An N-qubit stabilizer state is a state that can be produced by starting from the $|0\rangle^{\otimes N}$ state and applying only H, CNOT, and S gates. How many N-qubit stabilizer states are there?

Because every stabilizer state can be represented as a graph states, which has an edge (or not) between each pair out of N nodes and also one of 24 possible Clifford operations on each node, there are at most $2^{(N^2)} 24^N$ stabilizer states over $N$ qubits. But a stabilizer state can have multiple graph state representations. What's a corresponding lower bound on the stabilizer state count, and what's the exact count?


There are $S(n) = 2^n \prod_{i=1}^n (2^i + 1)$ $n$-qubit stabilizer states, as per Corollary 21 of D. Gross, Hudson's Theorem for finite-dimensional quantum systems, J. Math. Phys. 47, 122107 (2006).

Here are some simple-to-state bounds on $S(n)$:

$2^{(n^2 + 3n)/2} \leq S(n) \leq 3 \cdot 2^{(n^2 + 3n)/2}$


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