4.3.1 Einstein’s $A$ and $B$ coefficients

Consider for simplicity a gas of “atoms” which can either be in their ground state or an excited state (excitation energy E), and let the numbers in each be $n_0$ and $n_1$. The atoms will interact with the black-body radiation field, emitting and absorbing quanta of energy, to reach thermal equilibrium. We will need Planck’s law for the energy density of the black-body radiation field at a given frequency:

where the temperature-independent prefactor $f\left(\omega \right)$ arises from the density of states, and $n\left(\omega ,T\right)$ is the Bose-Einstein expression for the average number of quanta of energy in a given mode. See section A.10 for more details about the Bose-Einstein distribution.

The rates of absorption and stimulated emission are proportional to the energy density in the field at $\omega =E∕\hslash$, and the coefficients are denoted $B_{01}$ and $B_{10}$, while the rate of spontaneous emission is just $A_{10}$. (We have seen that $B_{01}=B_{10}$, but we won’t assume that here.) Then the rate of change of $n_0$ and $n_1$ is

At thermal equilibrium, ${ṅ}_0={ṅ}_1=0$ and $n_1∕n_0={\text{e}}^{-E∕kBT}$. Using the Planck law for $\rho \left(E∕\hslash \right)$, with some rearrangement we get

Now this has to be true for any temperature, so we can equate coefficients of ${\text{e}}^{E∕kBT}$ to give $A_{10}=B_{01}f\left(\omega \right)$ and $A_{10}=B_{10}f\left(\omega \right)$. So we recover $B_{01}=B_{10}$, which we already knew, but we also get a prediction for $A_{10}$. Thus we have

We see that the total emission probability corresponds to replacing $n\left(\omega ,T\right)$ with $n\left(\omega ,T\right)+1$. This result is confirmed by a full calculation with quantised radiation fields, where the factor arises from the fact that the creation operator for quanta in a mode of the EM field has the usual normalisation $a_{\omega }^{†}\left|\right.n_{\omega }⟩=\sqrt{n_{\omega }+1}\left|\right.n_{\omega }+1⟩$.

• Gasiorowicz Supplement 1

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