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diff --git a/source/know/concept/electric-dipole-approximation/index.md b/source/know/concept/electric-dipole-approximation/index.md index 35cf00c..06f0f45 100644 --- a/source/know/concept/electric-dipole-approximation/index.md +++ b/source/know/concept/electric-dipole-approximation/index.md @@ -13,20 +13,22 @@ layout: "concept" Suppose that an [electromagnetic wave](/know/concept/electromagnetic-wave-equation/) is travelling through an atom, and affecting the electrons. -The general Hamiltonian of an electron in such a wave is given by: +The general Hamiltonian of an electron in an electromagnetic field is: $$\begin{aligned} \hat{H} - &= \frac{(\vu{P} - q \vb{A})^2}{2 m} + q \varphi + &= \frac{(\vu{P} - q \vb{A})^2}{2 m} + q \Phi \\ - &= \frac{\vu{P}{}^2}{2 m} - \frac{q}{2 m} (\vb{A} \cdot \vu{P} + \vu{P} \cdot \vb{A}) + \frac{q^2 \vb{A}^2}{2m} + q \varphi + &= \frac{\vu{P}{}^2}{2 m} - \frac{q}{2 m} (\vb{A} \cdot \vu{P} + \vu{P} \cdot \vb{A}) + \frac{q^2 \vb{A}^2}{2m} + q \Phi \end{aligned}$$ -With charge $$q = - e$$, -canonical momentum operator $$\vu{P} = - i \hbar \nabla$$, -and magnetic vector potential $$\vb{A}(\vb{x}, t)$$. -We reduce this by fixing the Coulomb gauge $$\nabla \cdot \vb{A} = 0$$, -so that $$\vb{A} \cdot \vu{P} = \vu{P} \cdot \vb{A}$$: +Where $$q < 0$$ is the electron's charge, +$$\vu{P} = - i \hbar \nabla$$ is the canonical momentum operator, +$$\vb{A}$$ is the magnetic vector potential, +and $$\Phi$$ is the electric scalar potential. +We start by fixing the Coulomb gauge $$\nabla \cdot \vb{A} = 0$$ +such that $$\vu{P}$$ and $$\vb{A}$$ commute; +let $$\psi$$ be an arbitrary test function: $$\begin{aligned} \comm{\vb{A}}{\vu{P}} \psi @@ -35,20 +37,29 @@ $$\begin{aligned} &= -i \hbar \vb{A} \cdot (\nabla \psi) + i \hbar \nabla \cdot (\vb{A} \psi) \\ &= i \hbar (\nabla \cdot \vb{A}) \psi - = 0 + \\ + &= 0 +\end{aligned}$$ + +Meaning $$\vb{A} \cdot \vu{P} = \vu{P} \cdot \vb{A}$$. +Furthermore, we assume that $$\vb{A}$$ is so small that $$\vb{A}{}^2$$ is negligible, +so the Hamiltonian is reduced to: + +$$\begin{aligned} + \hat{H} + &\approx \frac{\vu{P}{}^2}{2 m} - \frac{q}{m} \vu{P} \cdot \vb{A} + q \Phi \end{aligned}$$ -Where $$\psi$$ is an arbitrary test function. -Assuming $$\vb{A}$$ is so small that $$\vb{A}{}^2$$ is negligible, we split $$\hat{H}$$ as follows, -where $$\hat{H}_1$$ can be regarded as a perturbation to $$\hat{H}_0$$: +We now split $$\hat{H}$$ like so, +where $$\hat{H}_1$$ can be regarded as a perturbation to the "base" $$\hat{H}_0$$: $$\begin{aligned} \hat{H} = \hat{H}_0 + \hat{H}_1 - \qquad \quad + \qquad\qquad \hat{H}_0 - \equiv \frac{\vu{P}{}^2}{2 m} + q \varphi - \qquad \quad + \equiv \frac{\vu{P}{}^2}{2 m} + q \Phi + \qquad\qquad \hat{H}_1 \equiv - \frac{q}{m} \vu{P} \cdot \vb{A} \end{aligned}$$ @@ -56,14 +67,16 @@ $$\begin{aligned} In an electromagnetic wave, $$\vb{A}$$ is oscillating sinusoidally in time and space: $$\begin{aligned} - \vb{A}(\vb{x}, t) = \vb{A}_0 \sin(\vb{k} \cdot \vb{x} - \omega t) + \vb{A}(\vb{x}, t) + = \vb{A}_0 \sin(\vb{k} \cdot \vb{x} - \omega t) \end{aligned}$$ Mathematically, it is more convenient to represent this with a complex exponential, whose real part should be taken at the end of the calculation: $$\begin{aligned} - \vb{A}(\vb{x}, t) = - i \vb{A}_0 \exp(i \vb{k} \cdot \vb{x} - i \omega t) + \vb{A}(\vb{x}, t) + = - i \vb{A}_0 \exp(i \vb{k} \cdot \vb{x} - i \omega t) \end{aligned}$$ The corresponding perturbative [electric field](/know/concept/electric-field/) $$\vb{E}$$ is then given by: @@ -75,10 +88,10 @@ $$\begin{aligned} \end{aligned}$$ Where $$\vb{E}_0 = \omega \vb{A}_0$$. -Let us restrict ourselves to visible light, -whose wavelength $$2 \pi / |\vb{k}| \sim 10^{-6} \:\mathrm{m}$$. -Meanwhile, an atomic orbital is several Bohr radii $$\sim 10^{-10} \:\mathrm{m}$$, -so $$\vb{k} \cdot \vb{x}$$ is negligible: +Light in and around the visible spectrum +has a wavelength $$2 \pi / |\vb{k}| \sim 10^{-7} \:\mathrm{m}$$, +while an atomic orbital is several Bohr radii $$\sim 10^{-10} \:\mathrm{m}$$, +so $$\vb{k} \cdot \vb{x}$$ is very small. Therefore: $$\begin{aligned} \boxed{ @@ -96,14 +109,13 @@ and the electron quantum-mechanically. Next, we want to rewrite $$\hat{H}_1$$ to use the electric field $$\vb{E}$$ instead of the potential $$\vb{A}$$. -To do so, we use that $$\vu{P} = m \: \idv{\vu{x}}{t}$$ +To do so, we use that momentum $$\vu{P} \equiv m \: \idv{\vu{x}}{t}$$ and evaluate this in the [interaction picture](/know/concept/interaction-picture/): $$\begin{aligned} \vu{P} - = m \idv{\vu{x}}{t} + &= m \dv{\vu{x}}{t} = m \frac{i}{\hbar} \comm{\hat{H}_0}{\vu{x}} - = m \frac{i}{\hbar} (\hat{H}_0 \vu{x} - \vu{x} \hat{H}_0) \end{aligned}$$ Taking the off-diagonal inner product with @@ -111,34 +123,37 @@ the two-level system's states $$\Ket{1}$$ and $$\Ket{2}$$ gives: $$\begin{aligned} \matrixel{2}{\vu{P}}{1} - = m \frac{i}{\hbar} \matrixel{2}{\hat{H}_0 \vu{x} - \vu{x} \hat{H}_0}{1} - = m i \omega_0 \matrixel{2}{\vu{x}}{1} + &= m \frac{i}{\hbar} \matrixel{2}{\hat{H}_0 \vu{x} - \vu{x} \hat{H}_0}{1} + \\ + &= m i \omega_0 \matrixel{2}{\vu{x}}{1} \end{aligned}$$ -Therefore, $$\vu{P} / m = i \omega_0 \vu{x}$$, -where $$\omega_0 \equiv (E_2 \!-\! E_1) / \hbar$$ is the resonance of the energy gap, +Where $$\omega_0 \equiv (E_2 \!-\! E_1) / \hbar$$ is the resonance of the energy gap, close to which we assume that $$\vb{A}$$ and $$\vb{E}$$ are oscillating, i.e. $$\omega \approx \omega_0$$. -We thus get: +Therefore, $$\vu{P} / m = i \omega_0 \vu{x}$$, so we get: $$\begin{aligned} \hat{H}_1(t) &= - \frac{q}{m} \vu{P} \cdot \vb{A} - = - (- i i) q \omega_0 \vu{x} \cdot \vb{A}_0 \exp(- i \omega t) \\ - &\approx - q \vu{x} \cdot \vb{E}_0 \exp(- i \omega t) - = - \vu{d} \cdot \vb{E}_0 \exp(- i \omega t) + &= - (- i i) q \omega_0 \vu{x} \cdot \vb{A}_0 \exp(- i \omega t) + \\ + &\approx - \vu{d} \cdot \vb{E}_0 \exp(- i \omega t) \end{aligned}$$ -Where $$\vu{d} \equiv q \vu{x} = - e \vu{x}$$ is +Where $$\vu{d} \equiv q \vu{x}$$ is the **transition dipole moment operator** of the electron, -hence the name **electric dipole approximation**. +hence the name *electric dipole approximation*. Finally, we take the real part, yielding: $$\begin{aligned} \boxed{ - \hat{H}_1(t) - = - \vu{d} \cdot \vb{E}(t) - = - q \vu{x} \cdot \vb{E}_0 \cos(\omega t) + \begin{aligned} + \hat{H}_1(t) + &= - \vu{d} \cdot \vb{E}(t) + \\ + &= - q \vu{x} \cdot \vb{E}_0 \cos(\omega t) + \end{aligned} } \end{aligned}$$ |