Zauner-symmetric Weyl–Heisenberg SIC fiducials
- Field
- Topic
Problem
Does every finite dimension admit a Weyl–Heisenberg SIC fiducial that is an eigenvector of a Zauner Clifford unitary? For \(d\geq2\), define the displacement operators \(D_{\mathbf p}:=D_{p,q}:=X_d^pZ_d^q\) for \(\mathbf p=(p,q)^{\mathsf T}\in\mathbb Z_d^2\), using
Equation (1) fixes the Weyl–Heisenberg orbit up to irrelevant phases. Let \(U_Z\) be a Clifford unitary whose action on displacement operators is
where indices are reduced modulo \(d\) and \(\doteq\) denotes equality up to phase. Equation (2) fixes the distinguished order-three Clifford symmetry. The question is whether, for every \(d\geq2\), there are a unit vector \(\lvert\phi\rangle\) and a phase \(e^{i\theta}\) such that
Equation (3) simultaneously imposes Zauner symmetry and the SIC overlap equations.
Source
Appleby states the all-dimensional existence of a Zauner-symmetric Weyl–Heisenberg SIC fiducial as Conjecture B [App05].
Progress
Appleby stated Eq. (3) as Conjecture B and verified it for the numerical fiducials then known, through \(d=45\) [App05].
Larger exact and numerical data sets continue to exhibit the Zauner symmetry, including Weyl–Heisenberg SIC solutions in every dimension through \(d=181\). Finite computations do not prove the universal quantifier in Eq. (3) [ABFG19].
Appleby, Flammia, and Kopp construct Zauner-symmetric SICs for all \(d>3\) under two number-theoretic conjectures. Those unproved assumptions leave the unconditional problem open [AFK25].
Comment
The open problem is the unconditional all-dimensional existence of a fiducial satisfying both conditions in Eq. (3). It is strictly stronger than Weyl–Heisenberg-covariant SIC existence in Problem 18.