As far as I understand, CV-QKD (also here) uses coherent states to send a key, that is encoded in amplitude and phase quadratures. When someone tries to eavesdrop, this will introduce an error and therefore the eavesdropper is detected. I think I can see that an intercept/resend attack of any kind would introduce such an error. What I do not understand, however:

How is the error introduced if and eavesdropper just taps a few of the photons?


1 Answer 1


In CV-QKD, we send the signal at low power. At that levels, the signal is masked by noise of quantum nature (e.g. Shot Noise i.e. uncertainty principle of the quadratures). This noise protects the communication.

In broad terms, each DV-QKD symbol sent becomes a bit of the secret key. In CV-QKD, we used the complete set of data to create a secret key. To do so, first we need to estimate some parameters and noises to assess the amount of info that Eve has. All this info will be removed from the secret key (privacy amplification). One of the parameters is the transmittance of the channel. Another one is the excess noise. Well, by measuring some of the photons (e.g. using a beam-splitter) Eve is changing these parameters.

Alice has a set of data, A = {I,Q} (I and Q are both quadratures). So does Bob, B; and Eve, E. A and B are partially correlated. E and B are also correlated in some way. As quantum information cannot be cloned, and Alice is the one sending the data to Bob, the info that Alice has about Bob, I(A,B) might be greater than the info that Eve has, I(E,B). If so, you can generate key: k = I(A,B) - I(E,B)

  • 1
    $\begingroup$ Thank you for the explanation! $\endgroup$
    – julian
    Commented May 9, 2023 at 8:52

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