Teleportation outcomes are equiprobable

For each $(a,b)$, let $P_{ab}$ be the diagonal projection onto the two basis vectors $\mathrm{\mid }ab0⟩,\mathrm{\mid }ab1⟩$ of the ordered three-wire register. These four projections are self-adjoint idempotents, their ranges are pairwise orthogonal, and their sum is the identity because the eight orthonormal computational basis vectors from [F8] are partitioned by their first two bits. Hence they form a projective measurement as in [F2].

Write the post-gate vector in [F1] as $\mathrm{\Omega }={\sum }_i{c}_i{e}_i$ in the orthonormal basis [F4]. For the matrix $\rho$ with entries ${\rho }_{ij}=c_i\overline{c_j}$, self-adjointness is immediate; for any coordinate vector $z$, $⟨\rho z,z⟩=\mathrm{\mid }{\sum }_j{z}_j\overline{c_j}{\mathrm{\mid }}^2\ge 0$. The matrix [F6] sends $(\alpha ,\beta )$ to $(\beta ,\alpha )$ under $X$ and to $(\alpha ,-\beta )$ under $Z$, so each preserves $\mathrm{\mid }\alpha {\mathrm{\mid }}^2+\mathrm{\mid }\beta {\mathrm{\mid }}^2$ and every $X^b{Z}^a$ preserves norm. The branch vectors in [F1] have orthogonal $AB$ labels, hence $\mathrm{tr}(\rho )={\sum }_i\mathrm{\mid }{c}_i{\mathrm{\mid }}^2=\mathrm{\parallel }\mathrm{\Omega }{\mathrm{\parallel }}^2=\mathrm{\parallel }\psi {\mathrm{\parallel }}^2=1$ by [A1] and [F7]. Thus $\rho$ is a density operator by [F9], since the preceding calculation establishes nonnegativity and trace one.

By [F2] and [F7], the probability weight of label $(a,b)$ is $\mathrm{tr}(\rho P_{ab})=\mathrm{\mid }{c}_{ab0}{\mathrm{\mid }}^2+\mathrm{\mid }{c}_{ab1}{\mathrm{\mid }}^2$, the sum of the two diagonal entries selected by $P_{ab}$. The branch vector in [F1] has these two coefficients equal to those of $\frac{1}{2}{X}^b{Z}^a\mathrm{\mid }\psi ⟩$, so this sum is $\frac{1}{4}\mathrm{\parallel }{X}^b{Z}^a\psi {\mathrm{\parallel }}^2$.

The matrices in [F6] are unitary: direct multiplication gives $X^{\ast }X=I$ and $Z^{\ast }Z=I$. Hence $\mathrm{\parallel }{X}^b{Z}^a\psi {\mathrm{\parallel }}^2=\mathrm{\parallel }\psi {\mathrm{\parallel }}^2=1$, so every weight in step 2.1 is $1\mathrm{/}4$. By [F3] these are the outcome probabilities, and the four weights sum to one.