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---
title: "Electric dipole approximation"
sort_title: "Electric dipole approximation"
date: 2021-09-14
categories:
- Physics
- Quantum mechanics
- Optics
- Electromagnetism
- Perturbation
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 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](/know/concept/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](/know/concept/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.
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