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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?

Craig Gidney
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3 Answers3

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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 $n$.

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

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

Tanmay Singal
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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.

shashvat
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