Standalone Global Phase Leaves Pure-State POVM Probabilities Unchanged

Positive definiteness gives $\mathrm{\parallel }x\mathrm{\parallel }>0$, so $u$ is defined. By [F1] and [F2], $\mathrm{\parallel }u{\mathrm{\parallel }}^2=⟨x\mathrm{/}\mathrm{\parallel }x\mathrm{\parallel },x\mathrm{/}\mathrm{\parallel }x\mathrm{\parallel }⟩=⟨x,x⟩\mathrm{/}\mathrm{\parallel }x{\mathrm{\parallel }}^2=1$. Again using linearity in the first argument and conjugate-linearity in the second, $\mathrm{\parallel }\lambda u{\mathrm{\parallel }}^2=⟨\lambda u,\lambda u⟩=\mathrm{\mid }\lambda {\mathrm{\mid }}^2⟨u,u⟩=1$. Taking nonnegative square roots gives both norms equal to one.

For every $z\in H$, the root-first conversion [L3] gives $\mathrm{\mid }\lambda u⟩⟨\lambda u\mathrm{\mid }z=⟨z,\lambda u{⟩}_{\mathrm{r}\mathrm{o}\mathrm{o}\mathrm{t}}\lambda u=\bar{\lambda }\lambda ⟨z,u{⟩}_{\mathrm{r}\mathrm{o}\mathrm{o}\mathrm{t}}u=\mathrm{\mid }u⟩⟨u\mathrm{\mid }z$. Hence the two rank-one operators are equal.

By [L4], the common operator is a density operator, so the outcome law [L2] defines its weight for each effect $E_a$. Since the density operator is identical for $u$ and $\lambda u$, those weights agree; [L5] proves the finite family of weights is a probability distribution.

Since $x=\mathrm{\parallel }x\mathrm{\parallel }u$ with $\mathrm{\parallel }x\mathrm{\parallel }$ a nonnegative real, [F1] and [L3] give $\mathrm{\mid }x⟩⟨x\mathrm{\mid }=\mathrm{\parallel }x{\mathrm{\parallel }}^2\mathrm{\mid }u⟩⟨u\mathrm{\mid }$. In any basis, its matrix is therefore $\mathrm{\parallel }x{\mathrm{\parallel }}^2$ times the matrix of $\mathrm{\mid }u⟩⟨u\mathrm{\mid }$; summing the diagonal entries in that basis as in [F3] gives $\mathrm{tr}(\mathrm{\mid }x⟩⟨x\mathrm{\mid })=\mathrm{\parallel }x{\mathrm{\parallel }}^2\mathrm{tr}(\mathrm{\mid }u⟩⟨u\mathrm{\mid })=\mathrm{\parallel }x{\mathrm{\parallel }}^2$ by [L4]. If $\mathrm{\parallel }x\mathrm{\parallel }\mathrm{\ne }1$, it is not trace one and hence is not a density operator by [L1].