Quantum Latin squares of order $6m$ with all possible cardinalities

TL;DR

The paper investigates the cardinality spectrum of quantum Latin squares of order $6m$, proving that for every $m\ge 2$ and every admissible cardinality $c\in [6m,36m^2]\setminus\{6m+1\}$ there exists a QLS$(6m)$ with cardinality $c$. The authors develop a constructive framework combining a fixed library of QLS$(6)$ with multiple cardinalities and a tensor-product-like block construction $W$ from row-quantum Latin rectangles to realize product cardinalities, then extend to $QLS$(6m) by tiling a classical Latin square of order $m$ with blocks $|a_{i,j}\rangle\otimes Y_{i,j}$ drawn from selected libraries. By carefully selecting libraries such as $H_0,H_1,\widetilde{W}_i,H_\ell,H_\ell'$, and leveraging infinite families of maximal-cardinality QLS$(6)$, the paper covers the entire target range except the single exception $6m+1$. This yields a broad, explicit catalog of QLS$(6m)$ with diverse cardinalities, offering constructive tools for quantum Sudoku and related quantum combinatorial designs.

Abstract

A quantum Latin square of order $n$ (denoted as QLS$(n)$) is an $n\times n$ array whose entries are unit column vectors from the $n$-dimensional Hilbert space $\mathcal{H}_n$, such that each row and column forms an orthonormal basis. Two unit vectors $|u\rangle, |v\rangle\in \mathcal{H}_n$ are regarded as identical if there exists a real number $θ$ such that $|u\rangle=e^{iθ}|v\rangle$; otherwise, they are considered distinct. The cardinality $c$ of a QLS$(n)$ is the number of distinct vectors in the array. In this note,we use sub-QLS$(6)$ to prove that for any integer $m\geq 2$ and any $c\in [6m,36m^2]\setminus \{6m+1\}$, there is a QLS$(6m)$ with cardinality $c$.

Theorems & Definitions (8)

• Lemma 1.1
• Theorem 1.2: ZZTS
• Theorem 1.3: ZWJPreprint
• Lemma 2.1
• Lemma 2.2
• Lemma 2.3
• Lemma 2.4
• Lemma 2.5