 Solution

Since $U$ is a unitary operator, its eigenstates form an orthonormal basis. Expand $|\xi ⟩={\sum }_j{c}_j|v_j⟩$, with eigenphases ${\varphi }_j=y_j/2^n$. By linearity, the unmeasured quantum phase estimation output isThis is generally an entangled state, not a phase label attached to an unchanged pure system state. The distinct eigenvalues give distinct phase labels, so a computational basis measurement yields $y_j$ with Born rule probability $|c_j{|}^2$ and leaves $|v_j⟩$ in the system register. Before measurement, all relative phases remain coherent; that is essential for the following spectral transformation. If phases were degenerate, a measured label would instead select the corresponding eigenspace component.