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Detailed balance says that in equilibrium, the total number of particles leaving a certain quantum state per unit time equals the particles arriving in that state per unit time.

Now, without the necessity of stimulated emission, we can imagine that the number of electrons reaching the upper level by stimulated absorption per unit time equals the number of electrons leaving that state by spontaneous emission per unit time.

So how did Einstein suggest the necessity of stimulated emission by detailed balance in equlibrium?

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The detailed balance says something stronger: each process in equilibrium is compensated by the time-reversed process. Your version of the balance needs to be strengthened to this form of balance whenever the fundamental laws are time-reversal-symmetric. Time reversal symmetric laws couldn't imply that in equilibrium, spontaneous emission should be balancing (stimulated) absorption because those two are not time reversals of each other. The strength of one of them depends on the number of photons around, the strength/probability of the other process doesn't, and so on.

In Einstein's version, using the modern – more comprehensible – quantum notation, the emission means that the number of particles (photons...) in a state goes from $N-1$ to $N$. The reverse process, absorption, means that the number of particles in a state goes from $N$ to $N-1$. These two processes must occur equally often or likely.

So it means that the probability of these two transitions must be the same. But the probability of absorption (the number of photons outside is reduced from $N$ to $N-1$) is proportional to $N$. It follows that the probability of emission needs to be proportional to $N$ as well, but in our convention for $N$, $N$ is the greater number among the two (initial and final numbers). $N$ is equal to $(N-1)+1$ where $(N-1)$ is the initial number of photons during emissions. If you call it $M=N-1$, then the probability must be proportional to $N=M+1$. The term $M$ is the stimulated emission (the more photons are around, the more likely the atom will drop) and $1$ is the spontaneous emission (independent of the number of photons already flying around).

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