From 42d409fa774efb8206ae5c701d5cbcc4ae1d9cad Mon Sep 17 00:00:00 2001 From: Prefetch Date: Tue, 14 Sep 2021 21:20:30 +0200 Subject: Expand knowledge base --- .../know/concept/einstein-coefficients/index.pdc | 189 +++++++-------------- 1 file changed, 57 insertions(+), 132 deletions(-) (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 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: $$\begin{aligned} \boxed{ - B_{21} = B_{12} = \frac{\pi q^2 |U|^2}{3 \varepsilon_0 \hbar^2} + B_{21} = B_{12} = \frac{\pi |\vec{d}|^2}{3 \varepsilon_0 \hbar^2} \qquad - A_{21} = \frac{\omega_0^3 q^2 |U|^2}{3 \pi \varepsilon \hbar c^3} + A_{21} = \frac{\omega_0^3 |\vec{d}|^2}{3 \pi \varepsilon \hbar c^3} } \end{aligned}$$ -- cgit v1.2.3