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.
I also got intrigued by this problem while listening to prof.Vazirani lectures on Quantum Computing.
I kind of explained to myself this "contradiction" by saying that you can "copy" the base states $|0\rangle$, $|1\rangle$, $|00\rangle$, $|11\rangle$, etc... but you cannot copy their amplitudes if they are in a superposition.
Therefore the no cloning theorem says that you cannot copy $|\psi\rangle = \alpha|0\rangle + \beta|1\rangle$ to another qubit that is now in the state $|\phi\rangle = |0\rangle$ BUT if $|\psi\rangle = |1\rangle$ you can make it became $|\psi\rangle = |0\rangle$ for instance.
Also, I think that when doing quantum computations, you are interested in the end if your Qubit is in the state $|0\rangle$ or $|1\rangle$ not in the amplitudes associated with the base states when in a superposition.
Any comments on this answer will be greatly appreciated, since I'm trying to teach myself Quantum Computing and sometimes I get stuck in some concept like this one.
