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Next: Measuring one photon Up: Attacks on two-photon pulses Previous: Attacks on two-photon pulses

Measuring both photons

First, we look at the case where she measures both photons. It is clear that she has an opportunity here to learn much more about the bit sent than in the single photon case. In fact, if she measures one photon in one of the two bases used for BB84 encoding and the other photon in the other basis, she will learn the bit with certainty after the basis has been revealed. Doing so, Eve will introduce noise on Bob's side. We want to find the minimum probability $p_e$ for Eve to introduce an error after measuring the two photons and having prepared a strong pulse according the outcome of her measurement.We call this a Double Intercept-Resend attack (DIR).

The maximum probability $p_b$ of guessing correctly the bit $b$ (for all possible measurements not only individual ones) acting on two identically polarized BB84 qubits can easily be obtained from the density matrices $\rho_0^{(2)}$ and $\rho_1^{(2)}$ for the encoding of bits 0 and 1 respectively using the formula $1-p_b \geq \frac{1}{2}-\frac{1}{4}\mbox{Tr$\vert\rho_0^{(2)}-\rho_1^{(2)}\vert$}$ [8]. Carrying out the calculations give $p_b\leq \cos^2{\frac{\pi}{8}}$. The maximum probability $p_\theta$ of guessing correctly the tranmission basis can be computed similarly giving $p_\theta\leq \frac{3}{4}$. Given those conditions on $p_b$ and $p_\theta$, we now find the best possible measurement for determining the state of the strong pulse that Eve resends to Bob.

Let $p_{b\vert\theta}$ be the probability of guessing the bit correctly given the guess on the basis is correct and let $\overline{p}_{b\vert\theta}$ be the probability of guessing the bit given the guess for the basis is wrong. We have that

\begin{displaymath}
\cos^2{\frac{\pi}{8}}\geq p_b = p_\theta p_{b\vert\theta} +
(1-p_{\theta})\overline{p}_{b\vert\theta}
\end{displaymath} (4)

since otherwise there would exist a measurement for the encoded bit succeding with probability better than $\cos^2{\frac{\pi}{8}}$ for a contradiction. Let $-\frac{\pi}{4}\leq \alpha \leq \frac{\pi}{4}$ be an angle relative to the state $\vert b\rangle_{\theta}$ for $b\in\{0,1\}$ and $\theta\in\{+,\times\}$ given the classical outcome $(b,\theta)$ for the measurement. We can write the resulting probability of error $p_e$ at Bob's site (given Alice's and Bob's bases match) as:
$\displaystyle 1-p_e$ $\textstyle =$ $\displaystyle p_{\theta}(p_{b\vert\theta}\cos^2{\alpha} +
(1-{p}_{b\vert\theta})\sin^2{\alpha})$  
    $\displaystyle +
(1-p_{\theta})(\overline{p}_{b\vert\theta}\cos^2{(\frac{\pi}{4}-\alpha)} +
(1-\overline{p}_{b\vert\theta})\sin^2{(\frac{\pi}{4}-\alpha}))$ (5)

for some choice of $\alpha$. Eve cannot do better than optimizing (5) subject to (4) and $p_\theta\leq \frac{3}{4}$ even if she was allowed to implement any measurement on two qubits. One can verify that the maximum of (5) occurs whenever $p_{b\vert\theta}=1$ and $p_{\theta}=\frac{3}{4}$ leading to 3

\begin{displaymath}p_e \geq 0.1243.
\end{displaymath}

The lower bound on $p_e$ holds even if Eve could perform any quantum measurement providing the state of the strong signal she then forwards to Bob. The bound is conservative since it would be surprising that a single quantum measurement could provide guesses for both the bit and the basis each reaching the maximum.


next up previous
Next: Measuring one photon Up: Attacks on two-photon pulses Previous: Attacks on two-photon pulses
Louis Salvail 2001-06-15