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The rough quantum mechanical explanation for linewidth is that the lifetime $\tau$ of an excited level is associated with an uncertainty $\Delta E$ in its energy satisfying $$\Delta E\tau=\hbar$$ and so there is a linewidth $$\Delta \omega=\frac{\Delta E}{\hbar}=\frac{1}{\tau}.$$ However, under stimulated emission, the rate of depopulation of the upper level increases and so its lifetime effectively decreases. Why does the linewidth not increase correspondingly?

Classically, it is obvious that stimulated emission does not contribute because it is coherent whereas spontaneous emission is incoherent. However, quantum mechanically, I can't see why the reduced lifetime would not be included in the uncertainty principle.

Ghorbalchov
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  • Can you explain, in detail, why you think the uncertainty relationship should have anything to do with linewidth in this problem? I think if you do that you'll find your own answer. – DanielSank Oct 19 '21 at 18:45
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    I can't really; everywhere I've found which uses the uncertainty relation to 'explain' linewidth just pulls it out of thin air. – Ghorbalchov Oct 19 '21 at 21:20
  • Why do you think the linewidth does not increase? Maybe it does, your argument sounds good. – Ján Lalinský Oct 19 '21 at 21:56
  • For example, with the Einstein coefficients you can show that the lineshape is the same for spontaneous and stimulated emission, with linewidth always limited by the spontaneous emission rate. – Ghorbalchov Oct 19 '21 at 21:59
  • @AlexGhorbal in that case, it may helpful for you to investigate a more sound argument about why linewidth is related to the lifetime of an excited level in the first place. If you understand that, you'll probably be able to answer your own question. I would suggest posting a new question focused on the linewidth of an excited transition, and link it here. – DanielSank Oct 20 '21 at 16:50
  • I already understand why linewidth is related to lifetime in terms of dephasing/coherence, but I clearly still don't know how to answer this question. The problem is I don't know how to relate those arguments to the uncertainty principle. Understanding other more sound arguments is not helping me with that. – Ghorbalchov Oct 20 '21 at 20:57
  • check out "power broadening" – Andrew Steane Nov 02 '21 at 22:46

2 Answers2

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The short answer: the relevant lifetime is not the one of an individual excitation, but the first-order coherence time, the timescale at which the phase diffuses.

The longer answer is very interesting. On the quantum level, both the photons coming in and out of the cavity are described by a Lindblad process. For the gain, the jump operators are $\sqrt{R}a^\dagger$ (leading to rate $Raa^\dagger=R(n+1) $) and for the losses they are $\sqrt{\gamma}a$ (leading to rate $\gamma a^\dagger a=\gamma n $). The +1 for the gain is what is typically attributed to the spontaneous emissions. Note that all the absorptions and emissions would invoke some kind of fluctuations, but they are much smaller.

The semiclassical picture, such as introduced in Theory of the linewidth of semiconductor lasers (Henry,'82) attributes all the fluctuations to the spontaneous emissions, but this makes them unphysically large. In this picture, each spontaneous emission doesn't increase the particle number deterministically with +1, but can change it either way up to the order of $\sqrt{n}$!

I puplished a paper related to this quantum-classical correspondence in a slightly different system, but should be related to this, especially sections III and IV.

Wouter
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  • Your link to IEEE.org leads to mostly empty page in Chromium and "Web address entered is invalid" in Firefox. – Ruslan Nov 10 '21 at 07:34
  • @Ruslan, thanks, corrected – Wouter Nov 10 '21 at 07:38
  • Also accessible through sci-hub.se . Note that the relevant mechanism of phase change here is the first one delta \phi' , \delta \phi'' can be disregarded if there is no optical nonlinearity – Wouter Nov 10 '21 at 07:59
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Emission means different things when talking about the lifetime of an excited state and about stimulated emission.

Finite lifetime of an excited state means that, if an atom in an excited state is left alone, it will eventually relax to the ground state, spontaneously emitting a photon. This results from the fact that we have an infinite number of states with equal energy:

  • one excited atomic state and photon vacuum: $|e, 0\rangle$
  • infinite number of states with a photon of wave vector $\mathbf{k}$ and polarization $\lambda$ and the atom in the ground state: $|g, 1_{\mathbf{k},\lambda}\rangle$.

The probability of finding the atom in the excited state is thus descreasing till it becomes negligeably small, approximately as $$ P_e(t)=e^{-t/\tau}, $$ where we call $\tau$ the lifetime. It is typically governed by the strength of the coupling between the atom and the photon field (See this answer for more detaild discussion.)

In some more complex situations the decay may be not exponential and we may want to include into the lifetime the effects related to the uncertainty of the atomic frequency (inhomogeneous broadening).

Stimulated emission is essentially formation of a coherent superposition between the excited states of atoms and a single photon mode. The effetctive strength of the coupling is proportional to the square root of the number of photons, as per well-known boson operators relations $$ a|n\rangle =\sqrt{n}|n-1\rangle, a^\dagger|n\rangle =\sqrt{n+1}|n+1\rangle. $$ The atoms and photons are never in a state with a definite number of photons or excited atoms, but rather in a state permanent Rabi oscillations (that is, till the light escapes the laser and we observe it). The most suitable model for this case is Dicke model, described as a critical phenomenon. There is no meaningful way to speak about an atom lifetime here, although one could meaningfully speak of the lifetime of the whole photon condensate as it leaks from the cavity and discussed the uncertainty of atomic frequencies due to various inhomogeneous effects (like random movement of atoms in a gas laser, random Stark shifts in a solid state, or random initial and final kinetic energies of the recombining electron and hole in a semiconductor laser.)

Roger V.
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  • Do we still have Rabi oscillations instead of irreversible decay if we consider open space with a (highly monochromatic) photon passing near the excited atom? I think the OP asked more about this scenario, rather than a laser. – Ruslan Nov 09 '21 at 19:30
  • @Ruslan "photon passing near" something is a poorly defined notion, since photons are not localized in space - unless we talk about a cavity or it is a broad wave packet. In ideal world, this system is described by Jaynes-Cummings model, where Rabi oscillations persist. Another way to view it: defining lifetime requires infinite number of modes, i.e., thermodynamic limit, whereas stimulated emission refers to the interaction with one selected mode. Spontaneous emission is an example of an irreversible process, stimulated emission is not. – Roger V. Nov 09 '21 at 21:35
  • Well, a superposition of single-photon states that makes up a wave packet is also often called a photon. This is what I meant in my comment. – Ruslan Nov 09 '21 at 21:37
  • @Ruslan Then we may also have many modes, but only some of them contain significant number of photons, so again the probability of the excited state will decay due to the continuum of the modes, which contain no photons. One could also consider interaction with the black body radiation, but that implies thermalization, i.e., existence of an implicit relaxation mechanism, which determines the lifetime - perhaps, in this case the lifetime indeed should be renormalized by the number of photons, i.e., the spectral density. But this probably takes us far from the OP hadin mind. – Roger V. Nov 09 '21 at 21:43
  • Your first sentence in the comment above seems to imply that if the atom is in free space, stimulated emission basically can't happen due to spontaneous emission happening before it. Do I understand it correctly? – Ruslan Nov 09 '21 at 21:46
  • @Ruslan a superposition that you talk about contains many states that are not resonant with the atom, and which are not the eigenmodes of the space/cavity. Stimulated emission is literally reflection of the action of the photon creation/annihilation operator. It should not be really treated on the same footing as spontaneous emission... as Einstein coefficients suggest - in fact these are hiding a lot more than basic processes. – Roger V. Nov 09 '21 at 21:51
  • @Ruslan the point is what we call emission: a) elementary process of interaction between an atom and a field, or b) a process in which the system irreversibly transitioned from one state to the other. – Roger V. Nov 09 '21 at 21:53
  • I think you took my superposition to be too wideband :) My idea was about a superposition of single-photon states such that any significant frequency in it is within the natural width of the atomic energy level. Corresponding wave packet will be quite wide, but still localized. The frequencies in such a superposition must be resonant with the atom. And of course, these constituent plane waves are eigenmodes of space. There's just a continuum of them in a superposition. – Ruslan Nov 09 '21 at 21:56
  • So the atom may emit a photon into the mode rhat already have photons, but then it will reabsorb it - this may happen many times, but eventually the photon leaks into the other modes and never comes back. Alternatively - the pumped mode may be leaking photons, and if this happens sufficiently fast, it will determine the decay time. – Roger V. Nov 09 '21 at 21:58
  • @Ruslan natural width is due to spontaneous emission - assuming natural width and then discussing the difference between spontaneous and stimulated emission is a bit circular. But it is also true that no photon has well-defined frequency, since it was emitted at some point... This is a very rich question, in my opinion – Roger V. Nov 09 '21 at 22:05
  • Right, it was sloppy of me to choose natural width as the indicator of resonance. Anyway, the point still remains for a "sufficiently narrow-band" wave packet that it's resonant with the atom. – Ruslan Nov 09 '21 at 22:08
  • @Ruslan This made me think of Landau-Zener - perhaps, one could replace deterministic time evolution by a quantized EM field. – Roger V. Nov 10 '21 at 08:10