How is the no-cloning theorem compatible with the fact that fan-out gates work?

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.

• It is perhaps worth adding that controlled not does perform fanout if your input is in the computational basis, but only if it is in the computational basis. Dec 9, 2019 at 6:18

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.

• You are correct. In general, any two orthogonal states can be cloned knowing prior which are the two orthogonal states we are dealing with. It doesn't just have to be basis states! Jan 17 at 23:59