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I am not able to understand an argument from Simple proof of Security of the BB84 QKD; I need your help. In page 2 it is mentioned that "Alice can measure her half of the encoded EPR pairs before or after transmission. If she measures them first, this is the same as choosing a random key $k$ and encoding $k$ using $Q_{x,z}$."

I am not getting how sharing maximally entangled EPR states is the same as sending non-entangled encoded (CSS) quantum states. Basically, it is proved that protocol is secure when the entangled pair is secure, but now they are replacing maximally entangled states with just encoded quantum states.

In the recently published book named "Quantum Key Distribution" by Dr. Ramona Wolf (and also in her lectures), she mentioned the same thing "... instead of preparing maximally entangled states, Alice can equivalently choose bit string $x,z,$ and $k$ at random, encode $| k \rangle$ in the code $Q_{x,z}$ and send the corresponding $n$-qubit codeword to Bob.

It's still unclear why they're the same.

If you could provide me with some understanding or a proper explanation for this equivalency, that would be extremely useful. (I have studied CSS error-correcting codes)

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I'm not sure that I would claim the two situations are entirely equivalent - if Alice and Bob share Bell pairs, there are extra things they could do (e.g. testing Bell inequalities), but in terms of the actual protocol applied, they're equivalent.

One way to see this is to think about the protocol (more or less) as described:

  • Alice first sends halves of Bell pairs to Bob
  • Alice measures her half of the Bell pair
  • Bob does some stuff with his halves

But now, does it matter exactly when Alice measures her qubits? Should she do it only once Bob has received his qubits (how would she know?), or could she do it a bit earlier? It makes no difference to what any other party knows or can do so long as Alice doesn't tell anybody her measurement results. So, take that to an extreme. Alice measures her half of the Bell pair at the moment that she sends the other half to Bob. Again, entirely equivalent. But, all that really means, is that for whatever measurement outcome she gets (which you know happens with 50:50 probability), this is just preparing some pure, separable state that she's sending to Bob.

So she doesn't have to prepare the Bell state and measure it. She can simulate the measurement result (after all, it's just choosing a measurement basis and tossing a fair coin) and, using those results, preparing a specific state to send to Bob. This is the standard BB84 protocol.

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  • $\begingroup$ Your answer does give me some intuition, but I think something is missing. If it made no difference that she measures her bell pair before or after sending it to Bob (ultimately, this is required), what is the need to introduce CSS code in between? Further "Bob acknowledges receipt of the qubits". $\endgroup$
    – IamKnull
    Oct 22 at 10:16
  • $\begingroup$ According to the text "Intuitively, the security of the protocol depends on the fact that for a sufficiently low error rate, a CSS code transmits the information encoded by it with very high fidelity, so that by the no-cloning principle very little information can leak to Eve." Perhaps I am missing something, the above statement from the text is not clear. $\endgroup$
    – IamKnull
    Oct 22 at 10:16
  • $\begingroup$ I won't pretend to have gone to read the specific documents you're linking to. A very basic BB84 protocol doesn't include error correction, which is what I was describing. I assume you can modify what I said by sending a bunch of Bell pairs together, and Alice making a joint measurement of them within the logical space of an error correcting code. That effective multi-qubit teleportation operation (minus corrective unitaries) must be just the same as physically sending the uncorrected state. $\endgroup$
    – DaftWullie
    Oct 22 at 11:17
  • $\begingroup$ To directly address your second comment: the point of using an error correcting code is that it can correct for errors up to some threshold. So, provided you're working below that threshold, the data arrives at Bob perfectly. If it arrives perfectly at Bob, it has not been cloned, and so Eve has no access to the information about the state (which would be on a clone). $\endgroup$
    – DaftWullie
    Oct 22 at 11:19
  • $\begingroup$ Thanks for your reply. I guess now I have a more clear picture than before. The paper is actually very interesting but requires a lot of work to understand the arguments. $\endgroup$
    – IamKnull
    Oct 22 at 13:16

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