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Suppose we have a single qubit mixed state described by a density matrix $\rho$, and we want to find a unitary transformation that brings $\rho$ to the pure state $|0\rangle\langle 0|$. Is there a matrix $U$ such that $U^{\dagger}U=I$ and that $U \rho U^\dagger=|0\rangle\langle 0|$, independent of the matrix elements of $\rho$?. That is, if for an arbitrary single qubit $\rho$, we can always find such projection with a fixed unitary $U$, with constant matrix elements, always the same $U$ regardless of the entries of $\rho$.

EDIT: Probably this is not possible since $\rho$ lies inside the unit sphere and a unitary transformation necessarily preserves the norm, so one needs a non-unitary operation in order to do that. Is there any good reference where I could look into this issue?

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What you're asking is basically what is forbidden by the no-deleting theorem.

If you want to work out things yourself, here's a (probably incomplete) reasoning. Suppose that $\rho$ is not pure. As such $\DeclareMathOperator{tr}{tr}\tr\left(\rho^2\right)<1$. Denote $\sigma=U\rho U^{\dagger}$. Then $\tr\left(\sigma^2\right)=\tr\left(U\rho^2U^\dagger\right)=\tr\left(\rho^2\right)<1$. Thus, $\sigma$ cannot be pure.

For the task of mapping an arbitrary state $\rho$ to $|0\rangle\!\langle0|$, this is forbidden by the fact that the transformation you're asking for is unitary, and as such invertible.

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