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I am interested in the existence of a set of vectors $\{ v_{ij} \}_{ij} \subseteq \mathbb{C}^N$ for $i \in \{1,\dots,N\}$, $j \in \{1,\dots,N+1\}$ such that $\left\vert v^*_{ij} v_{ij'} \right\vert = 1/N$ for $j \ne j'$ and $v^*_{ij} v_{i'j} = \delta_{ii'}$.

For real vectors, the condition $\left\vert v^*_{ij} v_{ij'} \right\vert = 1/N$ is satisfied if $\{ v_{ij} : i \}$ is a regular simplex for each $j$. I have verified numerically the existence of such solutions for $N \le 5$. For $N=3$ a solution is given by the Compound of three tetrahedra.

For $N=p^k$ with $p$ prime, a solution is given by a set of mutually unbiased bases. These are not known to exist when $N \ne p^k$ but my conditions are looser than the MUB conditions.

Numerical experiments show that such vectors exist for $N \le 6$. Do they exist for all $N$?

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up vote 1 down vote accepted

Vern Paulsen has provided me an answer, and I reproduce it here with his permission. For $j \in \{1,\dots,N+1\}$ define the matrix with columns $U_j = (v_{1j};\dots;v_{Nj})$. The condition $v^*_{ij} v_{i'j} = \delta_{ii'}$ is equivalent to $U_j$ being a unitary matrix. The other condition, $\lvert v^*_{ij} v_{ij'} \rvert=1/N$ for $j \ne j'$, is equivalent to the matrix product $U_j^* U_{j'}$ having values of modulus $1/N$ along the diagonal.

Define $\omega = e^{2\pi i/N}$ and $\gamma = e^{2\pi i/(N+1)}$. Let $W$ be the Fourier transform matrix $W_{ij} = \omega^{ij} / \sqrt{N}$. Define $V_j = \mathrm{diag}(\gamma^j, \gamma^{2j}, \dots, \gamma^{Nj})$ and $U_j = W V_j W^*$. Since $W$ and $V_j$ are unitary, so is $U_j$. And it can be seen that the diagonal entries of $U^*_j U_{j'} = W V^*_j V_{j'} W^*$ are equal to $\mathrm{Tr}(V^*_j V_{j'})/N = \sum_{i=1}^N \gamma^{(j'-j)N}/N = -1/N$ when $j \ne j'$.

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