Phase-estimation finite outcome law

Before the controlled powers, the state is $Q^{-1\mathrm{/}2}{\sum }_{x=0}^{Q-1}\mathrm{\mid }x⟩\otimes \psi$ by [F2]. Write $x={\sum }_j{c}_j{2}^j$, where $c_j$ is the bit at wire $q_{m-1-j}$ by [F1]. The controlled operation on that wire applies $U^{2^j}$ exactly when $c_j=1$; multiplying the eigenvalues gives ${\prod }_j\exp (2\pi i\theta c_j{2}^j)=\exp (2\pi i\theta x)$. Hence the state after the controlled-power stage is $Q^{-1\mathrm{/}2}{\sum }_x\exp (2\pi i\theta x)\mathrm{\mid }x⟩\otimes \psi$. Reading this gate sequence as an operator on the ordered register uses [F9].

The inverse QFT matrix coefficient from input $x$ to output $y$ is $Q^{-1\mathrm{/}2}\exp (-2\pi ixy\mathrm{/}Q)$, by [F3] and the adjoint convention. Applying it to step 1.1 gives $\chi ={\sum }_{y=0}^{Q-1}{a}_y\mathrm{\mid }y⟩\otimes \psi$, where $a_y=Q^{-1}{\sum }_{x=0}^{Q-1}\exp (2\pi ix(\theta -y\mathrm{/}Q))$. The sign is positive in the eigenphase and negative in the inverse-QFT coefficient, yielding the displayed difference $\theta -y\mathrm{/}Q$. The operator ordering is the one in [F9].

The vector $\chi$ is normalized. Expand the finite squared moduli using [F7]. For every real $t$, [F10] gives $\overline{\exp (it)}=\exp (-it)$, and the exponential addition law combines each conjugate pair of phases; interchanging the finite sums then gives ${\sum }_y\mathrm{\mid }{a}_y{\mathrm{\mid }}^2=Q^{-2}{\sum }_{x,x^{\mathrm{\prime }}}\exp (2\pi i(x-x^{\mathrm{\prime }})\theta ){\sum }_y\exp (-2\pi i(x-x^{\mathrm{\prime }})y\mathrm{/}Q)$. By [F6], the inner sum is $Q$ when $x=x^{\mathrm{\prime }}$ and $0$ otherwise, because $0\le x,x^{\mathrm{\prime }}<Q$; therefore ${\sum }_y\mathrm{\mid }{a}_y{\mathrm{\mid }}^2=Q^{-2}Q\cdot Q=1$. Since $\mathrm{\parallel }\psi \mathrm{\parallel }=1$ and the $\mathrm{\mid }y⟩$ are orthonormal, $\mathrm{\parallel }\chi \mathrm{\parallel }=1$.

The ray-projector theorem [F5] makes $\rho =\mathrm{\mid }\chi ⟩⟨\chi \mathrm{\mid }$ a density operator. The phase-register projective outcome has projector $P_y$ from [F4], so it extracts $a_y\mathrm{\mid }y⟩\otimes \psi$. By [F5] and [F8], $\mathrm{tr}(\rho P_y)=\mathrm{tr}(P_y\rho )=\mathrm{tr}(\mathrm{\mid }{P}_y\chi ⟩⟨\chi \mathrm{\mid })=⟨P_y\chi ,\chi {⟩}_{\mathrm{r}\mathrm{o}\mathrm{o}\mathrm{t}}=\mathrm{\parallel }{P}_y\chi {\mathrm{\parallel }}^2=\mathrm{\mid }{a}_y{\mathrm{\mid }}^2$, where the penultimate equality uses $P_y=P_y^{\ast }=P_y^2$. The outcome law in [F4] therefore gives this probability to the classical label $y$. Substituting the value of $a_y$ from step 2.1 gives exactly the displayed $P_{\theta }(y)$.

The finite sum in the displayed law is defined for every $y,\theta$, including when $Q\theta$ is an integer. For its geometric-sum specialization, the denominator is zero exactly when $z=\exp (2\pi i(\theta -y\mathrm{/}Q))=1$, equivalently $Q\theta -y\in Q\mathbb{Z}$; in every other case Finite geometric sum for phase-estimation amplitudes gives the nonzero-denominator quotient. Thus the stated squared-sum law handles all phases without an undefined quotient.