2
$\begingroup$

This is a repost from a question I posted in the physics stack exchange

I am having a hard time understanding why MDI-QKD is truly measurement device independent. My current vision is that Charlie (the one who performs the Bell state measurements) is simply sampling the qubits Alice and Bob sent: the ones for which the measurement does not fail should have certain correlations.

But why is this schema imune against all detector side channels?

$\endgroup$

1 Answer 1

1
$\begingroup$

Actually, Charlie measures not the qubits themself, but the result of its interference.

As there are currently no reliable quantum information medium other than photons, lets speak about photons:

  1. Alice prepares photon $a$
  2. Bob prepares photon $b$
  3. They send this photons to Charlie
  4. Inside the Charlie's setup the photons interfere and Charlie measures only the result of the interference.

The qubits are prepared in bases, so Charlie can't reliably measure them (he will chose the wrong basis with 50% probability). The result of the interference does not have any correlations with individual qubits. Meanwhile, for those, who exactly knows one of the qubit states, interference unambiguously points to the state of another qubit.

A classical analogy:

Imagine, Alice and Bob send classical bits $a'$ and $b'$ to Charlie, but for some reason it is impossible to measure them directly. Charlie is only able to measure $c = a'$ XOR $b'$.

Alice knows $a'$ and $c$, so she can find $b'$. Bob knows $b'$ and $c$, so he can find $c'$. Charlie knows only $c$, so he can't cfind neither $a'$, not $b'$.

$\endgroup$
3
  • $\begingroup$ I loved the analogy, and now I understand why Charlie cannot obtain information about the generated key. However one doubt remains: why is this protocol imune to all attacks against the detector, such as the dector blinding attack? $\endgroup$ Nov 27, 2021 at 0:13
  • $\begingroup$ It is immune to such attacks, because them does not allow the attacker to gather information about Alice's and Bob's qubits. The only thing, that the attacker can achieve is to interrupt the key exchange (if we speak about attacks against the detector). $\endgroup$
    – totikom
    Nov 27, 2021 at 8:42
  • $\begingroup$ However, the attacker still can perform probbing attacks against either Alice's or Bob's state preparation. $\endgroup$
    – totikom
    Nov 27, 2021 at 8:43

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.