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Earlier quoted context omitted.
Heh - I recall back in 94 when I was one of the first handful of webmasters at BT we also had a tube map centered round Martlesham (Think UK Bell Labs) with of shoots to LSEC TSEC(my lonely I486 that ran occasionally when I don't need all the horse power of my pc) AND BSEC
We are going back to our ‘BT Labs’ brand soon :)
In particular messages (bits) are sent as photons with certain polarisations. The space of possible polarisations is roughly rays through the origin in the plane (ignoring circular polarisation). The only way to observe polarisation is to take two possible polarisations and ask which one the photon is. The answer depends probabilistically on the photon’s state and the polarisations chosen. For example a +45° polarisation when observed in the basis of {0°, 90°} corresponds to 50% chance of 0° and 50% chance of 90°. If a certain polarisation p is sent and then one observes with a basis of p or q, there is a 100% chance of observing state p. When a photon is observed it’s polarisation becomes whatever polarisation was observed (this it can’t be measured twice)
The algorithm used works roughly as:
0. Four polarisations are known in advance, made of two orthogonal bases, (but the four states are together orthogonal or linearly independent) and parties can send any polarisation down some channel[2], and observe any photon they receive with one of the bases, chosen before the photon arrives. A 0-state and 1-state is decided for each basis.
1. Alice produces some random data to send to Bob
2. For each bit she wishes to send, she chooses a basis at random and sends the corresponding 0-state or 1-state
3. For each photon he receives, bob chooses a basis at random and observes the photon, recording his basis choice and whether he got the 0-state or the 1-state.
4. Alice and bob compare notes on the bases they chose over a classical insecure channel. They should have picked the same basis 50% of the time.
5. Bob and Alice keep the bits where the basis matched. In these cases, bob should have perfectly received the data.
6. Some [random I think] subset is compared over the classical channel. Where these don’t match up, the photons must have been observed (potentially eavesdropped). An eavesdropper can get a 75% probability [I think] that they were eavesdropped but not noticed by Bob/Alice as an eavesdropper would have had to pick a basis and could have either got it right and sent on an equivalent photon or got it wrong and had Bob probabilistically get the right bit out.
7. If the error rate is suitably low then Alice and bob assume their data was safely transmitted and the rest of it can be used to make a random key.
Please correct me if I’m wrong on this as I’m not an expert and I’m not sure I’ve understood it correctly.
[1] https://en.m.wikipedia.org/wiki/Quantum_key_distribution
[2] optical fibre is this channel. I don’t really understand the physics of it but I thought reflection (inside the fibre) could cause polarisation to change or just loss of photon. I don’t know this is dealt with but I’d love to find out.