From bd13537ee2fb704b02b961b5d06dd4f406f19a71 Mon Sep 17 00:00:00 2001
From: Prefetch
Date: Sat, 21 Oct 2023 14:21:59 +0200
Subject: Improve knowledge base

---
 .../concept/electric-dipole-approximation/index.md | 89 +++++++++++++---------
 1 file changed, 52 insertions(+), 37 deletions(-)

(limited to 'source/know/concept/electric-dipole-approximation/index.md')

diff --git a/source/know/concept/electric-dipole-approximation/index.md b/source/know/concept/electric-dipole-approximation/index.md
index 35cf00c..06f0f45 100644
--- a/source/know/concept/electric-dipole-approximation/index.md
+++ b/source/know/concept/electric-dipole-approximation/index.md
@@ -13,20 +13,22 @@ layout: "concept"
 
 Suppose that an [electromagnetic wave](/know/concept/electromagnetic-wave-equation/)
 is travelling through an atom, and affecting the electrons.
-The general Hamiltonian of an electron in such a wave is given by:
+The general Hamiltonian of an electron in an electromagnetic field is:
 
 $$\begin{aligned}
     \hat{H}
-    &= \frac{(\vu{P} - q \vb{A})^2}{2 m} + q \varphi
+    &= \frac{(\vu{P} - q \vb{A})^2}{2 m} + q \Phi
     \\
-    &= \frac{\vu{P}{}^2}{2 m} - \frac{q}{2 m} (\vb{A} \cdot \vu{P} + \vu{P} \cdot \vb{A}) + \frac{q^2 \vb{A}^2}{2m} + q \varphi
+    &= \frac{\vu{P}{}^2}{2 m} - \frac{q}{2 m} (\vb{A} \cdot \vu{P} + \vu{P} \cdot \vb{A}) + \frac{q^2 \vb{A}^2}{2m} + q \Phi
 \end{aligned}$$
 
-With charge $$q = - e$$,
-canonical momentum operator $$\vu{P} = - i \hbar \nabla$$,
-and magnetic vector potential $$\vb{A}(\vb{x}, t)$$.
-We reduce this by fixing the Coulomb gauge $$\nabla \cdot \vb{A} = 0$$,
-so that $$\vb{A} \cdot \vu{P} = \vu{P} \cdot \vb{A}$$:
+Where $$q < 0$$ is the electron's charge,
+$$\vu{P} = - i \hbar \nabla$$ is the canonical momentum operator,
+$$\vb{A}$$ is the magnetic vector potential,
+and $$\Phi$$ is the electric scalar potential.
+We start by fixing the Coulomb gauge $$\nabla \cdot \vb{A} = 0$$
+such that $$\vu{P}$$ and $$\vb{A}$$ commute;
+let $$\psi$$ be an arbitrary test function:
 
 $$\begin{aligned}
     \comm{\vb{A}}{\vu{P}} \psi
@@ -35,20 +37,29 @@ $$\begin{aligned}
     &= -i \hbar \vb{A} \cdot (\nabla \psi) + i \hbar \nabla \cdot (\vb{A} \psi)
     \\
     &= i \hbar (\nabla \cdot \vb{A}) \psi
-    = 0
+    \\
+    &= 0
+\end{aligned}$$
+
+Meaning $$\vb{A} \cdot \vu{P} = \vu{P} \cdot \vb{A}$$.
+Furthermore, we assume that $$\vb{A}$$ is so small that $$\vb{A}{}^2$$ is negligible,
+so the Hamiltonian is reduced to:
+
+$$\begin{aligned}
+    \hat{H}
+    &\approx \frac{\vu{P}{}^2}{2 m} - \frac{q}{m} \vu{P} \cdot \vb{A} + q \Phi
 \end{aligned}$$
 
-Where $$\psi$$ is an arbitrary test function.
-Assuming $$\vb{A}$$ is so small that $$\vb{A}{}^2$$ is negligible, we split $$\hat{H}$$ as follows,
-where $$\hat{H}_1$$ can be regarded as a perturbation to $$\hat{H}_0$$:
+We now split $$\hat{H}$$ like so,
+where $$\hat{H}_1$$ can be regarded as a perturbation to the "base" $$\hat{H}_0$$:
 
 $$\begin{aligned}
     \hat{H}
     = \hat{H}_0 + \hat{H}_1
-    \qquad \quad
+    \qquad\qquad
     \hat{H}_0
-    \equiv \frac{\vu{P}{}^2}{2 m} + q \varphi
-    \qquad \quad
+    \equiv \frac{\vu{P}{}^2}{2 m} + q \Phi
+    \qquad\qquad
     \hat{H}_1
     \equiv - \frac{q}{m} \vu{P} \cdot \vb{A}
 \end{aligned}$$
@@ -56,14 +67,16 @@ $$\begin{aligned}
 In an electromagnetic wave, $$\vb{A}$$ is oscillating sinusoidally in time and space:
 
 $$\begin{aligned}
-    \vb{A}(\vb{x}, t) = \vb{A}_0 \sin(\vb{k} \cdot \vb{x} - \omega t)
+    \vb{A}(\vb{x}, t)
+    = \vb{A}_0 \sin(\vb{k} \cdot \vb{x} - \omega t)
 \end{aligned}$$
 
 Mathematically, it is more convenient to represent this with a complex exponential,
 whose real part should be taken at the end of the calculation:
 
 $$\begin{aligned}
-    \vb{A}(\vb{x}, t) = - i \vb{A}_0 \exp(i \vb{k} \cdot \vb{x} - i \omega t)
+    \vb{A}(\vb{x}, t)
+    = - i \vb{A}_0 \exp(i \vb{k} \cdot \vb{x} - i \omega t)
 \end{aligned}$$
 
 The corresponding perturbative [electric field](/know/concept/electric-field/) $$\vb{E}$$ is then given by:
@@ -75,10 +88,10 @@ $$\begin{aligned}
 \end{aligned}$$
 
 Where $$\vb{E}_0 = \omega \vb{A}_0$$.
-Let us restrict ourselves to visible light,
-whose wavelength $$2 \pi / |\vb{k}| \sim 10^{-6} \:\mathrm{m}$$.
-Meanwhile, an atomic orbital is several Bohr radii $$\sim 10^{-10} \:\mathrm{m}$$,
-so $$\vb{k} \cdot \vb{x}$$ is negligible:
+Light in and around the visible spectrum
+has a wavelength $$2 \pi / |\vb{k}| \sim 10^{-7} \:\mathrm{m}$$,
+while an atomic orbital is several Bohr radii $$\sim 10^{-10} \:\mathrm{m}$$,
+so $$\vb{k} \cdot \vb{x}$$ is very small. Therefore:
 
 $$\begin{aligned}
     \boxed{
@@ -96,14 +109,13 @@ and the electron quantum-mechanically.
 
 Next, we want to rewrite $$\hat{H}_1$$
 to use the electric field $$\vb{E}$$ instead of the potential $$\vb{A}$$.
-To do so, we use that $$\vu{P} = m \: \idv{\vu{x}}{t}$$
+To do so, we use that momentum $$\vu{P} \equiv m \: \idv{\vu{x}}{t}$$
 and evaluate this in the [interaction picture](/know/concept/interaction-picture/):
 
 $$\begin{aligned}
     \vu{P}
-    = m \idv{\vu{x}}{t}
+    &= m \dv{\vu{x}}{t}
     = m \frac{i}{\hbar} \comm{\hat{H}_0}{\vu{x}}
-    = m \frac{i}{\hbar} (\hat{H}_0 \vu{x} - \vu{x} \hat{H}_0)
 \end{aligned}$$
 
 Taking the off-diagonal inner product with
@@ -111,34 +123,37 @@ the two-level system's states $$\Ket{1}$$ and $$\Ket{2}$$ gives:
 
 $$\begin{aligned}
     \matrixel{2}{\vu{P}}{1}
-    = m \frac{i}{\hbar} \matrixel{2}{\hat{H}_0 \vu{x} - \vu{x} \hat{H}_0}{1}
-    = m i \omega_0 \matrixel{2}{\vu{x}}{1}
+    &= m \frac{i}{\hbar} \matrixel{2}{\hat{H}_0 \vu{x} - \vu{x} \hat{H}_0}{1}
+    \\
+    &= m i \omega_0 \matrixel{2}{\vu{x}}{1}
 \end{aligned}$$
 
-Therefore, $$\vu{P} / m = i \omega_0 \vu{x}$$,
-where $$\omega_0 \equiv (E_2 \!-\! E_1) / \hbar$$ is the resonance of the energy gap,
+Where $$\omega_0 \equiv (E_2 \!-\! E_1) / \hbar$$ is the resonance of the energy gap,
 close to which we assume that $$\vb{A}$$ and $$\vb{E}$$ are oscillating, i.e. $$\omega \approx \omega_0$$.
-We thus get:
+Therefore, $$\vu{P} / m = i \omega_0 \vu{x}$$, so we get:
 
 $$\begin{aligned}
     \hat{H}_1(t)
     &= - \frac{q}{m} \vu{P} \cdot \vb{A}
-    = - (- i i) q \omega_0 \vu{x} \cdot \vb{A}_0 \exp(- i \omega t)
     \\
-    &\approx - q \vu{x} \cdot \vb{E}_0 \exp(- i \omega t)
-    = - \vu{d} \cdot \vb{E}_0 \exp(- i \omega t)
+    &= - (- i i) q \omega_0 \vu{x} \cdot \vb{A}_0 \exp(- i \omega t)
+    \\
+    &\approx - \vu{d} \cdot \vb{E}_0 \exp(- i \omega t)
 \end{aligned}$$
 
-Where $$\vu{d} \equiv q \vu{x} = - e \vu{x}$$ is
+Where $$\vu{d} \equiv q \vu{x}$$ is
 the **transition dipole moment operator** of the electron,
-hence the name **electric dipole approximation**.
+hence the name *electric dipole approximation*.
 Finally, we take the real part, yielding:
 
 $$\begin{aligned}
     \boxed{
-        \hat{H}_1(t)
-        = - \vu{d} \cdot \vb{E}(t)
-        = - q \vu{x} \cdot \vb{E}_0 \cos(\omega t)
+        \begin{aligned}
+            \hat{H}_1(t)
+            &= - \vu{d} \cdot \vb{E}(t)
+            \\
+            &= - q \vu{x} \cdot \vb{E}_0 \cos(\omega t)
+        \end{aligned}
     }
 \end{aligned}$$
 
-- 
cgit v1.2.3