Given a quantum state, can you generate a uniform superposition over its computational basis vectors with nonzero amplitudes?

Question

Given an arbitrary $|\psi\rangle=\sum_{i=0}^n\alpha_i|i\rangle$, $K=\{i\mid \alpha_i\not=0\}$, and $k=\vert K\vert$, is it possible to generate the state $\frac{1}{\sqrt k}\sum_{i\in K}|i\rangle$? I feel like this isn't allowed by no-cloning, but I haven't made progress either way.

Answer 1

TL;DR: No. Assuming we only have a single copy of $|\psi\rangle$, the transformation is prohibited by linearity of quantum mechanics. That said, it can be done if we have multiple copies of $|\psi\rangle$, e.g. using state tomography followed by preparation.

No-go result

Suppose that $\mathcal{E}:L(\mathbb{C}^2)\to L(\mathbb{C}^2)$ is a quantum operation that achieves the desired transformation. In particular, $$ \begin{align} \mathcal{E}(|0\rangle\langle 0|)=|0\rangle\langle 0|\quad \mathcal{E}(|{+}\rangle\langle{+}|)=|{+}\rangle\langle{+}|\tag1\\ \mathcal{E}(|1\rangle\langle 1|)=|1\rangle\langle 1|\quad \mathcal{E}(|{-}\rangle\langle{-}|)=|{+}\rangle\langle {+}|\tag2. \end{align} $$ Now, consider the action of $\mathcal{E}$ on the maximally mixed state. On one hand, $$ \mathcal{E}\left(\frac{I}{2}\right)=\frac{\mathcal{E}(|0\rangle\langle 0|)+\mathcal{E}(|1\rangle\langle 1|)}{2}=\frac{I}{2}\tag3 $$ and on the other $$ \mathcal{E}\left(\frac{I}{2}\right)=\frac{\mathcal{E}(|{+}\rangle\langle {+}|)+\mathcal{E}(|{-}\rangle\langle{-}|)}{2}=|{+}\rangle\langle {+}|.\tag4 $$ The contradiction means that the desired map is not linear and hence not physical.

Pitfalls of overspecifying a linear map

In general, a linear map is uniquely determined by its action on a basis. If we specify a desired action of a linear map on a set that is strictly larger than a basis (as is done in the question and also in the no-cloning theorem), then there is a possibility that no linear map is consistent with our requirements.