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


2 Answers 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:

just a cnot

  • $\begingroup$ 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. $\endgroup$ Commented Nov 15, 2019 at 22:56
  • $\begingroup$ @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. $\endgroup$ Commented Nov 16, 2019 at 0:37
  • $\begingroup$ 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. $\endgroup$ Commented Nov 16, 2019 at 2:29
  • $\begingroup$ 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. $\endgroup$
    – czwang
    Commented Nov 16, 2019 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!


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