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Error rate

The total transmission of the line is the product of two contributions:

F = F_{fiber}F_{Bob} = 10^{-\frac{\beta l+c}{10}}
\end{displaymath} (1)

where $\beta$ is the absorption coefficient of the fiber, $l$ the length of it and $c$ is the constant losses at Bob's station. The two contributions to the transmission are separated because obviously only one of them depends on the fiber length and therefore determines the maximal distance for the secure communication.

The photon statistics for the pulses generated by our laser is close to the Poissonian distribution described by the parameter $\lambda $ which is an average photon number in a pulse sent by Alice.

The probability to detect a photon at Bob's side, with no dark counts taken into account, is then

P^{signal}_{det} = \eta_B F\lambda
\end{displaymath} (2)

The detector is not capable of measuring the number of photons in the pulse. It produces a click independently on this number. However, Eve is assumed to be able to measure the photon number in every pulse (see below) and therefore the probabilities to have two or more photons in a pulse are vital for the security analysis. The probability to obtain two photons per pulse is at most $\frac{1}{2}\lambda^2$ whereas the probability to have three or more photons is equal to $\frac{1}{6}\lambda^3$.

Since it is impossible to distinguish between dark counts and regular counts, they both contribute to the probability of a detection event $
P_{det} \approx \kappa P_{det}^{signal} + P_{det}^{dark}
$. The coincidence term $P_{det}^{signal}P_{det}^{dark}$ is ignored here. $\kappa$ is the implementation efficiency, which is equal to the exploitation of the pulses, that is $\kappa = 1$ for BB84, $\kappa = \frac{1}{2}$ for BB84 implemented with only one detector1 and $\kappa = \frac{1}{4}$ for B92.

The contributions to the errors are the same as the contributions to the counts, i.e. ignoring the coincidence term, we get for the error probability $ e \approx \kappa e^{signal} + e^{dark} = \kappa
P_{det}^{signal}P_{e}^{signal} + \frac{_1}{^2}P_{det}^{dark} $, where $P_{e}^{signal}$ and $\frac{1}{2}$ is the probability of an error due to the signal and dark counts respectively. The error rate $\epsilon$ is given by the probability of an error divided by the probability of a detection event:

\epsilon = \frac{e}{P_{det}} \approx
\frac{\kappa P_{det}...
{\kappa P_{det}^{signal} + P_{det}^{dark}}
\end{displaymath} (3)

In the following section we consider, one, two and many photon attacks separately. Finally the results are combined to obtain optimal conditions on the pulse strength $\lambda $, quantum efficiency and the dark count probability $P^{dark}_{det}$ necessary to maximize the secure bit rate.

next up previous
Next: Restricted Adversaries Up: Security Previous: Security
Louis Salvail 2001-06-15