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


1 Answer 1


It depends on what you mean.

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.

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