What is the projective measurement operator for a stabilizer?

Question

I am learning about the stabilizer. And for a stabilizer $A$, its eigenspace with $\lambda=1$ corresponds to no-error condition while the eigenspace with $\lambda=-1$ corresponds to errored condition.

And I read it that we can define the projective operators regarding $A$ as follows:

$P_+^A=\frac{I+A}{2}$
$P_-^A=\frac{I-A}{2}$

For an arbitrary state $|\Psi>$, there is $|\Psi> = P_+^A|\Psi>+P_-^A|\Psi>$

I think the completeness $|\Psi> = P_+^A|\Psi>+P_-^A|\Psi>$ is clear based on the definition of $P_+^A$ and $P_-^A$.

But I don't know why $P_+^A$ and $P_-^A$ project $|\Psi>$ into states with eigenvalue 1 and -1. For example, from the definition of $P_+^A$, the eigenvalue of the projected state could be either 1 or 0, and so does $P_-^A$.

Some explanation about how to define the projectors regarding a stabilizer and how projected states relate to no-error and error conditions would really help!

Answer 1

What you want to prove is that $P^A_{\pm}|\Psi\rangle$ is a $\pm 1$ eigenvalue of $A$. Once stated like that, hopefully it's clear how to proceed: $$ AP^A_{\pm}|\Psi\rangle=(P^A_+-P^A_-)P^A_{\pm}|\Psi\rangle $$ Now from the properties of projectors, $(P^A_+-P^A_-)P^A_{\pm}=\pm P^A_{\pm}$ ($P_+^2=P_+$ and $P_+P_-=0$). Hence, we see that $$ AP^A_{\pm}|\Psi\rangle=\pm P^A_{\pm}|\Psi\rangle, $$ as required.