1
$\begingroup$

This is related to this but a little different, so bear with me.

When is a fanout not violating the no-cloning theorem? It seems accepted that a fanout (or at least what seems to be one in my novice understanding of computing) can be used in certain instances despite this seemingly cloning the state, which I know is impossible. For example, in a CNOT gate, to use the XOR gate, which takes two inputs and outputs one, the top qubit must be duplicated. How is this possible?

$\endgroup$
  • $\begingroup$ Does it help if I say that if $|\psi\rangle=\alpha|0\rangle+\beta|1\rangle$, then $|\psi\rangle|\psi\rangle$ is not the same as $\alpha|00\rangle+\beta|11\rangle$? $\endgroup$ – Mark S Jun 30 at 20:27
  • $\begingroup$ That I get. What I don't understand is not the final product of the gate as a whole, but more what must be done to perform the operation. $\endgroup$ – SpaceChicken Jun 30 at 21:50
  • 1
    $\begingroup$ The top qubit isn’t “duplicated”. It’s the same qubit. I think of it as a “controlled NOT” much more than an XOR. The top qubit controls the negation of the bottom qubit. After controlling, it’s done its job. An XOR classically takes two inputs and outputs 1. But the CNOT takes two and outputs two. We think of it as fanout because the bottom qubit could initially be zero; afterward it’s entangled with the top. $\endgroup$ – Mark S Jul 1 at 2:46
  • $\begingroup$ That makes a lot more sense to me. I think I just learned it wrong. Thank you. $\endgroup$ – SpaceChicken Jul 1 at 3:38

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.