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 $n \approx 8$

ReplyOptionsa.Eve is measuring in a random basis at each step. In order to know a bit value "for sure", i.e., with 100% confidence, she would have to measure in exactly the same basis as the one in which Alice prepared the qubit. But there are infinitely many possible measurement bases, and so the probability of doing so is exactly zero. She will have full confidence about $\boxed{0\%}$ of the bits.

b.We interpret the "final key" here as the classical bit string agreed upon by Alice and Bob after they have thrown away the bits for which they used different bases. Of course, in the BB84 protocol, Alice and Bob would then compare their a portion of their classical bit string and throw it away if they detected Eve's presence, but for the purposes of this problem, we ignore that portion.

Suppose Eve measures in the random basis $\{\ket{v}, \ket{v^\perp}\}$, and Alice has prepared the qubits in the basis $\{\ket{a}, \ket{a^\perp}\}$. When the qubit reaches Eve, it must be in either state $\ket{a}$ or $\ket{a^\perp}$. Without loss of generality, assume that Alice prepared $\ket{a}$, and assume that $\ket{v}$ is the “closer” of Eve's basis vectors to $\ket{a}$, i.e., the one such that the angle between $\ket{a}$ and $\ket{v}$ is $0 \le \theta \le \pi/2$. Then, the probability that Eve's measurement produced the same (classical) bit as Alice's is given by $\left|\braket{v}{a}\right|^2 = \cos^2 \theta$. To find the average probability that Eve obtains the correct bit, we integrate this as a probability density over the range of possible angles, $0 \le \theta \le \pi/2$:

(1)So on average, $\boxed{50\%}$ of Eve's bits are correct, as compared to Alice's original classical bit string. (Note that after Alice and Bob throw away the bits for which they used different bases, Eve will still have on average only 50% of the bits correct, since her measurement basis is random with respect to either possible basis that Alice and Bob have used, and the same calculation applies.)

(2)c.Remember that Alice and Bob have agreed on their two choices of bases ahead of time, and they will discard any bits in which they used different bases. So we can assume that, for each bit they choose to keep, Alice and Bob have both measured in the same basis. Call this basis $\{\ket{a}, \ket{a^\perp}\}$. Now, if Eve has made a measurement in her random basis $\{\ket{v}, \ket{v^\perp}\}$, then the qubit Bob receives will be in either state $\ket{v}$ or $\ket{v^\perp}$. Let's assume Alice sent $\ket{a}$, and again let $\theta$ be the angle between $\ket{a}$ and $\ket{v}$. What is the probability that Bob will still measure the correct value? We have two cases to consider:

1. Eve measures $\ket{v}$, and then Bob measures $\ket{a}$. Probability = $\left| \braket{v}{a} \right|^2 \left| \braket{a}{v} \right|^2 = \cos^4 \theta$.

2. Eve measures $\ket{v^\perp}$, and then Bob measures $\ket{a}$. Probability = $\left| \braket{v^\perp}{a} \right|^2 \left| \braket{a}{v^\perp} \right|^2 = \sin^4 \theta$.

So the probability that Bob measures the correct bit is $\cos^4 \theta + \sin^4 \theta$. Averaging this over the range of possible angles:

This means that, on average, Eve will go undetected $\frac{3}{4}$ of the time. In order for Alice and Bob to have at least a $90\%$ chance of detecting Eve, we need $(\frac{3}{4})^n \le 0.1$, which means that they need to compare $\boxed{n > 8}$ bits.

ReplyOptions