Global Phase Under Control

The diagonal blocks of $C(\lambda U)$ are $I_H$ and $\lambda U$. Multiplying the blocks of $(P_{\lambda }\otimes I_H)C(U)$ gives $I_H$ and $\lambda U$, so the two full matrices are equal exactly.

Under the whole-output update specified in the Given, $(\lambda I)\rho (\lambda I{)}^{\ast }=\lambda \bar{\lambda }\rho =\rho$ because $\mathrm{\mid }\lambda \mathrm{\mid }=1$. Hence the whole-output scalar leaves the density operator unchanged.

By [F2], $U$ and $\lambda U$ are phase-equivalent as target matrices. If $\lambda \mathrm{\ne }1$, $P_{\lambda }$ is not a scalar multiple of the identity: its diagonal entries are $1$ and $\lambda$. Thus step 1.1 places a phase on only the active control block. Left multiplication by $P_{\bar{\lambda }}\otimes I_H$ changes the blocks back to $I_H,U$, so it restores $C(U)$ exactly. For $\lambda =1$ no correction is needed.

By [F4], each defined measurement weight is $\mathrm{tr}(\rho E_a)$. The equality of density operators in step 1.2 makes these weights identical for every outcome $a$. Under the normalized-law conclusion [F5], the two probability distributions therefore agree. The global scalar in step 1.2 preserves the whole measurement law, while the controlled target scalar in step 1.1 does not become a whole-output scalar.