I'm currently grappling with a challenge in comprehending the inverse channel of the Pauli 4 POVM. The POVM elements are defined as follows: $ M_0 = \frac{1}{3} |0\rangle\langle0|, \quad M_1 = \frac{1}{3} |1 \rangle\langle1|, \quad M_2 = \frac{1}{3} |l \rangle\langle l|, \quad M_3 = \frac{1}{3} (|1 \rangle\langle1| + |- \rangle\langle-| + |r \rangle\langle r|). $ In their paper [1], the authors claim that due to the rank-2 property of $M_3$, we can deduce the following: "When the outcome corresponds to index 3, we consider the eigenvector $|t\rangle$ associated with the eigenvalue $\frac{1}{2} (1 + \frac{1}{\sqrt{3}})$ instead of the other eigenvector associated with $\frac{1}{2} (1 - \frac{1}{\sqrt{3}})$". However, I'm uncertain about the reasoning behind this assertion. I would greatly appreciate some clarification on this matter. In addition, I am not sure how they did find the eigenvalue above.

Thank you. Reference: [1] https://arxiv.org/abs/2105.05992


1 Answer 1


The authors say earlier

If a particular POVM element is not rank one, $|\psi_a\rangle$ can be taken as the eigenvector corresponding to the highest eigenvalue of $M_a$.

Here, the authors are saying that $M_3$ has two eigenvalues, $(1\pm 1/\sqrt{3})/2$. The larger eigenvalue is the one with the $+$ sign, so they use the eigenvector corresponding to that eigenvalue in their construction of the measurement channel.


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