Exercise 2.10

$\def\i{\mathbf {i}}\def\ket#1{|{#1}\rangle}\def\bra#1{\langle{#1}|}\def\braket#1#2{\langle{#1}|{#2}\rangle}\mathbf{Exercise\ 2.10}$

Analyze Eve's success in eavesdropping on the BB84 protocol if she does not even know which two bases to choose from so chooses a basis at random at each step.

a) On average, what percentage of bit values of the final key will Eve know for sure after listening to Alice and Bob's conversation on the public channel?

b) On average, what percentage of bits in her string are correct?

c) How many bits do Alice and Bob need to compare to have a $90\%$ chance of detecting Eve's presence?

page revision: 0, last edited: 12 Nov 2011 05:34

a) Eve only learns the correct basis for measurement on the classical line after she has a chance to measure the qubits, so those that she knows for sure are those that by chance she was able to measure exactly in the basis Alice encoded them. When Eve doesn't know which two bases to choose from, there are an infinite number of possibilities for her to choose from, so that chance is approximately zero. Thus, Eve doesn't know any of the final key bit values for sure.

b) The angle between bases determines the change in the probability that Eve measures the bit correctly. Suppose the basis Eve measures in is $\lvert v \rangle = cos(\theta) \lvert 0 \rangle + sin(\theta) \lvert 1 \rangle$, where $\theta$ is the angle between the basis chosen by Alice, and that chosen by Eve. Then, if we take the instance where the \levert 0 \rangle is the what Alice encodes, the probability that Eve will correctly measure this, is $\vert \langle 0 \vert v \rangle \vert ^2 = cos^2(\theta)$. Expanding to the the more general case, the average number of bits that Eve measures correctly will be $\frac{\pi}{2} \int_0^{\frac{\pi}{2}} cos^2 (\theta) d\theta = \frac{1}{2}$. So, on average, Eve is able to correctly measure 50% of the bits encoded by Alice.

c) If Alice encodes the bit as $\vert 0 \rangle$, Eve measures it as such in her randomly chosen basis, and sends it on in that basis, the probability that Bob will still measure the value originally encoded by Alice is represented by $|\langle v \vert 0 \rangle \langle v \vert 0 \rangle|^2 + |\langle v_{\bot} \vert 0 \rangle \langle v_{\bot} \vert \rangle|^2 = cos^4 (\theta) + sin^4 (\theta)$, where Eve's chance of correctly measuring the bit Alice encoded, and Bob's chance of measuring the bit resent by Eve as the same as Alice originally encoded, are represented in the two inner products in each term of the expression. Averaging this expression yields $\frac{\pi}{2} \int_0^{\frac{\pi}{2}} cos^4 (\theta) + sin^4 (\theta) d\theta = \frac{3}{4}$, which represents the chance that Eve will correctly measure a bit encoded by Alice, and go undetected. So, in order for Alice and Bob to have a 90% chance of detecting Eve's presence, they need to compare $n$ bits s.t. $(\frac{3}{4})^n \lt 0.1$, so $n \approx 8$

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