Categories: Electromagnetism, Optics, Perturbation, Physics, Quantum mechanics.

# Electric dipole approximation

Suppose that an electromagnetic wave is travelling through an atom, and affecting the electrons. 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 \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 \Phi \end{aligned}

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 &= (\vb{A} \cdot \vu{P} - \vu{P} \cdot \vb{A}) \psi \\ &= -i \hbar \vb{A} \cdot (\nabla \psi) + i \hbar \nabla \cdot (\vb{A} \psi) \\ &= i \hbar (\nabla \cdot \vb{A}) \psi \\ &= 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}

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\qquad \hat{H}_0 \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}

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) \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) \end{aligned}

The corresponding perturbative electric field $\vb{E}$ is then given by:

\begin{aligned} \vb{E}(\vb{x}, t) = - \pdv{\vb{A}}{t} = \vb{E}_0 \exp(i \vb{k} \cdot \vb{x} - i \omega t) \end{aligned}

Where $\vb{E}_0 = \omega \vb{A}_0$. 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{ \vb{E}(\vb{x}, t) \approx \vb{E}_0 \exp(- i \omega t) } \end{aligned}

This is the electric dipole approximation: we ignore all spatial variation of $\vb{E}$, and only consider its temporal oscillation. Also, since we have not used the word “photon”, we are implicitly treating the radiation classically, 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 momentum $\vu{P} \equiv m \: \idv{\vu{x}}{t}$ and evaluate this in the interaction picture:

\begin{aligned} \vu{P} &= m \dv{\vu{x}}{t} = m \frac{i}{\hbar} \comm{\hat{H}_0}{\vu{x}} \end{aligned}

Taking the off-diagonal inner product with 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} \end{aligned}

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. $\omega \approx \omega_0$. 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 - \vu{d} \cdot \vb{E}_0 \exp(- i \omega t) \end{aligned}

Where $\vu{d} \equiv q \vu{x}$ is the transition dipole moment operator of the electron, hence the name electric dipole approximation. Finally, we take the real part, yielding:

\begin{aligned} \boxed{ \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}

If this approximation is too rough, $\vb{E}$ can always be Taylor-expanded in $(i \vb{k} \cdot \vb{x})$:

\begin{aligned} \vb{E}(\vb{x}, t) = \vb{E}_0 \Big( 1 + (i \vb{k} \cdot \vb{x}) + \frac{1}{2} (i \vb{k} \cdot \vb{x})^2 + \: ... \Big) \exp(- i \omega t) \end{aligned}

Taking the real part then yields the following series of higher-order correction terms:

\begin{aligned} \vb{E}(\vb{x}, t) = \vb{E}_0 \Big( \cos(\omega t) + (\vb{k} \cdot \vb{x}) \sin(\omega t) - \frac{1}{2} (\vb{k} \cdot \vb{x})^2 \cos(\omega t) + \: ... \Big) \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.