Phase estimation on a superposition of eigenspaces

By [F1], write $H={⨁}_{\theta \in \mathrm{\Theta }}{E}_{\theta }$ orthogonally with projections ${\mathrm{\Pi }}_{\theta }$. By [F2], every eigenvalue has modulus one, and [F3] supplies its unique phase $\theta \in [0,1)$. Therefore ${\sum }_{\theta }{\mathrm{\Pi }}_{\theta }=I_H$ and $\psi ={\sum }_{\theta }{\mathrm{\Pi }}_{\theta }\psi$.

On a phase-register basis branch $\mathrm{\mid }j⟩$, the controlled powers act as $U^j$. On each eigenspace, $U^j{\mathrm{\Pi }}_{\theta }=\exp (2\pi ij\theta ){\mathrm{\Pi }}_{\theta }$. Applying the inverse-QFT matrix in [F4], the coefficient of $\mathrm{\mid }y⟩$ on $E_{\theta }$ is $a_y(\theta )$; hence the post-readout Kraus operator on $H$ is $K_y={\sum }_{\theta }{a}_y(\theta ){\mathrm{\Pi }}_{\theta }$. Interpreting the chronological gate list as this operator uses [F14].

Orthogonality in [F6] gives $K_y^{†}{K}_y={\sum }_{\theta }\mathrm{\mid }{a}_y(\theta ){\mathrm{\mid }}^2{\mathrm{\Pi }}_{\theta }$. Summing over $y$ and using [F5] yields ${\sum }_y{K}_y^{†}{K}_y={\sum }_{\theta }{\mathrm{\Pi }}_{\theta }=I_H$. Consequently, for a unit vector $\psi$, $\mathrm{\parallel }{K}_y\psi {\mathrm{\parallel }}^2={\sum }_{\theta }\mathrm{\mid }{a}_y(\theta ){\mathrm{\mid }}^2\mathrm{\parallel }{\mathrm{\Pi }}_{\theta }\psi {\mathrm{\parallel }}^2$; all cross terms vanish because the eigenspaces are orthogonal. The weights $\mathrm{\parallel }{\mathrm{\Pi }}_{\theta }\psi {\mathrm{\parallel }}^2$ are nonnegative and sum to one, so this is the eigenspace-weighted mixture law.

If every $\theta \in \mathrm{\Theta }$ has $Q\theta =t_{\theta }\in \{0,\dots ,Q-1\}$, the finite geometric sum in [F10] and cyclic-character orthogonality give $a_y(\theta )=1$ for $y=t_{\theta }$ and $0$ otherwise. Distinct phases have distinct labels, so $K_y$ is exactly the spectral projection for the eigenspace with label $y$, or zero if there is none.

By linearity of the circuit, its pre-readout map from $H$ to the phase register tensored with $H$ is $V:v\mapsto {\sum }_z\mathrm{\mid }z⟩\otimes K_{z}v$, with the $K_z$ from step 2.1. For any operator $X$ on $H\otimes R$, the joint output is $(V\otimes I_R)X(V^{†}\otimes I_R)$, using [F15] for the tensor adjoint. Applying the readout branch in [F13] gives $\mathrm{\mid }y⟩⟨y\mathrm{\mid }\otimes (K_y\otimes I_R)X(K_y^{†}\otimes I_R)$. Reorder the phase factor to the last position using [F11], then take its partial trace by the coordinate formula [F9]; the sole nonzero diagonal phase coordinate is $y$, so the result is exactly $(K_y\otimes I_R)X(K_y^{†}\otimes I_R)$. In particular this holds for the density operator $\rho$ in the statement. To prove complete positivity, take any finite reference factor $S$ and any nonnegative operator $X$ on $H\otimes S$. Its identity extension is $X\mapsto (K_y\otimes I_S)X(K_y^{†}\otimes I_S)$ by [F16]. Writing $A=K_y\otimes I_S$, its quadratic form is $⟨AX{A}^{†}v,v⟩=⟨X{A}^{†}v,A^{†}v⟩\ge 0$, so every finite extension is positive and each branch is completely positive by [F7].

If some phase $\theta$ is off-grid, then for every $y$, $z=\exp (2\pi i(\theta -y\mathrm{/}Q))\mathrm{\ne }1$ and $z^Q=\exp (2\pi iQ\theta )\mathrm{\ne }1$ by [F10]. The geometric formula gives $a_y(\theta )\mathrm{\ne }0$ for every $y$. Since $Q\ge 2$ and [F5] says the $Q$ squared moduli sum to one, each is strictly less than one. Thus $K_y$ acts on this eigenspace by a scalar of modulus strictly between zero and one, so $K_y^{†}{K}_y$ is not a projection; the induced system instrument is not the spectral projection instrument.

For any finite-dimensional reference $R$ (including $R=\mathbb{C}$) and any operator $X$ on $H\otimes R$, [F8] and step 2.2 give ${\sum }_y\mathrm{tr}{\mathcal{I}}_y^{(R)}(X)=\mathrm{tr}((({\sum }_y{K}_y^{†}{K}_y)\otimes I_R)X)=\mathrm{tr}X$. Taking $R=\mathbb{C}$ proves that the system maps form an instrument on $H$. For a density $\rho$ with arbitrary reference $R$, cyclicity and linearity also give $p_y=\mathrm{tr}{\mathcal{I}}_y^{(R)}(\rho )={\sum }_{\theta }\mathrm{\mid }{a}_y(\theta ){\mathrm{\mid }}^2\mathrm{tr}(({\mathrm{\Pi }}_{\theta }\otimes I_R)\rho )$. By [F12], the weights $\mathrm{tr}(({\mathrm{\Pi }}_{\theta }\otimes I_R)\rho )$ are nonnegative and sum to one. Therefore the reference outcome law is the claimed mixture. When $p_y>0$, ${\mathcal{I}}_y^{(R)}(\rho )\mathrm{/}{p}_y$ is nonnegative of trace one, and when $p_y=0$ no conditional density is defined. For a pure input the conditional vector is $K_y\psi \mathrm{/}\sqrt{p_y}$.

To witness coherence for a permitted $m\ge 1$, let $Q=2^m$ and choose the witness space $H_{\mathrm{w}\mathrm{i}\mathrm{t}}={\mathbb{C}}^2$ with orthonormal basis $u_0,u_1$. Define $U$ on this entire space by $U{u}_0=u_0$ and $U{u}_1=\exp (\pi i\mathrm{/}Q)u_1$, extending linearly. Each basis coefficient is multiplied by a unit scalar by [F3], so $U$ preserves the norm and is unitary by For an endomorphism in finite dimension, preserving lengths, preserving inner products, carrying orthonormal bases to orthonormal bases, and $T^{\ast }T=I$ are equivalent. Set $\psi =(u_0+u_1)\mathrm{/}\sqrt{2}$, which is unit. At output $y=0$, $a_0(0)=1$, while [F10] and [F3] give $a_0(1\mathrm{/}(2Q))=Q^{-1}{\sum }_{j<Q}\exp (\pi ij\mathrm{/}Q)=2\mathrm{/}[Q(1-\exp (\pi i\mathrm{/}Q))]\mathrm{\ne }0$: [F10] makes the denominator nonzero since $Q(1\mathrm{/}(2Q))=1\mathrm{/}2\mathrm{\notin }Q\mathbb{Z}$, and [F3] gives $\exp (\pi i)=-1$ for the numerator. Thus $K_0\psi =(u_0+a_0(1\mathrm{/}(2Q))u_1)\mathrm{/}\sqrt{2}$, and its rank-one operator has the nonzero cross terms $a_0(1\mathrm{/}(2Q))\mathrm{\mid }{u}_1⟩⟨u_0\mathrm{\mid }\mathrm{/}2$ and $\overline{a_0(1\mathrm{/}(2Q))}\mathrm{\mid }{u}_0⟩⟨u_1\mathrm{\mid }\mathrm{/}2$. The branch probability is positive because its $u_0$ coefficient is nonzero, so conditioning preserves these cross terms.