Minimum Power Required to Maintain a Population Inversion According to many sources, the minimum power required to maintain a population N in the upper level of a two level laser system is $$P_{min}=fA_{21}Nh\nu$$ where $f$ accounts for the lack of 100% efficiency, $A_{21}$ is the Einstein A coefficient and $\nu$ is the frequency of the transition. This is just the energy required to counteract the spontaneous decay in the system.
However I don't really understand why this is a minimum energy? A possibility is that if some external radiation field is applied we should also get an imbalance in the rates of spontaneous emission and absorption and so is this the extra contribution that we are missing in the above calculation (if not, why not)? Thanks in advance!
 A: It is not possible to have a population inversion in a system with just two states.
The trick with any laser is that you need three possible states - let's call them 1, 2 and 3 (in increasing order of energy - see image from wikipedia):
 
If we can make it so that the 2->1 transition is much slower than the 3->2 transition, then you can create a population inversion, but you need quite a lot of power (since you are exciting from the ground state, you need to excite more than half of all the atoms).
For this reason, in practice, most lasers actually are four-level: this means that the two levels between which transition occurs contain a minority of the atoms, and it's possible to create a population inversion between these two levels with minimal power.  At that point, "minimal" is however much power is needed to counter the spontaneous decay taking place (which is the expression given in your question) multiplied by some factor describing losses (for example, this will include the energy lost in going from state 4->3 and from 2->1)
