- #26

Cthugha

Science Advisor

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1) When I turn the knob on my power supply and increase the pump rate, I increase the number of photons in the cavity, and hence the loss. But the gain is already saturated at this point, so how is clamping satisfied?

Yes, the loss rate is constant, so the losses increase as you pump harder. Now the gain is defined as the amplification of the light field inside per length unit or per round trip. Let me just take some arbitrary exaggerated numbers. If you had a loss rate of 10 % per round trip and 10 photons inside the cavity you would lose one per round trip. Accordingly the steady state gain must also provide one photon per round trip, so you get a gain of 0.1 per round trip. Now you pump harder and have a mean photon number of 100 inside your cavity. The loss rate stays the same and you now lose 10 photons per round trip. The gain in the steady state regime will now also provide 10 photons per round trip. So 10 provided photons divided by the 100 photons inside still give you a gain of 0.1 per round trip. The gain stays the same. The number of photons provided by the gain of course differs.

2) So gain = loss isonlyvalid for cavities where loss depends on intensity, right?

I am not really aware of any cavities where losses do not depend on intensity. If you had an efficient way to introduce a loss to a cavity that does not depend on the intensity inside, you could extract a fixed number of photons from a cavity regardless of the photon number inside. People are chasing this kind of photon blockade nowadays and you could certainly publish a high-impact publication if you realized such a loss which does not depend on intensity.

3) Say a mirror in my cavity has a tunable transmission, such that we can get an output. Now I increase the transmission of the mirror, because I want to measure something. This will increase the loss of my cavity or - equivalently - make the gain smaller. But using the same line of throught as chrisbaird we can argue that the gain will just increase to the value of the new losses, right?

No, increasing loss does not mean that the gain will get smaller.

Ok, another simplified example. You have a cavity with a loss rate of 10% per round trip and 100 photons inside. You lose 10 photons per round trip and gain also provides 10 photons per round trip (so gain is also 0.1). Now you detune the cavity to reduce the quality factor and get a loss rate of 50%. So now you will lose 50 photons on the next round trip. Assuming a simplified system where the inversion does not really depend on what happens inside the cavity (think for example QD lasers), the gain will depend mostly on what the pump rate offers and now still provide 10 photons. As there are now only 50 photons inside, adding 10 photons with a photon number of 50 present already means that the gain increases to 10/50=0.2. Nevertheless the losses in the next round trip will still be larger than the photons added by gain. This game continues until the mean photon number reaches 20. Now the loss rate of 50% will cause 10 photons to leave the cavity and 10 will also enter the cavity due to gain. Now the gain is 10/20=0.5 and again equals the losses. We reached the steady state.

In reality of course the number of photons added is not strictly constant, but may depend on what is happening inside the cavity. However, I hope the basic idea became clear.