# How many N-qubit stabilizer states are there?

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's a simple approximation of $$S(n)$$:

$$S(n) \approx \sqrt{2}^{(n+1)(n+2)}$$

This approximation is within a factor of 2 of the real anwer for all $$n$$. It starts too high but by $$n=3$$ it's a bit too low and as $$n$$ goes to infinity it limits to being about 20% too low. Multiplying it by 1.192115 gets a much better estimate for large $$n$$.

For arbitrary $$d$$ (modular qudits), the number of stabilizer states is seen to scale similarly. Corollary 1 says that for $$c=2.17$$

$$$$\label{eq:nkd_count_order_states_intro} d^{\frac{n(n+3)}{2}} \ \le S(n,d) \ < \ \ d^{\frac{n(n+3)}{2}+c}.$$$$

In the paper for Aaronson-Gottesman algorithm, the expression for the number of stabiliser states is given with an accessible proof.

$$2^n \prod_{k=0}^{n-1} (2^{n-k}+1)$$

Thanks to Oscar Higgott for bringing my attention to it.