---
title: "Electric dipole approximation"
firstLetter: "E"
publishDate: 2021-09-14
categories:
- Physics
- Quantum mechanics
- Optics
- Electromagnetism
- Perturbation

date: 2021-09-14T13:11:54+02:00
draft: false
markup: pandoc
---

# Electric dipole approximation

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:

$$\begin{aligned}
    \hat{H}
    &= \frac{(\vu{P} - q \vb{A})^2}{2 m} + q \varphi
    \\
    &= \frac{\vu{P}{}^2}{2 m} - \frac{q}{2 m} (\vb{A} \cdot \vu{P} + \vu{P} \cdot \vb{A}) + \frac{q^2 \vb{A}^2}{2m} + q \varphi
\end{aligned}$$

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 $\vb{A} \cdot \vu{P} = \vu{P} \cdot \vb{A}$:

$$\begin{aligned}
    \comm*{\vb{A}}{\vu{P}} \psi
    &= -i \hbar \vb{A} \cdot (\nabla \psi) + i \hbar \nabla \cdot (\vb{A} \psi)
    \\
    &= i \hbar (\nabla \cdot \vb{A}) \psi
    = 0
\end{aligned}$$

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 $\hat{H}_0$:

$$\begin{aligned}
    \hat{H}
    = \hat{H}_0 + \hat{H}_1
    \qquad \quad
    \hat{H}_0
    \equiv \frac{\vu{P}{}^2}{2 m} + q \varphi
    \qquad \quad
    \hat{H}_1
    \equiv - \frac{q}{m} \vu{P} \cdot \vb{A}
\end{aligned}$$

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 $\vb{E}_0 = \omega \vb{A}_0$.
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 $\sim 10^{-10} \:\mathrm{m}$,
so $\vb{k} \cdot \vb{x}$ is negligible:

$$\begin{aligned}
    \boxed{
        \vb{E}(\vb{x}, t)
        \approx \vb{E}_0 \exp\!(- i \omega t)
    }
\end{aligned}$$

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 $\vu{P} = m \: \dv*{\vu{x}}{t}$
and evaluate this in the [interaction picture](/know/concept/interaction-picture/):

$$\begin{aligned}
    \vu{P}
    = m \dv*{\vu{x}}{t}
    = m \frac{i}{\hbar} \comm*{\hat{H}_0}{\vu{x}}
    = m \frac{i}{\hbar} (\hat{H}_0 \vu{x} - \vu{x} \hat{H}_0)
\end{aligned}$$

Taking the off-diagonal inner product with
the two-level system's states $\ket{1}$ and $\ket{2}$ gives:

$$\begin{aligned}
    \matrixel{2}{\vu{P}}{1}
    = m \frac{i}{\hbar} \matrixel{2}{\hat{H}_0 \vu{x} - \vu{x} \hat{H}_0}{1}
    = m i \omega_0 \matrixel{2}{\vu{x}}{1}
\end{aligned}$$

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,
close to which we assume that $\vb{A}$ and $\vb{E}$ are oscillating, i.e. $\omega \approx \omega_0$.
We thus get:

$$\begin{aligned}
    \hat{H}_1(t)
    &= - \frac{q}{m} \vu{P} \cdot \vb{A}
    = - (- i i) q \omega_0 \vu{x} \cdot \vb{A}_0 \exp\!(- i \omega t)
    \\
    &\approx - q \vu{x} \cdot \vb{E}_0 \exp\!(- i \omega t)
    = - \vu{d} \cdot \vb{E}_0 \exp\!(- i \omega t)
\end{aligned}$$

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

$$\begin{aligned}
    \boxed{
        \hat{H}_1(t)
        = - \vu{d} \cdot \vb{E}(t)
        = - q \vu{x} \cdot \vb{E}_0 \cos\!(\omega t)
    }
\end{aligned}$$

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.