Skip to main content
\(\require{cancel}\)
Physics LibreTexts

8.9: Absorption and Stimulated Emission of Radiation

Let us use some of the results of time-dependent perturbation theory to investigate the interaction of an atomic electron with classical (i.e., non-quantized) electromagnetic radiation.

The unperturbed Hamiltonian is

 

$\displaystyle H_0 = \frac{p^2}{2 \,m_e} + V_0(r).$ (862)

 

 

The standard classical prescription for obtaining the Hamiltonian of a particle of charge $ q$ in the presence of an electromagnetic field is

 

$\displaystyle {\bf p}$ $\displaystyle \rightarrow {\bf p} - q\,{ \bf A},$ (863)
$\displaystyle H$ $\displaystyle \rightarrow H-q\,\phi,$ (864)

 

 

where $ {\bf A}(\bf r)$ is the vector potential and $ \phi({\bf r})$ is the scalar potential. Note that

 

$\displaystyle {\bf E}$ $\displaystyle = - \nabla\phi - \frac{\partial {\bf A}}{\partial t},$ (865)
$\displaystyle {\bf B}$ $\displaystyle = \nabla\times {\bf A}.$ (866)

 

 

This prescription also works in quantum mechanics. Thus, the Hamiltonian of an atomic electron placed in an electromagnetic field is

 

$\displaystyle H = \frac{\left\vert{\bf p} + e\, {\bf A}\right\vert^{\,2} }{2\,m_e}- e \,\phi + V_0(r),$ (867)

 

 

where $ {\bf A}$ and $ \phi$ are real functions of the position operators. The above equation can be written

 

$\displaystyle H = \frac{ \left(p^2 +e \,{\bf A}\cdot {\bf p} +e \,{\bf p}\cdot{\bf A} + e^2 A^2\right)}{2\,m_e}- e \,\phi + V_0(r).$ (868)

 

 

Now,

 

$\displaystyle {\bf p}\cdot {\bf A} = {\bf A}\cdot {\bf p},$ (869)

 

 

provided that we adopt the gauge $ \nabla\cdot{\bf A} = 0$ . Hence,

 

$\displaystyle H = \frac{p^2}{2\,m_e} -\frac{e\,{\bf A}\cdot{\bf p}}{m_e} +\frac{ e^2 A^2}{2\,m_e}- e\, \phi + V_0(r).$ (870)

 

 

Suppose that the perturbation corresponds to a monochromatic plane-wave, for which

 

$\displaystyle \phi$ $\displaystyle = 0,$ (871)
$\displaystyle {\bf A}$ $\displaystyle = 2\, A_0 \,$$\displaystyle \mbox{\boldmath$\epsilon$}$$\displaystyle \,\cos\left (\frac{\omega}{c} \,{\bf n}\cdot{\bf x} - \omega\, t\right),$ (872)

 

 

where $ \epsilon$ and $ {\bf n}$ are unit vectors that specify the direction of polarization and the direction of propagation, respectively. Note that $ \epsilon$ $ \cdot{\bf n} = 0$ . The Hamiltonian becomes

 

$\displaystyle H = H_0 + H_1(t),$ (873)

 

 

with

 

$\displaystyle H_0 = \frac{p^2}{2\,m_e} + V(r),$ (874)

 

 

and

 

$\displaystyle H_1 \simeq \frac{e\,{\bf A}\cdot{\bf p}}{m_e},$ (875)

 

 

where the $ A^2$ term, which is second order in $ A_0$ , has been neglected.

The perturbing Hamiltonian can be written

 

$\displaystyle H_1 = \frac{e \,A_0\, \mbox{\boldmath$\epsilon$}\cdot{\bf p} }{m_...
...\exp[-{\rm i}\,(\omega/c)\, {\bf n}\cdot{\bf x} + {\rm i}\, \omega\, t]\right).$ (876)

 

 

This has the same form as Equation (850), provided that

 

$\displaystyle V = - \frac{e \,A_0\, \mbox{\boldmath$\epsilon$}\cdot{\bf p} }{m_e}\, \exp[-{\rm i}\,(\omega/c)\, {\bf n}\cdot{\bf x}\,]$ (877)

 

 

It is clear, by analogy with the previous analysis, that the first term on the right-hand side of Equation (876) describes the absorption of a photon of energy $ \hbar\,\omega$ , whereas the second term describes the stimulated emission of a photon of energy $ \hbar\,\omega$ . It follows from Equations (859) and (860) that the rates of absorption and stimulated emission are

 

$\displaystyle w_{i\rightarrow n} = \frac{2\pi}{\hbar} \frac{e^2}{m_e^{\,2}}\, \...
...\,2}\, \vert\langle n\vert\, \exp[\,{\rm i}\,(\omega/c)\,{\bf n}\cdot{\bf x}]\,$   $\displaystyle \mbox{\boldmath$\epsilon$}$$\displaystyle \cdot{\bf p} \,\vert i\rangle\vert^{\,2}\, \delta(E_n-E_i -\hbar\,\omega),$ (878)

 

 

and

 

$\displaystyle w_{i\rightarrow n} = \frac{2\pi}{\hbar} \frac{e^2}{m_e^{\,2}}\, \...
...{\,2}\, \vert\langle n\vert\, \exp[-{\rm i}\,(\omega/c)\,{\bf n}\cdot{\bf x}]\,$   $\displaystyle \mbox{\boldmath$\epsilon$}$$\displaystyle \cdot{\bf p} \,\vert i\rangle\vert^{\,2}\, \delta(E_n-E_i +\hbar\,\omega),$ (879)

 

 

respectively.

Now, the energy density of a radiation field is

 

$\displaystyle U = \frac{1}{2}\left(\frac{\epsilon_0\,E_0^{\,2}}{2}+ \frac{B_0^{\,2}}{2\,\mu_0} \right),$ (880)

 

 

where $ E_0$ and $ B_0=E_0/c= 2\,A_0\,\omega/c$ are the peak electric and magnetic field-strengths, respectively. Hence,

 

$\displaystyle U = 2\,\epsilon_0\,c\,\omega^2\,\vert A_0\vert^{\,2},$ (881)

 

 

and expressions (878) and (879) become

 

$\displaystyle w_{i\rightarrow n} = \frac{\pi}{\hbar} \frac{e^2}{\epsilon_0\,m_e...
...\, U\, \vert\langle n\vert\, \exp[\,{\rm i}\,(\omega/c)\,{\bf n}\cdot{\bf x}]\,$   $\displaystyle \mbox{\boldmath$\epsilon$}$$\displaystyle \cdot{\bf p} \,\vert i\rangle\vert^{\,2}\, \delta(E_n-E_i -\hbar\,\omega),$ (882)

 

 

and

 

$\displaystyle w_{i\rightarrow n} = \frac{\pi}{\hbar} \frac{e^2}{\epsilon_0\,m_e...
...}\, U\, \vert\langle n\vert\, \exp[-{\rm i}\,(\omega/c)\,{\bf n}\cdot{\bf x}]\,$   $\displaystyle \mbox{\boldmath$\epsilon$}$$\displaystyle \cdot{\bf p} \,\vert i\rangle\vert^{\,2}\, \delta(E_n-E_i +\hbar\,\omega),$ (883)

 

 

respectively. Finally, if we imagine that the incident radiation has a range of different frequencies, so that

 

$\displaystyle U = \int d\omega\,u(\omega),$ (884)

 

 

where $ d\omega\,u(\omega)$ is the energy density of radiation whose frequency lies in the range $ \omega$ to $ \omega+d\omega$ , then we can integrate our transition rates over $ \omega$ to give

 

$\displaystyle w_{i\rightarrow n} = \frac{\pi}{\hbar^2} \frac{e^2}{\epsilon_0\,m...
..., \vert\langle n\vert\, \exp[\,{\rm i}\,(\omega_{ni}/c)\,{\bf n}\cdot{\bf x}]\,$   $\displaystyle \mbox{\boldmath$\epsilon$}$$\displaystyle \cdot{\bf p} \,\vert i\rangle\vert^{\,2}$ (885)

 

 

for absorption, and

 

$\displaystyle w_{i\rightarrow n} = \frac{\pi}{\hbar^2} \frac{e^2}{\epsilon_0\,m...
...\, \vert\langle n\vert\, \exp[-{\rm i}\,(\omega_{in}/c)\,{\bf n}\cdot{\bf x}]\,$   $\displaystyle \mbox{\boldmath$\epsilon$}$$\displaystyle \cdot{\bf p} \,\vert i\rangle\vert^{\,2}$ (886)

 

for stimulated emission. Here, $ \omega_{ni} = (E_n-E_i)/\hbar>0$ and $ \omega_{in} = (E_i-E_n)/\hbar>0$ . Furthermore, we are assuming that the radiation is incoherent, so that intensities can be added.

Contributors