From ea12abd73dd1e624367935353605a3c1327b5281 Mon Sep 17 00:00:00 2001
From: Prefetch
Date: Tue, 7 Sep 2021 09:47:25 +0200
Subject: Expand knowledge base

---
 .../know/concept/einstein-coefficients/index.pdc   | 409 +++++++++++++++++++++
 1 file changed, 409 insertions(+)
 create mode 100644 content/know/concept/einstein-coefficients/index.pdc

(limited to 'content/know/concept/einstein-coefficients/index.pdc')

diff --git a/content/know/concept/einstein-coefficients/index.pdc b/content/know/concept/einstein-coefficients/index.pdc
new file mode 100644
index 0000000..37141f2
--- /dev/null
+++ b/content/know/concept/einstein-coefficients/index.pdc
@@ -0,0 +1,409 @@
+---
+title: "Einstein coefficients"
+firstLetter: "E"
+publishDate: 2021-07-11
+categories:
+- Physics
+- Optics
+- Quantum mechanics
+
+date: 2021-07-11T18:22:14+02:00
+draft: false
+markup: pandoc
+---
+
+# Einstein coefficients
+
+The **Einstein coefficients** quantify
+the emission and absorption of photons by a solid,
+and can be calculated analytically from first principles
+in several useful situations.
+
+
+## Qualitative description
+
+Suppose we have a ground state with energy $E_1$ containing $N_1$ electrons,
+and an excited state with energy $E_2$ containing $N_2$ electrons.
+The resonance $\omega_0 \equiv (E_2 \!-\! E_1)/\hbar$
+is the frequency of the photon emitted
+when an electron falls from $E_2$ to $E_1$.
+
+The first Einstein coefficient is the **spontaneous emission rate** $A_{21}$,
+which gives the probability per unit time
+that an excited electron falls from state 2 to 1,
+so that $N_2(t)$ obeys the following equation,
+which is easily solved:
+
+$$\begin{aligned}
+    \dv{N_2}{t} = - A_{21} N_2
+    \quad \implies \quad
+    N_2(t) = N_2(0) \exp\!(- t / \tau)
+\end{aligned}$$
+
+Where $\tau = 1 / A_{21}$ is the **natural radiative lifetime** of the excited state,
+which gives the lifetime of an excited electron,
+before it decays to the ground state.
+
+The next coefficient is the **absorption rate** $B_{12}$,
+which is the probability that an incoming photon excites an electron,
+per unit time and per unit spectral energy density
+(i.e. the rate depends on the frequency of the incoming light).
+Then $N_1(t)$ obeys the following equation:
+
+$$\begin{aligned}
+    \dv{N_1}{t} = - B_{12} N_1 u(\omega_0)
+\end{aligned}$$
+
+Where $u(\omega)$ is the spectral energy density of the incoming light,
+put here to express the fact that only photons with frequency $\omega_0$ are absorbed.
+
+There is one more Einstein coefficient: the **stimulated emission rate** $B_{21}$.
+An incoming photon has an associated electromagnetic field,
+which can encourage an excited electron to drop to the ground state,
+such that for $A_{21} = 0$:
+
+$$\begin{aligned}
+    \dv{N_2}{t} = - B_{21} N_2 u(\omega_0)
+\end{aligned}$$
+
+These three coefficients $A_{21}$, $B_{12}$ and $B_{21}$ are related to each other.
+Suppose that the system is in equilibrium,
+i.e. that $N_1$ and $N_2$ are constant.
+We assume that the number of particles in the system is constant,
+implying that $N_1'(t) = - N_2'(t) = 0$, so:
+
+$$\begin{aligned}
+    B_{12} N_1 u(\omega_0) = A_{21} N_2 + B_{21} N_2 u(\omega_0) = 0
+\end{aligned}$$
+
+Isolating this equation for $u(\omega_0)$,
+gives following expression for the radiation:
+
+$$\begin{aligned}
+    u(\omega_0)
+    = \frac{A_{21}}{(N_1 / N_2) B_{12} - B_{21}}
+\end{aligned}$$
+
+We assume that the system is in thermal equilibrium
+with its own black-body radiation, and that there is no external light.
+Then this is a [canonical ensemble](/know/concept/canonical-ensemble/),
+meaning that the relative probability that an electron has $E_2$ compared to $E_1$
+is given by the Boltzmann distribution:
+
+$$\begin{aligned}
+    \frac{\mathrm{Prob}(E_2)}{\mathrm{Prob}(E_1)}
+    = \frac{N_2}{N_1}
+    = \frac{g_2}{g_1} \exp\!(- \hbar \omega_0 \beta)
+\end{aligned}$$
+
+Where $g_2$ and $g_1$ are the degeneracies of the energy levels.
+Inserting this back into the equation for the spectrum $u(\omega_0)$ yields:
+
+$$\begin{aligned}
+    u(\omega_0)
+    = \frac{A_{21}}{(g_1 / g_2) B_{12} \exp\!(\hbar \omega_0 \beta) - B_{21}}
+\end{aligned}$$
+
+Since $u(\omega_0)$ represents only black-body radiation,
+our result must agree with Planck's law:
+
+$$\begin{aligned}
+    u(\omega_0)
+    = \frac{A_{21}}{B_{21} \big( (g_1 B_{12} / g_2 B_{21}) \exp\!(\hbar \omega_0 \beta) - 1 \big)}
+    = \frac{\hbar \omega_0^3}{\pi^2 c^3} \frac{1}{\exp\!(\hbar \omega_0 \beta) - 1}
+\end{aligned}$$
+
+This gives us the following two equations relating the Einstein coefficients:
+
+$$\begin{aligned}
+    \boxed{
+        A_{21} = \frac{\hbar \omega_0^3}{\pi^2 c^3} B_{21}
+        \qquad \quad
+        g_1 B_{12} = g_2 B_{21}
+    }
+\end{aligned}$$
+
+Note that this result holds even if $E_1$ is not the ground state,
+but instead some lower excited state below $E_2$,
+due to the principle of *detailed balance*.
+Furthermore, it turns out that these relations
+also hold if the system is not in equilibrium.
+
+A notable case is **population inversion**,
+where $B_{21} N_2 > B_{12} N_1$ such that $N_2 > (g_2 / g_1) N_1$.
+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
+
+In fact, we can analytically calculate the Einstein coefficients,
+if we make a mild approximation.
+Consider the Hamiltonian of an electron with charge $q = - e$:
+
+$$\begin{aligned}
+    \hat{H}
+    &= \frac{\vec{P}{}^2}{2 m} - \frac{q}{2 m} (\vec{A} \cdot \vec{P} + \vec{P} \cdot \vec{A}) + \frac{q^2 \vec{A}{}^2}{2m} + q \phi
+\end{aligned}$$
+
+With $\vec{A}(\vec{r}, t)$ the magnetic vector potential,
+and $\phi(\vec{r}, t)$ the electric scalar 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:
+
+$$\begin{aligned}
+    \hat{H}
+    &= \frac{\vec{P}{}^2}{2 m} - \frac{q}{m} \vec{P} \cdot \vec{A} + q \phi
+\end{aligned}$$
+
+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 $\hat{H}_1$:
+
+$$\begin{aligned}
+    \hat{H}
+    = \hat{H}_0 + \hat{H}_1
+    \qquad \quad
+    \hat{H}_0
+    \equiv \frac{\vec{P}{}^2}{2 m} + q \phi
+    \qquad \quad
+    \hat{H}_1
+    \equiv - \frac{q}{m} \vec{P} \cdot \vec{A}
+\end{aligned}$$
+
+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 $\vec{E}$ 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:
+
+$$\begin{aligned}
+    \hat{H}_1(t)
+    &= - \frac{q}{m} \vec{P} \cdot \vec{A}
+    = - i q \omega_0 \vec{r} \cdot \vec{A}_0 \exp\!(- i \omega t)
+    \\
+    &= - q \vec{r} \cdot \vec{E}_0 \exp\!(- i \omega t)
+    = - \vec{p} \cdot \vec{E}_0 \exp\!(- i \omega t)
+\end{aligned}$$
+
+Where $\vec{p} \equiv q \vec{r} = - e \vec{r}$ is the electric dipole moment of the electron,
+hence the name *electric dipole approximation*.
+
+
+## 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:
+
+$$\begin{aligned}
+    P_{ab}
+    = \frac{\big|\!\matrixel{a}{H_1}{b}\!\big|^2}{\hbar^2} \frac{\sin^2\!\big( (\omega_{ba} - \omega) t / 2 \big)}{(\omega_{ba} - \omega)^2}
+\end{aligned}$$
+
+If the location of the nucleus of the atom has $z = 0$,
+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:
+
+$$\begin{gathered}
+    \matrixel{1}{H_1}{2} = - q E_0 V
+    \qquad
+    \matrixel{2}{H_1}{1} = - q E_0 V^*
+    \\
+    \matrixel{1}{H_1}{1} = \matrixel{2}{H_1}{2} = 0
+\end{gathered}$$
+
+Where $V \equiv \matrixel{1}{z}{2}$ is a constant.
+The chance of an upward jump (i.e. absorption) is:
+
+$$\begin{aligned}
+    P_{12}
+    = \frac{q^2 E_0^2 |V|^2}{\hbar^2} \frac{\sin^2\!\big( (\omega_0 - \omega) t / 2 \big)}{(\omega_0 - \omega)^2}
+\end{aligned}$$
+
+Meanwhile, the transition probability for stimulated emission is as follows,
+using the fact that $P_{ab}$ is a sinc-function,
+and is therefore symmetric around $\omega_{ba}$:
+
+$$\begin{aligned}
+    P_{21}
+    = \frac{q^2 E_0^2 |V|^2}{\hbar^2} \frac{\sin^2\!\big( (\omega_0 - \omega) t / 2 \big)}{(\omega_0 - \omega)^2}
+\end{aligned}$$
+
+Surprisingly, the probabilities of absorption and stimulated emission are the same!
+In practice, however, the relative rates of these two processes depends heavily on
+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 $t$.
+
+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:
+
+$$\begin{aligned}
+    u = \frac{1}{2} \varepsilon_0 E_0^2
+    \qquad \implies \qquad
+    E_0^2 = \frac{2 u}{\varepsilon_0}
+\end{aligned}$$
+
+Putting this in the previous result gives the following transition probability:
+
+$$\begin{aligned}
+    P_{12}
+    = \frac{2 u q^2 |V|^2}{\varepsilon_0 \hbar^2} \frac{\sin^2\!\big( (\omega_0 - \omega) t / 2 \big)}{(\omega_0 - \omega)^2}
+\end{aligned}$$
+
+For a continuous light spectrum,
+this $u$ turns into the spectral energy density $u(\omega)$:
+
+$$\begin{aligned}
+    P_{12}
+    = \frac{2 q^2 |V|^2}{\varepsilon_0 \hbar^2}
+    \int_0^\infty \frac{\sin^2\!\big( (\omega_0 - \omega) t / 2 \big)}{(\omega_0 - \omega)^2} u(\omega) \dd{\omega}
+\end{aligned}$$
+
+From here, we the derivation is similar to that of
+[Fermi's golden rule](/know/concept/fermis-golden-rule/),
+despite the distinction that we are integrating over frequencies rather than states.
+
+At sufficiently large $t$, the integrand is sharply peaked at $\omega = \omega_0$
+and negligible everywhere else,
+so we take $u(\omega)$ out of the integral and extend the integration limits.
+Then we rewrite and look up the integral,
+which turns out to be $\pi t$:
+
+$$\begin{aligned}
+    P_{12}
+    = \frac{q^2 |V|^2}{\varepsilon_0 \hbar^2} u(\omega_0) \int_{-\infty}^\infty \frac{\sin^2\!\big(x t \big)}{x^2} \dd{x}
+    = \frac{\pi q^2 |V|^2}{\varepsilon_0 \hbar^2} u(\omega_0) \:t
+\end{aligned}$$
+
+From this, the transition rate $R_{12} = B_{12} u(\omega_0)$
+is then calculated as follows:
+
+$$\begin{aligned}
+    R_{12}
+    = \pdv{P_{2 \to 1}}{t}
+    = \frac{\pi q^2 |V|^2}{\varepsilon_0 \hbar^2} u(\omega_0)
+\end{aligned}$$
+
+Using the relations from earlier with $g_1 = g_2$,
+the Einstein coefficients are found to be as follows
+for a polarized incoming light spectrum:
+
+$$\begin{aligned}
+    \boxed{
+        B_{21} = B_{12} = \frac{\pi q^2 |V|^2}{\varepsilon_0 \hbar^2}
+        \qquad
+        A_{21} = \frac{\omega_0^3 q^2 |V|^2}{\pi \varepsilon \hbar c^3}
+    }
+\end{aligned}$$
+
+
+## Unpolarized light
+
+We can generalize the above result even further to unpolarized light.
+Let us return to the matrix elements of the perturbation $\hat{H}_1$,
+and define the polarization unit vector $\vec{n}$:
+
+$$\begin{aligned}
+    \matrixel{1}{\hat{H}_1}{2}
+    = - q \matrixel{1}{\vec{r} \cdot \vec{E}_0}{2}
+    = - q E_0 \matrixel{1}{\vec{r} \cdot \vec{n}}{2}
+    = - q E_0 W
+\end{aligned}$$
+
+The goal is to obtain the average of $|W|^2$,
+where $W \equiv \matrixel{1}{\vec{r} \cdot \vec{n}}{2}$.
+In [spherical coordinates](/know/concept/spherical-coordinates/),
+we integrate over all possible orientations $\vec{n}$ for fixed $\vec{r}$,
+using that $\vec{r} \cdot \vec{n} = |\vec{r}| \cos\!(\theta)$:
+
+$$\begin{aligned}
+    \expval{|W|^2}
+    = \frac{1}{4 \pi} \int_0^\pi \int_0^{2 \pi} |\matrixel{1}{\vec{r}}{2}|^2 \cos^2(\theta) \sin\!(\theta) \dd{\varphi} \dd{\theta}
+\end{aligned}$$
+
+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 $\vec{p}$.
+Evaluating the integrals yields:
+
+$$\begin{aligned}
+    \expval{|W|^2}
+    = \frac{2 \pi}{4 \pi} |V|^2 \int_0^\pi \cos^2(\theta) \sin\!(\theta) \dd{\theta}
+    = \frac{|V|^2}{2} \Big[ \!-\! \frac{\cos^3(\theta)}{3} \Big]_0^\pi
+    = \frac{|V|^2}{3}
+\end{aligned}$$
+
+With this additional constant factor $1/3$,
+the transition rate $R_{12}$ is modified to:
+
+$$\begin{aligned}
+    R_{12}
+    = \frac{\pi q^2 |V|^2}{3 \varepsilon_0 \hbar^2} u(\omega_0)
+\end{aligned}$$
+
+From which it follows that the Einstein coefficients for unpolarized light are given by:
+
+$$\begin{aligned}
+    \boxed{
+        B_{21} = B_{12} = \frac{\pi q^2 |V|^2}{3 \varepsilon_0 \hbar^2}
+        \qquad
+        A_{21} = \frac{\omega_0^3 q^2 |V|^2}{3 \pi \varepsilon \hbar c^3}
+    }
+\end{aligned}$$
+
+
+
+## References
+1.  M. Fox,
+    *Optical properties of solids*, 2nd edition,
+    Oxford.
+2.  D.J. Griffiths, D.F. Schroeter,
+    *Introduction to quantum mechanics*, 3rd edition,
+    Cambridge.
-- 
cgit v1.2.3