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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!

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    $\begingroup$ 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
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    $\begingroup$ 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

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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).

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  • $\begingroup$ 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

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