$ qubits into the channel, keeping the reference $R'$ . A perfect protocol would return a state on which $R'$ is still maximally entangled with the decoded output, so $I_c(\Phi,\mathrm{id}) = S(R') = nR$ : the coherent information of the noiseless channel on a maximally entangled input is exactly the qubit count. Why: this anchors the rate $R$ to a coherent information, converting an operational claim about codes into an entropic inequality.

Step 2. Interpose the real channel and the decoder. The actual protocol is $\mathcal{D}\circ\mathcal{N}^{\otimes n}$ applied to the input, and by the data-processing inequality (4) the decoder $\mathcal{D}$ cannot increase coherent information: $I_c(\rho^{(n)},\mathcal{D}\circ\mathcal{N}^{\otimes n}) \le I_c(\rho^{(n)},\mathcal{N}^{\otimes n})$ . Why: this is where "post-processing never helps" does the heavy lifting — the decoder is just another channel, so it is powerless to manufacture entanglement the channel already lost.

Step 3. Close the gap with continuity. Because the decoded state is $\delta_n$ -close to the perfect one with $\delta_n\to 0$ , Alicki–Fannes continuity gives $nR \le I_c(\rho^{(n)},\mathcal{N}^{\otimes n}) + 1 + 2n\delta_n R$ . Dividing by $n$ and taking $n\to\infty$ yields $R \le \frac{1}{n}Q^{(1)}(\mathcal{N}^{\otimes n}) + o(1) \to Q(\mathcal{N})$ . Why: the near-perfect protocol's coherent information cannot be far from the ideal $nR$ , so the ideal rate is pinned below the regularized optimum — exactly the LSD converse.

Go deeper · Derivation C · 1 minThe dephasing channel capacity from degradability

Derivation CThe dephasing channel capacity from degradability

The goal is to compute the exact quantum capacity of the qubit dephasing (phase-damping) channel $\mathcal{N}_p(\rho) = (1-p)\rho + pZ\rho Z$ , obtaining the closed form $Q = 1 - H_2(p)$ , and to use degradability to certify that the single-letter coherent information is the true capacity with no regularization. This is the cleanest nontrivial capacity in the theory: it shows how a symmetry argument fixes the optimal input and how the environment's binary record produces the binary-entropy penalty.

Step 1. Fix the optimal input by symmetry. The channel commutes with $Z$ -rotations, and $I_c$ is concave (degradability), so the maximizer is the fixed point of that symmetry — the maximally mixed input $\rho = \tfrac12 I$ , purified by a Bell state shared with the reference. Why: concavity plus a symmetry group means the optimum sits at the symmetric point, sparing us any explicit optimization.

Step 2. Compute the receiver's entropy. Dephasing leaves the diagonal of $\rho$ untouched, so the maximally mixed input passes to a maximally mixed output and $S(B) = 1$ bit. Why: the receiver sees no loss of population, only loss of coherence, so its marginal is as disordered as it can be.

Step 3. Compute the environment's entropy. The complementary channel records only whether a phase flip occurred — a classical bit that is $1$ with probability $p$ — so the environment carries entropy $S(E) = H_2(p) = -p\log_2 p - (1-p)\log_2(1-p)$ . Why: dephasing leaks exactly one classical "which-phase" bit to the environment, and its entropy is the price the receiver pays.

Step 4. Subtract and certify. The coherent information is $I_c = S(B) - S(E) = 1 - H_2(p)$ , and because the channel is degradable this single-letter value is the capacity: $Q(\mathcal{N}_p) = 1 - H_2(p)$ . It equals $1$ at $p = 0$ , falls to $0$ at $p = 1/2$ , and rises back to $1$ at $p = 1$ (a deterministic $Z$ is reversible). Why: degradability guarantees additivity, so the regularization $\lim_n\frac1n Q^{(1)}(\mathcal{N}_p^{\otimes n})$ collapses onto this one number — the capacity is genuinely computable.

Limiting cases

Coherent information specializes sharply across channel families, and the pattern is always the same: whoever wins the entropy competition, receiver or environment, sets the sign.

Case	Result	Meaning
Noiseless channel	$Q(\mathrm{id}_d) = \log_2 d$	A perfect $d$ -level channel carries $\log_2 d$ qubits; the maximally mixed input gives $S(B) = \log_2 d$ , $S(E) = 0$ .
Dephasing (prob. $p$ )	$Q = 1 - H_2(p)$	Degradable, exactly single-letter; $Q = 0.531$ at $p = 0.1$ , vanishes at $p = 1/2$ , revives to $1$ at $p = 1$ .
Erasure (prob. $p$ )	$Q = \max\{0,\,1 - 2p\}$	Degradable for $p \le 1/2$ ; the exact capacity is $1 - 2p$ with no single-letter gap, zero at $p = 1/2$ .
Antidegradable / very noisy	$Q = 0$	Environment can simulate the receiver, so $I_c \le 0$ for all inputs; e.g. depolarizing channel above its noise threshold.
Two zero-capacity channels	$Q(\mathcal{N}_1) = Q(\mathcal{N}_2) = 0$ , $Q(\mathcal{N}_1\otimes\mathcal{N}_2) > 0$	Superactivation (Smith–Yard): a PPT channel plus a symmetric-erasure channel combine to positive capacity — impossible classically.

Common trapAssuming the capacity is just the single-letter maximum of the coherent information

The formula $Q(\mathcal{N}) = \max_\rho I_c(\rho,\mathcal{N})$ is false in general. It holds only for degradable channels (and a few special cases like erasure). Coherent information is superadditive: $I_c$ of two channel uses can strictly exceed twice $I_c$ of one use, so a single-use optimization underestimates the true rate. The gap can be as extreme as it gets — with superactivation, two channels each with a single-letter (and true) capacity of exactly zero combine to a strictly positive joint capacity. The consequence is that the true $Q$ requires the $n\to\infty$ regularization, and for the workhorse depolarizing channel no finite algorithm is known to compute it; only bounds like the hashing rate exist. Never quote $\max_\rho I_c$ as "the quantum capacity" unless you have first checked degradability.

Worked examples

1 · The dephasing channel, fully numeric (confidence builder)

A qubit passes through the phase-damping channel $\mathcal{N}_p(\rho) = (1-p)\rho + pZ\rho Z$ with $p = 0.1$ . Find the quantum capacity, and confirm both endpoints of the entropy competition.

Step 1. Choose the optimal input. Because the channel is degradable and $Z$ -symmetric, the maximizer is the maximally mixed input $\rho = \mathrm{diag}(\tfrac12,\tfrac12)$ , purified by a Bell state. This fixes the whole calculation; there is nothing left to optimize.

Step 2. Receiver entropy. The maximally mixed input is a fixed point of dephasing, so the output is maximally mixed and $S(B) = 1$ bit. The receiver's marginal is maximally disordered, $S(B) = 1$ .

Step 3. Environment entropy. The complementary channel records the binary "did a phase flip happen" event, with flip probability $p = 0.1$ , so $S(E) = H_2(0.1) = -0.1\log_2 0.1 - 0.9\log_2 0.9 = 0.3322 + 0.1368 = \mathbf{0.4690}$ bits.

Step 4. Coherent information and capacity. Subtract: $I_c = S(B) - S(E) = 1 - 0.4690 = \mathbf{0.531}$ qubits per use. Degradability makes this exact, so $Q(\mathcal{N}_{0.1}) = 0.531$ qubits per channel use. As a sanity check the two symmetric limits hold: $H_2(0) = 0 \Rightarrow Q = 1$ (a noiseless channel) and $H_2(1/2) = 1 \Rightarrow Q = 0$ (a completely dephasing channel destroys all coherence). Q(𝒩ₚ) = 1 − H₂(0.1) = 0.531 qubits/use.

2 · The erasure channel and its exact threshold

The qubit erasure channel transmits the input with probability $1 - p$ and replaces it with a flag state $|\bot\rangle$ orthogonal to the input space with probability $p$ . Find $Q$ at $p = 0.2$ and at $p = 0.5$ , and explain why no regularization is needed.

Step 1. Reduce the coherent information to one line. Because the receiver knows whether an erasure happened (the flag is orthogonal to any real input), the output splits into a "got it" branch of weight $1-p$ carrying $\rho$ and an "erased" branch carrying nothing. The complementary channel is another erasure channel with roles swapped, so $I_c(\rho,\mathcal{N}^{\mathrm{er}}_p) = (1-p)S(\rho) - pS(\rho) = (1 - 2p)\,S(\rho)$ .

Step 2. Maximize over the input. The coefficient $(1-2p)$ is positive for $p < 1/2$ , so we maximize $S(\rho)$ by taking the maximally mixed qubit input, $S(\rho) = 1$ . Hence $Q^{(1)} = \max\{0,\,1-2p\}$ , and because the erasure channel is degradable for $p \le 1/2$ this single-letter value is the exact capacity.

Step 3. Evaluate. At $p = 0.2$ : $Q = 1 - 2(0.2) = \mathbf{0.60}$ qubits per use — a fifth of transmissions are lost, yet more than half a qubit still gets through. At $p = 0.5$ : $Q = \mathbf{0}$ . The threshold is sharp and physical: at exactly $p = 1/2$ the channel turns antidegradable — the environment now receives the input at least as often as the receiver — so no-cloning forces the capacity to zero. Q(𝒩ᵉʳ) = max{0, 1 − 2p}: 0.60 at p = 0.2, exactly 0 at p = 1/2.

Physics of the portrait

Coherent information is where quantum Shannon theory found what classical theory lacks — a capacity with no single-letter formula — and it took a decade and three people to pin down what it measures.

Origins · 1 mincoherent information and how this was discovered

coherent information

the entanglement-preservation measure I_c = S(B) − S(E) · Lloyd–Shor–Devetak quantum capacity

Coherent information was introduced in 1996 by Benjamin Schumacher and Michael Nielsen as the quantum analogue of mutual information, together with the quantum data-processing inequality that would later drive every converse in the field. Schumacher had already given quantum information its unit — the qubit — and its source-coding theorem; the coherent information was the natural next object, the quantity that decreases under processing and so must bound any capacity. The conjecture that its regularized maximum equals the quantum capacity was assembled in three complementary stages, which is why the theorem carries three names. Seth Lloyd gave the first achievability argument in 1997, physically motivated but not fully rigorous. Peter Shor supplied a complete proof of achievability in celebrated 2002 lecture notes at MSRI that were never formally published. Igor Devetak closed the problem in 2005 with a rigorous proof built on a surprising bridge: he showed the quantum capacity coincides with the regularized private classical capacity, so that codes protecting classical messages from an eavesdropper could be made "coherent" to protect qubits from the environment. The story did not end with a clean formula. In 2008 Graeme Smith and Jon Yard demonstrated superactivation — two channels of exactly zero quantum capacity that, used together, transmit qubits — proving that quantum capacity is not an additive, intrinsic property of a channel at all. A quantity defined to look like Shannon's mutual information turned out to obey none of Shannon's additivity, encoding instead the non-clonable structure of quantum information.

History

1996

— Schumacher and Nielsen introduce coherent information and the quantum data-processing inequality

1997

— Lloyd gives the first heuristic achievability argument for the coherent-information rate

2002

— Shor proves achievability in his unpublished MSRI lecture notes

2005

— Devetak completes the rigorous LSD proof via the private classical capacity

2008

— Smith and Yard discover superactivation of the quantum capacity

Going further

Texts. Wilde, Quantum Information Theory, 2nd ed., Ch. 13 (coherent information), 23 (private capacity), 24 (quantum capacity) · Nielsen & Chuang, Quantum Computation and Quantum Information, §12.4 · Watrous, The Theory of Quantum Information, Ch. 8 · Preskill, Lecture Notes on Quantum Computation, Ch. 10 · Devetak, IEEE Trans. Inf. Theory 51, 44 (2005) · Schumacher & Nielsen, Phys. Rev. A 54, 2629 (1996) · Smith & Yard, Science 321, 1812 (2008). The key numerical anchors are: the identity $I_c = S(B) - S(E) = -S(R|B)$ follows from purity and Schmidt symmetry of the pure tripartite state; the data-processing inequality (4) follows from strong subadditivity; the dephasing capacity $1 - H_2(p)$ satisfies $H_2(0.1) = 0.4690 \Rightarrow Q = 0.531$ , $H_2(0) = 0 \Rightarrow Q = 1$ , $H_2(1/2) = 1 \Rightarrow Q = 0$ ; and the erasure capacity $\max\{0,1-2p\}$ follows from $I_c = (1-2p)S(\rho)$ , giving $0.60$ at $p = 0.2$ and $0$ at $p = 1/2$ .

Prerequisites. Von Neumann Entropy · Quantum Conditional Entropy and Mutual Information · Quantum Channels and CPTP Maps · Stinespring Dilation · Complementary Channels and Degradability · Strong Subadditivity. Leads to. Data-Processing Inequality · Complementary Channels and Degradability · Holevo Bound.

Ask the tutor: ask me to walk through why two channels of exactly zero quantum capacity can superactivate into a positive one — the sharpest way to see why coherent information refuses to add.

Why is coherent information directional when classical mutual information is symmetric?

Classical mutual information $I(X{:}Y)$ is symmetric because a classical correlation can be read from either end — copying is free. Quantum information cannot be copied, so the receiver $B$ and the environment $E$ cannot both hold the input's quantum content; one gains only what the other loses. Coherent information $I_c = S(B) - S(E)$ measures precisely which of the two won, and swapping their roles flips the sign. On the pure state it is $-S(R|B)$ , a conditional entropy with a definite direction. The asymmetry is the fingerprint of no-cloning, and it is exactly why the single symmetric single-letter formula that classical Shannon theory enjoys cannot exist here.

What does negative coherent information mean physically?

Negative $I_c = S(B) - S(E) < 0$ means the environment holds more of the input's quantum information than the receiver, equivalently $S(R|B) > 0$ : the reference and the receiver are no longer entangled enough to reconstruct an arbitrary state. For that particular input, reliable quantum transmission at any positive rate is impossible. This is not the end of the road, though. A different input may make $I_c$ positive, and even if every single-use input gives $I_c \le 0$ , entangled inputs spread across many uses — or tensoring with a second channel, as in superactivation — can lift the regularized rate above zero. Negative coherent information is a statement about one input, not a verdict on the channel.

What exactly makes a channel degradable, and why does it matter so much?

A channel $\mathcal{N}$ is degradable if some further channel $\mathcal{D}$ turns the receiver's output into the environment's: $\mathcal{D}\circ\mathcal{N} = \mathcal{N}^c$ . The receiver is "ahead" of the environment — it can always simulate what the environment sees, so it never holds less. This ordering forces the coherent information to be concave in the input and additive across uses, collapsing the $n\to\infty$ regularization to a single letter. The payoff is decisive: for degradable channels the capacity is a finite, computable optimization ( $\max_\rho I_c$ ), which is why dephasing, erasure and amplitude damping below threshold, and the pure-loss bosonic channel all have closed-form capacities, while the depolarizing channel — neither degradable nor antidegradable — does not.

How does the no-cloning theorem force some capacities to be exactly zero?

If a channel is antidegradable, the environment can reconstruct the receiver's output: $\mathcal{N} = \mathcal{D}\circ\mathcal{N}^c$ for some $\mathcal{D}$ . Suppose such a channel had positive capacity. Then the legitimate receiver could recover the transmitted qubit — and because the environment can simulate the receiver, so could it, yielding two copies of an unknown state. That is forbidden by no-cloning. The contradiction forces the coherent information non-positive for every input, and since that survives regularization, the capacity is exactly zero — not small, exactly zero. This is why the erasure channel dies precisely at $p = 1/2$ , where it turns antidegradable.

Is coherent information the same as the quantum mutual information of a channel?

No, and the difference sets unassisted from entanglement-assisted capacity apart. The channel's quantum mutual information $I(\rho,\mathcal{N}) = S(\rho) + I_c$ includes the input entropy $S(\rho)$ , and its half-maximum is the entanglement-assisted capacity $Q_E = \tfrac12\max_\rho I(\rho,\mathcal{N})$ — the rate when sender and receiver share free prior entanglement. That quantity is single-letter and additive for every channel, no regularization required. The unassisted capacity uses $I_c$ alone, without the $S(\rho)$ bonus, and it is the one that misbehaves — superadditive, generally uncomputable. Free entanglement is exactly what tames the capacity.

Why can't we just compute the depolarizing channel's capacity numerically?

Because there is no known finite object to compute. The LSD capacity is a limit, $\lim_{n\to\infty}\frac1n\max_\rho I_c(\rho,\mathcal{N}^{\otimes n})$ , and for the depolarizing channel the coherent information is strictly superadditive: the optimum keeps rising as entangled inputs span more uses, and even highly entangled codes on large blocks beat all smaller ones. Since the channel is neither degradable (which would make the sequence constant) nor antidegradable (which would make it zero), no finite $n$ is known to saturate the limit. In practice one brackets it: the hashing bound $1 - H_2(p) - p\log_2 3$ from below, and no-cloning or degradable-extension bounds from above.