gpt 6 astra ultra attempt on Phys-19

Secret key from all entangled states

Definitions and mathematical statement

For a bipartite state $\rho_{AB}$, a secret-key protocol applies local quantum operations and public classical communication to $n$ copies. Eve may hold a purification and the public transcript. An $\ell_n$-bit key is secure when the final state approaches, in trace distance, \[ 2^{-\ell_n}\sum_{z\in\{0,1\}^{\ell_n}}|zz\rangle\langle zz|_{K_AK_B}\otimes\rho_E. \] The distillable key $K_D(\rho)$ is the supremum of asymptotic rates $\liminf\ell_n/n$ with error tending to zero. The question is whether $K_D(\rho)>0$ for every entangled bipartite state, where entangled means not separable into a convex mixture of product states. Secret classical key and maximally entangled quantum pairs are distinct output resources; their distillation rates must not be identified by definition.

Formulation and concept references: [S1].

Short English statement

Does every bipartite entangled state contain a strictly positive asymptotic rate of secret classical key extractable by local operations and public communication?

Research attempt

The resource and the quantifiers. The output is secret classical correlation decoupled from Eve and the public transcript. It is not a collection of Bell pairs. The 2026 revision of [S1] studies key cost and related formation quantities; in its discussion, absence of entangled zero-key states is an assumption for certain conclusions, not a proved theorem covering this question. A formation cost or an entanglement witness is not itself a distillation protocol. This attempt proves positivity for pure entangled states, proves zero key for finite-dimensional separable states, and isolates why the usual obstruction to entangled-pair distillation does not settle the mixed-state key question.

All finite-dimensional matrix arguments below use the standard model with arbitrary local ancillas and authenticated public communication. They do not assume that Eve lacks a purification. Any additional infinite-dimensional technical extension would require its own continuity and approximation arguments and is not claimed here.

A direct positive-rate protocol for a pure entangled state. Let \[ |\psi\rangle=\sum_{i=1}^r\sqrt{\lambda_i}\,|i\rangle_A|i\rangle_B, \qquad \lambda_i>0,\quad\sum_i\lambda_i=1, \] with $r\geq2$. Since $AB$ is pure, every purification factors as $|\psi\rangle_{AB}|e\rangle_E$. Choose two Schmidt coefficients, $\lambda_1,\lambda_2$, and put $\mu=\min(\lambda_1,\lambda_2)>0$. Alice applies the successful filter \[ F=\sqrt{\mu/\lambda_1}|1\rangle\langle1| +\sqrt{\mu/\lambda_2}|2\rangle\langle2|, \] with zero action on the remaining basis vectors. Since $F^*F\leq I$, this is a valid local measurement branch, completed by the failure operator $(I-F^*F)^{1/2}$. Its success probability is $2\mu$, and the normalized successful state is $(|11\rangle+|22\rangle)/\sqrt2$. Alice publicly announces which copies succeed. On independent copies the number of successes is binomial. For any $0<R<2\mu$, the probability of fewer than $\lfloor Rn\rfloor$ successes tends to zero by the law of large numbers. Keep that many successes and measure them in the Schmidt basis. The output bits are uniform, equal for Alice and Bob, and independent of Eve and the success transcript. On the rare failure event output an arbitrary fixed string; its probability bounds the trace-distance error. Hence $K_D(\psi)\geq2\mu>0$ after taking the supremum over $R<2\mu$. The same two-coefficient argument applies to a pure state with more Schmidt coefficients whenever the indicated Schmidt decomposition and filter exist.

A stronger pure-state rate without identifying key and entanglement. For finite Schmidt rank, both parties can measure every copy immediately in the Schmidt basis. Their resulting length-$n$ strings are identical and have law $\prod_{j=1}^n\lambda_{i_j}$, still independent of Eve. Publicly announce the type $t=(n_1,\ldots,n_r)$. Conditional on $t$, the common string is uniform over a type class of size \[ N_t=\frac{n!}{n_1!\cdots n_r!}. \] The law of large numbers and the elementary Stirling estimate imply that, with probability tending to one, \[ \log_2N_t\geq n(H(\lambda)-\varepsilon),\qquad H(\lambda)=-\sum_i\lambda_i\log_2\lambda_i. \] Fix $M=2^{\lfloor n(H(\lambda)-2\varepsilon)\rfloor}$. On each such type class, order its strings and keep the first $M\lfloor N_t/M\rfloor$ strings. Partition these equally among $M$ labels. The rejected fraction is less than $M/N_t\leq2^{-n\varepsilon}$. Given the publicly announced type and acceptance, the label is exactly uniform. Both parties compute the same label locally. Atypical types and rejections have vanishing total probability, so this proves \[ K_D(\psi)\geq H(\lambda)>0. \] This is directly a classical-secret-key construction. No equality between distillable key and distillable entanglement for general mixed states was used or inferred. The protocol's rate may depend on the state, as the question allows. Both parties retain any unused local measurement records in their laboratories; the protocol does not hand a copy of the raw secret string to Eve.

Why a finite-dimensional separable state has no secret key. Write a separable state as a finite convex combination of pure product states, \[ \rho_{AB}=\sum_jp_j|a_j\rangle\langle a_j| \otimes|b_j\rangle\langle b_j|. \] Eve may hold the purification $\sum_j\sqrt{p_j}|a_j\rangle|b_j\rangle|j\rangle_E$ and measure the orthogonal label register. Conditional on $J=j$, the two parties' state is a product. For $n$ copies use the sequence of such labels.

Under any finite-round local protocol, conditional on Eve's labels and the public transcript, the parties' states remain a product. This follows by induction over messages: a local instrument changes only one tensor factor, and conditioning on its announced outcome normalizes only that factor. Private outcomes can remain in that party's local register. Local private randomness is likewise part of a local tensor factor. Thus, after final key measurements, $K_A$ and $K_B$ are independent conditioned on $W=(J,\text{public transcript})$.

For each $w$, write their conditional laws as $p_z(w)$ and $q_z(w)$. The probability of agreement is $\sum_zp_z(w)q_z(w)\leq\max_zp_z(w)$. Eve can guess Alice's most probable value, so after averaging \[ P_{\rm guess}(K_A\mid W)\geq\mathbb P(K_A=K_B). \] If the output is within trace distance $\delta$ of an ideal equal uniform key on $M$ values independent of Eve, contractivity under measurement gives \[ \mathbb P(K_A=K_B)\geq1-\delta,\qquad P_{\rm guess}(K_A\mid W)\leq M^{-1}+\delta. \] Hence $1-\delta\leq M^{-1}+\delta$. With $\delta\to0$, a growing key alphabet is impossible. The argument extends to limits of finite-round protocols by trace-distance continuity of these probabilities. Security against a purification implies security against the particular Eve measurement used here, so weakening Eve in the proof creates no loophole. This establishes $K_D=0$ for this finite-dimensional separable class and clarifies why the question begins at entanglement.

Private states show what the secret output actually needs. Let $A'B'$ be auxiliary local registers, called the shield, let $\sigma$ be a state on them, and let $U_i$ be unitaries on the joint shield. The private state form is \[ \gamma=\frac1d\sum_{i,j=1}^d|ii\rangle\langle jj|_{AB} \otimes U_i\sigma U_j^*. \] It is positive with trace one: purify $\sigma$ as $|\chi\rangle_{A'B'E}$, and use the purification $d^{-1/2}\sum_i|ii\rangle(U_i\otimes I_E)|\chi\rangle$. Measuring $A,B$ in the displayed basis gives the same uniform label $i$. Eve's conditional state is \[ \operatorname{Tr}_{A'B'}(U_i\otimes I_E) |\chi\rangle\langle\chi|(U_i^*\otimes I_E) =\operatorname{Tr}_{A'B'}|\chi\rangle\langle\chi|, \] independent of $i$. Thus the state supplies $\log_2d$ secret bits per copy (or the corresponding asymptotic binary rate). The twisting unitaries need not be local operations across $A':B'$; their appearance describes the given state and does not claim that arbitrary such states can be prepared by LOCC. Nor is the shield handed to Eve. Eve holds a purification of all registers retained by Alice and Bob, exactly as required by the security definition.

The PPT obstruction stops at entangled pairs. Partial transpose positivity is preserved by LOCC: a product Kraus branch $A\otimes B$ sends $\rho^{\Gamma_B}$ to $(A\otimes\overline B)\rho^{\Gamma_B}(A^*\otimes B^T)$, and sums and conditional normalization preserve positivity. Tensor products also preserve it. If $\tau$ is PPT on $\mathbb C^d\otimes\mathbb C^d$, then its overlap with the maximally entangled state satisfies \[ \operatorname{Tr}(\tau\Phi_d) =\frac1d\operatorname{Tr}(\tau^{\Gamma_B}F)\leq\frac1d, \] where $F$ is the swap operator and $F\leq I$. This prevents convergence to Bell pairs from PPT inputs and proves zero distillable entanglement for them. The target ideal classical key state $d^{-1}\sum_i|ii\rangle\langle ii|$ is itself separable and PPT. The overlap argument therefore has no analogous obstruction to that target with decoupling from Eve. The source discusses known positive-key bound entangled examples; this separation is not merely terminological. A proof that a mixed state cannot yield Bell pairs is insufficient to show $K_D=0$.

The unresolved mixed-state step. For a mixed entangled state, Eve can correlate with every proposed measurement outcome. A witness detecting entanglement supplies a separating linear functional, whereas a key protocol must simultaneously give high agreement, uniformity, and decoupling after public communication on arbitrarily many copies. No direct implication between those tasks has been proved above. Likewise, a particular one-way key lower bound being nonpositive only shows that this chosen protocol fails to certify a rate. It is not an upper bound on all multi-copy, two-way protocols. To disprove the conjecture one would need an entangled state and a valid zero upper bound for the regularized operational quantity; to prove it one needs a positive protocol for every mixed entangled state. Neither follows from a finite-copy search.

Verification and outcome. The offline diagnostic checks the two-Schmidt-coefficient filter exactly for a rational example, balanced type-class partitions by exact integer counting, the conditional-independence guessing inequality for rational distributions, and Bell partial-transpose eigenvalues numerically. It also checks a small twisted private-state purification's conditional Eve marginals numerically. These checks validate the explicit subcases and resource distinction. The general positive-rate assertion for arbitrary mixed entangled states remains unresolved by this attempt.

Sources

• S1 K. Horodecki, J. Sikorski, S. Das, and M. M. Wilde, Cost of quantum secret key (2026 revision). https://arxiv.org/html/2402.17007. Formulation source.