Why is it impossible to make a mixed-state qubit into a pure-state qubit?

Question

How is it impossible to make a mixed-state qubit (having Bloch vector of length $l < 1$) into a pure-state qubit (having Bloch vector of length $l' = 1$) via a quantum operation that is invertible?

Can anyone give a proof of it?

Answer 1

It's possible to have an operation sending a specific mixed input into a specific pure output. Trivial case in point: just throw away the input state and replace it with $|0\rangle$ (or any other state). Mathematically, that's the replacement channel $\Phi(\rho)=\operatorname{tr}(\rho)|0\rangle\!\langle 0|$.

It's not possible to have an operation sending any mixed input into a pure state, with the only exception of a replacement channel. One way to see it easily is to assume $\Phi$ is such an operation. Then it sends $\mathbb{P}_0\equiv |0\rangle\!\langle0|$ and $\mathbb{P}_1$ into pure states, meaning $\Phi(\mathbb{P}_0)$ and $\Phi(\mathbb{P}_1)$ are pure states. But then $\Phi(\frac12(\mathbb{P}_0+\mathbb{P}_1))=\frac12(\Phi(\mathbb{P}_0)+\Phi(\mathbb{P}_1))$, and thus the mixed state $\frac12(\mathbb{P}_0+\mathbb{P}_1)$ is sent to a mixture of two pure states. Therefore, unless $\Phi(\mathbb{P}_0)=\Phi(\mathbb{P}_1)$, you built a mixed state that is not sent to a pure state. And if $\Phi(\mathbb{P}_0)=\Phi(\mathbb{P}_1)$, then repeat the above reasoning with any other pair of pure states that are not sent to the same output, and you got another mixed state that is not sent to a pure state. The only remaining possibility is that $\Phi$ sends every pure state into the same output, in which case it's a replacement channel of the form $\Phi(\rho)=\operatorname{tr}(\rho)\mathbb{P}_\psi$.

Answer 2

Imagine a quantum operation that takes any state with a Bloch vector of length $l<1$ and converts it into a state with a Bloch vector pointing in the same direction but with $l=1$. This operation would not know what to do with a vector of length $l=0$, because it points in all directions, so the operation cannot be defined.