SIC-POVM

Remarks

The equal overlaps force $\sum_j \lvert \psi_j \rangle \langle \psi_j \rvert = d\, I$, so $\sum_j E_j = I$ and the $E_j$ do form a POVM. The $E_j$ are linearly independent and span all operators on $\C^d$, so the probabilities $\tr(E_j\rho)$ determine every state $\rho$.

Example: for $d = 2$, let $r_1, \dots, r_4$ be the vertices of a regular tetrahedron on the unit sphere of $\R^3$, so $r_j \cdot r_k = -1/3$ for $j \ne k$. Let $\psi_j$ be the qubit state with Bloch vector $r_j$. Then $\abs{\ip{\psi_j}{\psi_k}}^2 = (1 + r_j \cdot r_k)/2 = 1/3 = 1/(d+1)$.

Non-example: the six eigenvectors of the three Pauli matrices in $\C^2$ are not a SIC-POVM. There are $6 \ne d^2 = 4$ of them, and their pairwise overlaps take the values $0$ and $1/2$, not $1/3$.

The phrase "SIC" also denotes the set of vectors $\psi_j$ or the lines they span.