# Laser Physics: Gain Clamping

Cthugha
Ok, let us check whether we can get this straight.

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 is only valid 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.

Thanks. I have a final piece that I can't fit into the puzzle. If you are getting fed up by now, its totally understandable if you stop reading from now on.

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It is related to something you said earlier:

Ok, to be precise one should distinguish between the case where the photon flux is increased as a consequence of the steady state pumping being increased and the case where the photon flux is increased as a consequence of a perturbation without changing the pump rate. In the latter case the gain will decrease and this is the case I referred to before.

So we have the expression 1 + flux/saturation_flux = small_signal_gain/gain. The small_signal_gain depends on the pump power, so above threshold where gain rate = loss rate, increasing the flux by increasing the pump increases the small_signal_gain. All good here, and it takes care of the first situation you refer to. But I am not quite sure what the other situation you mention refers to during our lasing process?

Cthugha
So we have the expression 1 + flux/saturation_flux = small_signal_gain/gain. The small_signal_gain depends on the pump power, so above threshold where gain rate = loss rate, increasing the flux by increasing the pump increases the small_signal_gain.

No, the small signal gain is still a constant. Why should it change with gain?

All good here, and it takes care of the first situation you refer to. But I am not quite sure what the other situation you mention refers to during our lasing process?

In a nutshell: If you increase the pump rate, you increase the possible number of photons that gain can add to the intracavity photon number per turn. In other words you open up the possibility to reach the same gain at a higher intracavity photon number.

If you have a higher intracavity photon number without increasing the pump rate, this means that the increase happened due to noise or you added some photons to the cavity or something like that. In that case the number of photons that gain can add to the intracavity photon number per turn does not increase, but stays constant as you do not increase the pump rate. As gain is the amplification per photon already present in the cavity and you will not add the same amount as before, but have more photons in the cavity, gain will decrease.

No, the small signal gain is still a constant. Why should it change with gain?

According to Eberly/Milonni the small-signal gain is given by

$$g_0(\nu)=\frac{\sigma(\nu)(P-\Gamma_{21})N_T}{P+\Gamma_{21}}$$

so it increases with increasing pump rate (but not with gain, so all OK here).

If you have a higher intracavity photon number without increasing the pump rate, this means that the increase happened due to noise or you added some photons to the cavity or something like that. In that case the number of photons that gain can add to the intracavity photon number per turn does not increase, but stays constant as you do not increase the pump rate. As gain is the amplification per photon already present in the cavity and you will not add the same amount as before, but have more photons in the cavity, gain will decrease.

I can't make any sense of that when I compare it to what Saleh/Teich write about it on page 576:

At the time of laser turn-on, the flux φ = 0 so that the gain γ = γ0. As the oscillation builds up in time, the increase in φ causes γ decrease through gain saturation. When γ reaches the losses, the photon-flux density ceases its growth and steady-state conditions are achieved. The smaller the loss, the greater the val ue of φ.

When they say the gain saturates, it doesn't seem like they are referring to photons added due to e.g. noise as you tell me. Honestly I don't believe Saleh/Teich make much sense, considering the things we have discussed in this thread.

Cthugha
According to Eberly/Milonni the small-signal gain is given by

$$g_0(\nu)=\frac{\sigma(\nu)(P-\Gamma_{21})N_T}{P+\Gamma_{21}}$$

so it increases with increasing pump rate (but not with gain, so all OK here).

I can't make any sense of that when I compare it to what Saleh/Teich write about it on page 576:

At the time of laser turn-on, the flux φ = 0 so that the gain γ = γ0. As the oscillation builds up in time, the increase in φ causes γ decrease through gain saturation. When γ reaches the losses, the photon-flux density ceases its growth and steady-state conditions are achieved. The smaller the loss, the greater the val ue of φ.

When they say the gain saturates, it doesn't seem like they are referring to photons added due to e.g. noise as you tell me. Honestly I don't believe Saleh/Teich make much sense, considering the things we have discussed in this thread.

Saleh/Teich are one of the prime references and here of course not really referring to noise added to the steady state, but changing the steady state by switching on the pump and wait until the system reaches the steady state. They basically just repeat what has been said in this thread.

I somewhat do not see your problem. Keep in mind that gain is a measure of the amplification describing the number of photons that are added to the intracavity field divided by the number that is already present, so the addition of the same number of photons to the light field at a low and a high intracavity photon number will give very different gains and the situation should be trivial.

What I said about photons added to noise was exactly that: the reaction of a steady state system when it is subject to external noise and slightly perturbed from the steady state. However, the process of switching a laser on is not fundamentally different. It is like adding a huge perturbation to the steady state by emptying the laser cavity and waiting for the system to recover to the steady state.

Also note that my last post you cite was about starting from the steady state. The steady state builds up and then there is some perturbation adding or removing photons and the system will recover to the steady state. If the photon number is perturbed and increased above the steady state value, things happen as described in my last post. If it is below the steady state photon number the situation is just the other way round: You get the same number of photons into the cavity as under steady state situations, but have a smaller intracavity photon number. Accordingly the gain is higher compared to the steady state gain.

What confuses me is that you say

If you have a higher intracavity photon number without increasing the pump rate, this means that the increase happened due to noise or you added some photons to the cavity or something like that. In that case the number of photons that gain can add to the intracavity photon number per turn does not increase, but stays constant as you do not increase the pump rate. As gain is the amplification per photon already present in the cavity and you will not add the same amount as before, but have more photons in the cavity, gain will decrease.

and then state

What I said about photons added to noise was exactly that: the reaction of a steady state system when it is subject to external noise and slightly perturbed from the steady state. However, the process of switching a laser on is not fundamentally different. It is like adding a huge perturbation to the steady state by emptying the laser cavity and waiting for the system to recover to the steady state.

As far as I understand from those two quotes, perturbing the laser by changing the pump during steady state will increase/decrease the gain?

So if I have a laser which is operating at steady state, and I suddenly increase the pump, then then gain will decrease because the intracavity flux increases -- see figure 4.6 (http://books.google.com/books?id=x5...DYQ6AEwAzge#v=onepage&q=gain clamping&f=false). But then gain will then return to its previous value because of what you said here previously

In fact, when you turn on a real laser, you will notice some small overshoot of the intracavity photon number above the steady state value. However, in this region losses are larger than gain (as seen to the right of the steady state in the figure), so the intracavity photon number will reduce again.

The gain returns to the loss value, and I understand that 100% physically. But I can't see that from figure 4.6, because according to that, the gain will decrease with increasing flux, but then the gain has to increase to the value of the loss. But this value is at a lower flux? So gain will always return the its value it has at threshold flux, but the point is that by increasing the pump we open up the possibility to reach the same gain at a higher intracavity photon number. Am I correct here?

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Cthugha
Two things:

So if I have a laser which is operating at steady state, and I suddenly increase the pump, then then gain will decrease because the intracavity flux increases

This is a bit tricky as two things are going on here while the laser has not already arrived at the steady state. A higher intracavity photon number decreases gain, but pumping harder also means more photons per cycle can be added, so this increases the possible gain. These effects balance and the system somewhen settles at the steady state.

So gain will always return the its value it has at threshold flux, but the point is that by increasing the pump we open up the possibility to reach the same gain at a higher intracavity photon number. Am I correct here?

Yes, exactly!

Thanks. I will have to study the replies in this thread for now and compare it to other references to see if I have truly understood it. I'll let you guys know later. Thanks for now.

Hi

I printed out the entire thread, and have read it since my last post. I believe there are still a few gaps missing. BTW, do you have a reference to all these things? Where do you all this from?

This is a bit tricky as two things are going on here while the laser has not already arrived at the steady state. A higher intracavity photon number decreases gain, but pumping harder also means more photons per cycle can be added, so this increases the possible gain. These effects balance and the system somewhen settles at the steady state.

1) So you say we have the two processes

*) "[...] pumping harder also means more photons per cycle can be added [...]"
**) "A higher intracavity photon number [...]"

Just to be perfectly clear: Doesn't (*) imply (**)? I mean, they are not two distinct processes.

2) The above process describes what happens when going from a nonzero rate R1 (at steady state) to a nonzero R2 (at steady state). But if we go from a rate 0 to R1, does the
gain also increase due to * and decrease due to ** until it ultimately reaches value of losses?

3) Why do people refer to the gain as being "saturated" when lasing occurs? I mean, it just attains the only value it can after small_signal_gain > losses, so its not like we have saturated it in any way.

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Cthugha
BTW, do you have a reference to all these things? Where do you all this from?

Good question. I read a few books on this topic, but to be honest I am not sure which one covers these questions best. If you have a good library you might just want to have a look at some books and pick one that you like best. Examples include Siegman's book on lasers, Boyd's Non-linear optics. Maybe Meschede's book If you are interested in some special lasers maybe also the semiconductor laser book by Chow and Koch.

1) So you say we have the two processes

*) "[...] pumping harder also means more photons per cycle can be added [...]"
**) "A higher intracavity photon number [...]"

Just to be perfectly clear: Doesn't (*) imply (**)? I mean, they are not two distinct processes.

Yes, (*) will usually induce (**).

2) The above process describes what happens when going from a nonzero rate R1 (at steady state) to a nonzero R2 (at steady state). But if we go from a rate 0 to R1, does the
gain also increase due to * and decrease due to ** until it ultimately reaches value of losses?

Yes, there is basically no difference between starting from any steady state and starting from 0 pumping. However, at 0 and below threshold the stimulated emission photon number under steady state conditions will be very small and in many cases 0.

3) Why do people refer to the gain as being "saturated" when lasing occurs? I mean, it just attains the only value it can after small_signal_gain > losses, so its not like we have saturated it in any way.

Well, it takes on a constant value and stays there if you pump harder. It is not far off to call this saturation. If I remember correctly, the term comes from amplifier theory in general and was also used for lasers, but I could be wrong there.

Thanks. I thought I understood it, until I read through the derivation of the gain in a 3-level system and saw the attached graph -- note that in the derivation they do not take into account a lossy cavity. But still I believe that the graph contradicts the main point we have been talking about: That the gain is clamped. Don't you agree with me on this one?

The gain increases with pump, until lasing starts. All good so far. But after lasing starts (gain > 0), then gain = loss, but I don't see that *anywhere* on the graph. According to them, gain increases linearly for low pump power, and saturates at some point. So they talk about it as if they have command over the specific value of the gain, we didn't in our discussion -- there it was either nothing or equal to losses.

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Cthugha
Thanks. I thought I understood it, until I read through the derivation of the gain in a 3-level system and saw the attached graph -- note that in the derivation they do not take into account a lossy cavity. But still I believe that the graph contradicts the main point we have been talking about: That the gain is clamped. Don't you agree with me on this one?

Well, I cannot say much without knowing under what assumptions the result was derived. Which book is the graph from? Just two comments: First, a lossless cavity is unphysical and therefore more of academic or pedagogical interest. Second, a lossless cavity usually does not attain a steady state as there is no real way to decrease the intracavity photon number (and thus also no gain clamping). At some point reabsorption becomes prominent and the growth of the intracavity photon number will decrease. You may get a steady state if the probability flow is interrupted for some interaction parameter (see e.g. chapter 23 (quantum theory of the laser) written by Bergou, Englert, LaxScully, Walther and Zubairy in handbook of optics. However this is a very non-standard issue usually not treated in introductory textbooks and also not treated in most detailed textbooks.

The gain increases with pump, until lasing starts. All good so far. But after lasing starts (gain > 0), then gain = loss, but I don't see that *anywhere* on the graph. According to them, gain increases linearly for low pump power, and saturates at some point. So they talk about it as if they have command over the specific value of the gain, we didn't in our discussion -- there it was either nothing or equal to losses.

Difficult to say. What is the definition of gain used here? Usually it is given in units of 1/m and denotes the amplification of a light field per distance traveled through the amplifying medium. Some textbooks simply use the intracavity power divided by the pump power and call this gain. This is all difficult to say without seeing the derivation and which factors are considered and which are not.

Well, I cannot say much without knowing under what assumptions the result was derived. Which book is the graph from? Just two comments: First, a lossless cavity is unphysical and therefore more of academic or pedagogical interest. Second, a lossless cavity usually does not attain a steady state as there is no real way to decrease the intracavity photon number (and thus also no gain clamping). At some point reabsorption becomes prominent and the growth of the intracavity photon number will decrease. You may get a steady state if the probability flow is interrupted for some interaction parameter (see e.g. chapter 23 (quantum theory of the laser) written by Bergou, Englert, LaxScully, Walther and Zubairy in handbook of optics. However this is a very non-standard issue usually not treated in introductory textbooks and also not treated in most detailed textbooks.

Difficult to say. What is the definition of gain used here? Usually it is given in units of 1/m and denotes the amplification of a light field per distance traveled through the amplifying medium. Some textbooks simply use the intracavity power divided by the pump power and call this gain. This is all difficult to say without seeing the derivation and which factors are considered and which are not.

It is from this book, chapter 5: http://books.google.com/books?id=uA...:"Emmanuel Desurvire"&source=gbs_similarbooks

But this is a generel criticism I have of this field: The textbooks contain so many different examples - both physical and unphysical - that it gets very confusing.

Cthugha
Ah, ok. That book is about Erbium doped fiber amplifiers. These are rather special as they are single pass amplifiers. You just put some signal and some pump laser into the fiber. The pump laser excites the dopant ions to a higher level from which stimulated emission to the signal wavelength can occur, so you get some amplified signal at the signal wavelength (usually 1550 nm). The amplification here is just single pass. You just send your signal through the fiber once and it gets amplified due to the presence of the pump beam pumping the fiber. As there are no mirrors, there are also no mirror losses (or better: you have 100% loss at the end of the fiber).

Erbium doped fiber amplifiers are basically lasers without mirrors and you have no feedback mechanism and it does not make sense to define a steady state in the same way you would do for lasers. Just to make the difference more clear: in lasers you just have a pump beam and the signal builds up due to spontaneous and stimulated emission caused by the pump beam and gets amplified. In EDFAs pump and signal are different beams and the signal just gets amplified.

Thanks for clarifying that. I have a few questions left, and I believe I know the answers. If you would like to, you are very welcome to confirm them (I lack confidence to believe I am correct).

1) In the fiber-book, they mention right below the figure of the gain that: "The experimentally determined saturation output power is defined as the signal output power at which the gain has been reduced (compressed) by 3dB". Just to be clear: When they say compressed, are they referring to the stagnation of the gain in the figure from my previous post? Because from that figure, we never see the gain decrease its size, and originally I thought this was the decrease they referred to.

2) A clarifying question regarding 1 + flux / fluxsat = g0/g: When we have steady state lasing, and increase the pump and wait for steady state to occur, then what has changed is the intracavity flux on the LHS. On the RHS g is still given by the same ratio, but g0 has increased, right?

3) In Milonni/Eberly, they derive the following equation for the gain of a three level laser (homogeneously broadened), which they ultimately write as (chapter 10.11)

$$g(\nu) = g(\nu_{21}) \frac{1}{(\nu_{21}-\nu)^2 / (\delta \nu_{21})^2 + 1 + \phi_\nu / \phi^{sat}_{21}}$$

Then they go on to show a figure of g/g0 as a function of (ν21-ν)/δν21 for different fluxes, and it gives a Lorentzian with decreasing ampltitude with increasing flux (they illustrate saturation). Just to be clear: That graph is nothing more than pedagogical interest, since in a physical laser we have a constant gain because of gain-clamping, right?

I really appreciate your help. I've said it a few times already, but I am sitting with 3-4 laser books around me, and none of them really give answers to the questions I have.

Best,
Niles.

Cthugha
1) In the fiber-book, they mention right below the figure of the gain that: "The experimentally determined saturation output power is defined as the signal output power at which the gain has been reduced (compressed) by 3dB". Just to be clear: When they say compressed, are they referring to the stagnation of the gain in the figure from my previous post? Because from that figure, we never see the gain decrease its size, and originally I thought this was the decrease they referred to.

Yes, gain compression usually refers to the differential slope ofd the gain curve.

2) A clarifying question regarding 1 + flux / fluxsat = g0/g: When we have steady state lasing, and increase the pump and wait for steady state to occur, then what has changed is the intracavity flux on the LHS. On the RHS g is still given by the same ratio, but g0 has increased, right?

If the quantities are defined as I think, that should be the case.

3) In Milonni/Eberly, they derive the following equation for the gain of a three level laser (homogeneously broadened), which they ultimately write as (chapter 10.11)

$$g(\nu) = g(\nu_{21}) \frac{1}{(\nu_{21}-\nu)^2 / (\delta \nu_{21})^2 + 1 + \phi_\nu / \phi^{sat}_{21}}$$

Then they go on to show a figure of g/g0 as a function of (ν21-ν)/δν21 for different fluxes, and it gives a Lorentzian with decreasing ampltitude with increasing flux (they illustrate saturation). Just to be clear: That graph is nothing more than pedagogical interest, since in a physical laser we have a constant gain because of gain-clamping, right?

Well, this is not necessarily only of pedagogical interest. The main point is to show that the saturation behavior will differ at different energies for broadened (and therefore realistic) transitions. Second, although the gain will be clamped in the lasing regime, g/g0 may still vary as g0 may differ.

Thanks, it was tough, but I learned a lot from this!

[...] The main point is to show that the saturation behavior will differ at different energies for broadened (and therefore realistic) transitions. [..]

I have been thinking about that for some time now, and I am actually not quite sure what different behaviors you refer to. At pump < pumpthreshold, g = g0 and it increases with pump. When g0>losses, lasing starts and g = losses. Where does the frequency come into play with regards to g?

I have been thinking about that for some time now, and I am actually not quite sure what different behaviors you refer to. At pump < pumpthreshold, g = g0 and it increases with pump. When g0>losses, lasing starts and g = losses. Where does the frequency come into play with regards to g?

OK, now I think I know what you were referring to earlier. Thanks. Regarding the gain saturation figure I posted earlier just above, then I believe there is a contradictory behavior. We have been talking about 1 + flux/flux_sat = g0/g quite alot. So when e.g. lasing begins in a cavity, the gain decreases its value (saturates) because of increased intracavity flux. On the other hand, you confirmed that when talking of e.g. a fiber, gain saturation is when the differential slope doesn't increase as much with pump (as we see in the figure).

So in the latter case, gain increases (but not as much), and in the first case the gain decreases its value. What is the difference here, why can't 1 + flux/flux_sat = g0/g be used in the case of single pass?

Best,
Niles.

OK, I just realized what the answer is. Formulating the question is actually quite useful at times! Thanks.