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I'm aware that this is basically a duplicate question, but I don't have any rep in this community so I can't comment on it, and I don't think I should "answer" that question with my own question:

No-cloning theorem and distinguishing between two non-orthogonal quantum states

Exercise 1.2: Explain how a device which, upon input of one of two non-orthogonal quantum states $|ψ⟩$ or $|ϕ⟩$ correctly identified the state, could be used to build a device which cloned the states $|ψ⟩$ and $|ϕ⟩$, in violation of the no-cloning theorem. Conversely, explain how a device for cloning could be used to distinguish non-orthogonal quantum states.

The first part isn't quite trivial to me. Since the device can distinguish both $|\psi\rangle$ and $|\phi\rangle$ with certainty, they are effectively orthogonal states, and thus can be cloned when the device measures in the "basis" $\{|\psi\rangle,|\phi\rangle\}$. Is this correct?

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  • $\begingroup$ Welcome to Quantum Computing SE! It's generally not a good idea to answer questions with another question, so a new question is better anyway - I'd argue they're not true duplicates (even if they're related) as you're asking for something that the other question takes for granted, so it's all good! $\endgroup$ – Mithrandir24601 Jun 21 at 6:52
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Forget that you know that non-orthogonal states cannot be distinguished, as this is what you're trying to prove. It's like a proof by contradiction.

The question is talking about a hypothetical device that could, if it existed, distinguish between $|\psi\rangle$ and $|\phi\rangle$. Assume it exists, and prove that it gives you cloning (I know which state I had because I could distinguish them, so I can make arbitrarily many copies). But that's in contradiction with no-cloning, so the hypothetical device must be impossible.

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No, $|\psi\rangle$ and $|\phi\rangle$ are non-orthogonal by assumption, they can't be effectively (whatever this means) orthogonal. The device just gives us the answer $\phi$ or $\psi$ (you can think it returns label and destroys the input state), so we can prepare $|\psi\rangle$ or $|\phi\rangle$ separately (or many copies of it). It is that trivial.

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  • $\begingroup$ Yeah, the "effectively" was pretty handwavey - I was grasping for straws. $\endgroup$ – Rourke Sekelsky Jun 21 at 21:09

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