Migrate from 'jekyll-katex' to 'kramdown-math-sskatex'

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.