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While reading through some lecture notes on quantum error correction, I read the statement:

"In particular, the syndrome doesn’t depend on the specific codeword, only on the Pauli error."

I'm trying to come up with a proof as to why this is the case. I know that codewords are vectors in the shared +1 eigenvalue subspace among the Pauli errors in the stabilizer group, $S$.

I suppose the question is whether given a Pauli error (not in the stabilizer group), $P$, does $P$ have the same error syndrome for all codewords? This seems like it would not be true because we only know how Pauli errors in $S$ act on the codespace.

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You can understand the syndrome by error propagation from the source of the error to the measurements. For example you can consider how an X error at a specific location affects the syndrome measurement. If it propagates to an X error right before a Z-measurement, for example, it flips the outcome. You see that this is completely independent of the state.

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  • $\begingroup$ I'm still a bit confused on your answer. Is there maybe a simple explicit example of what you are saying? $\endgroup$ Commented Jun 1 at 20:32
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    $\begingroup$ I think I understand now. A measurement of a stabilizer after an error occurs will be determined by whether or not that stabilizer commutes or anti-commutes with the error (Pauli operators must always commute/anticommute). Therefore applying an error E to the entire +1 subspace just translates it a subspace defined by the error syndrome. $\endgroup$ Commented Jun 1 at 22:04
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If the syndrome would depend on the codeword, this would imply that the syndrome measurement reveals information about the encoded qubit, which would necessarily destroy the encoded information (otherwise, no-cloning would be violated).

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