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Dicke model is a collection of two level atoms prepared in the excited state, when the wavelength of the light absorbed or emitted is very large compared to the distance between the atoms, the rate at which radiation emitted is proportional to $N^2$. My question is if we have a collection of $N$ atoms, if they independently emit radiation the maximum radiated intensity goes as $N$. In Dicke model, there is no interaction between atoms, all $N$ atoms interact with common light field only. How does the $N$-atoms emit collectively?

FearlessVirgo
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I have a bit of intuition for some of these concepts, so maybe this will help a bit.

First let's start with collective absorption. If you have a cluster of atoms that are close together, and you send a single photon into this collection. Only a single photon can get absorbed a single atom. But the photons wavefunction spatially overlaps with multiple atoms. What happens is that you end up with a wavefunction of all the different possibilities of exactly one atom each absorbing a photon. so you a superposition of all of the configurations of individual states where exactly one atom absorbed a photon: $|0\rangle |0\rangle |0\rangle |0\rangle ... |1\rangle ... |0\rangle$

If you send a second photon into this, there's a chance that the 1 states will emit a photon through stimulated emission. In the circumstance that stimulated emission is created because the photon that triggered the emission talks to all of the atoms simultaneously, and they each have a component of having absorbed a previous photon - they can all emit their probability simultaneously.

Now why is the collective emission enhanced when more atoms are added? I think it's just that it increases the chance of interaction (more atoms means you have more shots to interact with an atom), and when they are bunched together, this behavior happens simultaneously and they interfere constructively.

Steven Sagona
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  • "emission talks to all of the atoms simultaneously", is it because of the spatial overlap of wavefunctions? – FearlessVirgo Dec 03 '20 at 09:47
  • its that the photon itself has a wavefunction which overlaps with the atoms. if you don't "check" which atom the photon absorbed to, there is a probability amplitude for each possible atom the photon got absorbed by – Steven Sagona Dec 03 '20 at 18:26
  • Yeah, now everything makes sense. – FearlessVirgo Dec 03 '20 at 18:28
  • Glad I could help. If you like the answer, you should upvote it and select the checkmark that accepts it as an answer. – Steven Sagona Dec 03 '20 at 18:57
  • I understand how this explains a $N$ times higher chance of interacting with a single photon. But how does it explain that the interaction is $N^2$ times faster? – A. P. Dec 04 '20 at 23:44
  • You can think interms of $N$ emitters having an effective dipole moment which is $N$ times the dipole moment of single emitter. The radiation rate is proportional to square of dipole moment, so it goes as $N^2$. – FearlessVirgo Dec 05 '20 at 04:08
  • @Muthumanimaran If your comment refers to the situation you were asking about (all atoms in the excited state) I understand it. But if it refers to the state $|0\rangle|0\rangle...|1\rangle...|0\rangle$ in Steven's answer I would disagree. – A. P. Dec 05 '20 at 23:47
  • @A.P., I think its a reasonable question and I'm not exactly sure what the answer is. If I were to guess, I would think that it might be that an atom being triggered to emit spontaneous emission increases the amount of local radiation which triggers more radiation. That is one photon can hit an atom which creates a second photon, which can trigger more photons. – Steven Sagona Dec 06 '20 at 03:06