1. Quick check §
What does P represent in the correction condition?
Hint 1
It selects the logical subspace.
Worked solution
P is the orthogonal projector onto the code subspace; the condition is restricted to that subspace.
Quantum Information Theory · Chapter 08
Recovery and Quantum Error Correction
What does it mean for noise to be reversible on information we care about?
In this chapter · 3 sections
Reading guide · preparation and approach
Starting point: Chapters 1–7; channel adjoints, reference systems, and conditional mutual information.
Anchor example: A three-qubit repetition code under a specified single-bit-flip error set.
Method: Express environmental ignorance as an operator condition on the code subspace.
Watch for: Correcting one error family does not mean correcting arbitrary physical noise.
This lesson includes explanations, worked examples, and problems with solutions. Advanced results whose proofs are outside the chapter are identified explicitly.
8.1 Exact correction
8.1.1 The Knill–Laflamme condition
Let P project onto a code subspace and let \(\{E_a\}\) be errors spanning the noise operators of interest. Exact correction is possible precisely when \(PE_a^\dagger E_bP=c_{ab}P\) for all a,b. The coefficient matrix is independent of the logical state. This is the Knill–Laflamme criterion. It covers the linear span of the specified errors, not arbitrary errors on an unspecified number of qubits.
To see the necessity, use a dilation whose environment labels the errors. An environmental matrix entry is a linear combination of terms \(\langle\psi|E_b^\dagger E_a|\psi\rangle\). If any such entry depends on the encoded logical input, the environment has learned logical information, contradicting the decoupling requirement for exact recovery. Requiring constancy for every superposition gives the operator condition on P, not just equality on selected basis states.
For sufficiency, diagonalize the positive matrix c and replace the errors by the corresponding linear combinations. Their nonzero actions on the code become scaled isometries with mutually orthogonal images. A recovery measures which error-image subspace occurred and reverses its isometry. The syndrome distinguishes errors without distinguishing the logical state. Degenerate codes allow different physical errors to act identically on the code, so correcting noise does not require identifying a unique physical event.
Worked check 8.1. Why must the criterion include a≠b, rather than only error probabilities Eₐ†Eₐ?
Reasoned solution
Off-diagonal products determine coherence between environmental error labels. Ignoring them can hide information leakage visible only for logical superpositions or coherent combinations of errors.
8.2 A limited but exact code
8.2.1 Bit flips and phase flips are different
Encode \(|0_L\rangle=|000\rangle\) and \(|1_L\rangle=|111\rangle\). Consider the error set \(\{I,X_1,X_2,X_3\}\). Each error maps the code into a distinct two-dimensional subspace, and all four subspaces are orthogonal. Thus \(PE_a^\dagger E_bP=\delta_{ab}P\). Measuring the commuting parity observables \(Z_1Z_2\) and \(Z_2Z_3\) identifies which of the four subspaces occurred without measuring the logical amplitudes.
The syndromes are (+,+) for I, (−,+) for X₁, (−,−) for X₂, and (+,−) for X₃. Applying the corresponding X reverses the error. If physical bit flips occur independently with probability p on each qubit, the majority decoder fails when two or three flips occur: \(p_L=3p^2(1-p)+p^3=3p^2-2p^3\). At p=0.1, this is 0.028. The code is exact for the zero-or-one-flip error family, but independent noise also includes uncorrectable multiple flips.
A phase flip Z₁ acts as logical Z inside the code. Indeed, \(PZ_1P\) is not a scalar multiple of P. It changes the relative sign of the logical amplitudes while leaving both parity syndromes unchanged. Repeating computational-basis symbols therefore does not correct general single-qubit errors. The distinction is a reason to study richer codes, not a defect to conceal in the example.
Worked check 8.2. What does Z₁ do to (|000⟩+|111⟩)/√2?
Reasoned solution
It gives (|000⟩−|111⟩)/√2, an orthogonal logical state. Both stabilizer parities remain +1, so the bit-flip recovery sees no error.
8.3 Recovering a family of states
8.3.1 Petz recovery and approximate Markov structure
Recovery also appears in data processing. For a reference state \(\sigma\) and channel \(\mathcal N\), define the Petz map on the support of \(\mathcal N(\sigma)\) by
\[\mathcal R_{\sigma,\mathcal N}(X)=\sigma^{1/2}\mathcal N^\dagger\!\left(\mathcal N(\sigma)^{-1/2}X\mathcal N(\sigma)^{-1/2}\right)\sigma^{1/2}.\]Inverses mean inverses on the support; an arbitrary channel extension handles the orthogonal complement. The map recovers \(\sigma\). Under the usual support condition, equality in relative-entropy data processing for \(\rho,\sigma\) is equivalent to this map recovering \(\rho\) too. We state this equality theorem rather than prove it; see the discussion of Petz’s theorem in Carlen and Vershynina, Recovery map stability for the Data Processing Inequality. It concerns recovery of a specified state pair, not a universal inverse of a noisy device.
For a tripartite state, a related approximate theorem guarantees a channel \(\mathcal R_{B\to BC}\) such that \(I(A:C|B)\ge-\log_2 F(\rho_{ABC},(\operatorname{id}_A\otimes\mathcal R)(\rho_{AB}))\), using our squared-fidelity convention. See Fawzi and Renner. This statement does not assert that the ordinary, unrotated Petz map always attains the bound. Small conditional mutual information therefore certifies existence of good recovery, while identifying an efficient recovery is an additional task.
Worked check 8.3. If I(A:C|B)=0.1 bits, what squared fidelity does the recovery theorem guarantee?
Reasoned solution
At least 2^(−0.1)≈0.933033 for some recovery channel. This is an existence lower bound, not a claim that every reconstruction procedure achieves it.
Summary
- Knill–Laflamme conditions test all cross products of the specified errors.
- The repetition code corrects single bit flips, not arbitrary single-qubit noise.
- Recovery theorems concern specified states, supports, and error criteria; a noisy channel need not have a universal inverse.
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 §
List the X₂ syndrome in the repetition code.
Hint 1
X₂ anticommutes with both parity operators.
Worked solution
The syndrome is (−1,−1). Applying X₂ reverses the error.
3. Quick check §
Evaluate the repetition-code logical error at p=0.01.
Hint 1
Use 3p²−2p³.
Worked solution
The result is 0.000298 under independent bit flips and ideal encoding, syndrome extraction, and recovery.
4. Worked problem §
Why does Z₁ violate the condition for errors {I,Z₁}?
Hint 1
Evaluate the cross product on the logical basis.
Worked solution
PZ₁P has logical eigenvalues +1 and −1, so it is not proportional to P. The pair of errors is not correctable by this code.
5. Worked problem §
Show the Petz map for identity noise is identity on the reference support.
Hint 1
Cancel the square roots and inverse square roots.
Worked solution
For \(\mathcal N=\operatorname{id}\), the expression becomes \(\sigma^{1/2}\sigma^{-1/2}X\sigma^{-1/2}\sigma^{1/2}=X\) on the support.
6. Worked problem §
For complete dephasing and σ=I/2, find the Petz map.
Hint 1
Dephasing is self-adjoint and fixes I/2.
Worked solution
The scalar square roots cancel, so the recovery is dephasing itself. It recovers every diagonal state but cannot restore lost off-diagonal elements.
7. Worked problem §
If conditional mutual information is one bit, what fidelity guarantee remains?
Hint 1
Exponentiate the bound with base two.
Worked solution
Some recovery achieves squared fidelity at least 1/2. The theorem does not promise near-perfect recovery in this case.
8. Misconception §
Does correcting X₁ imply correcting every single-qubit operation?
Hint 1
Include phase errors.
Worked solution
No. An arbitrary single-qubit operator has components I, X, Y, Z. A code must satisfy the criterion for their relevant span, including cross terms.
9. Misconception §
Does the recovery inequality specify an efficient decoding algorithm?
Hint 1
Separate existence from construction and complexity.
Worked solution
No. It guarantees a suitable channel exists. Efficient implementation and experimentally available operations require further analysis.
10. Extension §
Show the bit-flip repetition code improves the error probability for 0<p<1/2.
Hint 1
Factor p minus the logical-error polynomial.
Worked solution
\(p-p_L=p-3p^2+2p^3=p(1-p)(1-2p)>0\). This conclusion assumes ideal recovery and only independent bit flips.