Is there a minimum limit for No. of photons that Alice would send to Bob in a QKD scheme like BB84? If not than what would be a desirable length for initial qubit string?

  • $\begingroup$ Many things have influence, like desired residual odds that an adversary taping the quantum link can go undetected, the error rate of the quantum link, synchronization, and the need to transmit enough bits to secure the next transmission. Related. $\endgroup$ – fgrieu Mar 30 '18 at 6:51

The minimum limit would be 1, which transmits a key of length 1 half of the time. Ideally, you would use hardware capable of conveying this information with a single photon per bit in the string, so one photon.

The string length is a practical consideration, not a cryptographic one. Sending fewer bits means you will have to re-key more often. Sending 1 bit at a time would basically be equivalent to creating a stream of key information. Size the qubit string to fit your application's practical needs.

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  • $\begingroup$ Thanks @Cort . It means that we can take any bit stream according to our choice depending upon the practical considerations? $\endgroup$ – Efs Jul 27 '17 at 14:38
  • $\begingroup$ Some QKDNs are tens of km, and I'd be extremely surprised if contemporary fibre (never mind fibreless) can do this. You loose photons just bouncing out at the bends. You'll easily get -5dB /km. Alice might have to transmit 10,000 for one to arrive in Bob's letter box. I expect that BB84 uses some sort of carrier modulation at the implementation level to mitigate this. And since individual photon loss is random, it would be impossible to know which photons exactly got through. They'll also arrive out of order. The answer therefore cannot be 1. $\endgroup$ – Paul Uszak May 29 '18 at 1:42
  • $\begingroup$ @PaulUszak The OP asked for a minimum limit without reference to any practical hardware. $\endgroup$ – Cort Ammon May 29 '18 at 2:53
  • $\begingroup$ Are you sure..? $\endgroup$ – Paul Uszak May 29 '18 at 11:40
  • $\begingroup$ @PaulUszak If you reduce it to a single photon, you basically end up with a single-photon quantum eraser. $\endgroup$ – Cort Ammon May 29 '18 at 14:27

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