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I am studying the BB84 protocol and I have some doubts regarding its security.

Consider the following example: Alice wants to send a message to Bob over a public quantum channel, where a third person (Eve) could try to detect the message. The first step for Alice is to compute a string of classical bits in qubits via orthogonal bases $X$ or $Z$. Then, she sends the encoded message over the channel. At this point, suppose Eve has intercepted the message.

It is known that the BB84 protocol is secure as Eve cannot intercept the message without being detected. This happens because she has to apply a set of bases in order to get Alice's message from qubits to bits; the fact is that Eve doesn't know the bases which have been used by Alice, as those are private. Based on that, when she tries to measure the qubits in her random selection of bases, if those are wrong, an altered message will be sent to Bob. We don't have to go further as my question lies in this part of the communication:

  1. As Eve doesn't know Alice's bases, almost certainly she will use different bases, leading to a different encoded message for Bob which will allow to detect the interception. What happens if she select exactly the same bases as Alice's? I suppose it's not impossible; even if the probability is so small, I still think it could happen.
  2. Based on 1., if Eve intercepts the whole message correctly because she used the same bases, doesn't this go against the non-cloning theorem of quantum computation, as it would imply creating the same message from unknown states.
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You are correct she could guess the bases correctly however each time she has a probability of $1/2$ of selecting the correct basis. So if Alice sends $n$ qubits then the probability she selects all of the correct bases is $1/2^n$. For any reasonable $n$ this is extremely unlikely.

For some perspective, if $n=100$ then (based on some rough estimates) Eve is more likely to select the same grain of sand as you from all possible grains of sand on the earth (choosing uniformly) than guess all the bases correctly. If $n=1000$ then Eve is more likely to choose the same atom as you from all atoms in the observable universe (choosing uniformly) than guess all the bases correctly.

Typical BB84 experiments run from $n=10^5$ to $n=10^9$ and possibly even higher. So guessing many bases is out of the question.

Regarding question 2, this does not go against no cloning as it is probabilistic, you can clone only with very small probability.

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  • $\begingroup$ Thank you for the answer. I understand the part about the probability of choosing the same bases being unlikely; I still don't get the point about the no cloning. This theorem states that "it is impossible to create identical copies of an unknown quantum state". If she (leave alone the fact that it couldn't happen, for the reasons you mentioned) would choose the same bases, she would get Alice's original message and send it to Bob; wouldn't this be "an identical copy from an unknown state"? Perhaps I am getting confused about what the teorem says. $\endgroup$
    – aghin
    Commented Feb 7 at 18:54
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    $\begingroup$ @aghin00 Eve cannot successfully clone the unknown state and concurrently know that she succeeded in cloning the unknown state, without knowing what the unknown state is, in which case it would no longer be an unknown state. $\endgroup$
    – Mark Spinelli
    Commented Feb 7 at 20:58
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    $\begingroup$ What exactly do you claim the no-cloning theorem to be? Write it out precisely as a mathematical statement and you'll find the above result is not in contradiction with the statement. $\endgroup$
    – Rammus
    Commented Feb 7 at 21:55
  • $\begingroup$ thank you @MarkSpinelli, I wasn't considering the "case it would no longer be an unkown state". I get it now. $\endgroup$
    – aghin
    Commented Feb 8 at 15:23

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