Information

Atoms light interaction outline

  • coupling matrix element (TODAY)
  • narrow band and broad band
  • quantized EM filed

Results of Classical EM and Lagrangian Formalism

Ref Goldstein Eq.(1.63) and Eq.(8.35).

$$\begin{align} \vec{p}_{\mathrm{canonical}} = \vec{p}_{\mathrm{kinetic}} + \frac{q}{c}\vec{A} \end{align}$$

for electron, charge $q = -e$

$$\begin{align} H = \underbrace{\frac{1}{2m} \left( \vec{p}_{\mathrm{canonical}} + \frac{e}{c}\vec{A} \right)^2}_{E_{\mathrm{kinetic}}} + V(r) \end{align}$$

where the first term is kinetic energy. Chose Coulomb gauge $\nabla\cdot \vec{A} = 0$ , so $[\vec{A}, \vec{p}] = 0$ , then

$$\begin{align} H = \underbrace{\frac{p^2}{2m} + V(r)}_{H_0} + \underbrace{\frac{e}{mc}\vec{p}\cdot \vec{A}}_{H_{\mathrm{int}}} + \underbrace{\frac{e^2}{2 mc^2}\vec{A}^2}_{H^{(2)}} \end{align}$$

Consider a plane wave

$$\begin{align} H_{\mathrm{int}} = \frac{e}{m c} \vec{p} \cdot\hat{e} A_0 e^{\mathrm{i}\vec{k}\cdot \vec{r} - \mathrm{i}\omega t} \end{align}$$

then matrix element

$$\begin{align} \langle b |H_{\mathrm{int}} | a\rangle = H_{ba} e^{-\mathrm{i}\omega t} \end{align}$$

where

$$\begin{align} H_{ba} = \frac{eA_0}{mc} \hat{e}\cdot \langle b | \vec{p} e^{\mathrm{i}\vec{k}\cdot \vec{r}} | a\rangle \end{align}$$

for long wave length approximation, $k r \ll 1$ , expand the exponential

$$\begin{align} e^{\mathrm{i}\vec{k}\cdot \vec{r}} = 1 + \mathrm{i} \vec{k}\cdot \vec{r} - \frac{1}{2} (\vec{k}\cdot \vec{r})^2 + \cdots \end{align}$$

Diople term

take the first term

$$\begin{align} H_{ba} = - \frac{\mathrm{i} e E_0}{m\omega} \hat{e} \langle b| \vec{p} | a\rangle \end{align}$$

where we have used (radiation field)

$$\begin{align} \vec{E} = - \frac{1}{c}\frac{\partial \vec{A}}{\partial t} = \frac{\mathrm{i}\omega}{c}A_0 \hat{e} \equiv E_0 \hat{e}. \end{align}$$

Then use

$$\begin{align} \vec{p} = \frac{m}{i\hbar} [\vec{r}, H_0] \end{align}$$

we get

$$\begin{align} H_{ba} = e E_0 \hat{e} \langle b| \vec{r} | a\rangle \frac{\omega_{ba}}{\omega} = \vec{E}\cdot \langle \vec{D}\rangle \frac{\omega_{ba}}{\omega} \end{align}$$

where $D$ is the dipole operator.

Higher-order radiation processes

assume the polarization is along the $z$ direction, and propagate along $x$ direction

$$\begin{align} A(\vec{r}) = A\hat{z} e^{\mathrm{i} kx} \end{align}$$ $$\begin{align} k r = \frac{\hbar \omega}{\hbar c} a_0 = \frac{e^2}{a_0} \frac{1}{\hbar c} a_0 = \frac{e^2}{\hbar c} = \alpha. \end{align}$$

The expansion

$$\begin{align} H_b = \frac{e A}{mc}\langle b| p_z \left[ 1 + \mathrm{i} kx -\frac{1}{2}(kx)^2 \right]| a\rangle, \end{align}$$

the 2nd term

$$\begin{align} p_z x = \underbrace{\frac{1}{2}(p_z x - z p_x)}_{- \frac{1}{2}\hbar L_y} + \frac{1}{2}(p_zx + z p_x) \end{align}$$

then

$$\begin{align} \mathrm{i} k \frac{e A}{mc}\langle b| \frac{1}{2}(p_z x - zp_x)| a\rangle = \underbrace{- \mathrm{i}A k}_{B_y} \underbrace{\frac{e\hbar}{2 mc}}_{\mu_B} \langle b| L_y |a\rangle = \vec{B}\mu_B \langle b| \underbrace{\vec{L}}_{-\vec{\mu}} | a\rangle \end{align}$$

where we have used $\vec{B} = \nabla\times \vec{A} = - \mathrm{i}k A \hat{y}$ .

the other term

$$\begin{align} \frac{1}{2}(p_zx + z p_x) = \frac{m}{2\mathrm{i}\hbar} \left( [z, H_0] x + z[x, H_0] \right) = \frac{m}{2\mathrm{i}\hbar} \left( -H_0 zx + zx H_0 \right) \end{align}$$

then (use $\vec{E} = - \frac{1}{c} \frac{\partial \vec{A}}{\partial t}, E = \mathrm{i}kA$)

$$\begin{align} \mathrm{i} k \frac{e A}{mc}\langle b| \frac{1}{2}(p_z x + zp_x)| a\rangle = -\frac{e A k}{2 c} \omega_{ba} \langle b|z x |a\rangle = \frac{\mathrm{i} e E \omega_{ba}}{2 c} \langle b|z x |a\rangle \end{align}$$

this is electric quadropole.

So,

$$\begin{align} H_{\mathrm{int}}^{(2)} = \underbrace{H_{\mathrm{int}}(M1)}_{\mathrm{real}} + \underbrace{H_{\mathrm{int}}(E2)}_{\mathrm{imaginary}} \end{align}$$

there are no interference, that is to say

$$\begin{align} |H_{\mathrm{int}}^{(2)}|^2 = |H_{\mathrm{int}}(M1)|^2 + |H_{\mathrm{int}}(E2)|^2 \end{align}$$

Summary

Operator Parity
Electric dipole E1 $-e\vec{r}$ -
Magnetic dipole M1 $-\mu_B (\vec{L} + g_s \vec{S})$ +
Electric dipole E2 Tensor $-e \vec{r}:\vec{r}$ +

M1, E2 are "Forbidden" transitions, $\alpha^2 \approx 5\times 10^{-5}$ weaker than an allowed E1 transition.

Reference

  • Goldstein, H., Poole, C. P. & Safko, J. Classical Mechanics. 665 (2011).