3
$\begingroup$

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?

$\endgroup$
1

2 Answers 2

11
$\begingroup$

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.

$\endgroup$
1
  • $\begingroup$ 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. $\endgroup$
    – DaftWullie
    Dec 9, 2019 at 6:18
2
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.