Dipole approximation

Question

I am studying the "dipole approximation" as is defined in quantum optics. (See, for example Introductory Quantum Optics by Gerry and Knight, pages 24–25.) In this situation, we have the operator of the electric field for a single-mode plane wave as $$\hat{E}(\textbf{r}, t)= i \mathcal{E}\textbf{e}_{x}\left[\hat{a}e^{i(\textbf{k}\cdot \textbf{r} - \omega t)} - \hat{a}^{\dagger}e^{-i(\textbf{k}\cdot \textbf{r} - \omega t)} \right], \tag{1}$$ where $\mathcal{E}$ is a real constant playing the role of an amplitude, $\textbf{e}_{x}$ is the polarization unit vector, $\omega$ is the frequency of the plane wave, and $\hat{a}$ ($\hat{a}^{\dagger}$) is the photon annihilation (creation) operator.

The book explains that in much of quantum optics, the spatial variation of the field over the dimensions of an atomic system may be negligible. For example, for optical radiation, the wavelength characterized by $\lambda=\frac{2\pi c}{\omega}$ is several times bigger than the dimensions of the atom; that is, $$\frac{\lambda}{2\pi}= \frac{1}{\left|\textbf{k} \right|}\gg \left|\textbf{r}_{\text{atom}} \right|, \tag{2}$$ where $\left|\textbf{r}_{\text{atom}} \right|$ is a length characteristic of the size of an atom. Under this assumption, the exponentials in eq. (1) containing the dot product $\textbf{k}\cdot \textbf{r}$ can be approximated as $$e^{\pm i(\textbf{k}\cdot \textbf{r}) }\approx 1 \pm i(\textbf{k}\cdot \textbf{r}). \tag{3} $$ The last equation of course comes from the Maclaurin series expansion of the exponential and taking $(\textbf{k}\cdot \textbf{r}) \ll 1$ (that is, very small), in order to have powers greater than one in the expansion be negligible; that is, $$e^{\pm i(\textbf{k}\cdot \textbf{r})}= 1 \pm i(\textbf{k}\cdot \textbf{r}) + \frac{1}{2!}(\pm i\textbf{k}\cdot \textbf{r})^2 \pm \frac{1}{3!} (\textbf{k}\cdot \textbf{r})^3 + \cdots \approx 1 \pm i(\textbf{k}\cdot \textbf{r}). \tag{4} $$ Then, taking this last equation, the electric field operator from eq. (1) can be written as $$ \hat{E}(\textbf{r}, t)= i \mathcal{E}\textbf{e}_{x}\left[\hat{a}e^{-i\omega t} - \hat{a}^{\dagger}e^{+i\omega t} \right]. \tag{5}$$ Then, my question is: Why, under the dipole approximation, can we take $(\textbf{k}\cdot \textbf{r}) \ll 1$?

Answer 1

By equation (2) $$ |\mathbf k||\mathbf r_{atom}|\ll1 $$

Noting that $$ \mathbf k\cdot\mathbf r =|\mathbf k||\mathbf r|\cos\theta $$

we have $$ |\mathbf k\cdot\mathbf r|\le|\mathbf k||\mathbf r| $$

Hence by (2) $$ |\mathbf k\cdot\mathbf r|\le |\mathbf k||\mathbf r|\ll1 $$

for $|\mathbf r|\approx|\mathbf r_{atom}|$.

In other words, near the atom, the field is a dipole.

Answer 2

The reason that $\lambda$ for optical photons is much bigger than the scale of a typical atom is that the electric forces that hold atoms together are (relatively) weak. The wavelength of a photon with energy $E$ is $$\lambda=\frac{2\pi\hbar c}{E}=\left(1.24\times 10^{-6}\,\text{m}\right)\left(\frac{1\,\text{eV}}{E}\right). \tag{1}$$ This puts the wavelengths for typical electronic transitions in atoms in the optical to near infrared ranges.

The size of the electron cloud in an atom is set by a different combination of quantities. The typical size of an atom is the Bohr radius, $$a_{0}=\frac{\hbar}{m_{e}c\alpha}=5.29\times10^{-11}\,\text{m}, \tag{2}$$ where $m_{e}$ is the electron mass and $\alpha$ is the fine structure constant $$\alpha\approx\frac{1}{137}\left(=\frac{e^{2}}{\hbar c}\right),$$ where the formula in parentheses applies in Gaussian (cgs) units. $\alpha$ represents the strength of the electromagnetic interaction, including how tightly atomic electrons are bound to their nuclei.

You might expect that the presence of the small quantity $\alpha$ in the denominator of (2) would make the $a_{0}$ scale larger than $\lambda$. However, it turns out the denominator in (1) is actually even smaller, making the wavelength the larger quantity! The reason is that the typical energy scale $E$ depends quadratically on $\alpha$; atomic energies are typically of the order of the Hartree energy $$E_{\text{Hartree}}=m_{e}c^{2}\alpha^{2};$$ for example, the Rydberg energy (the ionization energy of atomic hydrogen) is $$R=\frac{1}{2}E_{\text{Hartree}}+\text{higher-order corrections}$$. That means that the typical scale for the wavelengths of electron transitions is actually $$\lambda\sim\frac{\hbar}{m_{e}c\alpha^{2}},$$ larger than (2) by two orders of magnitude.