Here, we'll question the existence of a quantum operation that maps every quantum state to an orthogonal state relative to its input. More specifically: is there a one-qubit unitary operation $U$ that maps each pure state $\ket{\psi}$ to some state $U\ket{\psi}=\ket{\psi'}$ such that $\ip{\psi}{\psi'}=0$.

I claim that there does not exist such a unitary!

Suppose, for the sake of contradiction, that such a unitary $U$ did exist. Consider an arbitrary one-qubit state $\ket{\psi}$. Let $U\ket{\psi}=\ket{\psi'}$ so that $\ip{\psi}{\psi'}=0$. Moreover, let $U\ket{\psi'}=\ket{\psi''}$ so that $\ip{\psi''}{\psi'}=0$ as well by the assumption of the existence of such a unitary. Now, consider the state $\ket{\phi}=\frac{1}{\sqrt{2}}(\ket{\psi}+\ket{\psi'})$, and let $U\ket{\phi}=\ket{\phi'}$ so that $\ip{\phi}{\phi'}=0$. Then it must also follow that

\[

U\ket{\phi}=\frac{1}{\sqrt{2}}(U\ket{\psi}+U\ket{\psi'})=\frac{1}{\sqrt{2}}(\ket{\psi'}+\ket{\psi''})=\ket{\phi'}.

\]

This implies that the inner product of $\ket{\phi}$ and $\ket{\phi}$ must also satisfy

\[\begin{align*}

\ip{\phi}{\phi'}&=\frac{1}{2}(\bra{\psi'}+\bra{\psi''})(\ket{\psi'}+\ket{\psi''}) \\

&=\frac{1}{2}(\ip{\psi'}{\psi'}+\ip{\psi'}{\psi''}+\ip{\psi''}{\psi'}+\ip{\psi''}{\psi''}) \\

&=\frac{1}{2}(1+0+0+1) \\

&=1,

\end{align*}\]

but then $1=\ip{\phi}{\phi'}=0$ which a contradiction. Therefore, no such unitary can exist.

I claim that there does not exist such a unitary!

Suppose, for the sake of contradiction, that such a unitary $U$ did exist. Consider an arbitrary one-qubit state $\ket{\psi}$. Let $U\ket{\psi}=\ket{\psi'}$ so that $\ip{\psi}{\psi'}=0$. Moreover, let $U\ket{\psi'}=\ket{\psi''}$ so that $\ip{\psi''}{\psi'}=0$ as well by the assumption of the existence of such a unitary. Now, consider the state $\ket{\phi}=\frac{1}{\sqrt{2}}(\ket{\psi}+\ket{\psi'})$, and let $U\ket{\phi}=\ket{\phi'}$ so that $\ip{\phi}{\phi'}=0$. Then it must also follow that

\[

U\ket{\phi}=\frac{1}{\sqrt{2}}(U\ket{\psi}+U\ket{\psi'})=\frac{1}{\sqrt{2}}(\ket{\psi'}+\ket{\psi''})=\ket{\phi'}.

\]

This implies that the inner product of $\ket{\phi}$ and $\ket{\phi}$ must also satisfy

\[\begin{align*}

\ip{\phi}{\phi'}&=\frac{1}{2}(\bra{\psi'}+\bra{\psi''})(\ket{\psi'}+\ket{\psi''}) \\

&=\frac{1}{2}(\ip{\psi'}{\psi'}+\ip{\psi'}{\psi''}+\ip{\psi''}{\psi'}+\ip{\psi''}{\psi''}) \\

&=\frac{1}{2}(1+0+0+1) \\

&=1,

\end{align*}\]

but then $1=\ip{\phi}{\phi'}=0$ which a contradiction. Therefore, no such unitary can exist.

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