So, in Laser theory I learned that a two-level laser is not possible, because it would violate Einsteins equations and the Boltzmann distribution, which in a nutshell say that I cannot cause population inversion with electromagnetic radiation. Yet, a 180 degree NMR pulse is doing just that: we have a two-level system of nuclear spins, and I cause an inversion of the population by applying a long enough pulse. Yet funnier, I can cause population equity when I prolong the spin (270 pulse) or return it to standard (360 pulse).

I know the vector model of NMR. I also know that it's an approximation and that it cannot explain everything (e.g. phase coherence after a 90 pulse). It explains these phenomena exorbitantly well, but it cannot disable the underlying laws of physics. So, how is this possible with Einsteins equation in place?


1 Answer 1


There isn't a contradiction. In both NMR and for optical transitions used in lasers, you can achieve perfect population inversion by putting in the right kind of pulse. It's just harder to do for optical transitions.

The reason that a two-level laser is impossible is more complicated. Einstein's relations/statistical mechanics tell us that you can never get population inversion in thermal equilibrium at any positive temperature. You can get population inversion from the right kind of pulse.

However, this doesn't achieve lasing, because once the atoms decay back to the ground state, you can at most just get the same number of photons that you put in to start with. In other words, during the decay you do get Light by the Stimulated Emission of Radiation, but you don't get Amplification; you're missing the A in "laser".

  • $\begingroup$ First of all, what then does constitute the right kind of pulse? Shape in time-intensity dimension? Shape in frequency distribution? Something related to it's phase? A magic combination of those? Second, so if I can go beyond equal populations, why is there an upper limit and why do I go below that limit with longer pulses? Thanks a lot! $\endgroup$
    – tifrel
    Commented Jun 19, 2020 at 18:02
  • $\begingroup$ @tillyboy It's just the same idea as a $\pi$ pulse in NMR -- every step of the math is exactly the same. The pulse just needs to be at the right frequency with the right total "area". $\endgroup$
    – knzhou
    Commented Jun 19, 2020 at 18:47

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