Stimulated Emission

Coherent Electromagnetic Waves

The interaction of an electron with a photon can be described by the atomic interaction with an electromagnetic wave.

Consider the simple case that we shine a light polarized light source along the \(z\) direction such that the wavelength is greater than the atomic size \(\lambda \gg \lambda_{dB}\). We can them simplify this electromagnetic wave as,

\[\begin{split} \begin{gather} \boldsymbol{E} = E_0\cos(\omega t)\hat z\\ V = \int{\boldsymbol{E}\; d\boldsymbol{r}} = -qE_0z\cos(\omega t) \end{gather} \end{split}\]

The perturbed Hamiltonian is then,

\[\begin{split} \begin{gather} H' = -qE_0 z \cos(\omega t)\\ H'_{ba} = -qE_0\cos(\omega t)z_{ab}\\ z_{ab} \equiv \bra{\psi_a}z\ket{\psi_b} \end{gather} \end{split}\]

Notice that \(z_{ab}\) is always integrate to zero since \(z|\psi|^2\) is always odd. Of course, this only applies to wavefunctions that can be considered even or odd with \(z\).

The probability of transition between \(a\) to \(b\) for the case that \(\omega \gg \omega_0\) (true for our light source) is,

\[\begin{split} \begin{gather} P_{a\rightarrow b}(t) = \left(\frac{qE_0}{\hbar}|z_{ab}|\right)^2\frac{\sin^2\left(\frac{\Delta \omega t}{2}\right)}{\Delta \omega^2}\\ \Delta \omega \equiv \omega - \omega_0 \end{gather} \end{split}\]

Now for an astonishing fact, the probability going the other way \(P_{b\rightarrow a}(t)\) is the exact same! Notice all you need to do is flip the sign of \(\omega_0\) to get the opposite transition.

\[ \begin{equation} \boxed{P_{a\rightarrow b}(t) = P_{b\rightarrow a}(t)} \end{equation} \]

If \(E_a < E_b\) then \(a \rightarrow b\) is called absorption and \(b\rightarrow a\) is called stimulated emission. This is amazing because you wouldn’t expect \(\psi_b\) to go to a lower energy level if an electromagnetic is contributed. Naturally you may have heard of an atom emitting light without any perturbation. This is called instead spontaneous emission.

Incoherent Electromagnetic Waves

The incoherent perturbation is best described by its energy density \(\mu\) which is composed of all frequencies. Recall that the energy density of the electromagnetic wave is,

\[ \mu = \frac{\epsilon_0}{2}E_0^2 \]

• \(E_0\) : electrical field amplitude

We may rewrite the monochromic stimualted emission as

\[ \begin{equation} P_{b\rightarrow a}(t) = \frac{2q\mu}{\epsilon_0 \hbar^2}\abs{z_{ba}}^2\frac{\sin^2\left(\frac{\Delta \omega t}{2}\right)}{\Delta \omega^2} \end{equation} \]

To consider the full range of frequency we integrate over \(\omega\). Recall,

\[ d\mu = \rho(\omega)\; d\omega \]

\[\begin{equation} \boxed{P_{b\rightarrow a}(t) = \frac{2q}{\epsilon_0 \hbar^2}\abs{z_{ba}}^2\int_{-\infty}^{\infty}{\rho(\omega)\frac{\sin^2\left(\frac{(\Delta \omega) t}{2}\right)}{(\Delta \omega)^2}\; d\omega}} \end{equation}\]

We may assume that energy density dominates at resonant frequency \(\omega = \omega_0\) so \(\rho(\omega) = \rho(\omega_0)\). Substituting \(x \equiv (\Delta \omega)t/2\) we can extend the integral to \(x=\pm\infty\) and find that,

\[\begin{equation} \boxed{P_{b\rightarrow a}(t) \simeq \frac{\pi q \abs{z_{ba}}^2}{\epsilon_0\hbar^2} \rho(\omega_0)t} \end{equation}\]

Isotropic Electromagnetic Waves

Let’s now introduced unpolarized light, actually light coming from every direction isotropically. The equation no long has the electromagnetic wave coming from the \(\hat z\) direction but now everywhere \(\abs{z_{ba}} \rightarrow \abs{r_{ba}}\).

The isotropic light will be polarized along \(\hat n\) thus we project \(\abs{r_{ba}}\) to \(\hat n\) and take the average value as the factor of the effective electric field.,

\[ \abs{r_{ba} \cdot \hat n}^2 = \int{\frac{1}{4\pi} \cos^2\theta\sin\theta \; d\theta\; d\phi} = \frac{1}{3}\abs{r_{ba}}^2 \]

\[\begin{equation} \boxed{P_{b\rightarrow a}(t) \simeq \frac{\pi q \abs{r_{ba}}^2}{3\epsilon_0\hbar^2} \rho(\omega_0)t} \end{equation}\]

Transition Rate

The transition rate has a good approximation considering statistically large numbers of photons (attempts to transition). In this approximation the transition rate is the rate of success,

\[ \begin{equation} R_{b\rightarrow a} \equiv \frac{dP_{b\rightarrow a}}{dt} \end{equation} \]

Fermi’s Golden Rule

The most simplest way to memorize all of these is Fermi’s golden rule. The rule states that the transitional probability given the energy density of state \(\rho\) given by such integral,

\[ R_{i \rightarrow f} = \frac{2\rho}{\hbar}\abs{H'_{fi}}^2\int_{-\infty}^{\infty} \frac{\sin \omega t}{\omega} \d \omega\]

This is simply equal to,

\[\boxed{ R_{i \rightarrow f} = \frac{2\pi}{\hbar}\abs{H_{fi}}^2\rho}\]