From 29b49508a751649310173e592b63415dbf563a2a Mon Sep 17 00:00:00 2001
From: Prefetch
Date: Thu, 3 Sep 2026 15:05:35 +0200
Subject: Improve knowledge base

---
 source/know/concept/quantum-teleportation/index.md | 17 ++++++++++-------
 1 file changed, 10 insertions(+), 7 deletions(-)

(limited to 'source/know/concept/quantum-teleportation/index.md')

diff --git a/source/know/concept/quantum-teleportation/index.md b/source/know/concept/quantum-teleportation/index.md
index 095c2c6..f57f981 100644
--- a/source/know/concept/quantum-teleportation/index.md
+++ b/source/know/concept/quantum-teleportation/index.md
@@ -22,7 +22,7 @@ $$\begin{aligned}
 
 She can only directly communicate with Bob over a classical channel.
 This is not enough: even if Alice did know $$\alpha$$ and $$\beta$$ exactly
-(which would need her having infinitely many copies to measure),
+(for which she would need infinitely many copies to measure),
 sending an arbitrary real number requires an infinite amount of classical data.
 
 However, between them, she and Bob also have an entangled [Bell state](/know/concept/bell-state/),
@@ -32,7 +32,7 @@ with $$A'$$ being Alice' qubit, $$A$$ her side of the Bell state, and $$B$$ Bob'
 
 $$\begin{aligned}
     \Ket{q}_{A'} \otimes \ket{\Phi^+}_{AB}
-    &= \frac{1}{\sqrt{2}} \Big( \alpha \Ket{0} + \beta \Ket{1} \Big)_{A'} \Big( \Ket{00} + \Ket{11} \Big)_{AB}
+    &= \frac{1}{\sqrt{2}} \Big( \alpha \Ket{0} + \beta \Ket{1} \Big)_{A'} \otimes \Big( \Ket{00} + \Ket{11} \Big)_{AB}
     \\
     &= \frac{1}{\sqrt{2}} \Big( \alpha \Ket{000} + \beta \Ket{100}
     + \alpha \Ket{011} + \beta \Ket{111} \Big)_{A'AB}
@@ -78,8 +78,10 @@ $$\begin{aligned}
     + \ket{\Psi^{-}}_{A'A} \Big( \alpha \Ket{1} - \beta \Ket{0} \Big)_{B} \bigg)
 \end{aligned}$$
 
-Thus, purely due to entanglement,
-Bob's qubit $$B$$ is in a superposition of the following states:
+Therefore, thanks to entanglement,
+Bob's qubit $$B$$ is in a superposition of the following states,
+where $$\hat{\sigma}_x$$ and $$\hat{\sigma}_z$$ are Pauli matrices
+(see [quantum gate](/know/concept/quantum-gate/)):
 
 $$\begin{aligned}
     \Ket{q}
@@ -95,8 +97,8 @@ $$\begin{aligned}
     = \alpha \Ket{1} - \beta \Ket{0}
 \end{aligned}$$
 
-Consequently, Alice and Bob are sharing (or, to be precise, seeing different sides of)
-the following entangled three-qubit state:
+Consequently, Alice and Bob are seeing different sides of
+this entangled three-qubit state:
 
 $$\begin{aligned}
     \Ket{q}_{A'} \ket{\Phi^+}_{AB}
@@ -123,7 +125,7 @@ who then either does nothing (for $$\Ket{q}$$),
 applies $$\hat{\sigma}_z$$ (for $$\hat{\sigma}_z \Ket{q}$$),
 applies $$\hat{\sigma}_x$$ (for $$\hat{\sigma}_x \Ket{q}$$),
 or applies $$\hat{\sigma}_z \hat{\sigma}_x$$ (for $$\hat{\sigma}_x \hat{\sigma}_z \Ket{q}$$).
-Then, due to the fact that $$\hat{\sigma}_x^2 = \hat{\sigma}_z^2 = \hat{I}$$,
+Then, thanks to the fact that $$\hat{\sigma}_x^2 = \hat{\sigma}_z^2 = \hat{I}$$,
 he recovers $$\Ket{q}$$ in his local qubit $$B$$.
 
 This is not violating the [no-cloning theorem](/know/concept/no-cloning-theorem)
@@ -140,6 +142,7 @@ Before receiving that, Bob only sees his side of the maximally entangled
 Bell state $$\ket{\Phi^{+}}_{AB}$$, which contains nothing of $$\Ket{q}$$.
 
 
+
 ## References
 1.  J.B. Brask,
     *Quantum information: lecture notes*,
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