# Is there a quantum gate that can turn any superposition $|\varphi \rangle$ into a unit column vector $|00\cdots 1\cdots 0\rangle$

that is $${{U}_{f}}|\varphi \rangle =|00\cdots 1\cdots 00\rangle$$,where $$|00\cdots 1\cdots 0\rangle$$ is a random unit vector

eg.$${{U}_{f}}(\sqrt{\frac{1}{3}}|0\rangle +\sqrt{\frac{1}{3}}|1\rangle +\sqrt{\frac{1}{3}}|2\rangle )=|4{{\rangle }_{10}}=|00000100{{\rangle }_{2}}$$or$${{U}_{f}}(\sqrt{\frac{1}{3}}|0\rangle +\sqrt{\frac{1}{3}}|1\rangle +\sqrt{\frac{1}{3}}|2\rangle )=|2{{\rangle }_{10}}=|00000010{{\rangle }_{2}}$$

• – glS
Jun 8 at 17:22

For a given state $$|\varphi\rangle$$, it is always possible to find a gate $$U_\varphi$$ such that $$U_\varphi|\varphi\rangle=|i\rangle_{10}$$ for some given $$i$$.
However, it is not possible to find a single gate $$U$$ such that $$U|\varphi\rangle=|i\rangle_{10}$$ for any $$|\varphi\rangle$$ and some $$i$$. There are several reasons for that:
• $$U$$ has to be bijective. This means that for a given $$i$$, there must be a single state $$|\varphi\rangle$$ such that $$U|\varphi\rangle=|i\rangle_{10}$$
• $$U$$ has to be unitary. This (amongst other things) means that since $$U|\varphi\rangle=|i\rangle_{10}$$ and $$U|\psi\rangle|j\rangle_{10}$$ are orthogonal (since $$i\neq j$$ as per the previous point), $$|\varphi\rangle$$ and $$|\psi\rangle$$ also have to be orthogonal. Thus, if $$U|0\rangle=|i\rangle_{10}$$ and $$U\left(\frac{|0\rangle+|1\rangle}{\sqrt{2}}\right)=|j\rangle_{10}$$, for $$i$$ and $$j$$ being part of the computational basis, this would mean that $$|0\rangle$$ and $$\frac{|0\rangle+|1\rangle}{\sqrt{2}}$$ are orthogonal, which isn't the case.