Bit flip error correction syndrome measurements

Question

I'm coming across some confusion in chapter 10.1.1 of Nielsen and Chaung. In terms of the 'recovery' procedure, how can the result of the syndrome measurement be 0, 2 or 3?

I am assuming that, for example, if the state is $a|010\rangle$ + $b|101\rangle$ (bit flip on the second qubit), then will we not see that $\langle\psi|P_2|\psi\rangle = 1$ and the rest of the projection operators will give a syndrome measurement of 0 (i.e $\langle\psi|P_1|\psi\rangle = 0$)? From this example, and working through the rest in the same manner, I am struggling to see where/how an error syndrome of 2 will be measured.

I hope my current working makes sense!

Answer 1

The outcome of measuring $P_2$ will not be 2, but rather the outcome of measuring the operator with a spectral decomposition given by the four syndromes and corresponding eigenvalues. Remember that $\langle \psi | P_n | \psi \rangle$ gives the probability of measuring the eigenvalue of $M$ corresponding to $P_n$, not the eigenvalue itself. As such, an outcome of $n$ is not saying that measuring $P_n$ will give $n$, but just refers to the fact that $\langle \psi | P_n | \psi \rangle = 1$, so we are certain that the syndrome is $P_n$.

As for how to perform this syndrome measurement in practice, look at page 430.