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When an atom “lases” it always gives up its energy in the same direction and phase as the incoming light. Why does this happen? How can this be explained? How does the photon generated because of stimulated emission, know which direction to take? What are the factors leading to this?

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  • $\begingroup$ This is a strong thing to say without any reference or any additional context. Could you clarify this part of your question: When an atom “lases” it always gives up its energy in the same direction and phase as the incoming light? $\endgroup$ – seb May 24 '13 at 7:11
  • $\begingroup$ during the lasing action...that is when the stimulated emission takes place and the released photon comes out...why does it go in same phase and direction. $\endgroup$ – Nix May 24 '13 at 8:33
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The word "stimulated" means that the emission of the photon is "encouraged" by the existence of photons in the same state as the state where the new photon may be added. The "same state" is one that has the same frequency, the same polarization, and the same direction of motion. Such a one-photon state may be described by the wave vector and the polarization vector, e.g. $|\vec k,\lambda\rangle$.

The physical reason why photons like to be emitted in the very same state as other photons is that they are bosons obeying the Bose-Einstein statistics. The probability amplitude for a new, $(N-1)$-st photon to be added into a one-photon state which already has $N$ photons in it is proportional to the matrix element of the raising operator $$ \langle N+1| a^\dagger|N\rangle = \sqrt{N+1}$$ of the harmonic oscillator between the $N$ times and $(N+1)$ times excited levels. Because the probability amplitude scales like $\sqrt{N+1}$, the probability for the photon to be emitted into the state goes like the squared amplitude i.e. as $N+1$. Recall that $N$ is the number of photons that were already in that state.

This coefficient $N+1$ may be divided to $1$ plus $N$. The term $1$ describes the probability of a spontaneous emission – that occurs even if no other photons were present in the state to start with – while the term $N$ is the stimulated emission whose probabilities scales with the number of photons that are already present.

But in all cases, we must talk about "exactly the same one-photon state" which also means that the direction of the motion is the same. It's because quantum field theory associates one quantum harmonic oscillator with each state i.e. with each information $\vec k$ about the direction of motion and wavelength; combined with a binary information about $\lambda$, the polarization (e.g.left-handed vs right-handed).

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  • $\begingroup$ "The word "stimulated" means that the emission of the photon is "encouraged" by the existence of photons in the same state as the state where the new photon may be added" The word stimulated means you are making it happen by stimulating it. So when u say the emission is "encouraged" by the existence of photons i agree with you.Why do u say the new photon will be added in the same state? Just because they are bosons and so tend to be in same state? $\endgroup$ – Nix May 25 '13 at 5:34
  • $\begingroup$ @Luboš, if I understand your point, the reason why the two photons are colinear is statistics. You need many photons to have a large probability of the same state. However, the photons should be emitted in the same direction even if there is only 1 atom and 1 photon. How do Bose statistics work for only 1 photon? $\endgroup$ – fffred May 25 '13 at 8:04
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I think it's deeply related to the fact that photons are bosons, ergo they follow the Bose-Einstein Statistics, or in this case they make a Bose-Einstein Condensate.

If you are not familiar with this exciting concept, I suggest you make a look at this Wikipedia article or any other statistical mechanics textbook you have around. Anyway, the two photons making a Bose-Einstein Condensate means that the two photons will have the same phase(and obviously wavelength) and will occupy the same point in space all the time.

Hope this was helpful.

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  • $\begingroup$ thank u..since i have done introductory BE statistics, i can understand ur answer. $\endgroup$ – Nix May 24 '13 at 8:32
  • $\begingroup$ So let me see if i get this right...you are saying when the photon comes out after stimulated emission, it "somehow" knows the incoming photon is in some particular dynamic state, and so it chooses to be in that dynamic state? I am not asking about the statistics or probability or anything-just that why do bosons(photons in this case) behave like this? $\endgroup$ – Nix May 25 '13 at 5:44
  • $\begingroup$ Particles usually know things which you do not expect! Although, this is not the answer of your question, but this brilliant series from Feynman might shed some light on it: QED $\endgroup$ – Ali May 25 '13 at 7:30
  • $\begingroup$ well yeah...i know that bit"particles knowing things". Was looking for perhaps something that i could..well...digest! $\endgroup$ – Nix May 25 '13 at 8:23
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This is a great question, and I believe one that deserves very deep thought!

Let me briefly review my understanding of a laser. For a laser, typically there are two channels the atoms can decay into by emitting photons: out into free space or into a cavity. Purely from spontaneous decay, the atom is more likely to emit photons into free space. However, because the cavity stores its photons before releasing them in a beam, photons begin to build up in the cavity. Then, these excess photons can stimulate emission into the cavity and the atoms begin to emit more and more of their photons into the cavity rather than free space. When this process runs away with a large number of photons then the output light from the cavity is laser light.

Finally, understanding why the photons are identical means understanding what a cavity does. The effect of the cavity is to dramatically modify the allowed states of the electromagnetic environment, specifically limiting the possible properties of the photons it stores. In particular, it turns out that the single-mode cavities used in many lasers can only store photons that have identical properties, and because photons are bosons it is possible to have many identical particles of light (photons) in the cavity mode.

On the other hand, if an excited atom were sitting in free space and interacted with a photon, then photon emission does not generally produce two photons with identical properties. This is actually an active area of research, i.e. computing what the state after stimulation looks like.

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