# Cloning quantum states with a device that distinguishes between two non-orthogonal quantum states

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?

• 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! – Mithrandir24601 Jun 21 '19 at 6:52

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.
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.