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Suppose I have a quantum error correcting code with two stabilizers. Does this mean that I can potentially correct at most 3 distinct errors using this code?
My reasoning is that for each error, we need a syndrome. With only two stabilizers, we have the syndromes $(00, 01, 10, 11)$. The first one is the case of no error so that leaves three more errors.
In general, does the number of stabilizer elements bound the number of correctable errors?
Yes, and no.
Yes, in the sense that for a non-degenerate code, it does bound this, and this is essentially the content of the Quantum Gilbert-Varshamov Bound. Non-degenerate codes are exactly those ones for which distinct correctable errors must map to distinct syndromes.
No in the sense that there are non-degenerate codes for which different errors have the same syndrome and the same correction. The Shor code is a good example of this: there are sets of 3 qubits for which a $Z$ error on any of the individual qubits gives the same syndrome. You perform any one of the $Z$s as a correction, and whichever one you do, and whichever the single error was, you've got back to the original state.
