Einstein coefficients for emission and absorption ($A_{21}$, $B_{12}$, $B_{21}$) are often derived from a consideration of thermodynamic equilibrium using Boltzmann statistics and comparison with Planck's law. This (among many others) is the typical approach: https://www.youtube.com/watch?v=4TOvjtovRXY

How can it be that intrinsic microscopic properties (which might be calculated by quantum mechanics on n-level systems including radiation fields) can be derived so "easily" based just on thermodynamics, which is itself of rather statistical nature?

I've always seen some kind of deep miracle behind this derivation: What is the right way to understand it? Einstein, at this time (1917), had no idea about how quantum mechanics would look like in about one decade in the future - how is it possible, that he got somehow a valid relation without doing complicated quantum mechanical calculus on the atomic level? Isn't that great?

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    $\begingroup$ Thermodynamics isn't statistical in nature. Technically speaking not even statistical mechanics is because it sums over all states, just like quantum mechanics sums over all paths, which in both cases is fully deterministic. Quantum mechanics alone can't give you such a prediction to begin with because it describes all quantum systems, not just systems in thermodynamic equilibrium. Unsurprisingly systems not in thermodynamic equilibrium do not obey these relationships. If all bodies did, then neither light bulbs nor lasers could exist. $\endgroup$ Commented Nov 1, 2022 at 9:47

4 Answers 4


How can it be that intrinsic microscopic properties (which might be calculated by quantum mechanics on n-level systems including radiation fields) can be derived so "easily" based just on thermodynamics, which is itself of rather statistical nature?

Actually the video (see at 24:10) shows, from thermodynamics and Planck's law you can derive only the ratios between the Einstein coefficients $A_{21}$, $B_{12}$ and $B_{21}$, but not their absolute values.

$$\frac{A_{21}}{B_{21}}=\frac{2h\nu^3}{c^2}$$ $$\frac{B_{21}}{B_{12}}=\frac{g_1}{g_2}$$

So, the best you can do is writing the coefficients as $$\begin{align} A_{21} &= g_1 C \frac{2h\nu^3}{c^2} & \text{ (spontaneous emission)}\\ B_{12} &= g_2 C & \text{ (absorption)} \\ B_{21} &= g_1 C & \text{ (stimulated emission)} \end{align}$$ where $C$ (for lack of a better name) is still an unknown constant. This $C$ (and also $g_1$ and $g_2$) actually is an intrinsic microscopic property of the atom. And indeed, for finding $C$ you need complicated quantum mechanical calculus on the atomic levels.

The fact that all three Einstein coefficients ($A_{21}$, $B_{12}$, $B_{21}$) ultimately are proportional to the same intrinsic atomic property ($C$), indicates there is a common mechanism to all three processes.
enter image description here
(image from LibreTexts - Electrical Engineering - 7.1: Absorption, Spontaneous Emission, Stimulated Emission)


Einstein's argument can be broken down into two parts.

The first part involving stimulated emission concludes that in order for a charged two state system to be in thermal equilibria the external field has to have an intensity profile obeying $$I(\omega) = \frac{A}{B_{21} e^{\hbar\beta\omega} - B_{12}}.$$

The second part of his argument begins with the realization that if our atom is in equilibria, then so should the surrounding EM field. Should this not be the case you'd probably be breaking one of the laws of thermodynamics, exactly which one I can't say at this moment.

Since a black body satisfies $$I(\omega) = \frac{\hbar}{\pi^2}\frac{\omega^3}{ e^{\hbar\beta\omega} - 1}$$ we conclude that $$B_{21} = B_{12}$$ and $$\frac{A}{B}=\frac{\hbar}{\pi^2}\omega^3.$$

If these constraints were not obeyed you would have a charged system that could not come to thermal equilibrium with the EM field, which is obscene.

  • $\begingroup$ Hmmm. That's more or less the line of reasoning I was talking about. Maybe I still don't understand what this is trying to tell me. At least the question isn't demystified... The fact that A/B is proportional to w³ could most likely be derived using "hard" quantum mechanics alone on a single atom interacting with a radiation field (using dipole moments...), without ever using thermodynamics. It sounds like this to me: I can initially derive a statement about intrinsic properties from thermodynamics, but in the end I no longer need thermodynamics at all. $\endgroup$
    – MichaelW
    Commented Nov 1, 2022 at 11:21
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    $\begingroup$ What this argument is saying is that any system like this has to obey these constraints in order to be consistent with thermaldynamics. $\endgroup$
    – AfterShave
    Commented Nov 1, 2022 at 11:55

Planck's law cannot be derived from classic mechanics, so it is quantum in nature. Yes, it is statistical, but it contains some footprints of low-level mechanical theory. The idea is to apply statistical methods to not-fully-known quantum mechanics and compare the result with Planck's law. Using this approach we can "extract" this "quantum footprints" from Planck's law.


Thermodynamics is well-established empirically. Einstein understood that this demanded that whatever the microscopic physics was, it had to be consistent with thermodynamics.

There's nothing miraculous here at all. The "intrinsic microscopic properties" are mathematical abstractions, products of human imagination. They are not fundamental: the phenomena we observe are fundamental. In physics, we select our abstractions to conform to the observations. That's what Einstein did. It is no coincidence that more elaborate abstractions also conform, since we would have rejected them as physics if they didn't.


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