1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
79
80
81
82
83
84
85
86
87
88
89
90
91
92
93
94
95
96
97
98
99
100
101
102
103
104
105
106
107
108
109
110
111
112
113
114
115
116
117
118
119
120
121
122
123
124
125
126
127
128
129
130
131
132
133
134
135
136
137
138
139
140
141
142
143
144
145
146
147
|
---
title: "Electric dipole approximation"
firstLetter: "E"
publishDate: 2021-09-14
categories:
- Physics
- Quantum mechanics
- Optics
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:
$$\begin{aligned}
\hat{H}
&= \frac{\vec{P}{}^2}{2 m} - \frac{q}{2 m} (\vec{A} \cdot \vec{P} + \vec{P} \cdot \vec{A}) + \frac{q^2 \vec{A}{}^2}{2m} + V
\end{aligned}$$
With charge $q = - e$
and electromagnetic vector potential $\vec{A}(\vec{r}, t)$.
We reduce this by fixing the Coulomb gauge $\nabla \cdot \vec{A} = 0$,
so that $\vec{A} \cdot \vec{P} = \vec{P} \cdot \vec{A}$,
and assume that $\vec{A}{}^2$ is negligible:
$$\begin{aligned}
\hat{H}
= \hat{H}_0 + \hat{H}_1
\qquad \quad
\hat{H}_0
\equiv \frac{\vec{P}{}^2}{2 m} + V
\qquad \quad
\hat{H}_1
\equiv - \frac{q}{m} \vec{P} \cdot \vec{A}
\end{aligned}$$
We have split $\hat{H}$ into $\hat{H}_0$
and a perturbation $\hat{H}_1$, since $\vec{A}$ is small.
In an electromagnetic wave,
$\vec{A}$ is oscillating sinusoidally in time and space as follows:
$$\begin{aligned}
\vec{A}(\vec{r}, t) = \vec{A}_0 \exp\!(i \vec{k} \cdot \vec{r} - i \omega t)
\end{aligned}$$
The corresponding perturbative
[electric field](/know/concept/electric-field/) $\vec{E}$
points in the same direction:
$$\begin{aligned}
\vec{E}(\vec{r}, t)
= - \pdv{\vec{A}}{t}
= \vec{E}_0 \exp\!(i \vec{k} \cdot \vec{r} - i \omega t)
\end{aligned}$$
Where $\vec{E}_0 = i \omega \vec{A}_0$.
Let us restrict ourselves to visible light,
whose wavelength $2 \pi / k \approx 10^{-6} \:\mathrm{m}$.
Meanwhile, an atomic orbital is on the order of $10^{-10} \:\mathrm{m}$,
so $\vec{k} \cdot \vec{r}$ is negligible:
$$\begin{aligned}
\boxed{
\vec{E}(\vec{r}, t)
\approx \vec{E}_0 \exp\!(- i \omega t)
}
\end{aligned}$$
This is the **electric dipole approximation**:
we ignore all spatial variation of $\vec{E}$,
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 convert $\hat{H}_1$
to use the electric field $\vec{E}$ instead of the potential $\vec{A}$.
To do so, we rewrite the momemtum $\vec{P} = m \: \dv*{\vec{r}}{t}$
and evaluate this in the [Heisenberg picture](/know/concept/heisenberg-picture/):
$$\begin{aligned}
\matrixel{2}{\dv*{\vec{r}}{t}}{1}
&= \frac{i}{\hbar} \matrixel{2}{[\hat{H}_0, \vec{r}]}{1}
= \frac{i}{\hbar} \matrixel{2}{\hat{H}_0 \vec{r} - \vec{r} \hat{H}_0}{1}
\\
&= \frac{i}{\hbar} (E_2 - E_1) \matrixel{2}{\vec{r}}{1}
= i \omega_0 \matrixel{2}{\vec{r}}{1}
\end{aligned}$$
Therefore, $\vec{P} / m = i \omega_0 \vec{r}$,
where $\omega_0 \equiv (E_2 - E_1) / \hbar$ is the resonance frequency of the transition,
close to which we assume that $\vec{A}$ and $\vec{E}$ are oscillating.
We thus get:
$$\begin{aligned}
\hat{H}_1(t)
&= - \frac{q}{m} \vec{P} \cdot \vec{A}
= - i q \omega_0 \vec{r} \cdot \vec{A}_0 \exp\!(- i \omega t)
\\
&= - q \vec{r} \cdot \vec{E}_0 \exp\!(- i \omega t)
= - \vec{d} \cdot \vec{E}_0 \exp\!(- i \omega t)
\end{aligned}$$
Where $\vec{d} \equiv q \vec{r} = - e \vec{r}$ is
the **transition dipole moment operator** of the electron,
hence the name **electric dipole approximation**.
Finally, since electric fields are actually real
(we let it be complex for mathematical convenience),
we take the real part, yielding:
$$\begin{aligned}
\boxed{
\hat{H}_1(t)
= - q \vec{r} \cdot \vec{E}_0 \cos\!(\omega t)
}
\end{aligned}$$
If this approximation is too rough,
$\vec{E}$ can always be Taylor-expanded in $(i \vec{k} \cdot \vec{r})$:
$$\begin{aligned}
\vec{E}(\vec{r}, t)
= \vec{E}_0 \Big( 1 + (i \vec{k} \cdot \vec{r}) + \frac{1}{2} (i \vec{k} \cdot \vec{r})^2 + \: ... \Big) \exp\!(- i \omega t)
\end{aligned}$$
Taking the real part then yields the following series of higher-order correction terms:
$$\begin{aligned}
\vec{E}(\vec{r}, t)
= \vec{E}_0 \Big( \cos\!(\omega t) + (\vec{k} \cdot \vec{r}) \sin\!(\omega t) - \frac{1}{2} (\vec{k} \cdot \vec{r})^2 \cos\!(\omega t) + \: ... \Big)
\end{aligned}$$
## 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.
|
