D.42 Derivation of the Einstein A coefficients

Einstein did not really derive the spontaneous emission rate from relativistic quantum mechanics. That did not exist at the time. Instead Einstein used a dirty trick; he peeked at the solution.

To see how, consider a system of identical atoms that can be in a low energy state $\psi_{\rm {L}}$ or in an excited energy state $\psi_{\rm {H}}$ . The fraction of atoms in the low energy state is $P_{\rm {L}}$ and the fraction in the excited energy state is $P_{\rm {H}}$ . Einstein assumed that the fraction $P_{\rm {H}}$ of excited atoms would evolve according to the equation

$\begin{displaymath} \frac{{\rm d}P_{\rm {H}}}{{\rm d}t} = B_{\rm {L\to{H}}} ... ...o(\omega_0)\; P_{\rm {H}} - A_{\rm {H\to{L}}}\; P_{\rm {H}} \end{displaymath}$

where $\rho(\omega)$ is the ambient electromagnetic field energy density, $\omega_0$ the frequency of the photon emitted in a transition from the high to the low energy state, and the

and

values are constants. This assumption agrees with the expression (7.46) given in chapter 7.8.

Then Einstein demanded that in an equilibrium situation, in which $P_{\rm {H}}$ is independent of time, the formula must agree with Planck’s formula for the blackbody electromagnetic radiation energy. The equilibrium version of the formula above gives the energy density as

$\begin{displaymath} \rho(\omega_0) = \frac{A_{\rm {H\to{L}}}/B_{\rm {H\to{L}... ...{\rm {L\to{H}}}P_{\rm {L}}/B_{\rm {H\to{L}}}P_{\rm {H}}) - 1} \end{displaymath}$

Equating this to Planck’s blackbody spectrum as derived in chapter 6.8 (6.11) gives

$\begin{displaymath} \frac{A_{\rm {H\to{L}}}/B_{\rm {H\to{L}}}} {(B_{\rm {L\t... ...\pi^2c^3} \frac{\omega_0^3}{e^{\hbar\omega_0/{k_{\rm B}}T}-1} \end{displaymath}$

The atoms can be modeled as distinguishable particles. Therefore the ratio $P_{\rm {H}}$ $\raisebox{.5pt}{$/$}$ $P_{\rm {L}}$ can be found from the Maxwell-Boltzmann formula of chapter 6.14; that gives the ratio as $e^{-(E_{\rm {H}}-E_{\rm {L}})/{k_{\rm B}}T}$ , or $e^{-\hbar\omega_0/{k_{\rm B}}T}$ in terms of the photon frequency. It then follows that for the two expressions for $\rho(\omega_0)$ to be equal,

$\begin{displaymath} B_{\rm {L\to{H}}} = B_{\rm {H\to{L}}} \qquad \frac{A_{... ...o{L}}}}{B_{\rm {H\to{L}}}} = \frac{\hbar\omega_0^3}{\pi^2c^3} \end{displaymath}$

That $B_{\rm {L\to{H}}}$ must equal $B_{\rm {H\to{L}}}$ is a consequence of the symmetry property mentioned at the end of chapter 7.7.2. But it was not self-evident when Einstein wrote the paper; Einstein really invented stimulated emission here.

The valuable result for this book is the formula for the spontaneous emission rate $A_{\rm {H\to{L}}}$ . With $B_{\rm {H\to{L}}}$ given by (7.47), it determines the spontaneous emission rate. So it has been obtained without using relativistic quantum mechanics. (Or at least not explicitly; there simply are no nonrelativistic photons.)