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$$ \newcommand{\ket}[1]{\left|#1\right\rangle} \newcommand{\bra}[1]{\left\langle#1\right|} $$

The beginning portion of Box 2.3 on page 87 of QCQI is as follows:

"Suppose such a measurement is possible. If the state $\ket{\psi_1}(\ket{\psi_2})$ is prepared then the probability of measuring $j$ such that $f(j) = 1(f(j) = 2)$ must be $1$. Defining $E_i = \sum_{j:f(j)=i}M_j^\dagger M_j,$ these observations may be written as

$$ \bra{\psi_1}E_1\ket{\psi_1} = 1; \bra{\psi_2}E_2\ket{\psi_2} = 1. $$

Since $\sum_i E_i = I ...$"

It's this last part I take issue with. To me it seems this statement is potentially incorrect since there is no accounting for $j$ where $f(j) \neq i, \forall i$. If we add $E_0 = I - \sum_{i=1} E_i$ then it is true.

Is this an actual issue with this proof or am I missing something?

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The requirement that $\sum_i E_i = I$ is the requirement that the probabilities of the two outcomes sum to one, i.e. that you will get two outcomes that perfectly distinguish the states. If you drop this requirement, by adding an extra $E_0$ as you suggest, then that can occur with some probability and you haven't reliably distinguished the non-orthogonal states. Another way to say it is that if you add your extra $E_0$ then you also need to require that it has $0$ probability of outcomes, and when you do this the argument will continue to hold.

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  • $\begingroup$ But the possibility must exist that the state is neither $\psi_1$ nor $\psi_2$, right? On the previous page, when distinguishing orthonormal states, we added an $M_0$ to account for this, but we still were able to reliably distinguish states. $\endgroup$ Jul 13, 2022 at 13:39
  • $\begingroup$ $\langle\psi_1|E_0|\psi_1\rangle = \langle\psi_2|E_0|\psi_2\rangle = 0$ but $\langle\phi|E_0|\phi\rangle = 1$ where $|\phi\rangle \neq |\psi_i\rangle$. In fact without adding $E_0$ the probabilities do not sum to one, $\sum_i \langle\psi|E_i|\psi\rangle \neq 1$ since there is no measurement operator describing the outcome for when the state is not $|\psi_1\rangle$ or $|\psi_2\rangle$ $\endgroup$ Jul 13, 2022 at 13:49

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