Entanglement without secret key and PPT channel composition

Source and verification

This article explains family 272 of the October 6, 2026 OpenAI collection. Research conclusions are attributed to its manuscripts. The review covers statements, introductions, and selected proof mechanisms; it is not an independent verification of the complete proofs.

🏷️ States, channels, and secrecy

A bipartite quantum state is a positive matrix $\rho$ of trace one on ${\mathbb{C}}^a\otimes {\mathbb{C}}^b$. It is separable if it is a finite sum of positive product matrices, and entangled otherwise. A purification gives a third system $E$ a pure joint state whose marginal on the two laboratories is $\rho$; tracing out $E$ recovers $\rho$.

Secret-key distillation allows each laboratory to process all of its copies jointly and exchange authenticated public classical messages. Eve holds a purification and the complete public record. The desired output is a pair of identical uniform strings independent of Eve. For one bit, the ideal state is

$${\gamma }_{\sigma }=\frac{1}{2}\sum _{i=0}^1|i,i⟩⟨i,i|\otimes {\sigma }_{E^{\prime }}.$$

Here $E^{\prime }$ includes the public messages. Distance will mean the unhalved trace norm $\Vert X{\Vert }_1=\mathrm{tr}\sqrt{X^{\ast }X}$. Conventional trace distance is half this norm. Distillable secret key $K_D(\rho )$ is the supremum of asymptotic lower rates $\mathrm{liminf}{\ell }_n/n$ for key outputs of length ${\ell }_n$ whose error tends to zero.

The manuscript’s model includes ordinary finite two-way local protocols, with the number of rounds allowed to grow with $n$. It also admits specified countable measurable histories and terminal readouts satisfying reference-compatibility conditions. The supplied state is the only shared private resource: auxiliary resources are local, and initial shared randomness is public, independent of the input, and retained by Eve. Eve also retains speaker choices and stopping information. A completed protocol produces its prescribed outputs almost surely on the actual input and has normalized output. The countable extension requires an explicit compatible terminal process; an unspecified infinite evolution is not a terminal measurement. (OpenAI, 2026)

Entanglement with zero key, as claimed by the source

There is an explicitly specified entangled state on ${\mathbb{C}}^{10}\otimes {\mathbb{C}}^{10}$ whose range contains no nonzero product vector. For every number of copies and every admitted completed bit protocol, its output has trace-norm distance at least $1/5$ from every ideal bit state. Thus its conventional trace distance is at least $1/10$, and $K_D(\rho )=0$. (OpenAI, 2026)

The uniform one-bit separation is stronger than a zero asymptotic rate. Retaining only the first bit is a local channel and cannot increase trace norm. Therefore a secure sequence with even one output bit for infinitely many input sizes would contradict the fixed separation along that subsequence.

🏷️ Positive partial transpose and the composition question

A linear map $F:M_a(\mathbb{C})\to M_b(\mathbb{C})$ is completely positive, or CP, if ${\mathrm{id}}_k\otimes F$ preserves positive matrices for every $k$. It can be written $F(X)={\sum }_r{A}_{r}X{A}_r^{\ast }$. A channel is CP and trace preserving. Write $T_d$ for coordinate transposition. A PPT map has both $F$ and $T_b\circ F$ CP; this definition alone imposes no trace condition.

Use the input-first, unnormalized Choi convention

$$\begin{array}{rl}J(F)& =\sum _{i,j=1}^a{E}_{ij}\otimes F(E_{ij}),\\ {\Omega }_a& =\sum _{i=1}^a{e}_i\otimes e_i.\end{array}$$

Thus $J(F)=({\mathrm{id}}_a\otimes F)({\Omega }_a{\Omega }_a^{\ast })$, with $\Vert {\Omega }_a{\Vert }^2=a$. Complete positivity is equivalent to $J(F)\ge 0$. The PPT condition says that transposing its output factor preserves positivity. A map is entanglement breaking if its action on one part of every positive bipartite input produces a separable output. Equivalently, $J(F)$ is separable; this equivalence works for the CP cone without trace preservation.

The source constructs two PPT maps on $M_{10}(\mathbb{C})$ whose composition is not entanglement breaking. These maps may differ and need not preserve trace. It separately constructs a trace-preserving PPT channel on $M_{21}(\mathbb{C})$ whose own square is not entanglement breaking. Neither dimension is claimed minimal. These distinguish a two-map composition statement from a same-channel statement. (OpenAI, 2026)

PPT alone does not imply zero distillable key. The proof uses a more specific representation, defined next, so its secrecy conclusion is not a claim about all PPT states.

🏷️ A class preserved under tensor powers

Let $\mathcal{C}$ consist of normalized states admitting

$$\rho =(\mathcal{L}\otimes \mathcal{M})(\chi {\chi }^{\ast }),$$

where $\chi$ is a possibly unnormalized bipartite vector and both local maps are PPT. Their input and output dimensions may differ, and the maps need not be channels. Equivalently, $\mathcal{L},\mathcal{L}\circ T,\mathcal{M},\mathcal{M}\circ T$ are CP, using transposes on their respective input spaces. This is a representation property; it does not prescribe a physical preparation protocol.

Tensoring the representing maps and vectors proves closure under tensor powers after grouping all of Alice’s and all of Bob’s systems. This lets the eventual estimate apply directly to collective processing of $n$ copies.

For CP maps $F:M_a\to M_b$ and $G:M_b\to M_c$, put $F^{♯}=T_a\circ F^{†}\circ T_b$, where $F^{†}$ is the Hilbert–Schmidt adjoint. Expansion in matrix units gives

$$J(G\circ F)=(F^{♯}\otimes G)({\Omega }_b{\Omega }_b^{\ast }).$$

PPT is preserved by this transpose-adjoint operation. Consequently, if $F,G$ are PPT and their composite Choi matrix $Z$ is nonzero, then $Z/\mathrm{tr}Z\in \mathcal{C}$. The representing vector is ${\Omega }_b/\sqrt{\mathrm{tr}Z}$, not automatically the normalized maximally entangled vector. Constructing an entangled member of $\mathcal{C}$ therefore reduces to making a PPT composition with a nonseparable Choi matrix. (OpenAI, 2026)

🏷️ Common local effects at a public transcript

Fix a finite public transcript $t$. In Alice’s laboratory, compose the local instrument branches consistent with $t$ and her final bit $i$, summing her private outcomes locally. This defines a CP map ${\mathcal{A}}_i^t$. Bob similarly obtains ${\mathcal{B}}_j^t$. Once the public record is fixed, compatible private histories sum independently within the two laboratories. The branch is therefore ${\mathcal{A}}_i^t\otimes {\mathcal{B}}_j^t$.

Its effect on the input has the form $D_{ij}(t)=X_i(t)\otimes Y_j(t)$, where $X_i=({\mathcal{A}}_i^t{)}^{†}(I)$ and $Y_j=({\mathcal{B}}_j^t{)}^{†}(I)$ are positive. Crucially, the same two pairs work for all four outputs $00,01,10,11$. Separate product decompositions for the four branches would not suffice.

For input $R={\rho }^{\otimes n}$, the canonical purification identifies Eve’s unnormalized branch with

$${\tau }_{ij}(t)=(\sqrt{R}{D}_{ij}(t)\sqrt{R}{)}^{\mathsf{T}}.$$

There is no pointwise measurement normalization at fixed $t$; normalization follows only after integrating transcripts and summing bits. Private outcomes remain private, rather than being silently disclosed to Eve. For measurable complete transcripts, the source proves the corresponding common-density statement using local product probes with untouched references and conditional independence of their local histories. This extension is a substantial separate input. An arbitrary purification is reduced to the canonical one by an Eve-side channel retaining the transcript. (OpenAI, 2026)

🏷️ The four-effect inequality

For arbitrary positive local matrices $X_0,X_1,Y_0,Y_1$, let $D_{ij}=X_i\otimes Y_j$ and $p_{ij}=\mathrm{tr}(D_{ij}\rho )$. The source proves, for $\rho \in \mathcal{C}$,

$$\begin{array}{rl}& \Vert \sqrt{D_{00}}\rho \sqrt{D_{11}}{\Vert }_1\\ & \le \sqrt{p_{00}{p}_{01}}+\sqrt{p_{10}{p}_{11}}\\ & +2\sqrt{p_{01}{p}_{10}}.\end{array}$$

No normalization of the four effects is assumed. This is what makes it usable at a single transcript.

The elementary starting point is the positive-block estimate: if $\left(\begin{array}{cc}H& Z\\ Z^{\ast }& J\end{array}\right)\ge 0$, then $\Vert Z{\Vert }_1\le \sqrt{\mathrm{tr}H\mathrm{tr}J}$. Decompose the block matrix into outer products $\left(\genfrac{}{}{0ex}{}{a_k}{b_k}\right){\left(\genfrac{}{}{0ex}{}{a_k}{b_k}\right)}^{\ast }$ and use the triangle inequality followed by Cauchy–Schwarz. This proof includes singular diagonal blocks.

Pull back Alice’s effects to $A_i={\mathcal{L}}^{†}(X_i)$. They can be made commuting without changing $\rho$: set $H=A_0+A_1$, use $H^{-1/2}$ on its support and the identity on its kernel, precompose $\mathcal{L}$ with that invertible congruence, and apply the inverse to Alice’s part of $\chi$. The new positive effects sum to the support projection, hence commute. The PPT conditions survive the congruence.

Split $\chi =v+w$ according to whether the first pulled-back effect is at least the second. Pulling back Bob’s effects similarly defines four nonnegative weights for each vector. The split gives $p_{11}(v)\le p_{01}(v)$ and $p_{00}(w)\le p_{10}(w)$, with additive total weights.

Complete positivity bounds each filtered cross term $Z(u,z)$ by $\sqrt{p_{00}(u)p_{11}(z)}$. Complete positivity after transposing both input factors gives the second bound $\sqrt{p_{00}(z)p_{11}(u)}$ for the same term. Apply the first bound to the two diagonal terms, and choose the appropriate orientation for each mixed term. Both mixed terms cost at most $\sqrt{p_{01}{p}_{10}}$. Adding the four bounds proves the inequality. (OpenAI, 2026)

🏷️ A quantitative conflict between agreement and secrecy

For positive matrices, define root fidelity by $f(H,J)=\Vert \sqrt{H}\sqrt{J}{\Vert }_1$. The purification identity and polar decomposition identify each transcript coherence with $f({\tau }_{00}(t),{\tau }_{11}(t))$. Let $P_{ij}$ be the integrated bit probabilities and let $\mathfrak{f}$ be the integral of that root fidelity.

Integrating the four-effect bound and applying scalar Cauchy–Schwarz gives

$$\begin{array}{rl}\mathfrak{f}& \le \sqrt{e(1-e)}+e,\\ e& =P_{01}+P_{10}.\end{array}$$

Indeed, the first two square-root terms are at most $\sqrt{(P_{00}+P_{11})(P_{01}+P_{10})}$, and twice the geometric mean of the two error probabilities is at most their sum.

Now suppose the output is at unhalved distance $\eta$ from an ideal bit. Work with one dominating transcript measure for the actual and ideal laws, so singular parts of either law are preserved. Let $a_i$ be the integrated distance of the $ii$ block from $\sigma /2$. The classical block structure gives $\eta =e+a_0+a_1$. Since the actual equal-bit mass is $1-e$, but the ideal equal-bit mass is one, trace conservation gives $a_0+a_1\ge e$. Thus $e\le \eta /2$.

The unnormalized fidelity bound $f(H,J)\ge (\mathrm{tr}H+\mathrm{tr}J-\Vert H-J{\Vert }_1)/2$, together with the triangle inequality between the equal-bit blocks, gives

$$\begin{array}{c}\frac{1-\eta }{2}\le \mathfrak{f}\le \sqrt{e(1-e)}+e,\\ e\le \frac{\eta }{2}.\end{array}$$

If $\eta <1/5$, then $e<1/10$. On that interval the upper bound is strictly below $2/5$, whereas the lower bound is strictly above $2/5$. This contradiction proves the uniform gap, independently of copy number, private-memory size, and the number of admitted communication steps. The factor two in $e\le \eta /2$ is essential for the stated constant. (OpenAI, 2026)

🏷️ The dimension-ten entanglement certificate

The construction starts from four explicit integer $6\times 4$ matrices defining a PPT map $L:M_4\to M_4$. Its matrix pencil has twenty complex directions $x$ with singular positive output $L(x{x}^{\ast })$. The key finite geometric property is that any ten vectors

$$\begin{array}{rl}(\hat{x}{)}_{ii}& =x_i^2(0\le i<4),\\ (\hat{x}{)}_{ij}& =\sqrt{2}{x}_i{x}_j(i<j).\end{array}$$

span the ten-dimensional space; the notation includes the four square coordinates and six cross coordinates. The source establishes this property through exact modular data and additional algebraic arguments. We use it as an explicit construction input rather than develop that algebra here.

Symmetric and antisymmetric tensor compression, followed by a complementary-pair operation, turn the singular outputs into kernel vectors $\hat{x}\otimes \hat{x}$ of a positive composite Choi matrix $Z$. A separate positive definite output, $L(e_0{e}_0^{\ast })=36I_4$, ensures $Z\ne 0$.

Suppose a nonzero product $u\otimes v$ belonged to $\mathrm{ran}Z=(\ker Z{)}^{\perp }$. Then $(u^{\ast }\hat{x})(v^{\ast }\hat{x})=0$ at all twenty directions. Each factor is a nonzero homogeneous quadratic in $x$. The spanning property says that each can vanish at at most nine directions. Their zero sets cannot cover twenty, since $9+9<20$. Thus the range contains no nonzero product vector.

Every nonzero separable positive matrix has such a vector in its range: refine its decomposition into positive rank-one products, whose vectors belong to the range because the kernel is their common orthogonal complement. Hence $Z$ is entangled. Normalizing by its actual trace gives the required entangled member of $\mathcal{C}$, and the secrecy obstruction applies. (OpenAI, 2026)

🏷️ A trace-preserving channel with an entangled square

To turn the dimension-ten pair ${\Phi }_1,{\Phi }_2$ into a single channel, use two ten-dimensional blocks and one absorbing flag. Set $A_i={\Phi }_i^{†}(I)$, $c_i=(1+\mathrm{tr}{A}_i{)}^{-1}$, and $F_i=I-c_i{A}_i>0$. The scaled maps lose trace exactly measured by the positive effects $F_i$.

Discard off-diagonal input blocks. Send block zero to block one using $c_1{\Phi }_1$, and block one back to block zero using $c_2{\Phi }_2$. Send all missing trace to the flag and leave the flag fixed. Each branch is CP, output transposition preserves complete positivity, and $c_i{A}_i+F_i=I$ makes the total map trace preserving.

After two applications, compression to block zero recovers $c_1{c}_2{\Phi }_2\circ {\Phi }_1$. The corresponding local Choi compression is a positive multiple of $Z$. Separability would survive that compression, so the full channel square cannot be entanglement breaking. Its full Choi range does contain a product contribution from the flag; the product-free assertion applies to the entangled corner, not to the entire enlarged Choi matrix. (OpenAI, 2026)

🏷️ Alternative construction and exact checks

The paper also gives a much larger construction based on real unit vectors and projection measurements. A Gram-matrix estimate supplies a lower bound on the sum of entries of a compressed Choi matrix. Zero projection overlaps restrict the possible supports of product vectors, forcing a conflicting upper bound under separability. A six-measurement parity construction and a restricted-intersection estimate supply the required data. The same rectangular trace-preserving PPT channel can then act on both halves of a maximally entangled state. Those parallel local actions are distinct from squaring the dimension-21 channel. (OpenAI, 2026)

We ran all three supplied verification programs in a temporary copy, with assertions enabled and exact integer arithmetic. All programs exited successfully, and all three outputs matched the supplied records byte-for-byte. They check the matrix-pencil identities and modular data at primes 41, 131, and 139; the binomial inequalities at parameter 257; and all 4096 choices of one outcome from each of six measurements, with no selection having every pairwise overlap positive.

These computations do not enumerate the twenty complex directions, verify all their ten-point determinants, instantiate the enormous alternative matrices, or prove the transcript theorem. The selected-source review covers the finite-effect argument, normalization and fidelity gap, the tensor construction’s range exclusion, and the channel completion. The complete measurable-history argument and the algebraic deductions from the modular certificate remain attributed to the manuscript.

🏷️ Key steps and proof outlines

Product probes recover common factors for a complete transcript

The finite-record argument above is transparent because one can sum finitely many local branches. The measurable-history theorem needs a replacement that does not condition on a singleton transcript of probability zero.

Attach untouched reference systems $A_0,B_0$, and run the protocol on a product of normalized local maximally entangled probes:

$$|{\Phi }_a{⟩}_{A{A}_0}\otimes |{\Phi }_b{⟩}_{B{B}_0}.$$

This input is a product across the two laboratories. The probes are mathematical tests of the protocol, rather than extra shared resources. Choose finite informationally complete measurements on the references, meaning that their effects span all Hermitian matrices.

Represent each laboratory’s private random choices by a local sampling tape, $U$ for Alice and $V$ for Bob. Conditional on a finite public prefix ${\mathcal{F}}_m$, their laws remain independent:

$$\mathbb{E}[u(U)v(V)\mid {\mathcal{F}}_m]=\mathbb{E}[u(U)\mid {\mathcal{F}}_m]\mathbb{E}[v(V)\mid {\mathcal{F}}_m].$$

The induction uses who sends the next message. An Alice message depends only on her tape and the existing public prefix; conditioning on it changes Alice’s conditional law while leaving Bob’s conditional law unchanged. A public random draw uses the specified kernel depending on the public record, so the same factorization survives that step. Speaker choices and stopping records must be included among these public updates.

For the full transcript sigma field ${\mathcal{F}}_{\infty }=\sigma ({\bigcup }_m{\mathcal{F}}_m)$, conditional expectations converge in $L^2$. Boundedness and Cauchy–Schwarz let the products converge in $L^1$, preserving the factorization at the complete transcript. This is the precise limiting step; it does not assume independence after arbitrary conditioning.

Let $W_{ij}(t)$ be the positive reference-matrix density of output bits $i,j$, with respect to the probe transcript law. Define its local marginal densities by

$$U_i(t)=\sum _j{\mathrm{Tr}}_{B_0}{W}_{ij}(t),V_j(t)=\sum _i{\mathrm{Tr}}_{A_0}{W}_{ij}(t).$$

Conditional independence gives

$$\mathrm{Tr}[(E_r\otimes G_s)W_{ij}(t)]=\mathrm{Tr}[E_r{U}_i(t)]\mathrm{Tr}[G_s{V}_j(t)]$$

for every pair of reference-test effects. Because these tests span the Hermitian matrices, this proves

$$W_{ij}(t)=U_i(t)\otimes V_j(t)$$

almost everywhere, with the same local pairs for all four bit branches. Private records remain private throughout this step (OpenAI, 2026).

A reference filter returns to the actual input without inversion

For the actual input $R$, apply the fixed reference filter

$$K=\sqrt{ab}(\sqrt{R}{)}^{\mathsf{T}}.$$

Acting on the untouched references of the product probes, it produces the canonical purification of $R$. The filter commutes with all laboratory instruments and public-source updates. Equality on finite history cylinders therefore extends to the complete history measure, and Eve’s bit density becomes

$$\begin{array}{rl}K{W}_{ij}(t)K^{\ast }& =(\sqrt{R}{D}_{ij}(t)\sqrt{R}{)}^{\mathsf{T}},\\ D_{ij}(t)& =a{U}_i(t{)}^{\mathsf{T}}\otimes b{V}_j(t{)}^{\mathsf{T}}.\end{array}$$

This proves the desired common-effect formula even when $R$ is singular: no inverse of $R$ is used.

One subtlety is termination. A protocol completed on $R$ may fail to return a bit on the auxiliary probes. The proof assigns a local default only to missing probe outputs. On the actual input the missing-output event has probability zero. Its filtered reference matrix is positive with zero trace, hence is the zero matrix; every subevent contributes zero too. Thus those defaults leave the actual output unchanged. Reference-compatible terminal readouts are needed to make this argument meaningful for the admitted infinite histories.

The operational chain is now explicit: tensor closure handles collective inputs; the probe-and-filter theorem supplies common effects at each transcript; the four-effect estimate bounds their coherence; and the fidelity calculation converts that bound into the fixed secret-bit separation. The finite entanglement certificate supplies a state to which this chain applies, but does not replace any of these protocol steps (OpenAI, 2026).

🏷️ Connections

Unbiased bases in dimension six also uses exact finite data to constrain quantum measurement geometry. Here the finite measurement checks support a nonseparability construction; they do not replace the operational secrecy proof. Condensation and depletion provides another example where information about a state must control a specified observable, rather than only a scalar energy or a broad structural property.

🏷️ Manuscripts and verification scope

• Entanglement with zero distillable secret key in local dimension ten — pinned manuscript and companion source.

The family scope page describes selected formalization coverage. The page selects the dimension-ten PPT pair and the dimension-21 channel whose square is not entanglement breaking. It explicitly places the zero-distillable-secret-key statement outside those two Comparator statements; the operational secrecy theorem is not claimed here to have that formal coverage. No Lean build, axiom audit, or manuscript-to-formal-statement equivalence check was performed for these notes.

OpenAI collection provides prerequisites and the reading sequence; coverage register records completed and pending families.

🐻  OpenAI 2026. Entanglement with zero distillable secret key in local dimension ten.