Gain saturation in laser cavity

1. May 4, 2007

neu

Im studying gain saturation for homogeneous and inhomogeneously broaded transistions.

Im a little confused bout the logistics of laser operation regarding the gain in the cavity.

So from ground state, the atoms are optically pumped to produce population inversion(4 level system) where stimulated emission occurs and we have lasing operation.

obviously here there is a optical gain, the input photons are amplified by multiple passes through medium.

Im told gain saturation occurs where there is so much stimulated emission that population inversion runs out and gain reduces to the threshold (=1).

So saturation occurs after a very short time? ie does the system go from steady state population inversion, where gain is >1, to steady state N1=N2, where gain=1?

2. May 4, 2007

neu

I read more and got it now. Well i hope

Gain saturation is laser operation right? I mean, gain saturates as laser operation starts and stays saturated.

3. May 5, 2007

las3rjock

As I recall, gain saturation is steady-state laser operation, but it's been a few years since the last time I read Siegman's book.

4. May 5, 2007

Manchot

You are correct. In steady-state laser operation, the gain will saturate at the cavity's threshold gain value. Think of it this way: Suppose that we start pumping electrons into a higher energy state. In fact, we do it so suddenly and quickly that, for a short time, we are able to achieve a gain higher than the threshold value. Now, because the gain is higher than threshold, the number of photons trapped within the cavity will begin to increase exponentially. Obviously, the exponential increase cannot occur indefinitely, because as the number of photons increases, so does the stimulated emission rate, decreasing the population inversion (and the gain). It will then decrease the gain to exactly the lasing threshold. Essentially, the lasing process is self-limiting.

5. May 5, 2007

neu

ok yeah got it. thanks

6. May 5, 2007

neu

I have a related question about Q-switching.

net gain $$\kappa$$ is equal to roud trip gain - losses $$\gamma$$

and net gain is equal to threshold $$\kapaa_{th}$$ in steady state lasing operation.

where:

$$N_{th}^*=\frac{8\pi\nu^2n^2\tau_{21}\kappa_{th}\delta\nu}{c^2}$$

and $$\kappa_{th}=\gamma-\frac{1}{2L}Ln[R_{1}R_{2}]$$

So $$N_{th}^*\propto\kappa_{th}\propto\gamma$$

so as losses ($$\gamma$$) increase, so does $$\kappa_{th}$$ and hence so does the threshold population inversion $$N_{threshold}^*$$, right?

So when losses are instantaneouly dropped from high to low then the population inversion is much greater than threshold and so have massive gain and high intensity pulse emmitted?

Is that right?

Last edited: May 5, 2007
7. May 6, 2007

Manchot

Yes, that's correct. If you want to think of it qualitatively (which is probably more important, since the above equations aren't very general), in some ways, Q-switching is analogous to a dam. If you kill the stimulated emission by making the loss high (building a dam), the only loss processes from the upper state to the lower state are the spontaneous transitions, which are (relatively) slow. As a result, your population inversion (water behind the dam) builds up. If you allow stimulated emission to occur (destroying the dam), a huge number of electrons with be available for downward transition, and there will be massive photon emission. After a while, the "flooding" will have died down, and the laser will reach its usual steady-state value (normal river flow).

8. May 6, 2007

neu

thats great thanks, appreciate it