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Suppose Alice transmits a qubit in either of two states $|\psi\rangle_{1}, |\psi \rangle_{2}$.

Bob has 3 Kraus operators: $\hat{E1}, \hat{E2}, \hat{E3}$ such that the average measurement value of $\hat{E1}$ on $|\psi\rangle_{1}$ yields 0, $\hat{E1}$ on $|\psi\rangle_{2}$ yields $E_{1}$; similarly, $\hat{E2}$ on $|\psi\rangle_{2}$ yields 0, $\hat{E2}$ on $|\psi\rangle_{2}$ yields $E_{2}$.

The action of $\hat{E3}$ on either states Alice sent yields an indeterministic results.

If this is the case why can't Bob stick with using $\hat{E1}, \hat{E2}$ and only $\hat{E1}, \hat{E2}$? My best guess is that it could be possible but the completeness relation cannot be satisfied with $\hat{E1}, \hat{E2}$ and only $\hat{E1}, \hat{E2}$. I.e, $\sum_{i=1}^{2} E_{i}^{\dagger}E_{i} \neq I$

Edit:

The above is in reference to Quantum Computation and Quantum Information by Michael A. Nielson and Isaac L. Chung, page 92.

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    $\begingroup$ Can you add some context? Why is it that Bob has these operators, particularly $\hat{E}3$? If you are reading a textbook or paper, please cite the exact spot in them. $\endgroup$ Commented Oct 19, 2023 at 3:00
  • $\begingroup$ @AbdullahKhalid I've edited the OP $\endgroup$
    – Physkid
    Commented Oct 19, 2023 at 10:35

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Saying "Bob has 3 Kraus operators $E_1,E_2,E_3"$ means, in this context, that he is performing a measurement described by the POVM $\{E_1^\dagger E_1, E_2^\dagger E_2, E_3^\dagger E_3\}$. Generally speaking, a POVM having elements $\{\mu_1,...,\mu_n\}$ means that it describes a measurement with $n$ possible outcomes. You cannot "choose" to just use a subset of the operators as measurement, because that doesn't give you a POVM, that is, it doesn't represent an actual measurement.

It's like saying, "instead of measuring in the computational basis, which corresponds to the POVM $\{|0\rangle\!\langle 0|, |1\rangle\!\langle1|\}$, why don't you just measure $\{|0\rangle\!\langle 0|\}$?" It just doesn't mean anything (unless you're talking about a projection rather than a measurement, but that's a different thing)

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  • $\begingroup$ Am I correct then to say that it is implicit in the author's description that each of the two states Alice sent to Bob has three measurement outcome values and each of the measurement outcome value (and it's associated post - measurement state) can be obtained using the action of the respective measurement operator. $\endgroup$
    – Physkid
    Commented Oct 19, 2023 at 14:08
  • $\begingroup$ I don't really fully understand what you're saying. Any measurement will have a fixed set of outcomes, by definition. Some outcomes might never occur for certain inputs, but that doesn't affect the general description of a POVM. If you want to know the probability of getting the $i$-th outcome when the measured state is $\rho$, you compute, with your notation, $\operatorname{tr}(E_i^\dagger E_i \rho)$. That's by definition. You can't decide to "not measure $E_3$", that doesn't mean anything. You measure a POVM with elements $\{E_i^\dagger E_i\}$, and the outcome is some $i$ $\endgroup$
    – glS
    Commented Oct 19, 2023 at 14:46
  • $\begingroup$ I think my confusion stems from the description provided by the authors. There is no reason for this setup to have 3 POVM elements, is there? It could have been 2 POVM elements associated with E1, E2, unless the authors are trying to illustrate a point. $\endgroup$
    – Physkid
    Commented Oct 19, 2023 at 15:16
  • $\begingroup$ @Physkid if you are talking about errorless state discrimination, then yes, you need three outcomes. The third outcome is neccessary to "collect" all the parts of the information that cannot be used to discriminate the states. If the states are not orthogonal, there is no possible two-outcome POVM that can distinguish them with zero error $\endgroup$
    – glS
    Commented Oct 19, 2023 at 17:18
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    $\begingroup$ @Physkid "clearer" depends on the context. For me, it's clearer to just talk about a POVM as a set of operators $E_1,..., E_n$ satisfying the usual properties. I don't quite see the point in writing them as the "square" of another matrix, unless you want to think of them as arising from the Kraus decomposition of some channel $\endgroup$
    – glS
    Commented Oct 20, 2023 at 12:31

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