Spontaneous emission and induced radiation

In Einstein A., Zur Quantentheorie der Strahlung, Phys.ZS., 18, 121-137 (1917) spontaneous emission is considered to occur together with induced radiation so that one can write the following condition for equilibrium (which also includes induced absorption): $$p_n e^{-\frac{\varepsilon_n} {kT}} B_n^m \rho = p_m e^{-\frac{\varepsilon_m} {kT}} \left( B_m^n \rho + A_m^n \right)$$ where $p_n$ and $p_m$ are statistical weights of the states $n$ and $m$, $\rho$ is radiation density of frequency $\nu$, $A_m^n$ is a constant characteristic of the spontaneous $m \rightarrow n$ transition (spontaneous emission), $B_m^n$ and $B_n^m$ are constants expressing the change of state under induced emission and absorption.

To arrive at Planck's radiation density law it is considered that at high temperatures the above equation becomes $$p_n B_n^m = p_m B_m^n .$$ What is the justification, however, to substitute $B_n^m$ expressed through the latter equality (valid for extreme temperatures) into the initial equation above (valid for lower temperatures)?

-

1 Answer

It's much like the justification for detailed balance. Only the whole system knows whether the temperature is high or low, any individual atom is just absorbing and emitting photons independent of whether the whole system is at a high temperature or a low one. So the A,B coefficients are independent of anything except the final and initial state.

As an aside, these coefficients are important precursers for Heisenberg's matrices. They were calculated explicitly by Kramers and Heisenberg in the old quantum theory in the early 1920s.

-
The principle of detailed balance is valid for a given temperature, right? This isn't the case here. Thus, in our case, if we agree with the above substitution it would mean we agree that at a given temperature there are two completely different equillibria for one and the same system -- one involving spontaneous emission, the other in absence of spontaneous emission. – ganzewoort Sep 22 '11 at 5:08
@gazenwoort: I meant it is like the philosophical justification for detailed balance, that you can ignore the rest of the system when looking at the statistics of transitions of any one part. In this case, you can ignore all the other photons when looking at the absorption/emission of an atom. The equilibrium distribution is Planckian with the proper A/B coefficients and require spontaneous emission. Without spontaneous emission, the low-temperature distribution is wrong. – Ron Maimon Sep 22 '11 at 6:04