A Controlled Global Phase Becomes a Relative Control Phase

By [F1] and [F2], $U$ satisfies $U{U}^{\ast }=U^{\ast }U=I$, hence is normal. The complex spectral theorem [F3] gives an orthonormal eigenbasis; since $H$ is nonzero, choose one unit eigenvector $\varphi$ and write $U\varphi =\mu \varphi$. Norm preservation and [F6], [F7] give $1=\mathrm{\parallel }U\varphi \mathrm{\parallel }=\mathrm{\mid }\mu \mathrm{\mid }\mathrm{\parallel }\varphi \mathrm{\parallel }=\mathrm{\mid }\mu \mathrm{\mid }$, so $\mathrm{\mid }\mu \mathrm{\mid }=1$. This chooses one vector from a finite nonempty basis and uses no choice axiom.

Choose an ordered orthonormal basis of $H$ using [L8]. Together with the control basis from [L5], it supplies the factor bases required by [L3], whose proof shows the pairing is independent of this choice. Put $P_{\pm }=\mathrm{\mid }\pm ⟩⟨\pm \mathrm{\mid }$ on the control and $E_{\pm }=P_{\pm }\otimes I_H$ on the joint register. By the Statement, $\mathrm{\mid }\pm ⟩=(\mathrm{\mid }0⟩\pm \mathrm{\mid }1⟩)\mathrm{/}\sqrt{2}$. Using [L3] and [L5], direct inner-product calculations give $⟨+,+⟩=⟨-,-⟩=1$ and $⟨+,-⟩=0$; also $\mathrm{\mid }0⟩=(\mathrm{\mid }+⟩+\mathrm{\mid }-⟩)\mathrm{/}\sqrt{2}$ and $\mathrm{\mid }1⟩=(\mathrm{\mid }+⟩-\mathrm{\mid }-⟩)\mathrm{/}\sqrt{2}$, so these vectors form an orthonormal basis. Every $v\in C\otimes H$ is a finite sum of elementary tensors; expanding each control factor in this basis and grouping terms writes $v=\mathrm{\mid }+⟩\otimes h_{+}+\mathrm{\mid }-⟩\otimes h_{-}$. The product pairing [L3] makes these two summands orthogonal, and the tensor-map definition [L4] gives $E_{+}v=\mathrm{\mid }+⟩\otimes h_{+}$ and $E_{-}v=\mathrm{\mid }-⟩\otimes h_{-}$. Hence $E_{+}+E_{-}=I$ and $⟨E_{\pm }v,v⟩=\mathrm{\parallel }{h}_{\pm }{\mathrm{\parallel }}^2\ge 0$. Applying the same decomposition to $v,w$ gives $⟨E_{\pm }v,w⟩=⟨v,E_{\pm }w⟩$; the adjoint characterization [F10] and finite-dimensional existence/uniqueness [F11] imply each effect is self-adjoint. By [F9], each effect is non-negative; together with the two-element outcome set and $E_{+}+E_{-}=I$, this is a finite POVM by [L2].

Set $c_{\mu }=(\mathrm{\mid }0⟩+\bar{\mu }\mathrm{\mid }1⟩)\mathrm{/}\sqrt{2}$. By [L5], [F6], and [F7], its squared norm is $\frac{1}{2}(⟨0,0⟩+\mathrm{\mid }\mu {\mathrm{\mid }}^2⟨1,1⟩)=1$; the cross terms vanish by orthogonality. Since $\varphi$ is unit, the product pairing [L3] makes $c_{\mu }\otimes \varphi$ unit. The controlled block rule [L1] gives $C(U)(c_{\mu }\otimes \varphi )=\frac{\mathrm{\mid }0⟩\otimes \varphi +\bar{\mu }\mathrm{\mid }1⟩\otimes U\varphi }{\sqrt{2}}=\mathrm{\mid }+⟩\otimes \varphi$.

Put $\alpha =e^{i\theta }$. By [F4], $\mathrm{\mid }\alpha \mathrm{\mid }=1$. For every $v\in H$, [F6] and [F7] give $\mathrm{\parallel }\alpha Uv{\mathrm{\parallel }}^2=\mathrm{\mid }\alpha {\mathrm{\mid }}^2\mathrm{\parallel }Uv{\mathrm{\parallel }}^2=\mathrm{\parallel }v{\mathrm{\parallel }}^2$; since $\alpha \mathrm{\ne }0$ and $U$ is invertible, $\alpha U$ is invertible, so it is unitary by [F1]. For $c_{\mu }=(\mathrm{\mid }0⟩+\bar{\mu }\mathrm{\mid }1⟩)\mathrm{/}\sqrt{2}$, [L5], [F6], [F7] and step 1.1 give $\mathrm{\parallel }{c}_{\mu }{\mathrm{\parallel }}^2=1$. The separately granted controlled block gives $C(e^{i\theta }U)(c_{\mu }\otimes \varphi )=\frac{\mathrm{\mid }0⟩+e^{i\theta }\mathrm{\mid }1⟩}{\sqrt{2}}\otimes \varphi$. Its control factor has squared norm $\frac{1}{2}(1+\mathrm{\mid }{e}^{i\theta }{\mathrm{\mid }}^2)=1$ by [F4], [F6], and [F7]. No access to this block is inferred from a call to $C(U)$; it is part of the explicit model assumption in the Given.

Let $\omega =\chi \otimes \varphi$ be either output in steps 2.1–2.2. Both factors are unit vectors, so the product norm [L3] makes $\mathrm{\parallel }\omega \mathrm{\parallel }=1$; [L6] makes ${\rho }_{\omega }=\mathrm{\mid }\omega ⟩⟨\omega \mathrm{\mid }$ a density operator by [L11]. Choose a finite orthonormal basis $(e_j)$ of $C\otimes H$ by [L8]. From [L7] and [F8], $\mathrm{tr}({\rho }_{\omega }{E}_{+})={\sum }_j⟨E_{+}{e}_j,\omega ⟩⟨\omega ,e_j⟩={\sum }_j⟨e_j,E_{+}\omega ⟩⟨\omega ,e_j⟩=⟨\omega ,E_{+}\omega ⟩$, where self-adjointness from step 1.2 gives the middle equality and Parseval [L9] gives the last equality. By the product pairing [L3] and tensor-map action [L4], $⟨\omega ,E_{+}\omega ⟩=⟨\chi ,P_{+}\chi ⟩⟨\varphi ,\varphi ⟩=\mathrm{\mid }⟨\chi ,+{⟩}_{\mathrm{r}\mathrm{o}\mathrm{o}\mathrm{t}}{\mathrm{\mid }}^2$. The finite weights normalize by [L10], so this trace value is the probability of the plus outcome.

For the output in step 2.1, $\chi =\mathrm{\mid }+⟩$, so $p_{+}=1$. For the output in step 2.2, $\chi =(\mathrm{\mid }0⟩+e^{i\theta }\mathrm{\mid }1⟩)\mathrm{/}\sqrt{2}$. Put $z=e^{i\theta \mathrm{/}2}$. By [F12], $z^2=e^{i\theta }$ and $z^{-1}=e^{-i\theta \mathrm{/}2}$; [F4] gives $\mathrm{\mid }z\mathrm{\mid }=1$ and $z=\cos (\theta \mathrm{/}2)+i\sin (\theta \mathrm{/}2)$, so $z^{-1}=\bar{z}$ and $z+z^{-1}=2\cos (\theta \mathrm{/}2)$. Hence $1+e^{i\theta }=z(z+z^{-1})=2z\cos (\theta \mathrm{/}2)$ and $p_{+}=\mathrm{\mid }1+e^{i\theta }{\mathrm{\mid }}^2\mathrm{/}4={\cos }^2(\theta \mathrm{/}2)$.

By [F5], $\theta \mathrm{\notin }2\pi \mathbb{Z}$ means $z=e^{i\theta }\mathrm{\ne }1$. Since $\mathrm{\mid }z\mathrm{\mid }=1$, $\mathrm{\mid }1+z{\mathrm{\mid }}^2=2+2\mathrm{Re}z<4$: equality would force $\mathrm{Re}z=1$ and then $\mathrm{Im}z=0$, hence $z=1$. Thus the probability from step 4.1 is strictly below one, while the other is one; the two outcome laws differ.

If $\theta \in 2\pi \mathbb{Z}$, [F5] gives $e^{i\theta }=1$, so the two controlled block maps are identical by [L1]. The operators therefore produce the same state for every common input and hence the same outcome law for the specified POVM.