Counting complete sets of mutually unbiased bases composed of stabilizer states Consider $N$ qubits. There are many complete sets of $2^N+1$ mutually unbiased bases formed exclusively of stabilizer states. How many?
Each complete set can be constructed as follows: partition the set of $4^N-1$ Pauli operators (excluding the identity) into $(2^N+1)$ sets of $(2^N-1)$ mutually commuting operators. Each set of commuting Paulis forms a group (if you also include the identity and "copies" of the Paulis with added phases $\pm 1$, $\pm i$). The common eigenstates of the operators in each such group form a basis for the Hilbert space, and the bases are mutually unbiased. So the question is how many different such partitions there exist for $N$ qubits. For $N=2$ there are six partitions, for $N=3$ there are 960 (as I found computationally).
The construction above (due to Lawrence et al., see below) may be an example of a structure common in other discrete groups - a partition of the group elements into (almost) disjoint abelian subgroups having only the identity in common. Does anyone know about this?
Reference:
Mutually unbiased binary observable sets on N qubits - Jay Lawrence, Caslav Brukner, Anton Zeilinger, http://arxiv.org/abs/quant-ph/0104012
 A: Here is an answer that should work. I do not currently have access to matlab to check this for anything other than the smallest cases, so you should do that.
First off, I find it easier personally to work in the reduced set of $3^N-1$ stabilizers for N qubits (generating the other from those). Entirely a personal preference, and doesn't change the result here.
So we want to divide the $3^N-1$ possible stabilizers into sets of $2^N-2$ commuting stabilizers, and find how many such divisions are possible.
Define 
$\alpha = 2^N-2$ = size of sets
$\beta = \frac{3^N-1}{2^N-2}$ = number of sets.
We now pick our sets from the available stabilizers. The first time, we can pick anything - $3^N-1$ choices. Then we have to pick a commuting set, which we'll come back to. After picking the set, there are $(3^N-1) - \alpha$ stabilizers remaining. We can pick any for the first of the next set. And so on. However, for the final set there is no choice: there will only be $\alpha$ stabilizers left. So the choices for the first entry of each set are
$F = \prod_{k=0}^{\beta-2}(3^N-1) - \alpha k$
Now to pick each set. On average, half the stabilizers remaining to chose from will commute with any given one. So picking the second one, half the remainder will do. So for the first set, we have $((3^N-1) - 1)/2$ choices. The next choice has to commute with both the previous ones, so we have $((3^N-1) - 2)/2^2$ choices. And so on. For the next set, we start with $(3^N-1) - (\alpha+1)$ remaining stabilizers to pick the second entry. So the choices for picking the sets are
$S = \prod_{m=1}^{\beta} \prod_{i=1}^{\alpha-1}( \frac{3^N - (\alpha m + 1) - i)}{2^{i+1}} ) $
So the number of possible partitions is $F.S$ divided by the number of possible ways to permute within the sets x number of sets (PR) and number of ways of permuting the sets themselves (PC):
$PR = \beta.\alpha !$ 
$PC = \beta !$
So the number of partitions is
$\frac{F.S}{PR.PC}$
A: For finite dimensional systems, R. Buniy and T Kephart in 1012.2630 quant-ph provide a tool for defining a set of equivalence classes for entanglement states based on their algebraic properties.  Your answer should be in there.
