Thirty-six Entangled Officers of Euler: Quantum Solution to a Classically Impossible Problem

The negative solution to the famous problem of 36 officers of Euler implies that there are no two orthogonal Latin squares of order six. We show that the problem has a solution, provided the officers are entangled, and construct orthogonal quantum Latin squares of this size. As a consequence, we find an example of the long-elusive Absolutely Maximally Entangled state AME(4,6) of four subsystems with six levels each, equivalently a 2-unitary matrix of size 36, which maximizes the entangling power among all bipartite unitary gates of this dimension, or a perfect tensor with four indices, each running from one to six. This special state deserves the appellation golden AME state, as the golden ratio appears prominently in its elements. This result allows us to construct a pure nonadditive quhex quantum error detection code $((3,6,2){)}_6$, which saturates the Singleton bound and allows one to encode a six-level state into a triplet of such states.

Figure 1

Four dice in the golden absolutely maximally entangled state of four-quhex, AME(4,6), corresponding to 36 entangled officers of Euler. Any pair of dice is unbiased, although their outcome determines the state of the other two. This is not possible if the four dice are replaced by four coins (qubits).

Figure 2

An example of a Greco-Latin square of order $d=3$, relevant to the AME(4,3) state. In the middle, greek and latin letters are replaced by ranks and suits of cards; on the right, they are replaced by pairs of numbers.

Figure 3

Nonvanishing coefficients $T_{ijk\ell }$ of the AME(4,6) state $|\mathrm{\Psi }⟩$. Indices $(i,j)$ are indicated in rows, with $(k,l)$ in columns. Treating each coordinate $(i,j)$ as a position (row and column) of Euler’s officer, its rank and regiment are in a superposition of two or four canonical ranks and regiments. Equivalently, the picture shows nonzero entries of a 2-unitary matrix ${\mathcal{U}}_{36}$ of size 36, where $j+6(i-1)$ labels the relevant row while $\ell +6(k-1)$ determines the column. Remarkably, all 36 officers, each represented by a single row of the matrix, are maximally entangled qubit states: 16 of them have minimal support, being bipartite Bell states (red rows), while the remaining 20 have maximal support (yellow and green rows); see Eq. (4) and the Supplemental Material [37]. The order of the blocks reflects an additional structure, as the numbering of columns increases by six as we proceed to a block below. The depicted matrix preserves its structure of nine unitary blocks of size four with respect to transformations of reshuffling $j\leftrightarrow k$, and partial transposition $j\leftrightarrow \ell$.

Figure 4

Visualization of entanglement between Euler’s officers, corresponding to the golden AME(4,6) state. The position of each officer ($i$th row and $j$th column) is presented in the appropriate row and column on the array. Ranks and regiments of officers (relevant to the indices $k$ and $\ell$ in $T_{ijk\ell }$) are represented in the form of cards with corresponding ranks and suits (in particular, , , , , , , for $k/\ell$, respectively). The grade of each officer is in a superposition of four basic grades: two ranks and two regiments. The font size of each ket corresponds to the amplitude of the related tensor element. Notice that a classical solution to Euler’s problem would correspond to an array with only one card in each entry. Note the structures arising in the array—e.g., aces are entangled only with kings, queens with jacks, and 10’s with 9’s. Moreover, the colors of the cards are not entangled with each other. Hence, officers are grouped into nine sets with four elements, each sharing the same pairs of figures and colors of the card. For example, officers in the positions (1,1),(4,2),(5,6),(6,3) share two black figures $A$, $K$ of a black suit (, ), which corresponds to the top-left clique in Fig. 3.