# Joey Marcellino comments on Viliam’s Shortform

• Just to (hopefully) make the distinction a bit more clear:

A true copying operation would take |psi1>|0> to |psi1>|psi1>; that’s to say, it would take as input one qubit in an arbitrary quantum state and a second qubit in |0>, and output two qubits in the same arbitrary quantum state that the first qubit was in. For our example, we’ll take |psi1> to be an equal superposition of 0 and 1: |psi1> = |0> + |1> (ignoring normalization).

If CNOT is a copying operation, it should take (|0> + |1>)|0> to (|0> + |1>)(|0> + |1>) = |00> + |01> + |10> + |11>. But as you noticed, what it actually does is create an entangled state (in this case, a Bell state) that looks like |00> + |11>.

So in some sense yes, the forbidden thing is to have a state copied and not entangled, but more importantly in this case CNOT just doesn’t copy the state, so there’s no tension with the no-cloning theorem.

• Thank you!

Some context: I am a “quantum autodidact”, and I am currently reading a book Q is for Quantum, which is a very gentle, beginner-friendly introduction to quantum computing. I was thinking how it relates to the things I have read before, and then I noticed that I was confused. I looked at Wikipedia, which said that CNOT does not violate the no-cloning theorem… but I didn’t understand the explanation why.

I think I get it now. |00> + |11> is not a copy (looking at one qubit collapses the other), |00> + |01> + |10> + |11> would be a copy (looking at one qubit would still leave the other as |0> + |1>).