0
$\begingroup$

I'm trying to teach myself Quantum Cryptography for an exam I have soon, and I came across the following question:

To establish a common key, Alice and Bob analyse a sequence of 20 photon pairs. Alice measures the photon sequence: $$1\:1\:0\:1\:0 \qquad 0\:0\:0\:1\:0 \qquad 1\:0\:1\:0\:1 \qquad 0\:1\:1\:0\:1$$ using the random series of detection bases: $$\otimes \oplus \oplus \otimes\otimes \qquad \otimes \oplus \oplus \otimes \otimes \qquad \oplus \otimes \oplus \oplus \oplus \qquad \oplus \oplus \otimes \otimes \oplus$$ while Bob is analysing the photons using the following sequence: $$⊗⊕⊗⊗⊗ \qquad ⊕⊗⊕⊕⊗ \qquad ⊗⊗⊕⊗⊕ \qquad ⊕⊗⊕⊗⊕$$ Hence find the sifted data set they get as an encryption key as in BB84.

Examining all the cases when the detection bases were the same and discarding the bits when they were not; we find that the encryption key is:

$$111\:000\:011\:101$$

Is this the right key?


Note: I apologise if this is in the wrong forum, I'm only really familiar with Math.SE, SO and EE.SE so feel free to migrate this if it's more appropriate elsewhere

$\endgroup$
1
  • 2
    $\begingroup$ 1. What have you tried? Where did you get stuck? Did you try working through the algorithm they taught you in class, by hand? 2. I don't think you've given us enough context. I'm guessing that in class you were taught a specific method for deriving the key from the measurement (i.e., a specific method for privacy amplification). Without knowing what specific method you were taught, it will be difficult for us to work through this by hand. $\endgroup$
    – D.W.
    Jun 14 '14 at 0:28
1
$\begingroup$

The first stage of the BB84 protocol involves throwing away any measurements where the measurements were performed with a different basis. After throwing those away, we are left with the following binary sequence:

$$11 10 0 0 01 10 01.$$

However, in the BB84 protocol as it was originally published, this is not the final key. There are several subsequent steps that are performed after this.

In the second stage, Alice and Bob check whether they both got approximately the same sequence, by comparing a randomly chosen subset of this sequence. If there are too many errors (differences between Alice's string and Bob's string) in those randomly selected positions, then they declare it a failure and go back to the beginning and start over.

In the third stage, a privacy amplification protocol is applied to the remaining bits, to derive a key that Alice and Bob can use.

Consequently, in the real BB84 protocol, the final key is not $111000011101$. That's just the output of the first stage. The final key is the output of the third stage.

However, I don't know what they taught you in your class. I'd hope that they taught you all three stages, but if they only taught you the first stage, then you'd get a different answer than if they taught you the full protocol. Therefore, we can't really tell what the right answer to your exam might have been. If you want to know about that, I suggest you contact your instructor.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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