$\def\i{\mathbf {i}}\def\ket#1{|{#1}\rangle}\def\bra#1{\langle{#1}|}\def\braket#1#2{\langle{#1}|{#2}\rangle}\mathbf{Exercise\ 2.11}$

*B92 quantum key distribution protocol*. In 1992 Bennett proposed the following quantum key distribution protocol. Instead of encoding each bit in either the standard basis or the Hadamard basis as is done in the BB84 protocol, Alice encodes her random string $x$ as follows

and sends them to Bob. Bob generates a random bit string $y$. If $y_i = 0$ he measures the $i$-th qubit in the Hadamard basis $\{\ket{+}, \ket{-} \}$, if $y_i = 1$ he measures in the standard basis $\{\ket 0, \ket 1 \}$. In this protocol, instead of telling Alice over the public classical channel which basis he used to measure each qubit, he tells her the results of his measurements. If his measurement resulted in $\ket +$ or $\ket 0$ Bob sends $0$, if his measurement indicates the state is $\ket 1$ or $\ket{-}$, he sends $1$. Alice and Bob discard all bits from strings $x$ and $y$ for which Bob's bit value from measurement yielded $0$, obtaining strings $x'$ and $y'$. Alice uses $x'$ as the secret key and Bob uses $y'$. Then, depending on the security level they desire, they compare a number of bits to detect tampering. They discard these check bits from their key.

a) Show that if Bob receives exactly the states Alice sends, then the strings $x'$ and $y'$ are identical strings.

b) Why didn't Alice and Bob decide to keep the bits of $x$ and $y$ for which Bob's bit value from measurement was $0$?

c) What if an eavesdropper Eve measures each bit in either the standard basis or the Hadamard basis to obtain a bit string $z$ and forwards the measured qubits to Bob. On average, how many bits of Alice and Bob's key does she know for sure after listening in on the public classical? If Alice and Bob compare $s$ bit values of their strings $x'$ and $y'$, how likely are they to detect Eve's presence?

a) Suppose Alice encodes a $0 \to \vert 0 \rangle$ and send it to Bob. There are two possibilities: 1) Bob generates a random string, $y_i=0$, so he measures in the Hadamard basis, and has a 50% chance of measuring $\vert + \rangle$, which will be discarded and the same chance of measuring $\vert - \rangle$, which will be kept. 2) Bob generates a random string, $y_i=1$, so he measures in the standard basis, and has a 100% chance of measuring $\vert 0 \rangle$. As such, all of these bits will be discarded. Thus, when Alice encodes $\vert 0 \rangle$, the remaining bits satisfy $x' = y'$. Now, suppose Alice encodes a $1 \to \vert + \rangle = \frac{1}{\sqrt{2}}(\vert 0 \rangle + \vert 1 \rangle)$. Again, there are two possibilities: 1) Bob generates a random string, $y_i=0$, measuring in the Hadamard basis, the result has a 100% chance of being $\vert + \rangle$, and so all bits from this instance are discarded 2) Bob generates a random string, $y_i=1$, measuring in the standard basis, the result has a 50% chance of being $\vert 0 \rangle$, which are discarded or $\vert 1 \rangle$, which are kept. From the 4 possible instances, the result after discarding whenever Bob sends a 0 bit, is that $x'=y'$.

b) When Bob's bit value from measurement is 0, there is some chance that what Alice encoded could be 0 or 1, but when Bob's bit value from measurement is 1, the bit encoded by Alice must be 0 in the case that Bob measures $\vert - \rangle$, and must be 1 in the case that Bob measures $\vert 1 \rangle$.

c) With the assumption that Eve knows Alice and Bob are using the B92 protocol, if Eve follows a similar procedure to Alice and Bob, and resorts to discarding all bits measured as $\vert 0 \rangle$ or $\vert + \rangle$, then in the cases where she measures $\vert - \rangle$ and $\vert 1 \rangle$, she can be certain that the bit encoded by Alice is 0 and 1 respectively. If she doesn't know what protocol Alice and Bob are using, then there is no chance for her to be certain of the results of her measurements since she doesn't know what to discard, though she will certainly have some chance of being correct.

Without Eve's presence, in the case that Alice encodes a 0, and Bob generates a 1, there is a 100% chance that he should measure 0, and when Alice encodes a 1, and Bob generates a 0, there is again a 100% chance that he should measure a 1. When Eve interferes however, there is a 25% chance for errors in each case, and as such, overall. The likelihood the likelihood that Alice and Bob will detect Eve's presence as a function of the number of compared bits s, is then: $1 - (\frac{3}{4})^s$

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