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Diffstat (limited to 'content/know/concept/einstein-coefficients')
| -rw-r--r-- | content/know/concept/einstein-coefficients/index.pdc | 66 |
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: $$\begin{aligned} \boxed{ - B_{21} = B_{12} = \frac{\pi q^2 |V|^2}{3 \varepsilon_0 \hbar^2} + B_{21} = B_{12} = \frac{\pi q^2 |U|^2}{3 \varepsilon_0 \hbar^2} \qquad - A_{21} = \frac{\omega_0^3 q^2 |V|^2}{3 \pi \varepsilon \hbar c^3} + A_{21} = \frac{\omega_0^3 q^2 |U|^2}{3 \pi \varepsilon \hbar c^3} } \end{aligned}$$ |