# Is the eavesdropper detected by chance in QKD?

So I was starting to familiarise myself with QKD when I came across this video on YouTube breaking down the process (I believe this is the E91 protocol but I could be mistaken) using an animation that was rather easy to follow. However I have one question regarding the section where the eavesdropper scenario was shown.

We see that Alice and Bob receive the following keys with the bold parts being the "intercepted" qubits

A: 1 1 1 0 0 0 1 1 0 1 0 0

B: 1 1 0 0 1 0 1 0 0 1 0 0

Now obviously, the difference in these keys is an indication of an eavesdropper. However, it is also evident that all not all the qubits show signs of interception. For example qubits 2,6,7,11,12 still match between the keys.

My question is this, is it possible that the eavesdropper could leave no trace of the interception (assuming he's lucky enough since the process are assumed to be completely random)? If so, what are the "rules of thumb" that one has to consider in order to maximise the likelihood that an eavesdropper leaves a trace on the keys (for example an optimal key length, detector sequence, etc)?

• The eavesdropper could also guess the whole message by accident! – Norbert Schuch Dec 25 '18 at 0:17
• @NorbertSchuch So it's a legitimate concern that one has to cater for if one were to design/construct such a system and not merely because I am misunderstanding something? – Jepsilon Dec 25 '18 at 0:29
• Not really, since you can make either probability arbitrarily small. My point was rather: There cannot be 100% security, since there is always a non-zero chance to guess the message ad hoc. But if you can make it as small as you want (with reasonable effort), you are good. – Norbert Schuch Dec 25 '18 at 0:44