From 5cacf4ffaf3a9621ab536195f6469f98a420f054 Mon Sep 17 00:00:00 2001
From: Prefetch
Date: Sat, 5 Sep 2026 21:55:33 +0200
Subject: Improve knowledge base

---
 source/know/concept/heisenberg-picture/index.md | 3 +--
 1 file changed, 1 insertion(+), 2 deletions(-)

(limited to 'source/know/concept/heisenberg-picture')

diff --git a/source/know/concept/heisenberg-picture/index.md b/source/know/concept/heisenberg-picture/index.md
index 54bf397..3ffe29a 100644
--- a/source/know/concept/heisenberg-picture/index.md
+++ b/source/know/concept/heisenberg-picture/index.md
@@ -118,7 +118,7 @@ This result is arguably more intuitive than the Schrödinger picture,
 because it allows us to think about observables (i.e. operators) in a more classical way.
 For example, inserting the position $$\hat{X}$$
 and momentum $$\hat{P} = - i \hbar \: \idv{}{\hat{X}}$$
-gives the following Newton-style relations:
+gives the following Newton-style relations (details omitted):
 
 $$\begin{aligned}
     \dv{\hat{X}}{t}
@@ -130,7 +130,6 @@ $$\begin{aligned}
     = - \dv{V(\hat{X})}{\hat{X}}
 \end{aligned}$$
 
-Where the commutators have been treated as known.
 These equations would not be valid in the Schrödinger picture,
 unless we took their expectation value
 to get [Ehrenfest's theorem](/know/concept/ehrenfests-theorem/).
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