greatitself.
Continue
advanced27 min studyLesson 430 in Physics

Quantum channel capacities and coding limits

Define communication rates with explicit assistance and use the flagged erasure channel to separate classical, quantum, and two-way-assisted transmission.

Before you begin
On this page

The question for today

What you’ll understand

  • Specify the operational task and asymptotic error criterion behind a capacity.
  • Distinguish Holevo information from coherent information and their regularized coding formulas.
  • Compute flagged erasure-channel capacities while separating finite-block counts from asymptotic rates.

The capacity of a quantum channel is not one universal number. A sender may wish to convey classical messages, preserve unknown quantum states and their reference correlations, or establish entanglement with classical assistance. Each task assigns different value to the same physical channel. We restrict attention to finite-dimensional, memoryless, completely positive trace-preserving channels used independently many times. Rates are normalized per channel use, and reliability means that a specified decoding error tends to zero along an appropriate sequence of growing coding blocks.

A channel must preserve correlations with a reference

Complete positivity means that the channel remains positive when tensored with an identity map on any untouched reference. A Kraus representation makes this requirement explicit, while the Kraus normalization ensures trace preservation. A channel need not be unitary on the transmitted system: coupling to an environment and discarding it is permitted. Describing only output populations can miss damage to input coherence or entanglement with a reference, which matters especially for quantum communication.

Unassisted classical capacity counts recoverable classical bits per use, optimizing message ensembles and long-block decoding. Quantum capacity counts logical qubits transmitted with asymptotically faithful preservation of arbitrary reference entanglement. Pure-state agreement on a chosen test basis is inadequate for the latter. Two-way classical communication, free shared entanglement, an energy constraint, or correlated channel noise changes the task. Each must appear in the capacity definition rather than being silently added to a numerical answer.

N(ρ)=∑jKjρKj†,∑jKj†Kj=I,Ic(ρ,N)=S(B)−S(RB)\mathcal N(\rho)=\sum_jK_j\rho K_j^\dagger,\qquad\sum_jK_j^\dagger K_j=I,\qquad I_c(\rho,\mathcal N)=S(B)-S(RB)
R purifies the input, and RB is the output of identity on R tensor N. Coherent information I_c is in bits; the unassisted capacity it helps characterize is in qubits per use.↔ Scroll wide equations horizontally.

Coding formulas may need arbitrarily large blocks

For classical transmission, optimize the Holevo information of the output ensemble over possible input ensembles, including states spanning an entire input block. For quantum transmission, optimize coherent information over block input density operators. Divide by block length and take the regularized asymptotic value. The resulting coding theorems characterize achievable rates, but they do not promise an efficient encoder or decoder for a specified laboratory error budget.

A one-use optimized quantity need not equal the general asymptotic capacity because collective coding can change the optimum. Special channel classes admit simpler expressions, but their assumptions must be established. Negative coherent information for one trial input is also not a proof of zero capacity: the optimization and block limit remain. Distinguish a demonstrated code rate, a lower bound, an upper bound, and an exact capacity theorem when reporting what a noisy channel can transmit.

C(N)=lim⁡n→∞1nmax⁡{px,ρx(n)}χ({px,N⊗n(ρx(n))}),Q(N)=lim⁡n→∞1nmax⁡ρ(n)Ic(ρ(n),N⊗n)C(\mathcal N)=\lim_{n\to\infty}\frac1n\max_{\{p_x,\rho_x^{(n)}\}}\chi\bigl(\{p_x,\mathcal N^{\otimes n}(\rho_x^{(n)})\}\bigr),\qquad Q(\mathcal N)=\lim_{n\to\infty}\frac1n\max_{\rho^{(n)}}I_c(\rho^{(n)},\mathcal N^{\otimes n})
These are unassisted memoryless finite-dimensional capacity formulas. C is in bits/use and Q in qubits/use; χ uses base-two entropy of the full output ensemble. Regularization is part of the general statement.↔ Scroll wide equations horizontally.

A visible erasure flag permits exact answers

The d-dimensional erasure channel either delivers the input intact or replaces it by an orthogonal flag with probability epsilon. The receiver knows which uses were erased, unlike an unflagged arbitrary replacement error. Its classical capacity is the surviving fraction times log base two of d. Its unassisted quantum capacity instead contains twice the erasure penalty, clipped at zero. Protecting unknown quantum information is constrained by the information leaking to the complementary environment and by no-cloning.

With unlimited two-way classical assistance, successful transmissions of halves of entangled pairs can be acknowledged, retained, and converted into faithful quantum transmission by teleportation. The two-way-assisted quantum capacity reaches the surviving fraction times log base two of d. This does not contradict the unassisted value because the communication resources differ. These formulas are exact asymptotic results for independently flagged erasures, not finite-block promises and not formulas for a depolarizing channel, whose errors are unflagged.

Work it through

Three capacities for a qubit erasure channel

For a memoryless qubit erasure channel with erasure probability 0.25 and a perfectly distinguishable erasure flag, find unassisted classical capacity C, unassisted quantum capacity Q, and quantum capacity Q₂ with unlimited two-way classical assistance.

  1. The input dimension is d=2, so log₂d=1. The surviving fraction is 1−0.25=0.75, yielding C=0.75 classical bits per channel use.

  2. The unassisted quantum formula gives max(0,1−2×0.25)×1=0.5 logical qubits per use. It is smaller than the surviving fraction because coherent reference information must be protected.

  3. Unlimited two-way classical acknowledgement gives Q₂=(1−0.25)×1=0.75 qubits per use. These are asymptotic rates: multiplying by 100 suggests comparison scales of 75 bits or 50 qubits, not a guaranteed zero-error 100-use code.

Eϵ(ρ)=(1−ϵ)ρ⊕ϵ∣e⟩⟨e∣,C=Q2=(1−ϵ)log⁡2d,Q=max⁡{0,1−2ϵ}log⁡2d\mathcal E_\epsilon(\rho)=(1-\epsilon)\rho\oplus\epsilon|e\rangle\langle e|,\qquad C=Q_2=(1-\epsilon)\log_2d,\qquad Q=\max\{0,1-2\epsilon\}\log_2d
The erasure flag is orthogonal to the d-dimensional input subspace, and 0≤ε≤1. C and Q₂ share a number but have different information units and allowed resources.↔ Scroll wide equations horizontally.
The result

C=0.75 bits/use, Q=0.5 qubits/use, and Q₂=0.75 qubits/use, each under its stated assistance convention.

The living notebook

Explore the model

Interactive lab

Interactive lab · markov

A population channel is a smaller question

For each fresh comparison, Reset and set both controls. Reset and set p=0.25 and q=0.25. Predict one Step gives state-one probability 0.25 from the fixed initial zero, and a second Step gives 0.375. Compare a fresh run with both controls p=0.5 and q=0.5, which reaches 0.5 after one Step. These are exact classical transition probabilities. The instrument has no erasure flag, coherence, reference system, code, or channel-capacity readout, so it cannot evaluate the quantum capacities in this lesson.

A population channel is a smaller questionExact finite-step probabilities for a two-state chain starting in state 0, not sampled trajectories. Playback advances two transitions per second. When p = q = 0 every distribution is stationary; when p = q = 1 the chain alternates forever. A single zero transition creates an absorbing state. Current readings: P(state 1) at step 0: 0%; Stationary P(state 1): 60%; Chain behavior: Monotone convergence. All parameters are adjustable with the labeled controls below.Two-state Markov distribution after 0 steps: state zero probability 1, state one probability 0.Initial distribution (1, 0) · step 0p = 0.3q = 0.201100%0%1−p1−qP(0)P(1)Monotone convergence
P(state 0)P(state 1)Stationary distribution
P(state 1) at step 0
0%
Stationary P(state 1)
60%
Chain behavior
Monotone convergence
t = 0 s

Exact finite-step probabilities for a two-state chain starting in state 0, not sampled trajectories. Playback advances two transitions per second. When p = q = 0 every distribution is stationary; when p = q = 1 the chain alternates forever. A single zero transition creates an absorbing state. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.

Look forPopulation evolution alone does not specify a channel’s ability to preserve quantum information.

Pause & reason

Check your understanding

A channel has positive classical capacity but zero unassisted quantum capacity. Which statement is consistent with those facts?

Your turn

Try it yourself

A d=4 erasure channel has erasure probability 0.6. Compute C, Q, and Q₂. Also find the expected number of unerased uses in a block of 50 and explain why it is not a guaranteed decoded payload.

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

Keep the factor log₂4=2 and the nonnegative clipping in Q. A count expectation is separate from a coding theorem.

Compare with the solution

The surviving fraction is 0.4 and log₂4=2, so C=0.8 bits/use and Q₂=0.8 qubits/use. Unassisted Q=max(0,1−1.2)×2=0 qubits/use. In 50 independent uses the mean unerased count is 20 with variance 50×0.4×0.6=12. Individual blocks fluctuate, and finite error tolerance requires code design and overhead. Neither the expected 20 surviving four-level systems nor a capacity multiplied by 50 guarantees exact recovery in every block.

Keep the idea

Take it with you

  • A capacity is tied to a task, error criterion, asymptotic limit, and allowed assistance.
  • General classical and quantum capacity theorems involve regularized output information quantities.
  • Flagged erasure channels separate classical, unassisted quantum, and two-way-assisted quantum rates.

Read further

Original sources and deeper explanations to follow at your own pace.

Optional progress, saved in this browser when available. Revisit any lesson at any time.

Lesson not marked complete.