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

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

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

$${\bf p} \rightarrow {\bf p} - q\,{ \bf A},$$ (863)

$$H \rightarrow H-q\,\phi,$$ (864)

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

$${\bf E} = - \nabla\phi - \frac{\partial {\bf A}}{\partial t},$$ (865)

$${\bf B} = \nabla\times {\bf A}.$$ (866)

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

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

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

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

Now,

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

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

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

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

$$\phi = 0,$$ (871)

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

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

with

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

and

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

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

The perturbing Hamiltonian can be written

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

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

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

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

and

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

respectively.

Now, the energy density of a radiation field is

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

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

and expressions (878) and (879) become

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

and

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

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

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

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

for absorption, and

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