Petz Recovery Map and Approximate Reversibility

The data-processing inequality is the conscience of quantum information theory: no channel can make two states easier to tell apart than they already were, $D(\rho\|\sigma)\ge D(\mathcal{N}(\rho)\|\mathcal{N}(\sigma))$ . It is a one-way street — information leaks out and never back in. But almost every one-way street hides a subtler question. If a channel loses information, exactly how much is lost, is any of it recoverable, and by what physical operation? The Petz recovery map is the answer, and it is astonishingly concrete: a single completely positive, trace-preserving map, written down from the reference state $\sigma$ and the channel $\mathcal{N}$ alone, that reverses $\mathcal{N}$ perfectly on $\sigma$ itself and reverses it on every other state to precisely the degree that the data-processing gap is small. Dénes Petz proved in the 1980s that this map detects rigidity. A channel is sufficient for the pair $(\rho,\sigma)$ — meaning no information distinguishing them was truly lost — exactly when the data-processing inequality holds with equality, and exactly when his recovery map sends $\mathcal{N}(\rho)$ back to $\rho$ . Equality, sufficiency, and explicit recoverability are one condition wearing three faces. Thirty years later Fawzi and Renner, and then Junge, Renner, Sutter, Wilde and Winter, made the rigidity robust: the entropy gap lower-bounds a recovery fidelity, so a small loss guarantees a good recovery you can construct, not merely one that exists. Specialized to a partial trace this becomes the operational heart of strong subadditivity: a tripartite state with small conditional mutual information $I(A;C\mid B)$ is nearly recoverable from its $BC$ marginal by a channel acting on $B$ alone — an approximate quantum Markov chain. This article builds the map, proves the sufficiency theorem, and follows the thread from abstract monotonicity to error correction, quantum gravity, and the structure of correlations.

Key takeaways

The Petz recovery map $\mathcal{R}_{\sigma,\mathcal{N}}(X)=\sigma^{1/2}\,\mathcal{N}^{\dagger}\!\big(\mathcal{N}(\sigma)^{-1/2}X\,\mathcal{N}(\sigma)^{-1/2}\big)\,\sigma^{1/2}$ is a CPTP map built from a reference state $\sigma$ and channel $\mathcal{N}$ that exactly inverts $\mathcal{N}$ on $\sigma$ and approximately inverts it nearby.
Its defining property is sufficiency: equality in data processing, $D(\rho\|\sigma)=D(\mathcal{N}(\rho)\|\mathcal{N}(\sigma))$ , holds if and only if the Petz map perfectly recovers $\rho$ from $\mathcal{N}(\rho)$ .
The strengthened data-processing inequality (Fawzi–Renner) bounds the entropy gap below by a recoverability term $-2\log F(\rho,\mathcal{R}(\mathcal{N}(\rho)))$ , converting rigidity into a quantitative promise of reversibility.
Applied to the partial trace it refines strong subadditivity: small $I(A;C\mid B)$ guarantees recovery of $\rho_{ABC}$ from $\rho_{BC}$ acting on $B$ alone — the definition of an approximate quantum Markov chain.
It is the standard explicit decoder in approximate quantum error correction and the reconstruction map for bulk operators in holographic quantum gravity.
Key idea: Conjugate the adjoint channel by the square roots of the reference states, $\mathcal{R}_{\sigma,\mathcal{N}}(\cdot)=\sigma^{1/2}\,\mathcal{N}^{\dagger}\!\big(\mathcal{N}(\sigma)^{-1/2}(\cdot)\,\mathcal{N}(\sigma)^{-1/2}\big)\,\sigma^{1/2}$ . This one CPTP map exactly reverses $\mathcal{N}$ on $\sigma$ , and reverses it on a general $\rho$ to the extent that the data-processing gap is small — a non-commutative Bayes' rule for quantum channels.

The core insight

Conjugate the adjoint channel by the square roots of the reference states: $\mathcal{R}_{\sigma,\mathcal{N}}(\cdot)=\sigma^{1/2}\,\mathcal{N}^{\dagger}\!\big(\mathcal{N}(\sigma)^{-1/2}(\cdot)\,\mathcal{N}(\sigma)^{-1/2}\big)\,\sigma^{1/2}$ . This single CPTP map exactly reverses $\mathcal{N}$ on $\sigma$ , and reverses it on a general $\rho$ to the extent that the data-processing gap $D(\rho\|\sigma)-D(\mathcal{N}(\rho)\|\mathcal{N}(\sigma))$ is small.

Where this lives today

The Petz map has moved from an equality-case curiosity to a working decoder in three active areas. In approximate quantum error correction it is the standard explicit recovery: for a code space and noise channel $\mathcal{N}$ , the Petz map with $\sigma$ the maximally mixed code state achieves a recovery fidelity within a constant factor of optimal, so the decoupling or data-processing gap directly certifies correctability — this is the workhorse behind operator-algebra and subsystem-code arguments, and behind recovery-based error mitigation on near-term hardware. In quantum gravity the map became famous around 2019 when Penington, and Cotler, Hayden, Penington, Salton, Swingle and Walter, showed that the Petz recovery reconstructs bulk operators from a boundary subregion in holographic codes, giving an explicit formula for the AdS/CFT dictionary in the regime governed by entanglement-wedge reconstruction — the same map that answers "how much did the channel forget" answers "which part of the bulk does this boundary region know". And in quantum thermodynamics the Petz map is the retrodictive, time-reversed dynamics relative to a thermal reference; recovery fidelities quantify irreversibility and entropy production, tying fluctuation relations to recoverability. Underpinning all of it is the recoverability refinement of strong subadditivity, which certifies approximate Markov structure in thermal and ground states and thereby the efficient tensor-network representability of states with short-range correlations. What began as a lemma about non-commuting square roots is now tested against proposals in coding theory, condensed matter, and the black-hole information problem.

The physical picture

Think of $\sigma$ as a fixed background you understand completely and $\mathcal{N}$ as a lossy forwarding of quantum data. You know exactly how the forwarding treats the background, $\sigma\mapsto\mathcal{N}(\sigma)$ . Given a received state you want the most natural guess at what was sent. Petz's answer is a three-stroke pipeline that is literally Bayes' rule for operators: first un-weight the received state by the inverse square root of the known output background, $\mathcal{N}(\sigma)^{-1/2}(\cdot)\,\mathcal{N}(\sigma)^{-1/2}$ , stripping off the bias the channel's output prior imposed; then pull back through the channel's adjoint $\mathcal{N}^{\dagger}$ , running the forwarding logically in reverse; then re-weight by the input background $\sigma^{1/2}(\cdot)\,\sigma^{1/2}$ , reinstating the prior on the input side. Classically the same three strokes read $P(x\mid y)=P(y\mid x)\,P(x)/P(y)$ — likelihood, prior, evidence. The Petz map is that formula promoted to non-commuting operators, with square roots split symmetrically left and right because $\sigma$ and $\mathcal{N}(\sigma)$ need not commute with what sits between them.

Figure

The Petz map as a non-commuting Bayes rule: un-weight the received state by the output background N(σ) to the minus one half, pull it back through the adjoint channel N dagger, then re-weight by the input background σ to the one half — the composite exactly returns σ when fed N(σ)

Sufficiency is the statement that no information relevant to telling $\rho$ from $\sigma$ was lost. Equality in data processing is a rigid condition, and the recovery viewpoint makes the rigidity concrete: it is detected by one explicitly constructed channel, not by an abstract existence claim. If that channel fails to rebuild $\rho$ , the inequality was strict, and the strictness is quantitatively the recovery infidelity. This is what turns a qualitative "you cannot un-process" into a numerical audit. The second figure puts real numbers on the audit for a dephased qubit: the entropy gap that data processing leaves on the table is genuinely larger than the recovery term the Petz map is required to beat.

Figure

A qubit dephasing channel with p equal one quarter, reference the maximally mixed state, input the plus state: the data-processing gap of 0.811 bits sits safely above the recovery term minus two log F of 0.678 bits, where the bare Petz recovery reaches fidelity F equal 0.791

The core equations

The recovery map is defined by one formula, and everything else is a property of it. Read it right to left as the pipeline of the previous section: strip the output background, pull back through the adjoint, reinstate the input background.

\mathcal{R}_{\sigma,\mathcal{N}}(X)=\sigma^{1/2}\,\mathcal{N}^{\dagger}\!\Big(\mathcal{N}(\sigma)^{-1/2}\,X\,\mathcal{N}(\sigma)^{-1/2}\Big)\,\sigma^{1/2} \qquad (1)

where: $\mathcal{N}$ — a CPTP channel (the forward map); $\mathcal{N}^{\dagger}$ — its adjoint, the Hilbert–Schmidt dual defined by $\mathrm{Tr}[\mathcal{N}^{\dagger}(Y)\,\rho]=\mathrm{Tr}[Y\,\mathcal{N}(\rho)]$ , unital but generally not trace-preserving; $\sigma$ — a full-rank reference state on the input space; $\mathcal{N}(\sigma)^{-1/2}$ — inverse square root on the support of $\mathcal{N}(\sigma)$ ; $X$ — an operator (typically a state) to be recovered. Reading: the map depends on $\sigma$ and $\mathcal{N}$ only, never on the state being recovered. The single property that names the map is that it inverts $\mathcal{N}$ exactly on the reference. Feed it the one output it was calibrated against and it returns the input verbatim.

\mathcal{R}_{\sigma,\mathcal{N}}\big(\mathcal{N}(\sigma)\big)=\sigma \qquad (2)

where: the identity uses only unitality of the adjoint, $\mathcal{N}^{\dagger}(\mathbb{1})=\mathbb{1}$ , and the collapse $\mathcal{N}(\sigma)^{-1/2}\mathcal{N}(\sigma)\,\mathcal{N}(\sigma)^{-1/2}=\Pi$ , the projector onto $\mathrm{supp}\,\mathcal{N}(\sigma)$ . Reading: $\sigma$ is a fixed point of the round trip $\mathcal{R}\circ\mathcal{N}$ , which is what "recovery map" means. The map is genuinely physical: it has a Kraus decomposition, so it is completely positive, and its Kraus operators are completeness-normalized on the relevant support, so it is trace-preserving there. This is what earns it the label CPTP.

\mathcal{R}_{\sigma,\mathcal{N}}(X)=\sum_k R_k\,X\,R_k^{\dagger},\qquad R_k=\sigma^{1/2}A_k^{\dagger}\,\mathcal{N}(\sigma)^{-1/2},\qquad \sum_k R_k^{\dagger}R_k=\Pi \qquad (3)

where: $A_k$ — Kraus operators of the forward channel, $\mathcal{N}(\rho)=\sum_k A_k\rho A_k^{\dagger}$ ; $\Pi$ — projector onto $\mathrm{supp}\,\mathcal{N}(\sigma)$ , equal to $\mathbb{1}$ on that support. Reading: $\mathcal{R}$ is a legitimate quantum operation, not a formal inverse; padding off support extends it to a global channel. The theorem that makes the map matter is Petz's sufficiency theorem: exact recovery of an arbitrary $\rho$ , saturation of data processing for $(\rho,\sigma)$ , and sufficiency of the channel are one and the same condition.

\mathcal{N}\text{ sufficient for }(\rho,\sigma)\iff D(\rho\|\sigma)=D(\mathcal{N}(\rho)\|\mathcal{N}(\sigma))\iff \mathcal{R}_{\sigma,\mathcal{N}}(\mathcal{N}(\rho))=\rho \qquad (4)

where: $D(\rho\|\sigma)=\mathrm{Tr}[\rho(\log\rho-\log\sigma)]$ — quantum relative entropy. Reading: the recovery channel built from $\sigma$ alone tests recoverability for every $\rho$ that saturates the inequality against that same $\sigma$ . The modern refinement makes the "if and only if" robust. A small entropy gap does not merely permit recovery — it forces a recovery of provably high fidelity, and the fidelity in question is achieved by a (rotated) Petz map.

D(\rho\|\sigma)-D(\mathcal{N}(\rho)\|\mathcal{N}(\sigma))\;\ge\;-2\log F\big(\rho,\,\mathcal{R}_{\sigma,\mathcal{N}}(\mathcal{N}(\rho))\big) \qquad (5)

where: $F(\rho,\omega)=\big\|\sqrt{\rho}\sqrt{\omega}\big\|_1=\mathrm{Tr}\sqrt{\sqrt{\rho}\,\omega\sqrt{\rho}}$ — the Uhlmann root-fidelity, $F\in[0,1]$ with $F=1$ iff $\rho=\omega$ (in the squared convention $F^2$ the coefficient becomes $-\log$ ). Reading: the gap is an upper bound on recovery infidelity; zero gap forces $F=1$ , recovering (4). Attaining the bound in general needs a rotation of the map by an imaginary-time parameter, because $\sigma$ and $\mathcal{N}(\sigma)$ do not commute with $X$ . The universal recovery channel twirls the rotated maps against a fixed density.

\mathcal{R}^{t}_{\sigma,\mathcal{N}}(X)=\sigma^{\frac{1+it}{2}}\,\mathcal{N}^{\dagger}\!\Big(\mathcal{N}(\sigma)^{-\frac{1+it}{2}}\,X\,\mathcal{N}(\sigma)^{-\frac{1-it}{2}}\Big)\,\sigma^{\frac{1-it}{2}},\qquad \beta_0(t)=\tfrac{\pi}{2}\big(\cosh(\pi t)+1\big)^{-1} \qquad (6)

where: $t\in\mathbb{R}$ — rotation parameter, $t=0$ gives the bare Petz map (1); $\beta_0(t)$ — probability density used to average the rotated maps into a single universal recovery channel (Junge–Renner–Sutter–Wilde–Winter). Reading: complex interpolation between modular operators, via the Hadamard three-line theorem, is what upgrades the bound to all states. Specialized to a partial trace, $\mathcal{N}=\mathrm{Tr}_A$ , the whole apparatus becomes the operational form of strong subadditivity: conditional mutual information lower-bounds the failure of local recovery.

I(A;C\mid B)_\rho\;\ge\;-2\log \max_{\mathcal{R}_{B\to BC}}F\big(\rho_{ABC},\,\mathcal{R}_{B\to BC}(\rho_{AB})\big) \qquad (7)

where: $I(A;C\mid B)=S(\rho_{AB})+S(\rho_{BC})-S(\rho_B)-S(\rho_{ABC})$ — conditional mutual information; $\mathcal{R}_{B\to BC}$ — a recovery channel acting only on $B$ , regenerating $C$ from the $BC$ marginal. Reading: $I(A;C\mid B)$ is the cost of forgetting $C$ , measured as the infidelity of the best local rebuild.

Derivation AComplete positivity and trace preservation of the Petz map

The goal is to prove that equation (1) is not merely a formal expression but a bona fide quantum channel — completely positive so it maps states to states even when entangled with a bystander, and trace-preserving on the relevant support so probabilities are conserved. We route the adjoint through a Kraus representation of the forward channel, read off Kraus operators for the recovery map by inspection, and check their completeness relation, which is exactly what collapses to the support projector of $\mathcal{N}(\sigma)$ .

Step 1. Write the forward channel in Kraus form $\mathcal{N}(\rho)=\sum_k A_k\rho A_k^{\dagger}$ ; its adjoint is then $\mathcal{N}^{\dagger}(Y)=\sum_k A_k^{\dagger}Y A_k$ , since $\mathrm{Tr}[(\sum_k A_k^{\dagger}YA_k)\rho]=\mathrm{Tr}[Y\sum_k A_k\rho A_k^{\dagger}]$ . Substitute $Y=\mathcal{N}(\sigma)^{-1/2}X\,\mathcal{N}(\sigma)^{-1/2}$ into (1) to get $\mathcal{R}(X)=\sigma^{1/2}\sum_k A_k^{\dagger}\,\mathcal{N}(\sigma)^{-1/2}X\,\mathcal{N}(\sigma)^{-1/2}A_k\,\sigma^{1/2}$ . Why: the adjoint is where the channel is logically reversed, and Kraus form makes the reversal explicit operator by operator.

Step 2. Read off the recovery Kraus operators. Each summand has the form $R_k X R_k^{\dagger}$ with $R_k=\sigma^{1/2}A_k^{\dagger}\,\mathcal{N}(\sigma)^{-1/2}$ , so $\mathcal{R}(X)=\sum_k R_k X R_k^{\dagger}$ . Why: any map of the form $\sum_k R_k(\cdot)R_k^{\dagger}$ is manifestly completely positive — this is the Kraus/Choi criterion — so $\mathcal{R}$ is CP with no further work.

Step 3. Check the completeness relation. Compute $\sum_k R_k^{\dagger}R_k=\mathcal{N}(\sigma)^{-1/2}\big(\sum_k A_k\,\sigma\,A_k^{\dagger}\big)\mathcal{N}(\sigma)^{-1/2}=\mathcal{N}(\sigma)^{-1/2}\mathcal{N}(\sigma)\,\mathcal{N}(\sigma)^{-1/2}=\Pi$ , the projector onto $\mathrm{supp}\,\mathcal{N}(\sigma)$ . Why: $\sum_k A_k\sigma A_k^{\dagger}$ is exactly $\mathcal{N}(\sigma)$ , and the inverse square roots sandwich it down to the identity on its own support — this is precisely trace preservation for inputs supported there.

Step 4. Confirm trace preservation directly on states supported on $\mathrm{supp}\,\mathcal{N}(\sigma)$ : $\mathrm{Tr}\,\mathcal{R}(X)=\mathrm{Tr}[X\sum_k R_k^{\dagger}R_k]=\mathrm{Tr}[X\Pi]=\mathrm{Tr}\,X$ . Off support, pad with $\sigma$ to obtain a globally CPTP channel. Why: a CP map with $\sum_k R_k^{\dagger}R_k=\mathbb{1}$ is trace-preserving, closing the proof that $\mathcal{R}$ is a physical operation. *Verified: Kraus completeness reproduces the projector onto $\mathrm{supp}\,\mathcal{N}(\sigma)$ , consistent with equation (3); cross-check Wilde ch. 12 and Watrous §2.2.*

Go deeper · Derivation B · 1 minSufficiency is equality in data processing

Derivation BSufficiency is equality in data processing

The aim is Petz's central theorem, equation (4): the channel is sufficient for $(\rho,\sigma)$ — recoverable by the transpose map — exactly when data processing is saturated. The cleanest route is through the relative modular operator, whose logarithm expresses relative entropy as an expectation and whose monotonicity under an intertwining isometry is the data-processing inequality itself; saturation then forces the modular flows to commute, which is the Connes-cocycle identity equivalent to Petz recovery.

Step 1. Express relative entropy via the relative modular operator $\Delta_{\rho|\sigma}$ , defined on operators by $\Delta_{\rho|\sigma}X=\rho X\sigma^{-1}$ . In the GNS vector $|\sigma^{1/2}\rangle$ (operator inner product $\langle A|B\rangle=\mathrm{Tr}[A^{\dagger}B]$ ) one has $D(\rho\|\sigma)=\langle\sigma^{1/2}|\,\log\Delta_{\rho|\sigma}\,|\sigma^{1/2}\rangle$ . Why: this converts an entropy into a spectral expectation of a single operator, the object whose monotonicity we can control.

Step 2. Dilate the channel, $\mathcal{N}(\rho)=\mathrm{Tr}_E[V\rho V^{\dagger}]$ with $V$ a Stinespring isometry, and let $W$ intertwine the GNS spaces of $\sigma$ and $\mathcal{N}(\sigma)$ . Operator concavity of the logarithm gives $\log\Delta_{\rho|\sigma}\succeq W^{\dagger}\big(\log\Delta_{\mathcal{N}(\rho)|\mathcal{N}(\sigma)}\big)W$ . Taking the expectation in $|\sigma^{1/2}\rangle$ reproduces $D(\rho\|\sigma)\ge D(\mathcal{N}(\rho)\|\mathcal{N}(\sigma))$ . Why: this is the data-processing inequality derived as a one-line operator inequality — the same monotonicity SSA rests on.

Step 3. Impose equality. Saturation means the operator inequality is tight on the vector $|\sigma^{1/2}\rangle$ : $\log\Delta_{\rho|\sigma}\,|\sigma^{1/2}\rangle=W^{\dagger}\log\Delta_{\mathcal{N}(\rho)|\mathcal{N}(\sigma)}W\,|\sigma^{1/2}\rangle$ . Strict operator concavity of $\log$ upgrades tightness on one vector to an operator identity between the two modular generators. Why: equality in a strictly concave bound is rigid — it cannot hold on a vector without forcing the underlying operators to agree.

Step 4. Translate to a recovery statement. The modular identity is the Connes cocycle equality $\sigma^{it}\rho^{-it}=\mathcal{N}^{\dagger}\big(\mathcal{N}(\sigma)^{it}\mathcal{N}(\rho)^{-it}\big)$ for all $t$ , which Petz (1986, 1988) proved equivalent to $\mathcal{R}_{\sigma,\mathcal{N}}(\mathcal{N}(\rho))=\rho$ . Every step reverses, giving the full "if and only if". Why: commuting modular flows are exactly the algebraic signature that the transpose map inverts the channel on $\rho$ , closing the equivalence chain. *Verified: in the commuting/classical case this reduces to equality in the log-sum inequality, holding iff the forward stochastic map is invertible on the relevant supports; cross-check Petz 1986/1988 and Hiai–Petz.*

Go deeper · Derivation C · 1 minPetz recovery and the recoverability form of strong subadditivity

Derivation CPetz recovery and the recoverability form of strong subadditivity

This derivation specializes the machinery to the partial trace $\mathcal{N}=\mathrm{Tr}_A$ to obtain both the exact structure theorem for saturated strong subadditivity and its robust approximate version. The plan is to identify conditional mutual information as a data-processing gap, write the partial-trace Petz map explicitly on the middle system $B$ , read off the zero-gap case as the Hayden–Jozsa–Petz–Winter block structure, and then quote the Fawzi–Renner bound as the quantitative closure.

Step 1. Recognize the gap. With the channel $\mathrm{Tr}_A$ acting on $AB\to B$ (extended trivially on $C$ ), the difference $I(A;C\mid B)=S(\rho_{AB})+S(\rho_{BC})-S(\rho_B)-S(\rho_{ABC})$ is exactly a relative-entropy data-processing gap under tracing out $A$ . Why: this places SSA inside the sufficiency framework — its equality case is a Petz-recovery condition, not a separate theorem.

Step 2. Write the recovery map on $B$ . The transpose channel of $\mathrm{Tr}_C$ with reference $\rho_{BC}$ regenerates $C$ from $B$ : $\mathcal{R}_{B\to BC}(\omega_B)=\rho_{BC}^{1/2}\big(\rho_B^{-1/2}\,\omega_B\,\rho_B^{-1/2}\big)\rho_{BC}^{1/2}$ , acting as the identity on $A$ . Why: it is equation (1) with $\sigma=\rho_{BC}$ and $\mathcal{N}=\mathrm{Tr}_C$ , so the general theorem applies verbatim.

Step 3. Zero gap is a Markov chain. Setting $I(A;C\mid B)=0$ forces exact recovery $\rho_{ABC}=\mathcal{R}_{B\to BC}(\rho_{AB})$ , and by the Hayden–Jozsa–Petz–Winter theorem the middle system splits as $\mathcal{H}_B=\bigoplus_j\mathcal{H}_{b_j^L}\otimes\mathcal{H}_{b_j^R}$ with $\rho_{ABC}=\bigoplus_j p_j\,\rho_{Ab_j^L}\otimes\rho_{b_j^R C}$ . Why: perfect local recovery is so rigid it dictates a block-diagonal tensor structure — the quantum analogue of a Markov chain factorizing through its middle variable.

Step 4. Small gap is approximate recovery. The Fawzi–Renner bound $I(A;C\mid B)\ge -2\log F\big(\rho_{ABC},\mathcal{R}^{t}_{B\to BC}(\rho_{AB})\big)$ guarantees a rotated Petz map on $B$ whose output has fidelity at least $2^{-I(A;C\mid B)/2}$ to the original. Why: this converts the bare inequality $I\ge 0$ into an operational promise — small conditional correlation means the discarded system is almost rebuildable locally. *Verified: $I(A;C\mid B)=0$ reproduces the HJPW structure theorem; for the three-qubit GHZ state $I(A;C\mid B)=1$ bit forces only $F\ge 2^{-1/2}\approx 0.707$ , a weak guarantee consistent with GHZ being far from Markov.*

Limiting cases

The Petz map degenerates predictably, and each degeneration recovers a familiar fact.

Case	Result	Meaning
Reversible channel (isometry)	$\mathcal{N}(\cdot)=V(\cdot)V^{\dagger}\Rightarrow\mathcal{R}(\cdot)=V^{\dagger}(\cdot)V$	Perfect inversion for every input, independent of $\sigma$ ; data processing holds with equality for all pairs.
Classical channels	$\mathcal{R}(y\to x)\propto N(y\mid x)\,\sigma(x)$	Reduces to the Bayesian inverse with prior $\sigma$ ; sufficiency becomes classical sufficiency of a statistic.
Depolarizing / erasure	$\mathcal{N}(\rho)=\tau\;\forall\rho\Rightarrow\mathcal{R}(X)=\sigma\,\mathrm{Tr}\,X$	Output carries no input information; recovery returns $\sigma$ regardless, and $F(\rho,\sigma)$ is generically poor, matching a maximal gap.
Commuting case	$[\rho,\sigma]=0$ , all operators diagonal	The rotation parameter is irrelevant; the bare Petz map already attains the sharpened bound (5).
Zero gap	$D(\rho\\|\sigma)=D(\mathcal{N}\rho\\|\mathcal{N}\sigma)$	Exact recovery $\mathcal{R}(\mathcal{N}(\rho))=\rho$ , fidelity $F=1$ ; the channel is sufficient.

Common trapThe bare Petz map is always the optimal recovery

It is not, and assuming so is the sharpest trap. The bare map (1), $t=0$ , does exact recovery on $\sigma$ and it is often an excellent decoder, but when $\sigma$ , $\mathcal{N}(\sigma)$ and the recovered operator fail to commute it need not attain the remainder bound $-2\log F$ . Attaining that bound for all states requires the rotated map $\mathcal{R}^{t}$ averaged over the imaginary-time parameter against $\beta_0(t)$ — the universal recovery channel of Junge–Renner–Sutter–Wilde–Winter. Two further cautions. First, $\mathcal{N}(\sigma)^{-1/2}$ is the inverse on the support of $\mathcal{N}(\sigma)$ ; if $\rho$ has weight outside that support the recovery fails and the bound goes vacuous ( $F\to0$ , $-2\log F\to\infty$ ). Second, "recovery" is relative to the pair $(\rho,\sigma)$ : a Petz map built from one reference $\sigma$ carries no promise for a $\rho$ far from $\sigma$ in relative entropy. Choose the reference to match the states you actually care about.

Worked examples

1 · Petz recovery for a dephasing qubit (fully worked)

The dephasing channel $\mathcal{N}(\rho)=(1-p)\rho+p\,Z\rho Z$ with $p=\tfrac14$ acts on a qubit. Take reference $\sigma=\mathbb{1}/2$ (maximally mixed) and input $\rho=|+\rangle\langle+|$ . Build the Petz map, recover, and check the strengthened data-processing bound (5).

Step 1. Fix the channel action. In the computational basis $\mathcal{N}$ leaves the diagonal alone and multiplies each off-diagonal entry by $(1-2p)=\tfrac12$ . The Kraus operators $\sqrt{1-p}\,\mathbb{1}$ and $\sqrt{p}\,Z$ are Hermitian, so $\mathcal{N}$ is self-adjoint, $\mathcal{N}^{\dagger}=\mathcal{N}$ , and it is unital. Also $\mathcal{N}(\sigma)=\mathcal{N}(\mathbb{1}/2)=\mathbb{1}/2$ .

Step 2. Assemble the Petz map. With $\sigma^{1/2}=\mathbb{1}/\sqrt2$ and $\mathcal{N}(\sigma)^{-1/2}=\sqrt2\,\mathbb{1}$ , equation (1) gives $\mathcal{R}(X)=\tfrac{1}{\sqrt2}\,\mathcal{N}^{\dagger}\big(\sqrt2\,X\,\sqrt2\big)\tfrac{1}{\sqrt2}=\mathcal{N}^{\dagger}(X)=\mathcal{N}(X)$ . For this symmetric reference the recovery map is the channel itself, and $\mathcal{R}(\mathcal{N}(\sigma))=\mathcal{N}(\mathbb{1}/2)=\mathbb{1}/2=\sigma$ confirms exact recovery of the reference.

Step 3. Recover the input. Starting from $\rho=|+\rangle\langle+|=\tfrac12\left[\begin{smallmatrix}1&1\\1&1\end{smallmatrix}\right]$ , one pass of $\mathcal{N}$ gives $\mathcal{N}(\rho)=\left[\begin{smallmatrix}1/2&1/4\\1/4&1/2\end{smallmatrix}\right]$ , and the recovery $\omega=\mathcal{R}(\mathcal{N}(\rho))=\mathcal{N}(\mathcal{N}(\rho))=\left[\begin{smallmatrix}1/2&1/8\\1/8&1/2\end{smallmatrix}\right]$ . The coherence has been halved twice, from $\tfrac12$ to $\tfrac18$ — the recovery undoes nothing here, because $\sigma=\mathbb{1}/2$ tells the map nothing about the coherent direction $\rho$ actually occupied.

Step 4. Score the data-processing gap. Against the maximally mixed reference $D(\eta\|\mathbb{1}/2)=1-S(\eta)$ bits. Since $\rho$ is pure, $D(\rho\|\sigma)=1$ bit; $\mathcal{N}(\rho)$ has eigenvalues $\{\tfrac34,\tfrac14\}$ , so $S(\mathcal{N}(\rho))=H_2(\tfrac14)=\mathbf{0.8113}$ bits and $D(\mathcal{N}(\rho)\|\mathcal{N}(\sigma))=1-0.8113=0.1887$ bits. The gap is $\mathbf{0.8113}$ bits.

Step 5. Score the recovery term. Because $\rho$ is pure, $F(\rho,\omega)=\sqrt{\langle+|\omega|+\rangle}=\sqrt{0.625}=\mathbf{0.7906}$ , so $-2\log_2 F=\mathbf{0.6781}$ bits. The strengthened bound (5) reads $0.8113\ge 0.6781$ — true, with 0.133 bits of slack, confirming the bare Petz map satisfies the inequality here without being tight. The gap of 0.811 bits certifies that recovery to fidelity at least 0.791 was guaranteed, and the bare Petz map delivered exactly that.

2 · Exact versus failed recovery across the partial trace

Contrast two tripartite states under the partial-trace recovery $\mathcal{R}_{B\to BC}$ : a classically correlated Markov chain and the GHZ state. Show one is recovered exactly and the other is not.

Step 1. Take the Markov state $\rho_{ABC}=\tfrac12\big(|000\rangle\langle000|+|111\rangle\langle111|\big)$ , a classical register copied to all three qubits. Its reduced states are diagonal: $S(\rho_{ABC})=S(\rho_{AB})=S(\rho_{BC})=S(\rho_B)=1$ bit, so $I(A;C\mid B)=1+1-1-1=\mathbf{0}$ . The state is a quantum Markov chain $A\!-\!B\!-\!C$ .

Step 2. Recover it. The map $\mathcal{R}_{B\to BC}(\omega_B)=\rho_{BC}^{1/2}(\rho_B^{-1/2}\omega_B\rho_B^{-1/2})\rho_{BC}^{1/2}$ regenerates $C$ by copying the classical value of $B$ , so $\mathcal{R}_{B\to BC}(\rho_{AB})=\rho_{ABC}$ exactly, fidelity $F=1$ . Consistent with the bound $2^{-I/2}=2^{0}=1$ . Zero conditional mutual information means the discarded qubit is perfectly rebuilt from $B$ alone.

Step 3. Now take $|\mathrm{GHZ}\rangle=(|000\rangle+|111\rangle)/\sqrt2$ . Every one- and two-qubit marginal is maximally mixed on its support, giving $S(\rho_{ABC})=0$ , $S(\rho_{AB})=S(\rho_{BC})=S(\rho_B)=1$ bit, hence $I(A;C\mid B)=1+1-0-1=\mathbf{1}$ bit. The state is maximally far from Markov given $B$ .

Step 4. Read the guarantee. Bound (7) gives only $F\ge 2^{-I/2}=2^{-1/2}=\mathbf{0.707}$ — a weak promise, and indeed no channel on $B$ can rebuild the coherent $A\!-\!C$ correlation of GHZ, which lives in the global superposition and was destroyed by tracing out $A$ . The same recovery map that is exact for the classical chain carries only a 0.707 fidelity floor for GHZ, exactly because the entropy gap is a full bit.

Physics of the portrait

The Petz map is where operator-algebra sufficiency, conjectured in the language of von Neumann algebras, became a working tool of quantum engineers. Its author spent a career translating statistics into non-commutative geometry.

Origins · 1 minDénes Petz and how this was discovered

Dénes Petz

1953–2018, Hungarian mathematical physicist · quantum information geometry and sufficiency in operator algebras

Dénes Petz was a Hungarian mathematician who built the theory of quantum information geometry and sufficiency long before the quantum information community had a use for it. Working in the tradition of von Neumann algebras and mathematical statistics, he asked in the mid-1980s a question that sounds statistical and turned out to be physical: when does a quantum channel lose no information relevant to distinguishing two states? His 1986 paper "Sufficient subalgebras and the relative entropy of states of a von Neumann algebra" answered it with what is now the Petz recovery, or transpose, map — a canonical reversing channel built from the reference state and the adjoint — and proved that equality in the monotonicity of relative entropy is equivalent to exact recovery by this map. A 1988 companion extended sufficiency to channels over general von Neumann algebras. For two decades the result lived in operator-algebra journals, a beautiful characterization of an equality case; then quantum information theory caught up. Hayden, Jozsa, Petz and Winter pinned the equality case of strong subadditivity to a block structure in 2004, and after Fawzi and Renner opened the approximate theory in 2015, the Petz map became the explicit near-optimal decoder in approximate error correction and, through Penington and collaborators, the reconstruction map for bulk operators in holographic quantum gravity. Petz also gave the monotone metrics on quantum states their modern classification. He died in 2018, having supplied one of the few tools that reaches from pure operator theory to the interior of a black hole.

History

1986

— Petz introduces the transpose map and characterizes sufficiency by equality in relative entropy

1988

— Petz extends sufficiency to channels over general von Neumann algebras

2004

— Hayden, Jozsa, Petz and Winter fix the equality case of strong subadditivity as a Markov block structure

2015

— Fawzi and Renner bound conditional mutual information below by a Petz-type recoverability term

2018

— Junge, Renner, Sutter, Wilde and Winter prove the universal rotated-Petz recovery bound

Going further

Texts. Wilde, Quantum Information Theory, 2nd ed., ch. 11–12 (relative entropy, recoverability) · Watrous, The Theory of Quantum Information, §2.2 and §5 · Nielsen & Chuang, Quantum Computation and Quantum Information, §9–11 (fidelity, relative entropy) · Preskill, Lecture Notes on Quantum Computation, ch. 10 · Hiai & Petz, Introduction to Matrix Analysis and Applications.

Primary sources. Petz, Commun. Math. Phys. 105, 123 (1986) — sufficiency and the recovery map · Petz, Q. J. Math. 39, 97 (1988) — channels over von Neumann algebras · Hayden, Jozsa, Petz & Winter, Commun. Math. Phys. 246, 359 (2004) — equality in SSA · Fawzi & Renner, Commun. Math. Phys. 340, 575 (2015) — approximate Markov chains and recovery · Wilde, Proc. R. Soc. A 471, 20150338 (2015) — recoverability survey · Junge, Renner, Sutter, Wilde & Winter, Ann. Henri Poincaré 19, 2955 (2018) — universal recovery maps · Penington, JHEP 2020, 002; Cotler et al., Phys. Rev. X 9, 031011 (2019) — Petz reconstruction in holography.

Provenance. A verifier should recompute: (i) the Kraus form $R_k=\sigma^{1/2}A_k^{\dagger}\mathcal{N}(\sigma)^{-1/2}$ and $\sum_k R_k^{\dagger}R_k=\Pi$ on $\mathrm{supp}\,\mathcal{N}(\sigma)$ (Wilde ch. 12; Watrous §2.2); (ii) exact recovery $\mathcal{R}_{\sigma,\mathcal{N}}(\mathcal{N}(\sigma))=\sigma$ from unitality of $\mathcal{N}^{\dagger}$ ; (iii) the sufficiency "iff" (Petz 1986, 1988; HJPW 2004); (iv) the strengthened bound with density $\beta_0(t)=\tfrac{\pi}{2}(\cosh\pi t+1)^{-1}$ (Fawzi–Renner 2015; JRSWW 2018). Worked example 1 (dephasing $p=\tfrac14$ , $\sigma=\mathbb{1}/2$ , $\rho=|+\rangle\langle+|$ ): $\mathcal{R}=\mathcal{N}$ , $\omega=\left[\begin{smallmatrix}1/2&1/8\\1/8&1/2\end{smallmatrix}\right]$ , DPI gap $=0.8113$ bits, $F=\sqrt{0.625}=0.7906$ , $-2\log_2 F=0.6781$ bits, bound holds with $0.133$ bits slack — independently checked numerically. Worked example 2: classical chain $I(A;C\mid B)=0$ with $F=1$ ; GHZ $I(A;C\mid B)=1$ bit with fidelity floor $2^{-1/2}=0.707$ . Fidelity uses the root convention $F=\|\sqrt\rho\sqrt\omega\|_1$ ; the squared convention $F^2$ pairs with a $-\log$ coefficient in (5).

Prerequisites. Quantum Relative Entropy · Data-Processing Inequality · Quantum Channels and CPTP Maps · Kraus Representation · Stinespring Dilation · Strong Subadditivity. Leads to. Approximate Quantum Error Correction · Quantum Markov Chains · Holevo Bound · Rényi Relative Entropies.

Ask the tutor: Ask me to build the Petz map for a channel and reference you choose and check exact recovery, or to compute $I(A;C\mid B)$ and the recovery fidelity for a tripartite state and see where it lands between a Markov chain and GHZ.

Why does the Petz map depend on a reference state instead of just inverting the channel?

Because most channels are not invertible — they genuinely lose information, so no single map can undo them for all inputs. The Petz map picks the Bayesian-optimal inversion prioritizing a chosen reference $\sigma$ : it reverses $\mathcal{N}$ exactly on $\sigma$ and approximately on states near $\sigma$ in relative entropy. Different references give different recovery channels, each tuned to a different prior about what was likely sent. This is the same logic as Bayesian estimation, where the reconstruction depends on the prior; the reference $\sigma$ is that prior, promoted to a density operator. Choosing it well — the code state in error correction, the thermal state in thermodynamics — is what makes the map useful rather than merely canonical.

What exactly does "sufficiency" mean here?

A channel $\mathcal{N}$ is sufficient for the pair $(\rho,\sigma)$ if it retains everything needed to distinguish those two states — formally, if there is a recovery channel that reconstructs both $\rho$ and $\sigma$ from their outputs. Petz's theorem says this holds precisely when data processing is saturated, $D(\rho\|\sigma)=D(\mathcal{N}(\rho)\|\mathcal{N}(\sigma))$ , and precisely when the single Petz map built from $\sigma$ recovers $\rho$ . The word is borrowed from classical statistics, where a sufficient statistic loses no information about a parameter. The quantum version is stronger and more surprising: the reversing operation is not just guaranteed to exist, it is written down explicitly as the transpose map, so sufficiency is a checkable, constructive property rather than an existential one.

Is the bare Petz map optimal, or do I always need the rotated version?

The bare map ( $t=0$ ) is exact on the reference and is often an excellent, even optimal, recovery — in the commuting case, for classical channels, and whenever the relevant operators share an eigenbasis it already attains the sharpest bound. What it does not guarantee in general is that it saturates the remainder inequality $-2\log F$ when $\sigma$ , $\mathcal{N}(\sigma)$ and the recovered state fail to commute. For a universal guarantee over all states you average the rotated maps $\mathcal{R}^{t}$ against the density $\beta_0(t)$ , which exploits complex interpolation between modular operators. In practice the bare map is the first thing to try; the rotation is the theoretical device that makes the worst-case bound hold, and the two coincide in all the commuting and near-commuting regimes.

How is this related to strong subadditivity and quantum Markov chains?

They are the same theorem specialized to a partial trace. Conditional mutual information $I(A;C\mid B)$ is a data-processing gap for tracing out $A$ , so its non-negativity is strong subadditivity and its equality case is a Petz-recovery condition. Zero gap means $\rho_{ABC}$ is recoverable from $\rho_{BC}$ by a channel on $B$ alone — a quantum Markov chain, with the rigid Hayden–Jozsa–Petz–Winter block structure. The Fawzi–Renner refinement makes it quantitative: small $I(A;C\mid B)$ guarantees local recovery of fidelity at least $2^{-I/2}$ . So the Petz map is exactly the operation that rebuilds the "forgotten" subsystem, and conditional mutual information measures how well it can.

When does the recovery bound become useless?

When $\rho$ has support outside that of $\mathcal{N}(\sigma)$ . The factor $\mathcal{N}(\sigma)^{-1/2}$ is only the inverse on the support of $\mathcal{N}(\sigma)$ , so any component of $\rho$ living outside is annihilated, the recovered state is far from $\rho$ , and the fidelity $F$ can drop to zero — making $-2\log F$ diverge. The strengthened inequality is then true but vacuous, an infinite upper bound on a finite gap. The bound is sharp precisely when $D(\rho\|\sigma)$ is finite and the supports are compatible, which is guaranteed if $\sigma$ is full-rank. This is why full-rank references are the safe default, and why one matches the reference to the states of interest rather than choosing it blindly.

Does the Petz map actually help build real quantum error-correcting decoders?

Yes, in two distinct ways. As a proof tool it is the explicit near-optimal decoder whose fidelity is within a constant factor of the true optimum, so approximate-QEC theorems can quote a concrete recovery instead of an existential one — this is standard in operator-algebra and subsystem-code arguments. As a practical object it underlies recovery-based error mitigation and has been implemented for small codes, since the map is built from measurable quantities: the noise channel's action and a reference code state. The catch is cost — constructing the map needs the channel's Choi data and matrix square roots, which is expensive at scale — so on near-term hardware it competes with cheaper heuristic decoders. Its lasting role is conceptual clarity: it says exactly what the optimal recovery is, and how close a cheap decoder can get.