Explicit Absolutely Maximally Entangled states in dimensions twice a prime power

The paper in detail

For \(N=2k+1\) parties of local dimension \(d\), AME\((N,d)\) means \(\rho_A=\mathbb I/d^k\) for every set \(A\) of \(k\) parties. The product rule makes composite dimensions easy once their factors are available, so dimensions \(d=2s\) with \(s\) odd are the hard case: no AME state of nine or more qubits exists. Moreover, for odd \(q\) a stabilizer AME\((N,2q)\) state in the cyclic Weyl convention would yield a stabilizer qubit AME state, so the new states cannot be stabilizer states in that convention.

The coupling identifies a qubit and a \(q\)-level system at each site with one \(2q\)-level system, \(\ket b\ket j\leftrightarrow\ket{qb+j}\), and sets \[\ket\Psi=\sum_{a=0}^{K-1}\sqrt{\lambda_a}\,\ket{u_a}\otimes\ket{v_a}.\] Reducing to \(A\) produces terms \(\mathrm{Tr}_{A^c}\ket{u_a}\bra{u_b}\otimes\mathrm{Tr}_{A^c}\ket{v_a}\bra{v_b}\). If the \(q\)-level cross blocks vanish for \(a\ne b\) and equal \(\mathbb I/q^k\) for \(a=b\), only the average \(\sum_a\lambda_a\mathrm{Tr}_{A^c}\ket{u_a}\bra{u_a}\) survives on the qubit side. That average can be maximally mixed even though no single qubit state is.

Qubit mixtures come from commuting Pauli groups: the normalised projector \(P/K\) onto the joint fixed space is \(k\)-uniform exactly when every nonidentity group element acts on more than \(k\) qubits. For \(N=15\) a cyclic code over \(\mathbb F_4\) gives 12 generators, rank \(K=8\) and minimum weight 8. Shortening gives \(N=13\) (\(K=16\), weight 7), and a quasicyclic \([18,6,10]_4\) code gives \(N=17\) (\(K=64\), weight 9). For any odd \(N\), the group generated by \(X^{\otimes2k}\otimes I\) and \(Z^{\otimes2k}\otimes I\) works with \(K=2^{N-2}\). At \(N=9,11\) the known Atlas and Kitamura mixtures have 4 and 2 branches.

Parents are Cauchy-matrix MDS graph states, which exist for prime powers \(q\ge N+1\), and in extended form for \(q=N\) when \(N\) is a prime power; special parents cover \(q=5\) at \(N=9,11\) and \(q=9\) at \(N=15\). A parent supplies at most \(q\) branches, so \(q\ge K\) is also needed, which fixes the thresholds. For nine parties, odd \(d\) and multiples of 4 follow from prime-power constructions. For \(d=2s\) with \(s\) odd, use AME\((9,6)\) when \(3\mid s\) and otherwise AME\((9,2p)\) for a prime \(p\ge5\) dividing \(s\), times an odd factor. Rains' shadow bound excludes \(d=2\).

Limits of the coupling: no more than \(q\) branches can have vanishing cross blocks, and a \(k\)-uniform mixture of \(N\) qubits has rank at least 4, 9, 8 and 23 for \(N=9,13,15,17\), with the required dimension growing at least exponentially in \(N\). Choosing which qubit branch pairs with which parent branch also lets the construction impose symmetry: for primes \(p\equiv1\pmod{15}\), starting at 31, there is an AME\((15,2p)\) state with a cyclic shift of all parties compensated by local unitaries. A concrete open target is a six-uniform 13-qubit mixture of rank 9 to 13, which would give AME\((13,26)\).

Transferable methods

1. Erasing interference with a parent's projected branches

   The qubit side can only be made uniform on average, so the interference between branches must be removed completely on every \(k\)-party cut. Projecting the extra party of an AME state with one more party gives states whose reduced cross terms vanish, because that party counts towards the cut.

   Editorial derivation of the projection step. Let \(\ket\Phi\) be AME\((N+1,q)\) with reference party \(R\), and \(\ket{v_a}=\sqrt q\,(\mathbb I\otimes\bra a_R)\ket\Phi\). For a set \(A\) of \(k=(N-1)/2\) parties, \(A\cup R\) has \(k+1=(N+1)/2\) parties, half of \(N+1\), so \(\rho_{AR}=\mathbb I/q^{k+1}\). Hence \[\mathrm{Tr}_{A^c}\ket{v_a}\bra{v_b}=q\,\langle a\vert_R\,\rho_{AR}\,\vert b\rangle_R=\delta_{ab}\,\frac{\mathbb I_A}{q^k},\] and the reduction of \(\ket\Psi\) is \((\sum_a\lambda_a\mathrm{Tr}_{A^c}\ket{u_a}\bra{u_a})\otimes\mathbb I/q^k=\mathbb I/(2q)^k\).

   Editorial check on the smallest case, \(N=3\), \(q=3\): branches \(\ket{u_b}=\tfrac{1}{\sqrt2}(\ket{00}+\ket{11})\ket b\) for \(b=0,1\), and the parent \(\tfrac13\sum_{i,j}\ket{i,j,i+j,i+2j}\), which is AME\((4,3)\). The coupled state in \((\mathbb C^6)^{\otimes3}\) has every single-party reduction equal to \(\mathbb I/6\) to machine precision, while a single branch \(\ket{u_0}\) has a reduction that differs from \(\mathbb I/2\) by 0.5 in one entry. AME\((3,6)\) is known anyway (a GHZ state); the example only exhibits the mechanism.

   How to reuse it

   When a product construction needs a factor that does not exist, ask whether only its average is needed. A second factor whose states have vanishing reduced cross terms, such as branches projected from a larger code or AME state, can convert an average condition into a pure-state one.

2. The weight test for uniform stabilizer mixtures

   The qubit ingredient must be a mixed state of low rank with maximally mixed \(k\)-party reductions. Pauli groups make both properties checkable from a weight enumerator: the rank is \(2^{N-r}\) and uniformity is a minimum-weight condition.

   For an abelian group \(S\) of Hermitian Pauli words with \(r\) independent generators and \(-\mathbb I\notin S\), \[\frac{P}{K}=2^{-N}\sum_{E\in S}E,\qquad K=2^{N-r}.\] Tracing out any qubit on which a word acts nontrivially kills that word, so \(\mathrm{Tr}_{A^c}(P/K)\) keeps only words supported inside \(A\). Distinct Pauli words are linearly independent, so the reduction is \(\mathbb I/2^k\) for every \(k\)-set exactly when no nonidentity word has weight at most \(k\).

   Editorial check of Proposition B.1 at \(N=15\): the word \(XYXIXYIZZXIIIII\), its cyclic shifts by \(0,\dots,5\) and their images under \(X\to Z\to Y\to X\) give 12 pairwise commuting generators of rank 12 over \(\mathbb F_2\). The 4096 group elements have weights 0, 8, 10, 12 and 14 with multiplicities 1, 405, 1260, 1890 and 540, as stated; the minimum weight 8 exceeds 7, so \(P/8\) is 7-uniform. With the rank bound of 8 at \(N=15\), the threshold \(q\ge9\) is optimal for this coupling.

   How to reuse it

   To find low-rank mixed states with uniform marginals, search additive codes over \(\mathbb F_4\) for large minimum weight rather than large distance: the fixed-space projector needs only heavy stabilizer elements, a weaker condition than for a pure AME state.

Source: Proposition II.1 and Fig. 1; Theorem II.3, Corollaries II.2 and II.4; Appendix B.1 (weight test) and B.2 (Proposition B.1); Theorem II.6, Proposition II.7. 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. The nine-party existence question is now settled for every integer local dimension; the first new case is \(d=14\). The construction also separates two tasks, finding a low-rank uniform qubit mixture and finding a parent, so progress on either transfers directly.

Limitations. The thresholds are sufficient, not necessary: within the coupling, \(q=9,11,13\) at \(N=13\) and \(23\le q\le61\) at \(N=17\) remain open, and other ways of cancelling interference could reach smaller dimensions, possibly AME\((13,10)\). The nine-party classification depends on Kitamura's external AME\((9,6)\) state, which this coupling cannot produce. Finite inputs (qubit groups, small-field parents) are certified by exact computation in a companion repository rather than in the text. The paper builds on a nine-qubit mixture from the AME\((9,10)\) state of the Atlas Lean library and on Kitamura's eleven-party construction; it contains no AI-use statement.

Description based on the full arXiv HTML of v1, read 2026-11-14: Sections I–III and Appendices B.1–B.2 (the qubit groups) and G.2 (the mixture for every odd \(N\)); Appendices A, C–F and G.1, G.3 were not read beyond what the main text cites. The weight enumerator of the fifteen-qubit group was recomputed, and the coupling was checked on a small explicit instance.