Is it correct to say that for an $n$-qubit system, we can clone all kinds of pure non-entangled states, without violating the no-cloning theorem?

That is, is the correct interpretation for the proof of cloning theorem shown here, to be that if such a cloning operator $U$ exists, such an operator $U$ would not be a linear operator, but nonetheless $U$ can still act on only non-entangled, mutually orthogonal states?

Entangled linear superpositions of those states in the set are not included instead, because then $U$ would not be a linear operator.

Thank you for your help!

  • 3
    $\begingroup$ Please clarify your specific problem or provide additional details to highlight exactly what you need. As it's currently written, it's hard to tell exactly what you're asking. $\endgroup$
    – Community Bot
    Sep 8, 2021 at 15:47
  • 2
    $\begingroup$ Hi 王天羿, I have tried editing your post for readability. Please let me know if this is the spirit of your question! $\endgroup$
    – Mark S
    Sep 8, 2021 at 21:55
  • $\begingroup$ @MarkS Yep! Thanks! $\endgroup$
    – QubitTy
    Sep 9, 2021 at 2:31
  • $\begingroup$ @Community Hi, is my question ready to be open to answer? $\endgroup$
    – QubitTy
    Sep 9, 2021 at 3:19
  • $\begingroup$ I don't think entanglement is important for no-cloning. Even with a single qubit and only pure states you have the no-cloning theorem. $\endgroup$
    – M. Stern
    Sep 9, 2021 at 7:04

1 Answer 1


It's not the non-entangled part that's important in your statement. The important part is that any mutually orthogonal states can be cloned (provided you know what the set of states is).

  • $\begingroup$ So does it mean that it's practically possible to clone mutually orthogonal states? If that's true, that cloning operator U is not a linear operator, right? $\endgroup$
    – QubitTy
    Sep 9, 2021 at 10:50
  • $\begingroup$ It depends what you mean by "practically possible". It is certainly theoretically possible. The cloning operator is linear. It is the linearity that prevents non-orthogonal states from being cloned. $\endgroup$
    – DaftWullie
    Sep 9, 2021 at 12:31

Your Answer

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

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