2
$\begingroup$

Given a superposition of states "B" which is a subset of the suoerposition "A" of all possible states of a set of qbits, is there a quantum operation that produces superposition $R= B^C$, the complementary subset of "A", and in the process necessarily destroys "B", to avoid violating the no cloning theorem? If not, is there an operation that can produce an approximate complement ${B^C}(Approx)$?

Edit 1: @tsgeorgios has demonstrated in a comment that the proposed operation is not unitary in the general case, so wouldn't be possible in general. However, this challenge may be analogous to the challenge of cloning a superposition, which can't be done in general, but can be closely approximated if the set A of states are linearly independent. That means if there are N qbits, there can only be N states in the superposition A. Though cumbersome, any method for approximating the complement of information encoded in B would be potentially useful.

$\endgroup$
6
  • $\begingroup$ @MarkS points out, in a comment to a related question, that the complement to the universal set is the null set, which might not have any meaning in reference to superpositions of states in a set of qbits. That's a good point. However, in this context maybe a single fully determined state would correspond to the null set. $\endgroup$
    – S. McGrew
    Sep 27 '20 at 23:05
  • $\begingroup$ B must be a proper subset then? If not then if you have a state which is a superposition of all possible states (says the uniform superposition state) then its complement would be nothing... which is not possible. $\endgroup$
    – KAJ226
    Sep 27 '20 at 23:06
  • $\begingroup$ In practice, B would always be a proper subset. $\endgroup$
    – S. McGrew
    Sep 27 '20 at 23:14
  • 1
    $\begingroup$ If i understand it correctly, the output for all states of the form $ \frac{1}{\sqrt{3}} (| 00 \rangle + | 01 \rangle + e^{i\phi} |10 \rangle) $ should be $ |11 \rangle $, right? But then, this is not a unitary operation. This means that you can't do what you ask for an arbitary superposition of an (arbitary) subset B. $\endgroup$
    – tsgeorgios
    Sep 27 '20 at 23:23
  • $\begingroup$ Why is it not a unitary operation? $\endgroup$
    – BlackHat18
    Sep 28 '20 at 4:11

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.