Improve knowledge base

1 files changed, 15 insertions, 8 deletions

diff --git a/source/know/concept/bb84-protocol/index.md b/source/know/concept/bb84-protocol/index.md
index 0f75930..44ea57d 100644
--- a/source/know/concept/bb84-protocol/index.md
+++ b/source/know/concept/bb84-protocol/index.md
@@ -48,7 +48,7 @@ $$\begin{aligned}
 \end{aligned}$$
 
 After Alice has sent all her qubits,
-the next step is **basis reconciliation**:
+the next step is *basis reconciliation*:
 over the classical channel, Bob announces, for each bit,
 which basis he chose, and Alice tells him if he was right or wrong.
 Bob discards all bits where he guessed wrongly.
@@ -56,6 +56,7 @@ If their quantum channel did not have any noise or eavesdroppers,
 Alice and Bob now have a perfectly correlated secret string of bits.
 
 
+
 ## Eavesdropper detection
 
 But what if there is actually an eavesdropper?
@@ -99,6 +100,7 @@ that as long as the error rate is below 11%,
 the BB84 protocol is fully secure, i.e. there cannot be any eavesdroppers.
 
 
+
 ## Error correction
 
 In practice, even without Eve, quantum channels are imperfect,
@@ -125,11 +127,14 @@ $$\begin{aligned}
 
 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 $$A \neq B$$, something went wrong,
+so all of $$a_n$$, $$b_n$$, $$a_{n+1}$$ and $$b_{n+1}$$ are discarded.
+Then Alice and Bob move on to $$n = 3$$, etc.
 
-Given that $$A = B$$, the probability that $$a_n = b_n$$,
-which is what we want, is given by:
+Given that $$A = B$$, the probability that $$a_n = b_n$$ is as shown below.
+Note that there are two possible explanations for $$A = B$$:
+either $$a_{n} = b_{n} \land a_{n+1} = b_{n+1}$$,
+or $$a_{n} \neq b_{n} \land a_{n+1} \neq b_{n+1}$$:
 
 $$\begin{aligned}
     P(a_{n} = b_{n} | A = B)
@@ -164,6 +169,7 @@ $$\begin{aligned}
 More efficient schemes exist, which do not consume so many bits.
 
 
+
 ## Privacy amplification
 
 Suppose that after the error correction step, $$p = 1$$,
@@ -177,10 +183,10 @@ $$\begin{aligned}
     q = P(e_n = a_n) > \frac{1}{2}
 \end{aligned}$$
 
-**Privacy amplification** is an optional final step of the BB84 protocol
-which aims to reduce Eve's $$q$$.
+*Privacy amplification* is an optional final step of the BB84 protocol
+that aims to reduce Eve's $$q$$.
 Alice and Bob use their existing strings to generate a new one
-$$\{a_1', ..., a_M'\}$$:
+$$\{a_1', ..., a_M'\}$$ where:
 
 $$\begin{aligned}
     a_1'
@@ -225,6 +231,7 @@ Eve would only know 50% of the bits,
 which is equivalent to her guessing at random.
 
 
+
 ## References
 1.  N. Brunner,
     *Quantum information theory: lecture notes*,