# What is the argument that Einstein's induced emission and induced absorption coefficients $B_{mn}=B_{nm}$ must be equal?

The following is a summary of my reading of https://www.feynmanlectures.caltech.edu/I_42.html#Ch42-S5

## Defintions

• $$N_{i}$$ Population of molecules in state $$i$$
• $$R_{i\to j}$$Transition rate from state $$i$$ to state $$j$$
• $$A_{mn}$$ Coefficient of spontaneous emission
• $$B_{mn}$$ Coefficient of induced emission
• $$B_{nm}$$ Coefficient of absorption
• $$E_{m}-E_{n}=\Delta E=\hbar\omega>0$$ Transition energy
• $$\mathcal{I}(\omega)$$ Radiation intensity profile
• $$N_{m}=N_{n}e^{-\frac{\Delta E}{\mathit{k}T}}$$Boltzmann relation

## Feynman's equation 42-12

$$\mathcal{I}(\omega)d\omega=\frac{\hbar\omega^{3}d\omega}{\pi^{2}c^{2}\left(e^{\frac{\hbar\omega}{\mathit{k}T}}-1\right)}.$$

## Derivation

Write the expressions for transition rates and set them equal using argument by footnote \begin{aligned} R_{n\to m}&=N_{n}\mathcal{I}(\omega)B_{nm}\\ R_{m\to n}&=N_{m}\left(A_{mn}+\mathcal{I}(\omega)B_{mn}\right)\\ R_{n\to m}&=R_{m\to n}. \end{aligned}

Combining expressions and applying basic algebra we get $$\mathcal{I}(\omega)=\frac{A_{mn}}{B_{nm}e^{\frac{\hbar\omega}{\mathit{k}T}}-B_{mn}}=\frac{\hbar\omega^{3}}{\pi^{2}c^{2}\left(e^{\frac{\hbar\omega}{\mathit{k}T}}-1\right)}.$$

Therefore we can deduce something: First, that $$B_{nm}$$ must equal $$B_{mn}$$, since otherwise we cannot get the $$(e^{\hbar\omega/kT} - 1)$$. So Einstein discovered some things that he did not know how to calculate, namely that the induced emission probability and the absorption probability must be equal.

Clearly, setting $$B_{nm}=B_{mn}$$ gives a compelling result, but I don't believe that follows from the algebra. Does the "necessity" of the result follow from a variation of $$\omega$$ or some other method of differential calculus?

If $$\omega$$ were a continuous real number parameter with all other terms constant, the result would be obvious. But in this case $$\omega$$ is a discrete value determined by the transition energy.

I also observe that in this recent and more detailed application of these ideas, the equation $$B_{nm}=B_{mn}$$ does not, in general, hold. See equation 14

https://doi.org/10.1155/2013/503727

The expression

$$\mathcal{I}(\omega)=\frac{A_{mn}}{B_{nm}e^{\frac{\hbar\omega}{\mathit{k}T}}-B_{mn}}=\frac{\hbar\omega^{3}}{\pi^{2}c^{2}\left(e^{\frac{\hbar\omega}{\mathit{k}T}}-1\right)}$$

is a function of $$\omega$$. If it is supposed to hold for more than one specific value of $$\omega$$ (and $$T$$), then the argument holds. You can easily see this from inspection, but you could also say e.g. that for small $$\omega$$ (specifically, $$\omega \ll kT/\hbar$$), the expression on the right becomes inversely proportional to $$\omega$$, while the expression on the left becomes inversely proportional to $$\omega + \frac{kT}{\hbar}(1-B_{mn}/B_{nm})$$; demanding the same low-frequency (or high-temperature) behavior requires that $$B_{nm}-B_{mn}$$ vanishes.

• Ok. I guess it depends on a variation of $T$. I didn't consider that. The reason I balked at a variation in $\omega$ is that in this case $\omega$ is a discrete value determined by the transition energy. – Steven Thomas Hatton Jun 3 at 3:37
• @StevenThomasHatton Ah, okay. That's not quite right. The idea is that we are taking $I(\omega)=\hbar\omega^3/\pi^2c^2(e^{\hbar\omega/kT}-1)$ as an input. We say that we know that this is the correct form for the radiation intensity profile for all $\omega$. On the other hand, by balancing the emission/absorption rates, we can express it in terms of the $A$'s and $B$'s. By comparing this to the formula we know, we can read off the fact that $B_{nm}=B_{mn}$ by inspection. – J. Murray Jun 3 at 17:36
• "by inspection" is a bit more rigorous than argumentum footnoteum. To me "by inspection" means "the result is obvious", but it still requires a step of reasoning. The $B_{mn}$ and $B_{nm}$ are specific to a species of molecule, and to a particular state transition. The blackbody equation involving them really (as I see it) only applies to the resonant peak near the transition frequency. On the other hand, it is temperature agnostic. Holding everything but the temperature constant requires $B_{nm}=B_{mn}.$ See feynmanlectures.caltech.edu/I_41.html#Ch41-F3 or ... – Steven Thomas Hatton Jun 4 at 3:17
• ... [continued] Figure 3 in dx.doi.org/10.1155/2013/503727 – Steven Thomas Hatton Jun 4 at 3:20