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1 files changed, 38 insertions, 28 deletions

diff --git a/content/know/concept/einstein-coefficients/index.pdc b/content/know/concept/einstein-coefficients/index.pdc
index 37141f2..bd8f76c 100644
--- a/content/know/concept/einstein-coefficients/index.pdc
+++ b/content/know/concept/einstein-coefficients/index.pdc
@@ -5,6 +5,7 @@ publishDate: 2021-07-11
 categories:
 - Physics
 - Optics
+- Electromagnetism
 - Quantum mechanics
 
 date: 2021-07-11T18:22:14+02:00
@@ -105,7 +106,7 @@ $$\begin{aligned}
 \end{aligned}$$
 
 Since $u(\omega_0)$ represents only black-body radiation,
-our result must agree with Planck's law:
+our result must agree with [Planck's law](/know/concept/plancks-law/):
 
 $$\begin{aligned}
     u(\omega_0)
@@ -143,31 +144,30 @@ Consider the Hamiltonian of an electron with charge $q = - e$:
 
 $$\begin{aligned}
     \hat{H}
-    &= \frac{\vec{P}{}^2}{2 m} - \frac{q}{2 m} (\vec{A} \cdot \vec{P} + \vec{P} \cdot \vec{A}) + \frac{q^2 \vec{A}{}^2}{2m} + q \phi
+    &= \frac{\vec{P}{}^2}{2 m} - \frac{q}{2 m} (\vec{A} \cdot \vec{P} + \vec{P} \cdot \vec{A}) + \frac{q^2 \vec{A}{}^2}{2m} + V
 \end{aligned}$$
 
-With $\vec{A}(\vec{r}, t)$ the magnetic vector potential,
-and $\phi(\vec{r}, t)$ the electric scalar potential.
+With $\vec{A}(\vec{r}, t)$ the electromagnetic vector potential.
 We reduce this by fixing the Coulomb gauge $\nabla \!\cdot\! \vec{A} = 0$,
 such that $\vec{A} \cdot \vec{P} = \vec{P} \cdot \vec{A}$,
-and by assuming that $\vec{A}{}^2$ is negligibly small.
-This leaves us with:
+and by assuming that $\vec{A}{}^2$ is negligible:
 
 $$\begin{aligned}
     \hat{H}
-    &= \frac{\vec{P}{}^2}{2 m} - \frac{q}{m} \vec{P} \cdot \vec{A} + q \phi
+    &= \frac{\vec{P}{}^2}{2 m} - \frac{q}{m} \vec{P} \cdot \vec{A} + V
 \end{aligned}$$
 
 The last term is the Coulomb interaction
 between the electron and the nucleus.
-We can interpret the second term, involving the weak $\vec{A}$, as a perturbation $\hat{H}_1$:
+We can interpret the second term,
+involving the weak $\vec{A}$, as a perturbation $\hat{H}_1$:
 
 $$\begin{aligned}
     \hat{H}
     = \hat{H}_0 + \hat{H}_1
     \qquad \quad
     \hat{H}_0
-    \equiv \frac{\vec{P}{}^2}{2 m} + q \phi
+    \equiv \frac{\vec{P}{}^2}{2 m} + V
     \qquad \quad
     \hat{H}_1
     \equiv - \frac{q}{m} \vec{P} \cdot \vec{A}
@@ -179,7 +179,9 @@ $$\begin{aligned}
     \vec{A}(\vec{r}, t) = \vec{A}_0 \exp\!(i \vec{k} \cdot \vec{r} - i \omega t)
 \end{aligned}$$
 
-The corresponding perturbative electric field $\vec{E}$ points in the same direction:
+The corresponding perturbative
+[electric field](/know/concept/electric-field/) $\vec{E}$
+points in the same direction:
 
 $$\begin{aligned}
     \vec{E}(\vec{r}, t)
@@ -231,6 +233,14 @@ $$\begin{aligned}
 
 Where $\vec{p} \equiv q \vec{r} = - e \vec{r}$ is the electric dipole moment of the electron,
 hence the name *electric dipole approximation*.
+Finally, because electric fields are actually real
+(we made it complex for mathematical convenience),
+we take the real part, yielding:
+
+$$\begin{aligned}
+    \hat{H}_1(t)
+    = - q \vec{r} \cdot \vec{E}_0 \cos\!(- i \omega t)
+\end{aligned}$$
 
 
 ## Polarized light
@@ -249,19 +259,19 @@ then generally $\ket{1}$ and $\ket{2}$ will be even or odd functions of $z$,
 such that $\matrixel{1}{z}{1} = \matrixel{2}{z}{2} = 0$, leading to:
 
 $$\begin{gathered}
-    \matrixel{1}{H_1}{2} = - q E_0 V
+    \matrixel{1}{H_1}{2} = - q E_0 U
     \qquad
-    \matrixel{2}{H_1}{1} = - q E_0 V^*
+    \matrixel{2}{H_1}{1} = - q E_0 U^*
     \\
     \matrixel{1}{H_1}{1} = \matrixel{2}{H_1}{2} = 0
 \end{gathered}$$
 
-Where $V \equiv \matrixel{1}{z}{2}$ is a constant.
+Where $U \equiv \matrixel{1}{z}{2}$ is a constant.
 The chance of an upward jump (i.e. absorption) is:
 
 $$\begin{aligned}
     P_{12}
-    = \frac{q^2 E_0^2 |V|^2}{\hbar^2} \frac{\sin^2\!\big( (\omega_0 - \omega) t / 2 \big)}{(\omega_0 - \omega)^2}
+    = \frac{q^2 E_0^2 |U|^2}{\hbar^2} \frac{\sin^2\!\big( (\omega_0 - \omega) t / 2 \big)}{(\omega_0 - \omega)^2}
 \end{aligned}$$
 
 Meanwhile, the transition probability for stimulated emission is as follows,
@@ -270,7 +280,7 @@ and is therefore symmetric around $\omega_{ba}$:
 
 $$\begin{aligned}
     P_{21}
-    = \frac{q^2 E_0^2 |V|^2}{\hbar^2} \frac{\sin^2\!\big( (\omega_0 - \omega) t / 2 \big)}{(\omega_0 - \omega)^2}
+    = \frac{q^2 E_0^2 |U|^2}{\hbar^2} \frac{\sin^2\!\big( (\omega_0 - \omega) t / 2 \big)}{(\omega_0 - \omega)^2}
 \end{aligned}$$
 
 Surprisingly, the probabilities of absorption and stimulated emission are the same!
@@ -295,7 +305,7 @@ Putting this in the previous result gives the following transition probability:
 
 $$\begin{aligned}
     P_{12}
-    = \frac{2 u q^2 |V|^2}{\varepsilon_0 \hbar^2} \frac{\sin^2\!\big( (\omega_0 - \omega) t / 2 \big)}{(\omega_0 - \omega)^2}
+    = \frac{2 u q^2 |U|^2}{\varepsilon_0 \hbar^2} \frac{\sin^2\!\big( (\omega_0 - \omega) t / 2 \big)}{(\omega_0 - \omega)^2}
 \end{aligned}$$
 
 For a continuous light spectrum,
@@ -303,7 +313,7 @@ this $u$ turns into the spectral energy density $u(\omega)$:
 
 $$\begin{aligned}
     P_{12}
-    = \frac{2 q^2 |V|^2}{\varepsilon_0 \hbar^2}
+    = \frac{2 q^2 |U|^2}{\varepsilon_0 \hbar^2}
     \int_0^\infty \frac{\sin^2\!\big( (\omega_0 - \omega) t / 2 \big)}{(\omega_0 - \omega)^2} u(\omega) \dd{\omega}
 \end{aligned}$$
 
@@ -319,8 +329,8 @@ which turns out to be $\pi t$:
 
 $$\begin{aligned}
     P_{12}
-    = \frac{q^2 |V|^2}{\varepsilon_0 \hbar^2} u(\omega_0) \int_{-\infty}^\infty \frac{\sin^2\!\big(x t \big)}{x^2} \dd{x}
-    = \frac{\pi q^2 |V|^2}{\varepsilon_0 \hbar^2} u(\omega_0) \:t
+    = \frac{q^2 |U|^2}{\varepsilon_0 \hbar^2} u(\omega_0) \int_{-\infty}^\infty \frac{\sin^2\!\big(x t \big)}{x^2} \dd{x}
+    = \frac{\pi q^2 |U|^2}{\varepsilon_0 \hbar^2} u(\omega_0) \:t
 \end{aligned}$$
 
 From this, the transition rate $R_{12} = B_{12} u(\omega_0)$
@@ -329,7 +339,7 @@ is then calculated as follows:
 $$\begin{aligned}
     R_{12}
     = \pdv{P_{2 \to 1}}{t}
-    = \frac{\pi q^2 |V|^2}{\varepsilon_0 \hbar^2} u(\omega_0)
+    = \frac{\pi q^2 |U|^2}{\varepsilon_0 \hbar^2} u(\omega_0)
 \end{aligned}$$
 
 Using the relations from earlier with $g_1 = g_2$,
@@ -338,9 +348,9 @@ for a polarized incoming light spectrum:
 
 $$\begin{aligned}
     \boxed{
-        B_{21} = B_{12} = \frac{\pi q^2 |V|^2}{\varepsilon_0 \hbar^2}
+        B_{21} = B_{12} = \frac{\pi q^2 |U|^2}{\varepsilon_0 \hbar^2}
         \qquad
-        A_{21} = \frac{\omega_0^3 q^2 |V|^2}{\pi \varepsilon \hbar c^3}
+        A_{21} = \frac{\omega_0^3 q^2 |U|^2}{\pi \varepsilon \hbar c^3}
     }
 \end{aligned}$$
 
@@ -375,9 +385,9 @@ Evaluating the integrals yields:
 
 $$\begin{aligned}
     \expval{|W|^2}
-    = \frac{2 \pi}{4 \pi} |V|^2 \int_0^\pi \cos^2(\theta) \sin\!(\theta) \dd{\theta}
-    = \frac{|V|^2}{2} \Big[ \!-\! \frac{\cos^3(\theta)}{3} \Big]_0^\pi
-    = \frac{|V|^2}{3}
+    = \frac{2 \pi}{4 \pi} |U|^2 \int_0^\pi \cos^2(\theta) \sin\!(\theta) \dd{\theta}
+    = \frac{|U|^2}{2} \Big[ \!-\! \frac{\cos^3(\theta)}{3} \Big]_0^\pi
+    = \frac{|U|^2}{3}
 \end{aligned}$$
 
 With this additional constant factor $1/3$,
@@ -385,16 +395,16 @@ the transition rate $R_{12}$ is modified to:
 
 $$\begin{aligned}
     R_{12}
-    = \frac{\pi q^2 |V|^2}{3 \varepsilon_0 \hbar^2} u(\omega_0)
+    = \frac{\pi q^2 |U|^2}{3 \varepsilon_0 \hbar^2} u(\omega_0)
 \end{aligned}$$
 
 From which it follows that the Einstein coefficients for unpolarized light are given by:
 
 $$\begin{aligned}
     \boxed{
-        B_{21} = B_{12} = \frac{\pi q^2 |V|^2}{3 \varepsilon_0 \hbar^2}
+        B_{21} = B_{12} = \frac{\pi q^2 |U|^2}{3 \varepsilon_0 \hbar^2}
         \qquad
-        A_{21} = \frac{\omega_0^3 q^2 |V|^2}{3 \pi \varepsilon \hbar c^3}
+        A_{21} = \frac{\omega_0^3 q^2 |U|^2}{3 \pi \varepsilon \hbar c^3}
     }
 \end{aligned}$$