# Inverse channel of the Pauli 4 POVM

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

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.