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
148
149
150
151
152
153
154
155
156
157
158
159
160
161
162
163
164
165
166
167
168
|
---
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 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 $$\vb{A} \cdot \vu{P} = \vu{P} \cdot \vb{A}$$:
$$\begin{aligned}
\comm{\vb{A}}{\vu{P}} \psi
&= (\vb{A} \cdot \vu{P} - \vu{P} \cdot \vb{A}) \psi
\\
&= -i \hbar \vb{A} \cdot (\nabla \psi) + i \hbar \nabla \cdot (\vb{A} \psi)
\\
&= i \hbar (\nabla \cdot \vb{A}) \psi
= 0
\end{aligned}$$
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 radii $$\sim 10^{-10} \:\mathrm{m}$$,
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 $$\vu{P} = m \: \idv{\vu{x}}{t}$$
and evaluate this in the [interaction picture](/know/concept/interaction-picture/):
$$\begin{aligned}
\vu{P}
= m \idv{\vu{x}}{t}
= m \frac{i}{\hbar} \comm{\hat{H}_0}{\vu{x}}
= m \frac{i}{\hbar} (\hat{H}_0 \vu{x} - \vu{x} \hat{H}_0)
\end{aligned}$$
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.
|
