3
$\begingroup$

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?

$\endgroup$
6
$\begingroup$

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 $S(n)$:

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

| improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.