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1 files changed, 57 insertions, 132 deletions

diff --git a/content/know/concept/einstein-coefficients/index.pdc b/content/know/concept/einstein-coefficients/index.pdc
index bd8f76c..80707c6 100644
--- a/content/know/concept/einstein-coefficients/index.pdc
+++ b/content/know/concept/einstein-coefficients/index.pdc
@@ -136,115 +136,31 @@ This situation is mandatory for lasers, where stimulated emission must dominate,
 such that the light becomes stronger as it travels through the medium.
 
 
-## Electric dipole approximation
+## Coherent light
 
-In fact, we can analytically calculate the Einstein coefficients,
-if we make a mild approximation.
-Consider the Hamiltonian of an electron with charge $q = - e$:
-
-$$\begin{aligned}
-    \hat{H}
-    &= \frac{\vec{P}{}^2}{2 m} - \frac{q}{2 m} (\vec{A} \cdot \vec{P} + \vec{P} \cdot \vec{A}) + \frac{q^2 \vec{A}{}^2}{2m} + V
-\end{aligned}$$
-
-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 negligible:
-
-$$\begin{aligned}
-    \hat{H}
-    &= \frac{\vec{P}{}^2}{2 m} - \frac{q}{m} \vec{P} \cdot \vec{A} + V
-\end{aligned}$$
-
-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 $\hat{H}_1$:
-
-$$\begin{aligned}
-    \hat{H}
-    = \hat{H}_0 + \hat{H}_1
-    \qquad \quad
-    \hat{H}_0
-    \equiv \frac{\vec{P}{}^2}{2 m} + V
-    \qquad \quad
-    \hat{H}_1
-    \equiv - \frac{q}{m} \vec{P} \cdot \vec{A}
-\end{aligned}$$
-
-Suppose that $\vec{A}$ is oscillating sinusoidally in time and space as follows:
-
-$$\begin{aligned}
-    \vec{A}(\vec{r}, t) = \vec{A}_0 \exp\!(i \vec{k} \cdot \vec{r} - i \omega t)
-\end{aligned}$$
-
-The corresponding perturbative
-[electric field](/know/concept/electric-field/) $\vec{E}$
-points in the same direction:
-
-$$\begin{aligned}
-    \vec{E}(\vec{r}, t)
-    = - \pdv{\vec{A}}{t}
-    = \vec{E}_0 \exp\!(i \vec{k} \cdot \vec{r} - i \omega t)
-\end{aligned}$$
-
-Where $\vec{E}_0 = i \omega \vec{A}_0$.
-Let us restrict ourselves to visible light,
-whose wavelength $2 \pi / k \approx 10^{-6} \:\mathrm{m}$.
-By comparison, the size of an atomic orbital is on the order of $10^{-10} \:\mathrm{m}$,
-so we can ignore the dot product $\vec{k} \cdot \vec{r}$.
-This is the **electric dipole approximation**:
-the radiation is treated classicaly,
-while the electron is treated quantum-mechanically.
-
-$$\begin{aligned}
-    \vec{E}(\vec{r}, t)
-    \approx \vec{E}_0 \exp\!(- i \omega t)
-\end{aligned}$$
-
-Next, we want to convert $\hat{H}_1$
-to use the electric field $\vec{E}$ instead of the potential $\vec{A}$.
-To do so, we rewrite the momemtum $\vec{P} = m \: \dv*{\vec{r}}{t}$
-and evaluate this in the [Heisenberg picture](/know/concept/heisenberg-picture/):
-
-$$\begin{aligned}
-    \matrixel{2}{\dv*{\vec{r}}{t}}{1}
-    &= \frac{i}{\hbar} \matrixel{2}{[\hat{H}_0, \vec{r}]}{1}
-    = \frac{i}{\hbar} \matrixel{2}{\hat{H}_0 \vec{r} - \vec{r} \hat{H}_0}{1}
-    \\
-    &= \frac{i}{\hbar} (E_2 - E_1) \matrixel{2}{\vec{r}}{1}
-    = i \omega_0 \matrixel{2}{\vec{r}}{1}
-\end{aligned}$$
-
-Therefore, $\vec{P} / m = i \omega_0 \vec{r}$,
-where $\omega_0 = (E_2 - E_1) / \hbar$ is the resonance frequency of the transition,
-close to which we assume that $\vec{A}$ and $\vec{E}$ are oscillating.
-We thus get:
+In fact, we can analytically calculate the Einstein coefficients in some cases,
+by treating incoming light as a perturbation
+to an electron in a two-level system,
+and then finding $B_{12}$ and $B_{21}$ from the resulting transition rate.
+We need to make the [electric dipole approximation](/know/concept/electric-dipole-approximation/),
+in which case the perturbing Hamiltonian $\hat{H}_1(t)$ is given by:
 
 $$\begin{aligned}
     \hat{H}_1(t)
-    &= - \frac{q}{m} \vec{P} \cdot \vec{A}
-    = - i q \omega_0 \vec{r} \cdot \vec{A}_0 \exp\!(- i \omega t)
-    \\
-    &= - q \vec{r} \cdot \vec{E}_0 \exp\!(- i \omega t)
-    = - \vec{p} \cdot \vec{E}_0 \exp\!(- i \omega t)
+    = - q \vec{r} \cdot \vec{E}_0 \cos\!(\omega t)
 \end{aligned}$$
 
-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:
+Where $q = -e$ is the electron charge,
+$\vec{r}$ is the position operator,
+and $\vec{E}_0$ is the amplitude of
+the [electromagnetic wave](/know/concept/electromagnetic-wave-equation/).
+For simplicity, we let the amplitude be along the $z$-axis:
 
 $$\begin{aligned}
     \hat{H}_1(t)
-    = - q \vec{r} \cdot \vec{E}_0 \cos\!(- i \omega t)
+    = - q E_0 z \cos\!(\omega t)
 \end{aligned}$$
 
-
-## Polarized light
-
 This form of $\hat{H}_1$ is a well-known case for
 [time-dependent perturbation theory](/know/concept/time-dependent-perturbation-theory/),
 which tells us that the transition probability from $\ket{a}$ to $\ket{b}$ is:
@@ -259,19 +175,20 @@ 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:
 
 $$\begin{gathered}
-    \matrixel{1}{H_1}{2} = - q E_0 U
+    \matrixel{1}{H_1}{2} = - E_0 d
     \qquad
-    \matrixel{2}{H_1}{1} = - q E_0 U^*
+    \matrixel{2}{H_1}{1} = - E_0 d^*
     \\
     \matrixel{1}{H_1}{1} = \matrixel{2}{H_1}{2} = 0
 \end{gathered}$$
 
-Where $U \equiv \matrixel{1}{z}{2}$ is a constant.
+Where $d \equiv q \matrixel{1}{z}{2}$ is a constant,
+namely the $z$-component of the **transition dipole moment**.
 The chance of an upward jump (i.e. absorption) is:
 
 $$\begin{aligned}
     P_{12}
-    = \frac{q^2 E_0^2 |U|^2}{\hbar^2} \frac{\sin^2\!\big( (\omega_0 - \omega) t / 2 \big)}{(\omega_0 - \omega)^2}
+    = \frac{E_0^2 |d|^2}{\hbar^2} \frac{\sin^2\!\big( (\omega_0 - \omega) t / 2 \big)}{(\omega_0 - \omega)^2}
 \end{aligned}$$
 
 Meanwhile, the transition probability for stimulated emission is as follows,
@@ -280,7 +197,7 @@ and is therefore symmetric around $\omega_{ba}$:
 
 $$\begin{aligned}
     P_{21}
-    = \frac{q^2 E_0^2 |U|^2}{\hbar^2} \frac{\sin^2\!\big( (\omega_0 - \omega) t / 2 \big)}{(\omega_0 - \omega)^2}
+    = \frac{E_0^2 |d|^2}{\hbar^2} \frac{\sin^2\!\big( (\omega_0 - \omega) t / 2 \big)}{(\omega_0 - \omega)^2}
 \end{aligned}$$
 
 Surprisingly, the probabilities of absorption and stimulated emission are the same!
@@ -289,8 +206,12 @@ the availability of electrons and holes in both states.
 
 In theory, we could calculate the transition rate $R_{12} = \pdv*{P_{12}}{t}$,
 which would give us Einstein's absorption coefficient $B_{12}$,
-for this particular case of coherent monochromatic light.
-However, the result would not be constant in time $t$.
+for this specific case of coherent monochromatic light.
+However, the result would not be constant in time $t$,
+so is not really useful.
+
+
+## Polarized light
 
 To solve this "problem", we generalize to (incoherent) polarized polychromatic light.
 To do so, we note that the energy density $u$ of an electric field $E_0$ is given by:
@@ -301,11 +222,12 @@ $$\begin{aligned}
     E_0^2 = \frac{2 u}{\varepsilon_0}
 \end{aligned}$$
 
-Putting this in the previous result gives the following transition probability:
+Where $\varepsilon_0$ is the vacuum permittivity.
+Putting this in the previous result for $P_{12}$ gives us:
 
 $$\begin{aligned}
     P_{12}
-    = \frac{2 u q^2 |U|^2}{\varepsilon_0 \hbar^2} \frac{\sin^2\!\big( (\omega_0 - \omega) t / 2 \big)}{(\omega_0 - \omega)^2}
+    = \frac{2 u |d|^2}{\varepsilon_0 \hbar^2} \frac{\sin^2\!\big( (\omega_0 - \omega) t / 2 \big)}{(\omega_0 - \omega)^2}
 \end{aligned}$$
 
 For a continuous light spectrum,
@@ -313,11 +235,11 @@ this $u$ turns into the spectral energy density $u(\omega)$:
 
 $$\begin{aligned}
     P_{12}
-    = \frac{2 q^2 |U|^2}{\varepsilon_0 \hbar^2}
+    = \frac{2 |d|^2}{\varepsilon_0 \hbar^2}
     \int_0^\infty \frac{\sin^2\!\big( (\omega_0 - \omega) t / 2 \big)}{(\omega_0 - \omega)^2} u(\omega) \dd{\omega}
 \end{aligned}$$
 
-From here, we the derivation is similar to that of
+From here, the derivation is similar to that of
 [Fermi's golden rule](/know/concept/fermis-golden-rule/),
 despite the distinction that we are integrating over frequencies rather than states.
 
@@ -329,8 +251,8 @@ which turns out to be $\pi t$:
 
 $$\begin{aligned}
     P_{12}
-    = \frac{q^2 |U|^2}{\varepsilon_0 \hbar^2} u(\omega_0) \int_{-\infty}^\infty \frac{\sin^2\!\big(x t \big)}{x^2} \dd{x}
-    = \frac{\pi q^2 |U|^2}{\varepsilon_0 \hbar^2} u(\omega_0) \:t
+    = \frac{|d|^2}{\varepsilon_0 \hbar^2} u(\omega_0) \int_{-\infty}^\infty \frac{\sin^2\!\big(x t \big)}{x^2} \dd{x}
+    = \frac{\pi |d|^2}{\varepsilon_0 \hbar^2} u(\omega_0) \:t
 \end{aligned}$$
 
 From this, the transition rate $R_{12} = B_{12} u(\omega_0)$
@@ -338,8 +260,8 @@ is then calculated as follows:
 
 $$\begin{aligned}
     R_{12}
-    = \pdv{P_{2 \to 1}}{t}
-    = \frac{\pi q^2 |U|^2}{\varepsilon_0 \hbar^2} u(\omega_0)
+    = \pdv{P_{12}}{t}
+    = \frac{\pi |d|^2}{\varepsilon_0 \hbar^2} u(\omega_0)
 \end{aligned}$$
 
 Using the relations from earlier with $g_1 = g_2$,
@@ -348,9 +270,9 @@ for a polarized incoming light spectrum:
 
 $$\begin{aligned}
     \boxed{
-        B_{21} = B_{12} = \frac{\pi q^2 |U|^2}{\varepsilon_0 \hbar^2}
+        B_{21} = B_{12} = \frac{\pi |d|^2}{\varepsilon_0 \hbar^2}
         \qquad
-        A_{21} = \frac{\omega_0^3 q^2 |U|^2}{\pi \varepsilon \hbar c^3}
+        A_{21} = \frac{\omega_0^3 |d|^2}{\pi \varepsilon \hbar c^3}
     }
 \end{aligned}$$
 
@@ -363,31 +285,33 @@ and define the polarization unit vector $\vec{n}$:
 
 $$\begin{aligned}
     \matrixel{1}{\hat{H}_1}{2}
-    = - q \matrixel{1}{\vec{r} \cdot \vec{E}_0}{2}
-    = - q E_0 \matrixel{1}{\vec{r} \cdot \vec{n}}{2}
-    = - q E_0 W
+    = - \vec{d} \cdot \vec{E}_0
+    = - E_0 (\vec{d} \cdot \vec{n})
 \end{aligned}$$
 
-The goal is to obtain the average of $|W|^2$,
-where $W \equiv \matrixel{1}{\vec{r} \cdot \vec{n}}{2}$.
+Where $\vec{d} \equiv q \matrixel{1}{\vec{r}}{2}$ is
+the full **transition dipole moment** vector, which is usually complex.
+
+The goal is to calculate the average of $|\vec{d} \cdot \vec{n}|^2$.
 In [spherical coordinates](/know/concept/spherical-coordinates/),
-we integrate over all possible orientations $\vec{n}$ for fixed $\vec{r}$,
-using that $\vec{r} \cdot \vec{n} = |\vec{r}| \cos\!(\theta)$:
+we integrate over all directions $\vec{n}$ for fixed $\vec{d}$,
+using that $\vec{d} \cdot \vec{n} = |\vec{d}| \cos\!(\theta)$
+with $|\vec{d}| \equiv |d_x|^2 \!+\! |d_y|^2 \!+\! |d_z|^2$:
 
 $$\begin{aligned}
-    \expval{|W|^2}
-    = \frac{1}{4 \pi} \int_0^\pi \int_0^{2 \pi} |\matrixel{1}{\vec{r}}{2}|^2 \cos^2(\theta) \sin\!(\theta) \dd{\varphi} \dd{\theta}
+    \expval{|\vec{d} \cdot \vec{n}|^2}
+    = \frac{1}{4 \pi} \int_0^\pi \int_0^{2 \pi} |\vec{d}|^2 \cos^2(\theta) \sin\!(\theta) \dd{\varphi} \dd{\theta}
 \end{aligned}$$
 
 Where we have divided by $4\pi$ (the surface area of a unit sphere) for normalization,
-and $\theta$ is the polar angle between $\vec{n}$ and $\vec{p}$.
+and $\theta$ is the polar angle between $\vec{n}$ and $\vec{d}$.
 Evaluating the integrals yields:
 
 $$\begin{aligned}
-    \expval{|W|^2}
-    = \frac{2 \pi}{4 \pi} |U|^2 \int_0^\pi \cos^2(\theta) \sin\!(\theta) \dd{\theta}
-    = \frac{|U|^2}{2} \Big[ \!-\! \frac{\cos^3(\theta)}{3} \Big]_0^\pi
-    = \frac{|U|^2}{3}
+    \expval{|\vec{d} \cdot \vec{n}|^2}
+    = \frac{2 \pi}{4 \pi} |\vec{d}|^2 \int_0^\pi \cos^2(\theta) \sin\!(\theta) \dd{\theta}
+    = \frac{|\vec{d}|^2}{2} \Big[ \!-\! \frac{\cos^3(\theta)}{3} \Big]_0^\pi
+    = \frac{|\vec{d}|^2}{3}
 \end{aligned}$$
 
 With this additional constant factor $1/3$,
@@ -395,16 +319,17 @@ the transition rate $R_{12}$ is modified to:
 
 $$\begin{aligned}
     R_{12}
-    = \frac{\pi q^2 |U|^2}{3 \varepsilon_0 \hbar^2} u(\omega_0)
+    = \pdv{P_{12}}{t}
+    = \frac{\pi |\vec{d}|^2}{3 \varepsilon_0 \hbar^2} u(\omega_0)
 \end{aligned}$$
 
 From which it follows that the Einstein coefficients for unpolarized light are given by:
 
 $$\begin{aligned}
     \boxed{
-        B_{21} = B_{12} = \frac{\pi q^2 |U|^2}{3 \varepsilon_0 \hbar^2}
+        B_{21} = B_{12} = \frac{\pi |\vec{d}|^2}{3 \varepsilon_0 \hbar^2}
         \qquad
-        A_{21} = \frac{\omega_0^3 q^2 |U|^2}{3 \pi \varepsilon \hbar c^3}
+        A_{21} = \frac{\omega_0^3 |\vec{d}|^2}{3 \pi \varepsilon \hbar c^3}
     }
 \end{aligned}$$