# Does a CNOT quantum gate violate no cloning theorem? [duplicate]

I am a curious quantum computing learner. :)

Once observe the CNOT gate: as you can see there it converts a |+> to |-> in the top or to say in another way it clones the |->

state. So does this violate "No Cloning Theorem".

Say for example when you apply a CNOT gate to a qubit in superposition then that is indeed as the

same as cloning the qubit in superposition.

## 1 Answer

Cloning means you have a unitary transforming $$U$$ that takes $$|\psi\rangle|0\rangle$$ to the $$|\psi\rangle|\psi \rangle$$ for all quantum states $$|\psi\rangle$$. That is,

$$U\big(|\psi\rangle|0\rangle \big) = |\psi\rangle|\psi \rangle \hspace{0.5 cm} \forall \ |\psi \rangle \in \mathcal{H}$$

where $$\mathcal{H}$$ is a Hilbert space. Thus, what you shown is not a violation of the no cloning theorem.

Note that the no-cloning theorem tells us that it is impossible to clone a specific unknown quantum state, but it does not preclude the construction of a known quantum state from a known quantum state.