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Consider the following scenario:

Alice and Bob run BB84 to each other.

Eavesdropper Eve is present in the middle. Here we assume that she has access to all channels: the quantum channel, in the classical control channel (where the BB84 protocol runs), and the classical payload channel (where the encrypted traffic is exchanged).

Let's assume for this scenario that the classical control channel (where the BB84 protocol runs) is NOT authenticated.

The lack of authentication allows Eve to perform a woman-in-the-middle attack as follows:

  • Eve can negotiate a key with Alice, where Alice thinks she is negotiating a key with Bob.

  • Eve can negotiate a key with Bob, where Bob thinks he is negotiating a key with Alice.

  • Eve can decrypt the payload traffic from Alice (using the Alice-Eve key) and re-encrypt it (using the Eve-Bob key) and send it to Bob.

  • Eve can decrypt the payload traffic from Bob (using the Bob-Eve key) and re-encrypt it (using the Eve-Alice key) and send it to Alice.

  • In general, the Alice-Eve and the Bob-Eve keys will be different.

My question is: is there any way for Eve to perform a woman-in-the-middle attack and force the Alice-Eve key and the Bob-Eve key to be the same?

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  • $\begingroup$ The basis of BB84 is that Alice and Bob's non-quantum communications are over public channels. There is no authentication or man-in-the-middle. $\endgroup$ Mar 18, 2023 at 0:09
  • $\begingroup$ @FrankYellin yes, the classic channel is allowed to be public, but it must be authenticated. If it is not authenticated, then BB84 is vulnerable to a man-in-the-middle attack. See quantumcomputing.stackexchange.com/questions/9015/… $\endgroup$ Mar 18, 2023 at 1:04
  • $\begingroup$ Your schema would work only in case Eve is a "woman in the middle" in the quantum chanel. Otherwise, she is not able to negotiate keys with Alice and Bob. $\endgroup$ Mar 18, 2023 at 9:57
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    $\begingroup$ @MartinVesely Yes, correct, Eve would have to be a woman-in-the-middle in both the quantum channel and the classical channel. I will clarify that. $\endgroup$ Mar 18, 2023 at 14:32

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