# Probability of an atom wanting to Excite/Deexcite in the state of Population Inversion

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!!

Thanks!

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.

• 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. May 27, 2020 at 20:00
• 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? May 27, 2020 at 20:03
• 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. May 27, 2020 at 21:20