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/bell-state/index.md | 20 +++++++++++---------
 1 file changed, 11 insertions(+), 9 deletions(-)

(limited to 'source/know/concept/bell-state')

diff --git a/source/know/concept/bell-state/index.md b/source/know/concept/bell-state/index.md
index fa289de..d4508e6 100644
--- a/source/know/concept/bell-state/index.md
+++ b/source/know/concept/bell-state/index.md
@@ -24,14 +24,14 @@ $$\begin{aligned}
     }
 \end{aligned}$$
 
-Where e.g. $$\ket{0}_A \ket{1}_B = \ket{0}_A \otimes \ket{1}_B$$
+Where e.g. $$\ket{0}_A \ket{1}_B \equiv \ket{0}_A \otimes \ket{1}_B$$
 is the tensor product of qubit $$A$$ in state $$\ket{0}$$ and $$B$$ in $$\ket{1}$$.
 These states form an orthonormal basis for the two-qubit
 [Hilbert space](/know/concept/hilbert-space/).
 
 More importantly, however,
-is that the Bell states are maximally entangled,
-which we prove here for $$\ket{\Phi^{+}}$$.
+is that all four Bell states are *maximally entangled*.
+For brevity, we will only show this for $$\ket{\Phi^{+}}$$ here.
 Consider the following pure [density operator](/know/concept/density-operator/):
 
 $$\begin{aligned}
@@ -40,7 +40,8 @@ $$\begin{aligned}
     &= \frac{1}{2} \Big( \ket{0}_A \ket{0}_B + \ket{1}_A \ket{1}_B \Big) \Big( \bra{0}_A \bra{0}_B + \bra{1}_A \bra{1}_B \Big)
 \end{aligned}$$
 
-The reduced density operator $$\hat{\rho}_A$$ of qubit $$A$$ is then calculated as follows:
+The *reduced* density operator $$\hat{\rho}_A$$ of qubit $$A$$
+is then calculated like so, using a partial trace:
 
 $$\begin{aligned}
     \hat{\rho}_A
@@ -54,12 +55,13 @@ $$\begin{aligned}
     = \frac{1}{2} \hat{I}
 \end{aligned}$$
 
-This result is maximally mixed, therefore $$\ket{\Phi^{+}}$$ is maximally entangled.
-The same holds for the other three Bell states,
-and is equally true for qubit $$B$$.
-
+The same holds for qubit $$B$$. This result is *maximally mixed*,
+therefore $$\ket{\Phi^{+}}$$ is maximally entangled.
 This means that a measurement of qubit $$A$$
-has a 50-50 chance to yield $$\ket{0}$$ or $$\ket{1}$$.
+has a 50-50 chance to yield $$\ket{0}$$ or $$\ket{1}$$,
+or in other words, no useful information can be gathered
+from measuring just one of the qubits.
+
 However, due to the entanglement,
 measuring $$A$$ also has consequences for qubit $$B$$:
 
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