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quantum cryptography

quantum key distribution

Basic order of events:

  1. generating a one time pad between Alice and Bob
  2. then checking if anybody was eavesdropping
  • after sharing the key a test is performed:
    • if anybody was listening the key is discarded
    • if not they now have a new key to communicate with
  • the reason why this works is because any measurement on a quantum object leaves a trace
    • it is impossible to read information traveling over a quantum channel without perturbing it

BB84 protocol

  • this scheme can use any quantum object with a two value observable
  • in practice the quantum object is the polarization of a photon (quantum optics)

prerequisites Alice

Alice has a source to generate photons with four possible polarizations:

  1. H, horizontal
  2. V, vertical
  3. L, left
  4. R, right

The four polarizations can be grouped into two sets. The HV set is located along the X and Y axis. The LR set 45 degrees rotated from that.

polarizations.jpg

prerequisites Bob

Bob has two polarization beam splitters with which he analyzes photons sent by Alice. One oriented at 0 degrees (vertical) and one at 45 degrees (diagonal).

The vertical beam splitter can analyze photons of the HV set. The diagonal beam splitter can analyze photons of the LR set.

Each beam splitter can produce two values.

  • Vertical beam splitter:
    1. H, horizontal
    2. V, vertical
  • Diagonal beam splitter:
    1. L, left
    2. R, right

beam-splitter.jpg

To obtain the original value of a photon (without fail) the orientations of Alice and Bob have to match up. Some examples:

  • inputting H into the vertical beam splitter will always return H (match)
  • inputting V into the vertical beam splitter will always return V (match)
  • inputting L or R into the vertical beam splitter will return a random value (H or V, mismatch)

procedure of events

For each (potential) bit of the key the following happens:

  • Alice generates a randomly polarized photon (H, V, L or R) and publicly sends it to Bob
  • Bob choses a random beam splitter to analyze the photon with
    • if Bob chose the correct beam splitter: he obtains the original value
    • if Bob chose the wrong beam splitter: he obtains a random value
  • Bob saves the orientation of the beam splitter and the resulting value
    • of which he still does not know if it's meaningful
    • there's a 50% chance it's just a random value due to orientation mismatch

After sending a bunch of photons we move on to the next step: reconciliation.

  • Bob publicly announces all his chosen beam splitter orientations
  • Alice publicly replies which were correct
  • They both drop all the values where Bob chose wrong
  • They now hold the key (which consists of all values where Bob chose correctly)

consequences of eavesdropping

If Eve intercepts a photon to measure it, Bob will not receive it and thus will not use it in the key. Alternatively, Eve can retransmit a photon, but Bob can detect this by sacrifing some Bits of the key and publicly sending them back to Alice!

This is where the quantum nature of the transmission medium kicks in:

  • Eve does not know the correct orientation (vertical or diagonal) and thus has to guess
    • if Eve guesses correctly she has no problem
      • the orientations match up and Bob will get the correct value
    • if Eve guesses wrong there is a 50% chance that Bob will receive the wrong value
      • if Bob randomly chooses this Bit to sacrifice (and send it back to Alice) she has a problem indeed
  • Alice sees that Bob sent back a wrong value despite the correct orientation! The key cannot be trusted!
    • any one Bit being wrong when eavesdropped will happen in about 25% of the cases
      • 50% chance that Eve chooses the wrong orientation set and 50% chance of that for a wrong value

This entire scheme is based on the fact that you can not find out the basis of polarization of a single photon.

A couple of example scenarios:

eve.jpg

  • To keep it simple Alice always send V.
  • In the first case Eve and Bob select a vertical splitter, eavesdropping takes place but is not detected.
  • In the second case Alice and Bob have an orientation mismatch, the value is discarded.
    • Eve does not matter.
  • In the third case Bob chose the correct and Eve the wrong orientation.
    • Eve is unlucky and the result is H instead of V.
    • Bob and Alice chose the same orientation but they still got differing values!
      • If Bob sacrifices this Bit and sends it to Alice they immediately know someone was eavesdropping.

sources and further reading