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

\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}}

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 $A$ and $B$ 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

\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}

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

\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}

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,

B_{\rm {L\to{H}}} = B_{\rm {H\to{L}}}
...o{L}}}}{B_{\rm {H\to{L}}}} = \frac{\hbar\omega_0^3}{\pi^2c^3}

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.)