# 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 $$\langle \psi | \phi \rangle = \langle \psi | \phi \rangle^2$$.)

The proof does not seem to rule out the case that there exists a specific U that can clone only the specific state |ψ⟩.

That's because you can clone specific states. Cloning is only impossible if the set of possible input states includes a pair of states that are not orthogonal.

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 '19 at 22:56
• @abrahimladha As long as the set of states is orthogonal, you can clones items from it. Bell states, superpositions, anything, just has to be orthogonal. Nov 16 '19 at 0:37
• 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 '19 at 2:29
• To "clone" a known state like |+> or its orthogonal state like |->, you can prepare the state from |0> with a unitary gate like Hadamard. To "clone" a bell state, you can prepare it directly. This will be a new state identical to the given one and they are disentangled from each other. Of course, one might argue it is not really a clone. Nov 16 '19 at 6:33

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!