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I was studying BB84 protocol from lecture notes by John Watrous. I had some doubts which i thought of asking. Here is the protocol. Alice prepares strings $x=(01110100)$ and $y=(11010001)$. based on the pair $(x_i,y_i)$ it prepares $|A_{00}\rangle=|0\rangle$,$|A_{10}\rangle=|1\rangle$, $|A_{01}\rangle=|+\rangle$,$|A_{11}\rangle=|-\rangle$. He distributes these to Bob. Bob genarates a random key $y'=01110110$. If $y_i=0$ Bob measures the $i-th$ qubit in $Z$ basis, else if $y_i=1$ Bob measures it in $X$ basis. He notes his measurement result as $x_i'$. Then the protocol says that whenever $y_i=y{_i'}$ then $x_i=x{_i'}$. But in the example for $x_5,y_5=00$ and Bob has $y_{5'}=0$ then his outcome $x_5'=0$. Whereas I think it should be $x_5'=1$. Am I missing something?

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  • $\begingroup$ Lecture notes by John Watrous. $\endgroup$
    – Upstart
    Dec 25, 2021 at 2:55

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Let us work this out for your particular example, and see why it works in the general case.


We have: $$ \begin{array}{c|c|c|c|c|c|c|c|c} \text{index}&1&2&3&4&5&6&7&8\\\hline x&0&1&1&1&0&1&0&0\\\hline y&1&1&0&1&0&0&0&1\\ \hline y'&0&1&1&1&0&1&1&0 \end{array} $$ For the fifth index, since $y_5=0$, Alice will encode $x_5$ using the $\{|0\rangle;|1\rangle\}$ basis. Since $x_5=0$, she will send $\left|A_{00}\right\rangle=|0\rangle$. Assuming there is nothing to disturb this state, this is what Bob will receive. Since $y'_5=0$, he will also use the $\{|0\rangle;|1\rangle\}$ basis to measure it, and will thus measure $|0\rangle$ with probability $1$. Thus, we have $x_5=x'_5=0$.


Now, in the general case. We know that we will only consider the indexes $i$ such that $y_i=y'_i$, so we can afford to only consider those cases. Assuming nothing disturbs the state sent by Alice, since Bob measures in the same basis as Alice encoded the state, he will always measure the right result with probability $1$.

Now, the point of the protocol is that Eve may disturb the state by measuring it in the wrong basis. Let us assume that Eve measures in the $|\pm\rangle$ basis, while Alice encoded the state in the $\{|0\rangle;|1\rangle\}$ basis. With probability $\frac12$ each, the state will then collapse to either $|+\rangle$ or $|-\rangle$. Subsequently, Bob will measure it and get $|0\rangle$ or $|1\rangle$ with probability $\frac12$ each. Thus, if this index is used to veridy the protocol, there is a probability $\frac12\times\frac12=\frac14$ (Eve measures in the wrong basis and Bob gets the wrong result) that Alice and Bob detects that the channel is noisy, or that Eve was eavesdropping.

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