Pauli measurement projectors and total probability

For $s=(-1{)}^b$, $P_b^{†}=P_b$ because $s$ is real and $M^{†}=M$. In any ordered basis write $M=(m_{ij})$. By [F2], $(P_b^2{)}_{ij}=\frac{1}{4}{\sum }_k({\delta }_{ik}+s{m}_{ik})({\delta }_{kj}+s{m}_{kj})=\frac{1}{4}({\delta }_{ij}+2s{m}_{ij}+(M^2{)}_{ij})$. Since $M^2=I_H$, this equals $(P_b{)}_{ij}$. Thus both $P_b$ are self-adjoint idempotents.

Direct multiplication gives $P_0+P_1=I$ and $P_0{P}_1=(I-M^2)\mathrm{/}4=0=P_1{P}_0$. Hence their ranges are orthogonal and sum to the full register.

Put $Q_b=P_b\rho P_b$. Self-adjointness of $P_b$ and $\rho$ gives $⟨Q_{b}u,v⟩=⟨\rho P_{b}u,P_{b}v⟩=⟨u,Q_{b}v⟩$, so $Q_b$ is self-adjoint. For every vector $v$, $⟨Q_{b}v,v⟩=⟨\rho P_{b}v,P_{b}v⟩\ge 0$, using the root-linear-first inner-product convention. Therefore $Q_b$ is positive semidefinite.

By cyclicity [F3] and $P_b^2=P_b$, $p_b=\mathrm{Tr}(\rho P_b)=\mathrm{Tr}(P_b\rho P_b)=\mathrm{Tr}(Q_b)$. In the orthonormal eigenbasis supplied by [F4], the diagonal entries of $Q_b$ are nonnegative; [F3] and [F5] therefore give $p_b=\mathrm{Tr}(Q_b)\ge 0$. Linearity of the finite diagonal sum and $P_0+P_1=I$ give $p_0+p_1=\mathrm{Tr}(\rho )=1$.