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authorPrefetch2022-10-20 18:25:31 +0200
committerPrefetch2022-10-20 18:25:31 +0200
commit16555851b6514a736c5c9d8e73de7da7fc9b6288 (patch)
tree76b8bfd30f8941d0d85365990bcdbc5d0643cabc /source/know/concept/bb84-protocol
parente5b9bce79b68a68ddd2e51daa16d2fea73b84fdb (diff)
Migrate from 'jekyll-katex' to 'kramdown-math-sskatex'
Diffstat (limited to 'source/know/concept/bb84-protocol')
-rw-r--r--source/know/concept/bb84-protocol/index.md50
1 files changed, 25 insertions, 25 deletions
diff --git a/source/know/concept/bb84-protocol/index.md b/source/know/concept/bb84-protocol/index.md
index 1091773..0f75930 100644
--- a/source/know/concept/bb84-protocol/index.md
+++ b/source/know/concept/bb84-protocol/index.md
@@ -26,8 +26,8 @@ because the later stages of the protocol involve revealing parts of the data
over the (insecure) classical channel.
For each bit, Alice randomly chooses a qubit basis,
-either $\{ \Ket{0}, \Ket{1} \}$ (eigenstates of the $z$-spin $\hat{\sigma}_z$)
-or $\{ \Ket{-}, \Ket{+} \}$ (eigenstates of the $x$-spin $\hat{\sigma}_x$).
+either $$\{ \Ket{0}, \Ket{1} \}$$ (eigenstates of the $$z$$-spin $$\hat{\sigma}_z$$)
+or $$\{ \Ket{-}, \Ket{+} \}$$ (eigenstates of the $$x$$-spin $$\hat{\sigma}_x$$).
Using the basis she chose, she then transmits the bits to Bob over the quantum channel,
encoding them as follows:
@@ -38,7 +38,7 @@ $$\begin{aligned}
\end{aligned}$$
Crucially, Bob has no idea which basis Alice used for any of the bits.
-For every bit, he chooses $\hat{\sigma}_z$ or $\hat{\sigma}_x$ at random,
+For every bit, he chooses $$\hat{\sigma}_z$$ or $$\hat{\sigma}_x$$ at random,
and makes a measurement of the qubit, yielding 0 or 1.
If he guessed the basis correctly, he gets the bit value intended by Alice,
but if he guessed incorrectly, he randomly gets 0 or 1 with a 50-50 probability:
@@ -67,7 +67,7 @@ Suppose that Eve is performing an *intercept-resend attack*
(not very effective, but simple),
where she listens on the quantum channel.
For each qubit received from Alice, Eve chooses
-$\hat{\sigma}_z$ or $\hat{\sigma}_x$ at random and measures it.
+$$\hat{\sigma}_z$$ or $$\hat{\sigma}_x$$ at random and measures it.
She records her results and resends the qubits to Bob
using the basis she chose, which may or may not be what Alice intended.
@@ -105,17 +105,17 @@ In practice, even without Eve, quantum channels are imperfect,
and will introduce some errors in the qubits received by Bob.
Suppose that after basis reconciliation,
Alice and Bob have the strings
-$\{a_1, ..., a_N\}$ and $\{b_1, ..., b_N\}$, respectively.
-We define $p$ as the probability that Alice and Bob agree on the $n$th bit,
+$$\{a_1, ..., a_N\}$$ and $$\{b_1, ..., b_N\}$$, respectively.
+We define $$p$$ as the probability that Alice and Bob agree on the $$n$$th bit,
which we assume to be greater than 50%:
$$\begin{aligned}
p = P(a_n = b_n) > \frac{1}{2}
\end{aligned}$$
-Ideally, $p = 1$. To improve $p$, the following simple scheme can be used:
-starting at $n = 1$, Alice and Bob reveal $A$ and $B$ over the classical channel,
-where $\oplus$ is an XOR:
+Ideally, $$p = 1$$. To improve $$p$$, the following simple scheme can be used:
+starting at $$n = 1$$, Alice and Bob reveal $$A$$ and $$B$$ over the classical channel,
+where $$\oplus$$ is an XOR:
$$\begin{aligned}
A = a_n \oplus a_{n+1}
@@ -123,12 +123,12 @@ $$\begin{aligned}
B = b_n \oplus b_{n+1}
\end{aligned}$$
-If $A = B$, then $a_{n+1}$ and $b_{n+1}$ are discarded to prevent
+If $$A = B$$, then $$a_{n+1}$$ and $$b_{n+1}$$ are discarded to prevent
a listener on the classical channel from learning anything about the string.
-If $A \neq B$, all of $a_n$, $b_n$, $a_{n+1}$ and $b_{n+1}$ are discarded,
-and then Alice and Bob move on to $n = 3$, etc.
+If $$A \neq B$$, all of $$a_n$$, $$b_n$$, $$a_{n+1}$$ and $$b_{n+1}$$ are discarded,
+and then Alice and Bob move on to $$n = 3$$, etc.
-Given that $A = B$, the probability that $a_n = b_n$,
+Given that $$A = B$$, the probability that $$a_n = b_n$$,
which is what we want, is given by:
$$\begin{aligned}
@@ -140,7 +140,7 @@ $$\begin{aligned}
&= \frac{P(a_{n} = b_{n}) \: P(a_{n+1} = b_{n+1})}{P(a_{n} = b_{n}) \: P(a_{n+1} = b_{n+1}) + P(a_{n} \neq b_{n}) \: P(a_{n+1} \neq b_{n+1})}
\end{aligned}$$
-We use the definition of $p$ to get the following inequality,
+We use the definition of $$p$$ to get the following inequality,
which can be verified by plotting:
$$\begin{aligned}
@@ -150,9 +150,9 @@ $$\begin{aligned}
\end{aligned}$$
Alice and Bob can repeat this error correction scheme multiple times,
-until their estimate of $p$ is satisfactory.
+until their estimate of $$p$$ is satisfactory.
This involves discarding many bits,
-so the length $N_\mathrm{new}$ of the string they end up with
+so the length $$N_\mathrm{new}$$ of the string they end up with
after one iteration is given by:
$$\begin{aligned}
@@ -166,11 +166,11 @@ More efficient schemes exist, which do not consume so many bits.
## Privacy amplification
-Suppose that after the error correction step, $p = 1$,
+Suppose that after the error correction step, $$p = 1$$,
so Alice and Bob fully agree on the random string.
However, in the meantime, Eve has been listening,
and has been doing a good job
-building up her own string $\{e_1, ..., e_N\}$,
+building up her own string $$\{e_1, ..., e_N\}$$,
such that she knows more that 50% of the bits:
$$\begin{aligned}
@@ -178,9 +178,9 @@ $$\begin{aligned}
\end{aligned}$$
**Privacy amplification** is an optional final step of the BB84 protocol
-which aims to reduce Eve's $q$.
+which aims to reduce Eve's $$q$$.
Alice and Bob use their existing strings to generate a new one
-$\{a_1', ..., a_M'\}$:
+$$\{a_1', ..., a_M'\}$$:
$$\begin{aligned}
a_1'
@@ -197,8 +197,8 @@ more efficient schemes exist, which consume less.
To see why this improves Alice and Bob's privacy,
suppose that Eve is following along,
-and creates a new string $\{e_1', ..., e_M'\}$
-where $e_m' = e_{2m - 1} \oplus e_{2m}$.
+and creates a new string $$\{e_1', ..., e_M'\}$$
+where $$e_m' = e_{2m - 1} \oplus e_{2m}$$.
The probability that Eve's result agrees with
Alice and Bob's string is given by:
@@ -209,7 +209,7 @@ $$\begin{aligned}
&= P(e_1 = a_1) \: P(e_2 = a_2) + P(e_1 \neq a_1) \: P(e_2 \neq a_2)
\end{aligned}$$
-Recognizing $q$ 's definition,
+Recognizing $$q$$ 's definition,
we find the following inequality,
which can be verified by plotting:
@@ -219,8 +219,8 @@ $$\begin{aligned}
< q
\end{aligned}$$
-After repeating this step several times, $q$ will be close to 1/2,
-which is the ideal value: for $q =$ 0.5,
+After repeating this step several times, $$q$$ will be close to 1/2,
+which is the ideal value: for $$q =$$ 0.5,
Eve would only know 50% of the bits,
which is equivalent to her guessing at random.