In the introduction of Eastin and Knill's paper on the no-go Theorem for universal and transversal quantum gate sets, they assert that an error E can only be detected if PEP∝P, where P is the projection operator onto the logical subspace. Why is this true, and is there a nice intuitive reason for this to be the case (or at least a slick geometric proof)?
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What the condition $PEP\propto P$ says is that the error, when projecting to the code space, does not change the quantum state of the code (up to an overall factor). To be more explicit, suppose $PEP=aP$. If $|\psi\rangle$ is a state in the code space (i.e. $P|{\psi}\rangle=|\psi\rangle$), then $E|\psi\rangle=PEP|\psi\rangle+(1-P)EP|\psi\rangle=a|\psi\rangle + (1-P)E|\psi\rangle$, where the second term $(1-P)E|\psi\rangle$ lives entirely outside the code space, which can be easily captured by measuring the projector and corrected.
answered Mar 19, 2022 at 19:29
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$\begingroup$ How is this consistent with, say, the surface code? A single X error takes us out of the code space, so $PEP=0$, but a single X error is certainly correctable. $\endgroup$ Apr 25, 2022 at 2:24
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$\begingroup$ @JahanClaes $a$ can be $0$, that's the surface code example. $\endgroup$ Apr 25, 2022 at 3:34