From cc391ce3b9867d88d124e147931d33be34e756fc Mon Sep 17 00:00:00 2001
From: Prefetch
Date: Mon, 14 Sep 2026 18:11:38 +0200
Subject: Improve knowledge base

---
 source/know/concept/ehrenfests-theorem/index.md | 75 ++++++++++++-------------
 1 file changed, 36 insertions(+), 39 deletions(-)

(limited to 'source/know/concept/ehrenfests-theorem')

diff --git a/source/know/concept/ehrenfests-theorem/index.md b/source/know/concept/ehrenfests-theorem/index.md
index fba0192..14d17fa 100644
--- a/source/know/concept/ehrenfests-theorem/index.md
+++ b/source/know/concept/ehrenfests-theorem/index.md
@@ -8,20 +8,19 @@ categories:
 layout: "concept"
 ---
 
-In quantum mechanics, **Ehrenfest's theorem** gives a general expression for the
-time evolution of an observable's expectation value $$\expval{\hat{L}}$$.
-
-The time-dependent Schrödinger equation is as follows,
+In quantum mechanics, **Ehrenfest's theorem** gives a general expression
+for the time evolution of an observable's expectation value $$\expval{\hat{L}}$$.
+Recall the time-dependent Schrödinger equation,
 where prime denotes differentiation with respect to time $$t$$:
 
 $$\begin{aligned}
     \Ket{\psi'} = \frac{1}{i \hbar} \hat{H} \Ket{\psi}
-    \qquad
+    \qquad \qquad
     \Bra{\psi'} = - \frac{1}{i \hbar} \Bra{\psi} \hat{H}
 \end{aligned}$$
 
 Given an observable operator $$\hat{L}$$ and a state $$\Ket{\psi}$$,
-the time-derivative of the expectation value $$\expval{\hat{L}}$$ is as follows
+the $$t$$-derivative of the expectation value $$\expval{\hat{L}}$$ is as follows
 (due to the product rule of differentiation):
 
 $$\begin{aligned}
@@ -43,28 +42,26 @@ $$\begin{aligned}
     }
 \end{aligned}$$
 
-In practice, since most operators are time-independent,
-the last term often vanishes.
-
-As a interesting side note, in the [Heisenberg picture](/know/concept/heisenberg-picture/),
-this relation proves itself,
-when one simply wraps all terms in $$\Bra{\psi}$$ and $$\Ket{\psi}$$.
+In practice, since most operators are time-independent, the last term often vanishes.
+Note that this relation is trivial to prove
+in the [Heisenberg picture](/know/concept/heisenberg-picture/),
+by wrapping all terms in $$\Bra{\psi}$$ and $$\Ket{\psi}$$.
 
-Two observables of particular interest are the position $$\hat{X}$$ and momentum $$\hat{P}$$.
-Applying the above theorem to $$\hat{X}$$ yields the following,
-which we reduce using the fact that $$\hat{X}$$ commutes
-with the potential $$V(\hat{X})$$,
-because one is a function of the other:
+Two observables of particular interest
+are position $$\hat{X}$$ and momentum $$\hat{P}$$.
+Applying the theorem to $$\hat{X}$$ yields the following,
+using $$\hat{H} = \hat{P}^2 / (2 m) + V(\hat{X})$$
+and a few basic properties of commutators:
 
 $$\begin{aligned}
     \dv{\expval{\hat{X}}}{t}
     &= \frac{1}{i \hbar} \Expval{[\hat{X}, \hat{H}]}
-    = \frac{1}{2 i \hbar m} \Expval{[\hat{X}, \hat{P}^2] + 2 m [\hat{X}, V(\hat{X})]}
-    = \frac{1}{2 i \hbar m} \Expval{[\hat{X}, \hat{P}^2]}
+    \\
+    &= \frac{1}{2 i \hbar m} \Expval{[\hat{X}, \hat{P}^2] + 2 m [\hat{X}, V(\hat{X})]}
     \\
     &= \frac{1}{2 i \hbar m} \Expval{\hat{P} [\hat{X}, \hat{P}] + [\hat{X}, \hat{P}] \hat{P}}
-    = \frac{2 i \hbar}{2 i \hbar m} \expval{\hat{P}}
-    = \frac{\expval{\hat{P}}}{m}
+    \\
+    &= \frac{2 i \hbar}{2 i \hbar m} \expval{\hat{P}}
 \end{aligned}$$
 
 This is the first part of the "original" form of Ehrenfest's theorem,
@@ -72,7 +69,8 @@ which is reminiscent of classical Newtonian mechanics:
 
 $$\begin{gathered}
     \boxed{
-        \dv{\expval{\hat{X}}}{t} = \frac{\expval{\hat{P}}}{m}
+        \dv{\expval{\hat{X}}}{t}
+        = \frac{\expval{\hat{P}}}{m}
     }
 \end{gathered}$$
 
@@ -82,33 +80,32 @@ gives us:
 $$\begin{aligned}
     \dv{\expval{\hat{P}}}{t}
     &= \frac{1}{i \hbar} \Expval{[\hat{P}, \hat{H}]}
-    = \frac{1}{2 i \hbar m} \Expval{[\hat{P}, \hat{P}^2] + 2 m [\hat{P}, V(\hat{X})]}
-    = \frac{1}{i \hbar} \Expval{[\hat{P}, V(\hat{X})]}
+    \\
+    &= \frac{1}{2 i \hbar m} \Expval{[\hat{P}, \hat{P}^2] + 2 m [\hat{P}, V(\hat{X})]}
+    \\
+    &= \frac{1}{i \hbar} \Expval{[\hat{P}, V(\hat{X})]}
 \end{aligned}$$
 
-To find the commutator, we go to the $$\hat{X}$$-basis and use a test
-function $$f(x)$$:
+To evaluate the commutator,
+we go to the $$\hat{X}$$-basis and use a test function $$f(x)$$:
 
 $$\begin{aligned}
     \Comm{- i \hbar \dv{}{x}}{V(x)} \: f(x)
+    &= - i \hbar \dv{}{x} \Big( V(x) \: f(x) \Big) - V(x) \Big( \!-\! i \hbar \dv{}{x} \Big) f(x)
+    \\
     &= - i \hbar \frac{dV}{dx} f(x) - i \hbar V(x) \frac{df}{dx} + i \hbar V(x) \frac{df}{dx}
-    = - i \hbar \frac{dV}{dx} f(x)
-\end{aligned}$$
-
-By inserting this result back into the previous equation, we find the following:
-
-$$\begin{aligned}
-    \dv{\expval{\hat{P}}}{t}
-    &= - \frac{i \hbar}{i \hbar} \Expval{\frac{d V}{d \hat{X}}}
-    = - \Expval{\frac{d V}{d \hat{X}}}
+    \\
+    &= - i \hbar \frac{dV}{dx} f(x)
 \end{aligned}$$
 
-This is the second part of Ehrenfest's theorem,
-which is also similar to Newtonian mechanics:
+By inserting this result back into the previous equation,
+we find the second part of Ehrenfest's original theorem,
+which is again reminiscent Newtonian mechanics:
 
 $$\begin{gathered}
     \boxed{
-        \dv{\expval{\hat{P}}}{t} = - \Expval{\pdv{V}{\hat{X}}}
+        \dv{\expval{\hat{P}}}{t}
+        = - \Expval{\pdv{V}{\hat{X}}}
     }
 \end{gathered}$$
 
@@ -121,7 +118,7 @@ $$\begin{gathered}
         \Expval{\pdv{\hat{H}}{\hat{P}}}
         = \dv{\expval{\hat{X}}}{t}
     }
-    \qquad \quad
+    \qquad \qquad
     \boxed{
         - \Expval{\pdv{\hat{H}}{\hat{X}}}
         = \dv{\expval{\hat{P}}}{t}
-- 
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