Q & A: fluorescence

As an element is exposed to a relatively narrow spectrum of light and the frequencies are increased, do emission bands appear at the lower emission/absorption lines when higher absorption bands are reached?
- Ed Parr (age 51)
Stamford, CT, USA

It sounds like what you're describing is fluorescence. In some cases, when energy is absorbed in one of the higher bands, putting a molecule (for instance) into an excited state, the molecule has ways of shedding some of the energy and slipping down to a lower excited state. Then when it emits the light it comes out at the frequency corresponding to that lower-frequency absorption band. Of course this is quite a common effect.

There are some related effects. In Raman scattering, the light comes out right away along with a vibrational excitation (in a solid, a phonon) so the Raman light is at a slightly lower frequency than the incoming light. There's also a weaker type of Raman scattering where the extra energy from the material gets absorbed and the light comes out a little higher frequency. In Raman scattering the emitted light is really part of the same band.

There's a special type of fluorescence, phosphorescence, in which the excited state is a very poor light emitter. Then there can be long delays before the light comes out. The origin of the long delays (the excited electron is in a spin triplet state) was first discovered by G. N. Lewis. (Personal aside: Unfortunately my dad didn't keep the letter Lewis set him about discovering the idea, only a follow-up letter describing how he and, I think, Condon figured out that a little spin-orbit coupling is what lets the light ever get out.)

Mike W.

(published on 07/30/2012)

Thanks for the answer, your first paragraph covered it nicely. In fact, mightn't this phenomenon be explained if subatomic particles have a wavelike property? (i.e, absorption bands correspond to "constructive interference", re-emission of lower bands at higher frequencies correspond to "destructive intereference"). If so, wouldn't that provide an alternative explanation to the current explanation wave-particle duality? (i.e, currently explained by particle-like properties of light)
- Ed Parr (age 51)
Stamford, CT, USA

You're certainly right that all particles have states which are described as waves and follow a wave equation. Your suggestion to rephrase the solution of the wave equation in terms of interference effects is a little reminiscent of the Feynman path integral formulation, in which quantum states follow all different possible paths, with the likely outcomes being those where there's constructive interference between the different paths.

As you guessed, "wave-particle duality" no longer plays a serious role in how quantum mechanics is understood. It served to patch things over for a while, and still shows up in many superficial presentations.

So far as we know, at all times the quantum state is a wave-like thing, never a dot-like thing. So what about its behavior suggests "particle-like" features? The key thing is there exists an observable (formally, a Hermitian operator) whose possible observed results (eigenvalues) are non-negative integers. For bosons, like photons, it's {0,1,2,...} but for fermions only {0,1}. So there's something about the state that we can count, and that reminds us of classical particles.

Mike W.

(published on 07/31/2012)

Hi Mike, thank you again for your patience. I admit much of what you wrote sailed right over my head, as my training in physics involved only high school and 1 year of undergrad and that was 25 years ago. My train of thought actually goes back to the origin of quantum physics. From my understanding (please correct me if I’m wrong), Einstein’s work (building on Planck’s) clearly showed the quantal nature of emitted light, and because also of limitations in modeling transfer of energy from waves to particles, this was interpreted as “particle-like” behavior of light that gave rise to modern quantum theory. So far as I understand it, this was also because it was assumed that, if light were waves, it must be emitted as a continuous spectrum, for which the math did not work. However, it isn’t clear to me why waves might also be emitted in a quantal fashion. If the behavior of particles such as electrons has wave-like properties (e.g., actual “spin” of asymmetric electromagnetic charge would be “wavelike”) and if the spin frequencies determined the interactions with other particles and with light (i.e., specific emission/absorption corresponding to specific actions/reactions), it might be expected that any emission/absorption would involve discrete wavelengths (even though relative motion would allow for existence of any wavelength). Has that ever been explored? And if so, was there a fatal flaw?
- Ed Parr (age 51)
Stamford, CT, USA

Hi Ed- My last answer was way too technical and short, a symptom of trying to work out my own understanding rather than trying to work out a good presentation for something I knew well.

Your question is pretty deep. Perhaps by "spin" you just mean the spatial part of the wave function. Spin is a very special property, which I should not have mentioned except for its role in the personal anecdote.

I think that this may be what you're working towards. The initial wave function of the system has some time dependence. In a slightly idealized case, it has a single frequency, i.e. the whole thing doesn't change except to rotate like e^-iωt in the complex plane. Now if anything physical is going to change, you need to have some slight spreading of the ω's (frequencies) of different components so that the interference patterns between them change. However, a very deep principle (Noether's theorem) says that the distribution of ω's itself doesn't change in a system where the rules are independent of time. By the way, those frequencies are just what we call "energies" after we multiply them by Planck's constant to put them in ordinary energy units.

Anyway, if the light and the  non-light part part of the system (say of a molecule) are nearly independent, the frequency (energy) is about the sum of their separate contributions. So the frequency of the outgoing light has to be the difference between the frequency the molecule initially had and the one it ends up with. I think that's your point. I guess you can view conservation of energy as an interference effect: a starting state and an ending state with different frequencies have a rapidly oscillating  overlap, destructively interfering over time.

What that doesn't say is why the light typically comes out as a single-lump photon, which then typically gets absorbed somewhere else as a single lump. Why not several photons, with the same net energy, which could then be absorbed separately in several places. This gets much deeper into the actual quantum details. Multi-photon effects do occur, but under most familiar circumstances are much rarer.

Mike W.

(published on 07/31/2012)

Hi Mike, again, thanks for your patience – hopefully I’m not taxing it. I really should have said “rotation around an internal axis” instead of spin, given the quantum meaning for that term. It had occurred to me that, unless I’m missing something (definitely possible!), rotation of an asymmetric charge would provide a standing wave. If the magnetic component (but not the electric component) of a light wave acted additively with the magnetic field of the particle, then the light wave and particle field could experience constructive or destructive interference when the frequencies were matched to increase or decrease the rotational velocity. That would predict the phenomenon of emission at lower frequency band(s) during absorption of higher frequencies, which is why I asked that question (couldn’t find an answer anywhere else, so thanks!). That would also suggest that it would not be necessary to invoke particle-like properties for light to explain the quantal behavior. I know it’s rather radical by contemporary standards in that it suggests no need to invoke a “photon”. Taking it a step further, if electrons rotate, it would stand to reason that protons might too (albeit at a much faster velocity, given the energies of nuclear reactions). That might explain why it was possible to determine a solution for the Schrodinger equation for the Bohr model for hydrogen, because with only one proton, the electron could be thought of as “orbiting”. Of course, this idea would be radical by the standards of classical physics because it would suggest there really is no such thing as electrostatic force because all electric charge is in motion. I know this is “out there” in terms of the current direction of physics, and I’d be glad to drop the whole thing if you could tell me of a hole in the logic, but it just seemed to me that this might be a way to interpret Einstein’s work that might have taken the field in a completely different direction. If nothing else, so long as there is not a serious problem with the logic, it might be publishable as a curiosity by someone with physics “chops”.
- Ed Parr (age 51)
Stamford, CT, USA

Hey Ed- First, a caveat. The standard current picture can correctly calculate many things well, including the magnetic moment of an electron, via a series of corrections to corrections, to one part in 10¹¹. No incorrect predictions are known. So the bar for any alternate picture is extremely high. On a minor note, the Bohr model, which never really made sense, was superseded by Schroedinger. Also, I'm not sure why the existence of electron motion would negate the existence of the electrostatic term in the forces.

Anyway, you certainly can have a rotating asymmetric charge. That comes from a state which is the superposition of two states with different energies and shapes, for example a 1s and a 2p state. You can see some beautiful graphical illustrations of this sort of behavior on the Falstad site, e.g..  A classical picture of the effects of a rotating charge cloud like that would include electromagnetic radiation. Nonetheless, such pictures really do a terrible job of capturing the full quantum behavior of  real things. A hydrogen atom can be prepared in an essentially pure 2p state, with no changing charge distribution. Yet this state will emit a photon by electric dipole radiation, falling down to 1s. The rule is interesting: if the superposition of the starting and ending state would have the right type of oscillating charge or current distribution, you can get the corresponding type of radiation out.

Mike W.

(published on 08/01/2012)