Quantum information · Theory
One-shot entanglement generation and distillation via Petz recovery and binary hypothesis testing
arXiv:2610.31501v1 ↗PDF ↗Submitted 22 October 2026
- Question
- How much entanglement can two laboratories obtain from one use of a noisy channel or one copy of a shared noisy state, and can the answer be expressed as a binary hypothesis-testing cost?
- Result
- An \((m,\varepsilon)\) entanglement generation or one-way distillation protocol exists whenever \(e_{d,m}(1-e_{d,m}/dm)\le\varepsilon\). Here \(e_{d,m}\) is the optimal error for testing \(\rho_{AB}\) against an explicit alternative \(B_{d,m}\propto dI_A\otimes\rho_B-\rho_{AB}\). Consequently \(\log_2m\le I_H^{\varepsilon-\nu}(A\rangle B)-\log_2(1/\nu)\) suffices, and repeated use gives \(nI_c+\sqrt{nV_c}\,\Phi^{-1}(\varepsilon)-\tfrac12\log_2n-c\) ebits.
The paper in detail
Three tasks share the target of an \(m\)-dimensional maximally entangled state with infidelity at most \(\varepsilon\). In entanglement generation, Alice prepares a state and sends one share through a channel \(\mathcal N\). In entanglement transmission, she encodes one share of a prescribed maximally entangled state. In one-way distillation, the parties already share \(\rho_{AB}\) and Alice sends a classical message. For classical communication, Cheng bounded the pretty-good measurement through binary tests; quantum decoding must also preserve coherence.
Write \(d=\dim A\) and define the testing cost \(\mathsf e(X,Y)=\min_{0\le T\le I}\{\operatorname{Tr}[(I-T)X]+\operatorname{Tr}[TY]\}=\tfrac12(\operatorname{Tr}(X+Y)-\|X-Y\|_1)\). Theorem 3.1: for every integer \(2\le m\le d\), \(e_{d,m}=\mathsf e(\rho_{AB},B_{d,m})\) with \(B_{d,m}=\tfrac{dm-1}{d^2-1}(dI_A\otimes\rho_B-\rho_{AB})\) certifies infidelity \(e_{d,m}(1-e_{d,m}/dm)\). The code projects Alice’s system onto a random \(m\)-dimensional subspace and Bob applies the Petz recovery for the resulting state.
The key step applies Cheng’s trace inequality, \(\operatorname{Tr}[XS^{-1/2}YS^{-1/2}]\le e(1-e/s)\) with \(S=X+Y\), before averaging over codes. The bound then depends linearly on a code-dependent comparison operator, whose Haar average is \(B_{d,m}\), so only a second moment of random subspaces is needed and no inverse of a random operator has to be averaged. Choosing an optimal test at tolerance \(\varepsilon-\nu\) gives the hypothesis-testing coherent-information bound with exact integer rounding. Lemma 4.4 converts an \((M,\varepsilon)\) generation code into a \((\lfloor M\sqrt{1-\varepsilon}\rfloor,\varepsilon)\) transmission code by local post-processing.
Checks and consequences. For a noiseless channel the test \(T=\Phi^d\) gives zero cost, certifying perfect recovery for every \(m\le d\). For an erasure channel with erasure probability \(p\), \(e_{d,m}=p\), while the actual Petz decoder has infidelity \(p(1-m^{-2})\), leaving room for channel-specific analysis. For \(n\) copies, the normal approximation to hypothesis testing gives a second-order expansion with coherent information \(I_c\) and its variance \(V_c\), for generation, distillation and transmission alike.
Transferable methods
Apply Cheng’s trace inequality before averaging over random codes
Bounding the recovery error by a binary testing cost while the code is fixed makes the bound linear in a comparison operator, so averaging over codes needs only a second moment.
For \(X,Y\ge0\), \(S=X+Y\), \(s=\operatorname{Tr}S\) and \(e=\mathsf e(X,Y)\): \[\operatorname{Tr}[XS^{-1/2}YS^{-1/2}]\le e\Bigl(1-\frac es\Bigr)\le\operatorname{Tr}[(I-T)X]+\operatorname{Tr}[TY]\quad\text{for every effect }T.\] With \(X=\rho_{AB}\) and \(Y=\Gamma-\rho_{AB}\) for a code-dependent \(\Gamma\ge\rho_{AB}\), the right side is linear in \(\Gamma\), and its average over Haar-random subspaces is \(\operatorname{Tr}[(I-T)\rho]+\operatorname{Tr}[TB_{d,m}]\).
Editorial checks. Over 200 random positive pairs in dimensions 2–5, the inequality never failed. For random \(\rho_{AB}\) with \(d=3,4\), \(B_{d,m}\) was positive semidefinite with trace exactly \(dm-1\). For an erasure channel (\(d=4\), \(p=0.3\)) the testing cost evaluated to \(p\), and the guarantee \(p(1-p/dm)\) exceeded the actual Petz infidelity \(p(1-1/m^2)\) for \(m=2,4\).
How to reuse it
In random-coding analyses where the decoder contains an inverse depending on the code, such as Petz or pretty-good measurements, use a trace inequality to replace the decoder error by a binary-test cost first. The average then involves only linear functionals of the code and is usually a simple moment computation.
From a binary test to an integer code dimension
An optimal hypothesis test at a slightly reduced tolerance turns the one-shot bound into a statement about hypothesis-testing coherent information, with a single logarithmic slack and exact rounding.
Let \(T\) attain \(D_H^\delta(\rho_{AB}\|I_A\otimes\rho_B)\), so \(\operatorname{Tr}[(I-T)\rho]\le\delta\) and \(\operatorname{Tr}[T(I_A\otimes\rho_B)]=2^{-I_H^\delta}\). Then \[e_{d,m}\le\delta+\frac{d(dm-1)}{d^2-1}2^{-I_H^\delta}-\frac{dm-1}{d^2-1}\operatorname{Tr}[T\rho]\le\delta+m\,2^{-I_H^\delta},\] since \(m-\tfrac{d(dm-1)}{d^2-1}=\tfrac{d-m}{d^2-1}\ge0\). With \(\delta=\varepsilon-\nu\), any integer \(m\le\nu2^{I_H^{\varepsilon-\nu}}\) gives error at most \(\varepsilon\).
For \(n\) copies, choosing \(\nu_n=n^{-1/2}\) and the normal approximation \(I_H^\delta\ge nI_c+\sqrt{nV_c}\,\Phi^{-1}(\delta-b/\sqrt n)\) yields the \(-\tfrac12\log_2n\) term while keeping the Gaussian quantile at the target error \(\varepsilon\).
How to reuse it
Whenever a one-shot coding error is bounded by a testing cost against an alternative of trace about \(m\) times a fixed operator, evaluate it at an optimal Neyman–Pearson test to get a hypothesis-testing rate with one slack term. Second-order expansions then follow from known normal approximations to hypothesis testing.
Context & limitations
Why it matters. Brings Cheng’s one-shot method for classical communication to quantum communication with a Petz decoder. The three central coding tasks (classical, entanglement-assisted, quantum) are now controlled by one kind of hypothesis-testing quantity, and a code-conversion lemma closes a gap in an earlier second-order argument.
Limitations. These are achievability bounds only: the \(-\tfrac12\log_2n\) term is a guarantee, not an optimal third-order coefficient. The proof shows that a good code exists but gives no efficient Petz circuit. The erasure example shows the bound is not tight for specific channels. The authors report AI assistance.