1. Quick check §
Write coherent information in terms of conditional entropy.
Hint 1
Use purity of RBE.
Worked solution
\(I_c=-S(R|B)=S(B)-S(RB)=S(B)-S(E)\).
Quantum Information Theory · Chapter 07
Quantum Communication and Decoupling
When can a noisy channel transmit a quantum state while preserving its reference correlations?
In this chapter · 3 sections
Reading guide · preparation and approach
Starting point: Chapters 1–6; purifications, complementary channels, and coding rates.
Anchor example: An entangled qubit sent through a flagged erasure channel.
Method: Compare output entropy with environmental entropy through a channel dilation.
Watch for: A positive coherent information is a rate lower bound; a single input does not determine a general capacity.
This lesson includes explanations, worked examples, and problems with solutions. Advanced results whose proofs are outside the chapter are identified explicitly.
7.1 Coherent information
7.1.1 Keep the reference in the calculation
Purify an input \(\rho_A\) as \(|\psi\rangle_{RA}\), apply an isometry \(V:A\to BE\), and write the resulting pure state as \(|\omega\rangle_{RBE}\). The receiver holds B, while E is inaccessible. Define coherent information \(I_c(\rho,\mathcal N)=S(B)-S(E)=-S(R|B)\). The equality follows from purity: \(S(E)=S(RB)\). Unlike mutual information, coherent information may be negative.
For the identity channel on a maximally mixed qubit, B has entropy one and E can be pure, giving \(I_c=1\). For a channel that replaces every input with a fixed pure state, B has entropy zero while E retains the input’s purification partner, so the same input gives \(I_c=-1\). The contrast captures the direction in which the input’s correlations have traveled.
The quantum capacity Q is the supremum of asymptotic qubit transmission rates with entanglement-transmission error tending to zero, using independent channel uses and no assistance. Its coding theorem is \(Q(\mathcal N)=\sup_{n\ge1}n^{-1}\max_{\rho_{A^n}} I_c(\rho_{A^n},\mathcal N^{\otimes n})\). We state this theorem without its full proof; the general regularized expression appears in Watrous, Chapter 8. Pure inputs ensure the maximized expression is at least zero, even when a particular mixed input gives a negative value.
Worked check 7.1. What coherent information does a pure input give through any channel?
Reasoned solution
The output BE is pure when the input is pure, so S(B)=S(E) and coherent information is zero. This does not imply that the channel’s optimized coherent information is zero.
7.2 Decoupling
7.2.1 Recovery is possible when the environment forgets
Suppose the reference and environment are exactly independent: \(\omega_{RE}=\omega_R\otimes\omega_E\). Any purification of this product can be chosen as the desired entangled reference–logical state tensored with a purification of E. Since purifications of the same state differ by an isometry on the purifying system, an isometry on the receiver’s B system can recover that logical system, with an extra local ancilla if necessary. The environment has no record of the logical information, so its correlations can be restored at the receiver.
This is the exact decoupling argument. Approximate product structure yields approximate recovery through continuity of purification, but turning it into coding rates requires quantitative estimates. A random encoder is not automatically a decoder, and small local noise alone is not a decoupling proof. The relevant comparison is the full joint RE state for the encoded source.
There is also a converse intuition with an exact statement behind it. If B can perfectly recover an arbitrary logical state while preserving R, then the recovered RL pair is pure. A pure subsystem cannot remain correlated with the rest, so E must be independent of R. This is the information–disturbance connection underlying error correction. Chapter 8 makes the requirement algebraic for a code subspace.
Worked check 7.2. Why is the marginal state of E alone insufficient to test decoupling?
Reasoned solution
Different joint RE states can have the same E marginal. A maximally entangled pair and the product of its marginals both have E=I/2, but only the product has no R–E correlation.
7.3 Erasure and assistance
7.3.1 The same channel has different operational capacities
For a d-dimensional erasure channel, a maximally mixed input gives receiver entropy \(h_2(p)+(1-p)\log_2d\). A complementary output receives the input on erased branches, giving entropy \(h_2(p)+p\log_2d\). Subtracting yields \(I_c=(1-2p)\log_2d\). The full unassisted quantum capacity is \(Q=\max\{0,1-2p\}\log_2d\). Its equality with this expression is a special theorem for the erasure channel, not a general license to remove regularization.
For \(p\le1/2\), the erasure channel is degradable: an additional erasure on B can reproduce the complementary channel. For \(p\ge1/2\), the environment can simulate B; such antidegradable channels have zero unassisted quantum capacity, consistent with the no-cloning constraint. See the erasure-capacity paper for the capacity result and assistance distinctions.
With unlimited two-way classical communication, the receiver can announce which halves of distributed Bell pairs survived. The surviving pairs support teleportation, and the erasure channel’s two-way assisted quantum capacity becomes \((1-p)\log_2d\). For qubits with p=0.2, unassisted classical capacity is 0.8 bits/use, unassisted quantum capacity is 0.6 qubits/use, and two-way assisted quantum capacity is 0.8 qubits/use. The numbers refer to different tasks and cannot be interchanged.
Worked check 7.3. At p=0.6, is the qubit erasure channel useless for distributing entanglement with classical feedback?
Reasoned solution
No. Unassisted Q is zero, but feedback identifies the surviving 40% of distributed Bell pairs. Two-way assistance achieves 0.4 ebits per use and supports that quantum transmission rate using teleportation.
Summary
- Coherent information compares the receiver with the environment.
- Decoupling concerns the reference–environment joint state, not only their marginals.
- Assistance changes capacities; special channel formulas must not be generalized without proof.
Questions and worked problems
These exercises draw on the whole chapter. Attempt each question before opening a hint, explain your reasoning, then compare it with the worked solution.
2. Quick check §
What is Q for a noiseless qubit channel?
Hint 1
The receiver gets the full state.
Worked solution
One qubit per use; a maximally mixed input has coherent information one.
3. Quick check §
Compute unassisted Q for p=0.3 qubit erasure.
Hint 1
Use the erasure-specific theorem.
Worked solution
Q=1−2(0.3)=0.4 qubits per use.
4. Worked problem §
Compute C, Q, and Q with two-way assistance for p=0.5 qubit erasure.
Hint 1
Keep the three models separate.
Worked solution
C=0.5 bits/use, Q=0, and two-way assisted Q=0.5 qubits/use.
5. Worked problem §
Compute Q for a qutrit erasure channel at p=0.25.
Hint 1
Dimension enters through log₂d.
Worked solution
Q=0.5 log₂3≈0.792481 qubits per use.
6. Worked problem §
Explain why exact recovery of a reference entanglement forces decoupling.
Hint 1
A pure recovered pair cannot be correlated with another system.
Worked solution
After a recovery isometry, RL is in the desired pure state and therefore factors from every remaining register. Tracing out local ancillas leaves R independent of E.
7. Worked problem §
A chosen input has coherent information −0.2. Does this imply negative capacity?
Hint 1
Compare with pure inputs and doing nothing.
Worked solution
No. Capacity is nonnegative, and pure inputs yield zero coherent information. Optimization and regularization may yield a positive rate.
8. Misconception §
Can equal environmental marginals prove correctability of a code?
Hint 1
Check superpositions and off-diagonal logical operators.
Worked solution
No. Basis states can have equal marginals while superpositions reveal logical coherence to the environment. Correctability requires independence for every logical input, including a reference.
9. Misconception §
Can the one-use coherent information always be called quantum capacity?
Hint 1
Recall the regularized coding theorem.
Worked solution
No. It is an achievable-rate lower bound after optimization; equality needs additional conditions such as degradability.
10. Extension §
For p<1/2, find the extra erasure probability that simulates E from B.
Hint 1
The simulated complement must survive with probability p.
Worked solution
If extra erasure is q, then \((1-p)(1-q)=p\), giving \(q=(1-2p)/(1-p)\). This is a valid probability exactly in the stated range.