Premise of the proof of the No-Cloning Theorem

I have seen two similar proofs of the no-cloning theorem. They assume (to the contrary) that there exists a unitary operator $$U$$ such that $$U |\psi\rangle |0 \rangle = | \psi \rangle | \psi \rangle$$, For any possible $$|\psi\rangle$$. The proof does not seem to rule out the case that there exists a specific $$U$$ that can clone only the specific state $$| \psi \rangle$$. Discussion of the no-cloning theorem implies that there cannot be a specific $$U$$, which can only clone a certain state, even when the proof only proves that there cannot be a general $$U$$ which can clone any state. Is there a proof of this specific case somewhere? Or maybe I am missing something from the original proof.

(I am referencing the one in Nielsen and Chuang which ends with the contradicition that $$\langle \psi | \phi \rangle = \langle \psi | \phi \rangle^2$$.)

For example, here is a circuit that performs $$|\psi⟩ \to |\psi⟩|\psi⟩$$ as long as $$|\psi\rangle$$ is promised to be exactly $$|0\rangle$$ or exactly $$|1\rangle$$ and never anything else:
• Thank you for the reply. I kind of assumed you could already clone $|0 \rangle$ or $|1 \rangle$, what about any other states? A bell state? Is there a list of possible clonable states, and is there anything on that list besides those two. Nov 15, 2019 at 22:56
• As you say, if you can clone some state, you can clone any state orthogonal state to it, right. But this is under the assumed premise that there is a state you can already clone. We know that $|0\rangle$ is clonable, so is everything orthogonal to it. But if there is there anything else clonable? Right its possible this set contains only $| 0 \rangle, | 1 \rangle$. so those are the only states clonable. To show that a bell state is clonable, we would have to show that a state orthogonal to it is clonable, but this might not be true. Nov 16, 2019 at 2:29
The real missing keyword in the stating the theorem is "arbitrary unknown state"! If you have some information about $$|\psi\rangle$$, i.e. specific state, then perhaps you can reconstruct that state!