I know that population inversion is achieved when the probability of stimulated emissions is greater than the probability of absorptions. But after the state is achieved, if $E_1 < E_2$ then what will be the probability of an atom wanting to excite from $E_{\text{1}}$ to $E_{\text{2}}$ versus the probability of an atom wanting to de-excite from $E_{\text{2}}$ to $E_{\text{1}}$?

I thought that if population inversion has been achieved because the probability of excitation is greater than the probability of de-excitation, then the same reasoning should apply in my question too. Is my reasoning correct? Any help whatsoever would be greatly appreciated!!



1 Answer 1


1- 50/50. At the point of inversion both probabilities are the same.

2-I think you are missing some key points here. You always need to include the energy structure. For example in a 4-level system: $E_0$ ground level, $E_1$ upper pump level, $E_2$ upper laser level, $E_3$ lower laser level. The inversion between ground state $E_0$ and upper-pump level $E_1$ is always close to 0. And the population of the lower laser level $E_3$ is always really low. This means that 1: you always have a high probability of exciting an electron from $E_0$ to $E_1$ because the occupancy of $E_1$ is low. And 2: that for virtually every electron decaying from $E_1$ to $E_2$ you achieve inversion. And after an electron falls back to $E_3$, it almost immediately goes down to $E_0$ again. This leaves the lower laser level practically free, and with lack of electrons to be excited back from $E_3$ to $E_2$.

What this entails is that you virtually achieve very high values of inversion with low effort. The probability of emission becomes much higher than that of absorption with "just a handful of electrons". Or in other words, if the only electrons of your system are in the upper laser level, even if just a few, the only probable thing is that they will de-excite and emit a photon.

This is of course simplifying things a bit, not going to extreme pumping or lasing, but should clear the confusion.

Decided to add another point:

What you describe above is also called the transparency point. At that point (lets call it 50/50) the net gain is 0. You sometimes absorb, sometimes emit. You need to picture all these processes with the energy level scheme though. You reach that point by pumping just a bit that the excitation gets enough electrons to match the number of electrons on the lower and not very populated level.

  • $\begingroup$ I'm not getting the logic behind the 50/50 probability, are you saying that if only one atom is left in the lower lasing level $E_{\text{1}}$ (hypothetically), then it has an equal probability of excitation as an atom in $E_{\text{2}}$ has of de-excitation? It's pretty counter intuitive tbh. $\endgroup$ May 27, 2020 at 20:00
  • $\begingroup$ Also, in a regular case of pumping, if $E_{\text{2}}$ is metastable state and if all the atoms truly are in $E_{\text{2}}$, then why would it undergo stimulated emissions? $\endgroup$ May 27, 2020 at 20:03
  • $\begingroup$ You specified the point of inversion in your question: i.e. population in upper and lower level are the same. As soon as population is higher on upper, you have real inversion. At the point where both have the same population though, in the presence of a photon for that transition, the probability for absorption or emission is the same. So yes, a single electron in the lower level has the same probability. Do not forget that in normal cases you have a much higher percentage of electrons in the upper level, and the net effect is of stimulated emission, not absorption. $\endgroup$ May 27, 2020 at 21:00
  • $\begingroup$ 2nd question- To answer why they undergo stimulated emission, we need a lot of quantum mechanics background. But I think you are confusing stimulated and spontaneous emission with your second question. Might it be? $\endgroup$ May 27, 2020 at 21:01
  • 1
    $\begingroup$ as an analogy, the transition $E_2$-$E_3$ is a resonance of the system. Now imagine the system as being a tuning fork for the note of A. If you bang it (you excite it) it will emit the note A. If you have it "quiet" and play and A on a violin, the tuning fork will also resonate and "absorb" part of the acoustic wave. Now the 50/50 part is when you "shook" the tuning fork the right amount, where it is now being transparent by emitting and absorbing the same amounts. $\endgroup$ May 27, 2020 at 21:20

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