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First of all, Heisenberg principle is usually invoked to say that two operators don't commute and therefore if your qubit is in $|0\rangle$ which is an eigenstate of $\sigma_z$, you will have uncertainty in the measurement of $\sigma_x$. However, let's check that Heisenberg principle is still holding even in this situation! Let's see Heisenberg's principle ...


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[0001] Regarding the OP's first paragraph and the comments therein, there is no protocol, call it $X$, that can be executed efficiently on classical computers, that has been proven to be secure against quantum computers. If we had such a proof then we would also know that P$\ne$NP. This follows because, as @Martin said, a quantum computer can efficiently ...


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