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Measuring one photon

If Eve measures one of the two photons, she may then from her measurement result produce a pulse that she forwards to Bob, possibly containing the photon she did not measure. Thus, this is actually a single photon intercept-resend attack, where however the pulse sent to Bob contains at most 1 photon in the correct state. Thus, from the above, Eve knows the bit sent with probability at most $0.85$.

Eve may choose to let only the unmeasured photon go to Bob. Bob's error probability will then only be the normal one ($\epsilon$, since Eve cannot influence Bob's apparatus), but Eve runs the risk that the photon is lost in Bob's apparatus. We call this the Split-off attack (SO). As we have seen, the probability that Alice emits a two-photon pulse is less than $\lambda^2/2$. Even if Eve transmits the unmeasured photon to Bob's apparatus through a perfect fiber, it will only contribute to the key, if it makes it through Bob's apparatus and is detected. So Eve is left with a fraction $\lambda^2F_{Bob}\eta_B/2$ of all pulses for which the SO attack could work. To increase the effect of this attack, Eve may choose to block some of the other pulses, typically one-photon pulses that she did not measure. This will increase the fraction of attacked pulses among those that Bob detects. However, there is limit to the number of pulses Eve can block: If Eve is not present, Bob expects to detect a certain number of pulses, more precisely, the fraction of sent pulses that are normally detected by Bob is $\lambda F \eta_B$. Eve must maintain approximately the number of pulses detected, since otherwise the raw key will be shorter than expected. Consequently, the fraction $f_{SO}$ of bits in the raw key coming from pulses subjected to SO can be at most

\begin{displaymath}f_{SO}\leq \frac{\lambda ^2F_{Bob}\eta_B}{2\lambda F\eta_B}=
\frac{\lambda}{2F_{fiber}}.\end{displaymath}

Eve may also inject one or more photons of her own into the pulse. If Bob happens to measure one of these instead of the correct one, the error probability will be at least 0.25 (again by the above considerations on intercept-resend). So, because Eve has no control over which of the photons Bob will measure, Bob's overall error probability in this case will be at least $Pr(\mbox{Bob measures one
of Eve's photons})\cdot 0.25 \geq 0.125$. On the other hand, there is now a larger chance that Bob will detect the pulse - for simplicity we will be generous to Eve and assume that Bob will always detect pulses attacked this way. We call this the Split-Off-Resend attack (SOR).


next up previous
Next: Three-or-more-photon pulses Up: Attacks on two-photon pulses Previous: Measuring both photons
Louis Salvail 2001-06-15