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I have some difficulty with understanding no-cloning theorem. Simply speaking, according to the theorem, it is not possible to copy a quantum state.
On the other hand, CNOT gate can be used as so-called fan-out gate which purpose is to copy one qubit to another one, previously in state $|0\rangle$.
It seems that these two facts negate each other.
My question: How is no-cloning theorem compatible with the fact that fan-out gate works?
By "copying a quantum state", we mean that we cannot take
$$|\psi⟩|0⟩=\alpha|00⟩+\beta|10⟩$$
into
$$|\psi⟩|\psi⟩=(\alpha|0⟩+\beta|1⟩)(\alpha|0⟩+\beta|1⟩)=\alpha^2|00⟩+\alpha\beta|01⟩+\beta\alpha|10⟩+\beta^2|11⟩$$
for arbitrary single qubit state $|\psi⟩=\alpha|0⟩+\beta|1⟩$. Notice that this resulting two-qubit state $|\psi⟩|\psi⟩$ is still separable.
But in the case of CNOT gate, it takes $|\psi⟩|0⟩$ to
$$CNOT|\psi⟩|0⟩=CNOT(\alpha|00⟩+\beta|10⟩)=\alpha|00⟩+\beta|11⟩.$$
As you can see, the results are different. The resulting two-qubit state from CNOT is now entangled (not separable).
CNOT gate does not copy qubit states; rather it creates entanglement between them.
As a complement to haoyu's answer, it is also worth noting that the no-cloning theorem does not forbid ANY cloning. Instead, cloning some states is ok.
In other words, what the no-cloning theorem says is that you cannot clone ARBITRARY states with a certain cloning device.
In fact, a cloning device can only clone states that are orthogonal to one another. For example, if you clone $|0\rangle$ and $|1\rangle$, then you cannot clone $a|0\rangle+b|1\rangle$ where $ab\neq$0.
