Quantum information · Theory

One-shot entanglement generation and distillation via Petz recovery and binary hypothesis testing

Matthew Can; Mark M. Wilde

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

  1. 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.

  2. 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.

Source: Sections 1.2 and 2.3 (Lemma 2.1); Section 3 (Theorem 3.1); Section 4.1 (Corollary 4.1); Section 5; Section 6. Based on the linked paper version. Derivations marked as editorial, and suggested uses in other problems, are editorial guidance rather than results of the paper.

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.

Description based on the full text of the linked arXiv version, read 2026-10-23; equations follow the paper’s notation.