Talk:Spontaneous emission

Nonradiative

How do they know it's nonradiative? lysdexia 02:55, 28 Oct 2004 (UTC)

I'm not sure that I understand your question. If you look at the object, you don't see any light coming out but the energy level changes so you know it lost energy. Since you didn't see any light, it must have lost the energy in a non-radiative (that is not causing ratiation or light) way. —Preceding unsigned comment added by Troyrock (talk • contribs) 14:35, November 4, 2004 UTC

Purcell effect

The following was recently added by MuthuKutty (talk • contribs):

“ In the 1950's E. Purcell, discovered the enhancement of spontaneous emission rates of atoms, when they are matched in a resonant cavity. See Purcell Effect. ”

The closest match I can find is the Smith-Purcell effect, which is completely different. The effect described does occur, but may be misnamed. References? --Christopher Thomas 02:22, 20 April 2007 (UTC)

-- Hi Mr Thomas, I found this citation on the 'French' Wikipedia! http://fr.wikipedia.org/wiki/Effet_Purcell which is what I wanted to say. You also point to the right article, but the article is not well written. --பராசக்தி 19:32, 22 April 2007 (UTC)

In the article it's stated that a photonic crystal can prohibit spontaneous emission. I do not believe this is correct. Evanescent modes still exist, and hence spontaneous emission, while inhibited, still occurs. —Preceding unsigned comment added by 67.188.224.196 (talk) 19:04, 22 December 2007 (UTC)

Dispute paragraph regarding most emission sources not being spontaneous emission

In the article it states:

"In spectroscopy one can frequently find that atoms or molecules in the excited states dissipate their energy in the absence of any external source of photons.People have a general tendency to use the term spontaneous emission to explain this phenomenon, but this is actually relaxation of the atoms or molecules caused by the fluctuation of the surrounding molecules present inside the bulk."

I believe this is practically the opposite of the truth, in addition to being unspecific about the "relaxation of .... surrounding molecules present inside the bulk". Actually most light from everyday sources is primarily due to spontaneous emission. When the brightness is very high (photon number per spatial-temporal mode not <<1, such as black-body radiation at longer wavelengths in the Rayleigh-Jeans regime) then there is some contribution due to stimulated emission amplifying the spontaneous emission. For visible light from common light sources (besides lasers) the latter is not the case. It could also be mentioned (maybe this was the source of the confusion) that the emission need not be from a single atom (or molecule) but can for instance come from a transition between energy levels in a crystal lattice (thus involving many atoms). However this is still spontaneous emission (no?).

I propose to correct this paragraph almost to its opposite. Interferometrist (talk) 13:42, 27 October 2010 (UTC)

Clarified the paragraph by changing to "nonradiative relaxation", (dissapating the energy as heat). Hopefully this should fix your objection. BruceThomson (talk) 07:17, 22 January 2011 (UTC)

Is "fluctuation" the best way to describe the cause of nonradiative transitions? In the case of a gas at STP, with a mean time between collisions of 70/.5 = 140 picoseconds, do any nonradiative transitions occur for any reason besides collision? --Vaughan Pratt (talk) 03:13, 14 September 2011 (UTC)

Here is the present state of this paragraph:

In spectroscopy one can frequently find that atoms or molecules in the excited states dissipate their energy in the absence of any external source of photons. This is not spontaneous emission, but is actually nonradiative relaxation of the atoms or molecules caused by the fluctuation of the surrounding molecules present inside the bulk.

Even with the corrections, I am puzzled by this. First of all, why emphasize that there is no external source of photons? For spontaneous emission there is already no external source of photons, so the emphasis does not explain anything. Should it be "without producing photons"?

Second (and more important): Does the energy move from one atom to another by direct coupling of the charges in different atoms (for example, collisions), without any intermediate production of photons? This should be explained in fundamental language that names the process, not just left to unspecified "fluctuations", since photon emission itself is a fundamental process. At this level of description, the number of possible types of interaction is quite limited and we can actually name them. 178.38.125.245 (talk) 07:49, 23 April 2015 (UTC)

Mistake in formula

Shouldn't the formula for the rate of spontaneous emission in terms of the dipole moment have n^2 rather than n in the numerator? I'm pretty sure it does; if so can someone (else) fix it please. Interferometrist (talk) 16:51, 30 March 2011 (UTC)

In the absense of perturbations

@Evgeny I see you added this qualifier but I would like to learn more. Do you have a reference for this exception? What kinds of perturbations are involved? Johnjbarton (talk) 17:27, 1 December 2024 (UTC)