Difference between Scattering and Emission of Photons

[JohnnyGui]

Hello,

Ok, so I’ve been searching about this for a while and there seems to be a difference in explanations that confuses me. What concerns me is the exact detailed physical mechanism that makes particles cause a scatter of a photon and an emission of a photon.

The Wiki page about scattering says the following about one type of scattering, Rayleigh Scattering:

“The particles may be individual atoms or molecules. It can occur when light travels through transparent solids and liquids, but is most prominently seen in gases. Rayleigh scattering results from the electric polarizability of the particles. The oscillating electric field of a light wave acts on the charges within a particle, causing them to move at the same frequency. The particle therefore becomes a small radiating dipole whose radiation we see as scattered light.”

To me, this explanation seems to be the same as when a particle meets a photon that has the same energy/frequency as a potential energy/frequency state of the particle, with the result of the particle absorbing this photon and emitting it again. This is the description found here: http://hyperphysics.phy-astr.gsu.edu/hbase/mod3.html#c3

However, several sources say that that absorption and emission is different from scattering. My question is then, in what way exactly? I am aware that, for example, Rayleigh scattering is elastic which means that the photon that left the particle has the same energy level as the one that hit the particle, but what happened exactly in between? Did it get absorbed and emitted in the same energy quantity?

Here’s a good discussion about the difference between emission (fluorescence) and Raman scattering:
http://physics.stackexchange.com/questions/38459/what-is-the-difference-between-raman-scattering-and-fluorescence

There seems to be 2 different opinions on this in that discussion:
1. That they’re the same in the sense of the photon being absorbed and then emitted again, with the only difference being the time between absorption and emission if you compare fluorescence with Raman scattering. So there is a time in which the particle is being excited by the photon.
2. That, in the case of Raman scattering, there is no absorption of the photon but a partial interaction of the photon with the particle which results in the particle taking a bit of energy away from the photon.So there is no excitation of the particle whatsoever by the photon.

Point 2 bothers me because I’m taught that a photon cannot be partially absorbed or interacted with, since it’s a discrete energy state.

So what is it exactly that separates the mechanism of emission from scattering? Preferably comparing it with every type of scattering.

 

[jambaugh]

The distinctions being made are distinct paradigms but physically you cannot distinguish between a photon interacting with another system and the photon being absorbed and re-emitted. In fact the free propagation can be seen as repeated "absorption and re-emissions" in the sense of Huygen's principle.

One can however distinguish types of absorption/re-emission and you should re-read the discussion with this in mind.

 

[Saw]

I think the question is very good (I had the same doubt!) and well deserves some elaborated answer, which unfortunately I don't have...

When one reads about transitions, you see a clear distinction between:

Case A, where the E of the incident photon matches a gap between E levels of the atom, so the atom is excited into a higher state (electron transition). This is often called "resonance absorption".
Case B, where the E of the incident photon is lower than such gap, so there is no such excitation / transition, and the photon goes out as it came in. This is called "Rayleigh scattering". When the E of the incident photon is higher than the gap, the photon loses energy to the particle but we don't say that there is partial absorption (?) and this is called "Raman scattering" (also Stokes scattering, but there is also an anti-Stokes effect, let us leave that aside...).

But then you learn that in case A the excitation lasts for an incredibly short time after which the photon is re-emitted with the same E as it came in, just as in case B. May it be that the duration of "excitation" in A is longer than "whatever happens" in B? (but it doesn't look like a great difference, given that the time is very short).

The distinctions being made are distinct paradigms but physically you cannot distinguish between a photon interacting with another system and the photon being absorbed and re-emitted. In fact the free propagation can be seen as repeated "absorption and re-emissions" in the sense of Huygen's principle.

One can however distinguish types of absorption/re-emission and you should re-read the discussion with this in mind.

Well, yes, given the difficulties, I would also prefer to call everything simply absorption / re-emission. But still we need to distinguish cases A and B, by both the causes and the effects. The causes are clearly different (case A, excitation/transition; case B, no such thing). But what about the effects, if in both cases the photon is emitted after a very short time with same E as it came in? Well, I am aware that if a second photon impinges upon an already excited atom, the phenomenon of stimulated emission arises, but any other difference? Well, it can be that in A, before re-emission, the E may be absorbed by the material via collisions as thermal E, but that looks more like a collective property of the material, rather than an individual feature of the atom.

On another note, I am reading here that Raman scattering is not a phenomenon of re-radiation due to an oscillating dipole, just to complicate things a little more.

 

[JohnnyGui]

The distinctions being made are distinct paradigms but physically you cannot distinguish between a photon interacting with another system and the photon being absorbed and re-emitted. In fact the free propagation can be seen as repeated "absorption and re-emissions" in the sense of Huygen's principle.

One can however distinguish types of absorption/re-emission and you should re-read the discussion with this in mind.

I tried rereading it again, but there are people who say that in case of scattering, there isn't any absorption at all. That's where the problem lies. Is there or isn't there any absorption in the case of scattering?

@Saw : Thanks for your well extensive explanation. I got some questions when reading your post.

Case B, where the E of the incident photon is lower than such gap, so there is no such excitation / transition, and the photon goes out as it came in. This is called "Rayleigh scattering".

I've also read another interaction in this scenario. That if an incident photon has a lower energy than the energy gap in a particle, the particle would let it just through instead of scattering it. I assume that's what happens in transparent objects if a material let's all the visible light through?

When the E of the incident photon is higher than the gap, the photon loses energy to the particle but we don't say that there is partial absorption (?) and this is called "Raman scattering" (also Stokes scattering, but there is also an anti-Stokes effect, let us leave that aside...).

Won't an incident photon with a higher energy than the gap lead to shooting the electron away and ionizing the atom? Or do you mean a photon with an energy level between 2 energy gaps of the particle?

Well, I am aware that if a second photon impinges upon an already excited atom, the phenomenon of stimulated emission arises, but any other difference? Well, it can be that in A, before re-emission, the E may be absorbed by the material via collisions as thermal E, but that looks more like a collective property of the material, rather than an individual feature of the atom.

Regarding interaction with the rest of the material, someone in the discussion link said the following about Raman scattering and fluorescence;

"Maybe the difference is really just terminology, but we normally think of Raman spectroscopy as the excited state decaying to a vibrationally excited state by emitting a photon of reduced frequency, and the vibrationally excited state then decays by interaction with the lattice. In fluorescence it's the other way around. The excited state decays to a lower energy state by interaction with the lattice and this lower excited state then decays by emission of a photon."

I'm not really sure what to conclude from that, other than there is absorption in both cases and that there's a time difference. But if this quote is true, then one could say that Raman scattering and fluorescence also depend on the collective property of the material.There's something else important that I noticed about scattering types. Rayleigh scattering, Mie scattering and geometrical scattering all depend on the size of the particle with respect to the wavelength, the particle being smaller (Rayleigh), equally large (Mie) or larger (geometrical scattering) than the wavelength. If scattering really involves absorption and re-emission of photons, then that means that the energy gaps of particles depend somehow on the size of the particle (and/or perhaps the mass as well?) which would seem weird to me. Are there partciles that are equally large/massive with different energy gaps?

Second to that, if we extend scattering to reflection and call that absorption and re-emission of photons as well, then how would one explain the same emission angle for every particle since reflection is characterized by an emission angle that is the same as the incident angle of a photon? If absorption and emission can't do that, then surely it must be nothing more than a "bounce" of the photon against the particle. Is there even a slight energy reduction of the photon in the case of reflection?

 

[Saw]

I've also read another interaction in this scenario. That if an incident photon has a lower energy than the energy gap in a particle, the particle would let it just through instead of scattering it. I assume that's what happens in transparent objects if a material let's all the visible light through?

The usual description does assume that the “lattice” (because it is a collective behavior, see this FAQ) briefly vibrates with a non-resonant frequency and then re-emits it. That brief stop and ensuing delay is what would explain that light travels more slowly in the medium. Whether that is "scattering" or not is another question. I would say, yes, the light is scattered/re-emitted in random directions. How is it that it keeps its original direction? What I read (classical explanation) is that the network of atoms acts cooperatively so that there is destructive interference except in the original direction, but someone objected once to that in a discussion...

Won't an incident photon with a higher energy than the gap lead to shooting the electron away and ionizing the atom? Or do you mean a photon with an energy level between 2 energy gaps of the particle?

Yes, I meant the second thing. Ionization would happen when the incident photon has enough E to make the electron jump beyond the highest level.

Regarding interaction with the rest of the material, someone in the discussion link said the following about Raman scattering and fluorescence;

"Maybe the difference is really just terminology, but we normally think of Raman spectroscopy as the excited state decaying to a vibrationally excited state by emitting a photon of reduced frequency, and the vibrationally excited state then decays by interaction with the lattice. In fluorescence it's the other way around. The excited state decays to a lower energy state by interaction with the lattice and this lower excited state then decays by emission of a photon."

I'm not really sure what to conclude from that, other than there is absorption in both cases and that there's a time difference. But if this quote is true, then one could say that Raman scattering and fluorescence also depend on the collective property of the material.

Interesting. What I had noted down is that fluorescence is for instance that the particle absorbs UV radiation through several E jumps but relaxes in steps emitting photons of several frequencies (jointly white light). But your quote implies that part of the E is dissipated (through interaction with the lattice) and then a single photon of lower frequency is emitted, which would be the same outcome as Raman scattering, although here the process would follow the inverted order…

I tend to believe that is true and see this site for a similar description of fluorescence.

There's something else important that I noticed about scattering types. Rayleigh scattering, Mie scattering and geometrical scattering all depend on the size of the particle with respect to the wavelength, the particle being smaller (Rayleigh), equally large (Mie) or larger (geometrical scattering) than the wavelength. If scattering really involves absorption and re-emission of photons, then that means that the energy gaps of particles depend somehow on the size of the particle (and/or perhaps the mass as well?) which would seem weird to me. Are there partciles that are equally large/massive with different energy gaps?

Yes, good question…

Second to that, if we extend scattering to reflection and call that absorption and re-emission of photons as well, then how would one explain the same emission angle for every particle since reflection is characterized by an emission angle that is the same as the incident angle of a photon? If absorption and emission can't do that, then surely it must be nothing more than a "bounce" of the photon against the particle. Is there even a slight energy reduction of the photon in the case of reflection?

In terms of reflection, scattering in random directions would be what happens when there are rugosities (bigger than light’s wavelength BTW…) and so there is diffuse reflection. If the principle angle of incidence = angle of emission is to be respected, then it will be because the surface is polished and so we have specular reflection, but I don’t know how to describe that at quantum level. We may have to read Feynman's description.

 

[PeterDonis]

Point 2 bothers me because I’m taught that a photon cannot be partially absorbed or interacted with, since it’s a discrete energy state.

I don't know where you were taught this, but it's not correct. What is correct is that EM radiation of a given frequency is quantized--heuristically, it comes in "packets" of energy $h \nu$, where $\nu$ is the frequency. But that does not mean it is impossible for EM radiation to undergo interactions that change its frequency. That is perfectly possible, and point 2 is just an example of it.

Is there or isn't there any absorption in the case of scattering?

You are assuming a binary distinction that does not exist. "Absorption" and "scattering" are not names for two distinct fundamental processes. They are just two different approximate descriptions of processes involving EM radiation, and the cases for which they are reasonable approximations overlap.

 

[JohnnyGui]

I don't know where you were taught this, but it's not correct. What is correct is that EM radiation of a given frequency is quantized--heuristically, it comes in "packets" of energy hνhνh \nu, where νν\nu is the frequency. But that does not mean it is impossible for EM radiation to undergo interactions that change its frequency. That is perfectly possible, and point 2 is just an example of it.

Impossible for EM radiation to undergo interactions that change its frequency, is not what I meant. What I'm saying is that in order to change a photons energy level, the incident photon has to be fully absorbed/destructed first. In the discussion link it is said that the incident photon is interacted with partially without the incident photon being wholly destructed/absorbed first. Someone in the link mentioned the Feynman Diagtam and according to that diagram, Compton scattering for example involves total destruction of the incident photon first.

You are assuming a binary distinction that does not exist. "Absorption" and "scattering" are not names for two distinct fundamental processes. They are just two different approximate descriptions of processes involving EM radiation, and the cases for which they are reasonable approximations overlap.

So they both involve the same mechanism? What is the exact process that I'm wrongly making a distinction in? For example, when it is mentioned that they both involve absorption, it is said that they differ in time duration between absorption and emission.

 

[PeterDonis]

What I'm saying is that in order to change a photons energy level, the incident photon has to be fully absorbed/destructed first.

And that is not correct. The photon has to interact, but "fully absorbed/destructed" is not the only possible interaction. At least, not if you're talking about the level of approximation you were talking about before, where we have EM radiation and electrons interacting with each other in a basically classical manner.

Someone in the link mentioned the Feynman Diagram and according to that diagram, Compton scattering for example involves total destruction of the incident photon first.

Yes, but this is a different approximation, the one in which we use Feynman diagrams to describe what is going on. In this approximation, there is no such thing as "scattering" in the sense of two particles interacting "at a distance" as in the classical scattering model. There is just one interaction, a vertex with one photon "leg" and two electron/positron "legs". Whether you call this vertex "photon emission" or "photon absorption" or just "electron-photon interaction" is a matter of convenience of terminology.

The reason this is still just an approximation is that it depends on perturbation theory, i.e., on trying to solve the equations of quantum electrodynamics approximately, by computing successive terms in a series expansion, because we don't know how to solve them exactly.

What is the exact process that I'm wrongly making a distinction in?

Whatever the exact solution (which we don't know, see above) of the equations of quantum electrodynamics is. But even in the Feynman diagram approximation described above, you can see that there is only one interaction, not two.

 

[JohnnyGui]

And that is not correct. The photon has to interact, but "fully absorbed/destructed" is not the only possible interaction. At least, not if you're talking about the level of approximation you were talking about before

If that's not the only possible interaction, then there may be a possible distinction between scattering and absorption like it is said. For example, giving the fully absorption/destruction of a photon the name "absorption" and if it's not fully absorbed/destructed the name "scatter". The discussion between people being that if it is possible for a photon to partially interact or not. If it is possible like you said then the distinction is that the photon isn't absorbed in the case of scatter. If it isn't possible, then perhaps they're in both cases absorbed but there's a time difference until re-emission. That's where my question lies.

Quantum electrodynamics

So there's no exact mechanism described? Only that scatter and absorption are overlapping approximations of the same mechanism? But there is apparently another way of photon interaction like you said (see above).

 

[PeterDonis]

The discussion between people being that if it is possible for a photon to partially interact or not.

And the answer to that is "it depends on which approximate model you are using". There is no single answer. If you use one approximate model, photons can either be "scattered" or "absorbed". If you use another approximate model, there is only one interaction, the vertex with one photon leg and two electron legs. Both models are approximations.

So there's no exact mechanism described?

The closest thing we have to an "exact mechanism" is the description in terms of quantum fields--there is an electron/positron field and a photon field. But we can't exactly solve the equations for this field. That's why we have to use approximations, and why different approximations are used in different scenarios.

there is apparently another way of photon interaction like you said (see above).

You have to pay more careful attention. I was talking about one particular approximation. See above.

 

[JohnnyGui]

And the answer to that is "it depends on which approximate model you are using". There is no single answer. If you use one approximate model, photons can either be "scattered" or "absorbed". If you use another approximate model, there is only one interaction, the vertex with one photon leg and two electron legs. Both models are approximations.

The closest thing we have to an "exact mechanism" is the description in terms of quantum fields--there is an electron/positron field and a photon field. But we can't exactly solve the equations for this field. That's why we have to use approximations, and why different approximations are used in different scenarios..

Ok, this clears things up a bit for me. Is there a limit to where these different approximations give different results? Like for example, seeing that the type of scattering is dependent on the particle size (Rayleigh, Mie, geometrical like reflection), can the approximation theory of total absorption/destruction with re-emission of a photon predict such a particles size/geometry-dependent behavior?

You have to pay more careful attention. I was talking about one particular approximation. See above.

I'm not sure if you've edited your last post but when I replied, your post didn't contain the explanation of other possible approximations.

 

[PeterDonis]

seeing that the type of scattering is dependent on the particle size (Rayleigh, Mie, geometrical like reflection), can the approximation theory of total absorption/destruction with re-emission of a photon predict such a particles size/geometry-dependent behavior?

In principle, yes. In practice, nobody does this because the math is too complicated. That's one of the main reasons we have different approximations for different scenarios--because using the most fundamental one (which at the current state of our knowledge is the Feynman diagram one) all the time would be beyond our practical capabilities.

your post didn't contain the explanation of other possible approximations.

Read my post #8.

 

[JohnnyGui]

In principle, yes. In practice, nobody does this because the math is too complicated. That's one of the main reasons we have different approximations for different scenarios--because using the most fundamental one (which at the current state of our knowledge is the Feynman diagram one) all the time would be beyond our practical capabilities.

Great. Thanks for this enlightment.

Read my post #8.

Yes, that's the post I was referring to being extended after I replied. I understand now what you mean.

 

[JohnnyGui]

@Saw : Just have been searching a bit about the quantum descriptions/approximations of different types of scattering and its dependence on the size of particles.

Regarding the dependency on the particle size, there are for example semiconductor atoms called "quantum dots". It seems that the widths between the available energy gaps in such atoms do depend on the size of the atom. This dependency is called quantum confinement effect. This effect explains that an electron in very large atoms usually have continuous available energy levels but as the atom size gets smaller, it would possesses increasingly more discrete steps of energy levels. The smaller an atom size gets, the larger the gaps between these available discrete energy states which would mean that a small atom would need a higher energy photon to get excited/(partially) absorb it and would therefore emit blue light instead of red for example. I'm not sure to what extent this theory would be applicable to non-semiconductor atoms but this might explain why very small particles scatter blue light much more strongly in the case of Rayleigh scattering. I wonder if fluoresence is also dependent on particle size?

Another quantum description regarding the difference between scattering and fluorescence is found here . It shows that scattering involves, not really the excitation of electrons to the higher available energy states, but more of a virtual energy state between these energy states that makes them go back immediately to their original state again. I think you gave a similar description on this in your post. This could explain the shorter time delay until emission compared to fluorescence. This description could be called an absorption but I think people don't prefer it to call it that since that term is reserved for if electrons are truly excited to higher energy levels. Reaching a virtual energy state and going back immediately to the original state is considered more of a "bounce" instead. The article further shows that there is a difference between scattering and fluorescence regarding how much the energy of the emitted photon differ from the absorbed one. An exception is resonant Raman scattering which seem to be the same as resonant fluorescence.

As for a quantum approximation of reflection, the atoms do indeed scatter/emit photons in all directions after absorption but in the case of reflection where the atoms are more or less at the same level, interference destructs the emission/scatter of photons at any angle except for the angle that is the same as the angle of incident. You've given a similar description on transparency. A good explanation and discussion about reflection and transparency is found here .

This is what I have so far. I'm planning to delve into the QED Lectures of Feynman about reflection and scatter based on a quantum description to get more details on all this. There are videos available.

 

[Saw]

Regarding the dependency on the particle size, there are for example semiconductor atoms called "quantum dots". It seems that the widths between the available energy gaps in such atoms do depend on the size of the atom. This dependency is called quantum confinement effect. This effect explains that an electron in very large atoms usually have continuous available energy levels but as the atom size gets smaller, it would possesses increasingly more discrete steps of energy levels. The smaller an atom size gets, the larger the gaps between these available discrete energy states which would mean that a small atom would need a higher energy photon to get excited/(partially) absorb it and would therefore emit blue light instead of red for example. I'm not sure to what extent this theory would be applicable to non-semiconductor atoms but this might explain why very small particles scatter blue light much more strongly in the case of Rayleigh scattering.

You've been doing excellent work, this "quantum dots" issue is really interesting.

I will study and comment later the rest of your post but wanted to share now the following, a sort of mechanical analogy with your findings on the above issue. In Conceptual Physics by Paul Hewitt, a simple but very enlightening book that I am lately enjoying, the author makes this comparison when discussing about Rayleigh scattering: "this is similar to the way that small bells ring with higher notes than larger bells".

See also this quote from here:

how do you adjust the speed at which a tuning fork vibrates? Well, first, you could adjust the length of your tuning fork. The smaller a tine, the less distance it has to move, and the faster it will be able to vibrate. It's the same principle as strings on a guitar. Without much room to wobble, a tight string vibrates quickly. A loose string, on the other hand, takes longer to shudder back and forth, resulting in a lower tone. The largest tuning fork in the world, by the way, is a 45-foot (13.7-meter) sculpture in Berkeley, Calif. [source: http://www.ci.berkeley.ca.us/ContentDisplay.aspx?id ]. If someone ever finds a hammer big enough to hit it, the sound would most likely be too low to be heard by human ears.

I don't know to what extent the analogy can be pushed further, but it is worth noting at least. Actually for a complete analogy with Rayleigh scattering we would need two things: 1/ the bell or fork should be much smaller than the wavelength of sound (diameter not larger than 1/10th of the wavelength) and 2/ then the bell or fork tends to vibrate more with wavelengths with regard to which it is not too small (higher notes...).

This is what I gather that your quantum confinement example guarantees: 1/ first you need an atom small enough to have discrete E gaps and 2/ once we are in the discrete domain, it appears that the small size of the atom entails small vibration radius, which calls for higher energy photons.

As to the reason for 1/ a hypothesis could be that the atom is simpler this way and suffers fewer interactions; instead a big one has many constituents, hence more interactions and thus becomes more like, ultimately, a network of atoms, which give off continuous spectra.

 

[jambaugh]

My apologies for not getting back to this discussion earlier and it appears you're in good hands with the other posts. I would add one qualifier to your thinking and that is that you should not think of photon absorption, emission, or scattering as instantaneous events. Especially when photons have highly resolved momenta (frequencies) the scattering process takes an extensive length of time and the "emission" isn't per se after the "absorption" but rather they happen together. This is true both classically and quantum mechanically.

 

[JohnnyGui]

Post

Thanks for this explanation. The theory does fit this confinement effect.

@jambaugh : No problem. I've read your description somewhere before indeed. The problem I had was that I didn't know where to place this explanation to clarify the behaviour of scattering.

@Saw

In the meantime, I've been doing quite some research and it seems that Geometrical scattering is the largest problem to be described purely on a quantum level basis.

When it comes to scattering in the case of individual atoms or molecules, like in Rayleigh or Raman, there are some quantum level approximations such as the quantum dots I talked about earlier, or electrons reaching a virtual energy state instead of a bounding one and then jumping back to the original state again, in the case of Raman the state the electron jumps back to being a tad bit different than the original.

But as soon as the behavior of scattering depends on a structure or shape of a complex of molecules or atoms, such as in reflection, refraction and transparency, people always seem to explain it with a mix of wavetheory. And I think the root cause being that there's a preference of a direction of photon scattering in the case of geometrical scattering for which people then use interference to explain it just like you did on transparency.

So far I couldn't find a very pure quantum approximation on geometrical scattering except for the one of Feynman which is a probability theory of how photons scatter in a preferred direction. Still, it doesn't really give me a relieved answer since it doesn't tell me HOW this probability dependency is created on the quantum level. And this is probably not possible as far as I know.

 

[Saw]

Yes, the takeaways that I am drawing from the thread are:
- A reinforcement of the idea, which I find very revealing, that direction of emission is random at quantum/microscopic level and that preferred direction is something that only emerges at macroscopic level as a consequence of cooperative behavior of the particles.
- As to why the cooperation leads to direction, I am also left with a feeling of incompleteness. There are as you have well noted the two parallel explanations: the classical wave theory based on constructive/destructive interference and the (as usual disconnected from the latter) probabilistic Feynman-style explanation, but... what one usually expects is to be given the key as to why a given structure favors a certain probability distribution... and it may be our fault, sure, but I did not grasp either such key when I browsed the ST of LandM time ago...
- I am correcting the idea that Rayleigh scattering be a non-resonant phenomenon. Latest readings indicate that the incident radiation must match an E gap of the particle for this scattering to take place.
- With fluorescence I am puzzled because I am still seeing the 2 explanations: (1) decay in 2 photons when de-exciting by steps versus (2) decay into the material plus into a photon.
- There is a lot of terminological confusion because authors use terms with different meanings. What I will try to do is at least agree with myself on a reasonable terminology and then adjust what I read to my own terminology. For example, I would say that an interaction where the atom is made to vibrate with more amplitude is "resonant" and, well, you can call it "absorption" as long as you understand that it will not last long and its only lasting effect may be, if applicable, "collective": thanks to this higher vibration the acquired energy may be converted via collisions in thermal energy, in which case, yes, the radiation will be absorbed by the material; otherwise it will be scattered. In conclusion, atomic absorption and re-emission or scattering can go together, although collective absorption frustrates scattering, at least if not on the first occasion, after a few occasions.

Thanks for the conversation. Will be glad to hear, if you have more comments on this subject!

 

[jambaugh]

Saw, I think that feeling of incompleteness you expressed is exactly the classical incompleteness of QM that Einstein so hated. Seeking that foundational, reductive, objective explanation of phenomena at the quantum level will lead you to great frustration. You instead have complementary and contradictory paradigms which are always only a partial explanation.

 

[Saw]

Saw, I think that feeling of incompleteness you expressed is exactly the classical incompleteness of QM that Einstein so hated. Seeking that foundational, reductive, objective explanation of phenomena at the quantum level will lead you to great frustration. You instead have complementary and contradictory paradigms which are always only a partial explanation.

Interesting... it may be so... :)!

 

[JohnnyGui]

Saw, I think that feeling of incompleteness you expressed is exactly the classical incompleteness of QM that Einstein so hated. Seeking that foundational, reductive, objective explanation of phenomena at the quantum level will lead you to great frustration. You instead have complementary and contradictory paradigms which are always only a partial explanation.

This answer is a bit of a relief for me in some way since apparently, the culprit of confusion and incompleteness isn't really at my end alone.

I am correcting the idea that Rayleigh scattering be a non-resonant phenomenon. Latest readings indicate that the incident radiation must match an E gap of the particle for this scattering to take place.

I'm interested in where you've read this.

With fluorescence I am puzzled because I am still seeing the 2 explanations: (1) decay in 2 photons when de-exciting by steps versus (2) decay into the material plus into a photon.

I've searched but I couldn't really find an answer to that either. An additional question that I have regarding fluorescence is the following; I know that an electron doesn't fully absorb the given photon because the atom or molecule itself absorbs a part of that photon energy. However, from thinking this a bit through, I concluded that this isn't the only reason that an electron emits a lower energy photon. There are 2 additional possible reasons, or so I think:

1. Seeing that the atom/molecule absorbs a part of that photon energy, it will have also a higher state energy itself, just like the electron. This should give the electron a bit of a higher ground state so that, when the electron after excitation jumps back, it should make a smaller jump back to that higher ground state. So in addition to the absorbed energy by the atom/molecule, this should thus result in an even a lower energy photon emission than if the ground state didn't get elevated.

2. If the excitation of the electron is long enough so that the partly absorbed photon energy by the atom/molecule gets lost due to collision with the rest of the lattice, then the electron would first make small jumps to lower sublevels due to that energy loss before jumping all the way back to the ground state. This might also be an additional factor as to why an electron emits a lower photon energy.

Are both of these conclusions true in the case of energy loss in fluorescence?One other question that came up is about transparency. If transparency at the quantum level is described as scattering of photons in all directions that interfere with each other so that only a constructive interference is in the same direction as the incident photons, can't I say that in the case of Rayleigh scattering the reddish wavelengths are transparent to air molecules? So that these air molecules DO scatter red in all directions but in such a way that there's only a constructive interference path in the same direction as the incident red photons?

 

[Saw]

I'm interested in where you've read this.

You make me doubt... The manual I mentioned before by Paul G. Hewitt, when he makes that comparison with tuning forks or bells, is assuming that the incident radiation / sound matches the natural frequencies of the air molecules / bells.

Also University Physics, by Young Hugh Ford Freedman, state that when you look at the sky you see light that has been "absorbed" and re-radiated.

Also this book and Serway discuss that Rayleigh scattering is a phenomenon of the air molecules acting as oscillating dipoles, like little antennae that re-radiate light...

Well, it is not that they expressly say that light matches an E gap in the particles, but clearly they imply it because they talk about resonance with the appropriate frequency of the particles.

But I have googled to find the source of my notes on transparency and found this: http://www.sparknotes.com/physics/optics/light/section3.rhtml

In the end, what do you make of it? Does the author, whoever it is, defend that blue sky scattering = Rayleigh scattering is related to non-resonance (as he initially implies) or the opposite (later text)?

One other question that came up is about transparency. If transparency at the quantum level is described as scattering of photons in all directions that interfere with each other so that only a constructive interference is in the same direction as the incident photons,

That is what by the way the above text defends, at least. But in my notes I also had another possibility, which is that the radiation goes through without interaction with the particles of the medium because it diffracts / bends round the obstacles inasmuch as the wavelength is long enough compared to the size of the object.

can't I say that in the case of Rayleigh scattering the reddish wavelengths are transparent to air molecules? So that these air molecules DO scatter red in all directions but in such a way that there's only a constructive interference path in the same direction as the incident red photons?

Air is certainly transparent to the frequencies that we see when we look straight to the sun. But in this case I think that the reason is the second one, bending round the obstacles, like when long radio waves bend round buildings and hills or low-pitched notes overcomes barriers or visible light (unlike X rays) passes by a virus without "seeing" it. This is because in the sky the molecules are sufficiently far apart from each other. If instead they were packed together as in glass, you would need the other explanation (cooperative behavior favoring constructive interference in the forward direction) for transparency to be there.

 

[JohnnyGui]

You make me doubt...

My apologies, had a temporary blackout about what you told me before.

This is because in the sky the molecules are sufficiently far apart from each other. If instead they were packed together as in glass, you would need the other explanation (cooperative behavior favoring constructive interference in the forward direction) for transparency to be there.

Ah I understand now. But isn't this wave theory in both cases; that waves bend in both spaced out air molecules and glass but that the bent waves in glass are so close that they interfere?

If that's the case then there isn't truly a quantum explanation on the selectiveness between red and blue photons in Rayleigh scattering. Because virtual energy states is, as I would think, available for any kind of wavelength to be scattered and yet it's not happening like that. Unless virtual energy states are discrete as well and determined by the size of a particle for example. If that's the case, then we have a quantum explanation for (selective) transparency, I think.

 

[PeterDonis]

isn't this wave theory in both cases; that waves bend in both spaced out air molecules and glass but that the bent waves in glass are so close that they interfere?

This is an approximation in which the light is treated as classical waves but the atoms are (or can be) treated quantum mechanically. In some cases the atoms can also be treated classically--basically as classical dipoles--but in other cases quantum properties of the atoms, such as discrete energy levels, have to be taken into account.

virtual energy states is, as I would think, available for any kind of wavelength to be scattered

Not if the quantum properties of the atoms are taken into account.

 

[Saw]

My apologies, had a temporary blackout about what you told me before.

Well, it is good you asked because on further searching what I have found is that my latest approach seems to be wrong, whereas the initial one would be correct. See http://www.thephysicsmill.com/2014/03/30/a-quantum-of-scattering/ where the author claims that the quantum explanation of Rayleigh scattering is that an atom absorbs a photon of the "wrong" energy (not the E gap between two levels), thus jumping into a virtual state, which can be done because it lasts very little (which is a surprising statement because absorption of a photon of the "right" energy is also said to last very little, so what is the difference...?). In any case, the site refers also to this other one, which even gives a math treatment of Rayleigh scattering and clearly states that "The frequency of visible radiation is much less than the typical emission frequencies of a Nitrogen or Oxygen atom (which lie in the ultra-violet band), so it is certainly the case that ω<<ω_o".

 

[JohnnyGui]

Well, it is good you asked because on further searching what I have found is that my latest approach seems to be wrong, whereas the initial one would be correct. See http://www.thephysicsmill.com/2014/03/30/a-quantum-of-scattering/ where the author claims that the quantum explanation of Rayleigh scattering is that an atom absorbs a photon of the "wrong" energy (not the E gap between two levels), thus jumping into a virtual state, which can be done because it lasts very little (which is a surprising statement because absorption of a photon of the "right" energy is also said to last very little, so what is the difference...?). In any case, the site refers also to this other one, which even gives a math treatment of Rayleigh scattering and clearly states that "The frequency of visible radiation is much less than the typical emission frequencies of a Nitrogen or Oxygen atom (which lie in the ultra-violet band), so it is certainly the case that ω<<ω_o".

Yes, that's indeed what I've understood from most of the sources as well. Absorption and emission in the case of scattering is way shorter than absorption of the photon of the "right" energy. We're talking femtoseconds vs nanoseconds. My explanation could be a bit too general but here it goes:

The reason for the difference in time duration is that there's this so called Heisenberg principle which explains that the more accurate you want to measure the exact position of something, the less accurate you'd measure its velocity and energy possession, and vice versa. In general, to be able to make a very accurate measure of a position of an object, in this case an electron at a particular state, you'd need a very short time duration to do so.
So according to the Heisenberg principle, if you did that, you'd not know the exact velocity or the exact energy possession of that electron. Here's when things get strange. Since your measuring accuracy of the velocity or energy of that electron is lowered, the electron can basically possesses ANY amount of energy as long as that range of possible energies suits the inacurracy of the short time duration you're trying to measure its position in. Hence, an electron can absorb photons of "wrong" energies in that very short time.
However, Heisenberg principle also said that if you start measuring the position of that electron for a longer time duration, then you're increasingly able to measure a more accurate velocity and energy level. The electron, that has already broken "the rules" by absorbing a photon of a "wrong" energy would thus have to release it as fast as it can to adapt to the more accurate possible energy levels that you're able to measure.
So the absorption and emission of a "wrong" photon energy must be happening in a very short time duration in which our ability to measure energylevels is inaccurate enough; that's scattering. You could say that the "official" S energy levels of the electron are energy levels according to the longer time duration (nanoseconds) that we're measuring the absorption and emission in; luminescence.

To make things even more confusing; I read that virtual energy states is merely a theory to make our understanding of possessing "wrong" energies a bit easier. It's theorized because, apparently, you can also excite an electron to an official S energy state by shooting two lower energy photons at it instead of just one photon with the right amount of energy, if the beam of photons has a high intensity. The electron must thus reach some kind of intermediate (virtual) state between the 2 photons to be able to reach the official S energy state. Virtual energy states comes from the Pertubation Theory that, as far as I understand, calculates the possible virtual energy levels by using superpositions of the "official" S energy states. So it's some kind of derivation from the available S energy states of an electron.

This is an approximation in which the light is treated as classical waves but the atoms are (or can be) treated quantum mechanically. In some cases the atoms can also be treated classically--basically as classical dipoles--but in other cases quantum properties of the atoms, such as discrete energy levels, have to be taken into account.

Yes, that's indeed what I have understood. What I was wondering is if there's a pure quantum level approximation on the selectivity between red and blue photons for example in Rayleigh Scattering (see below).

Not if the quantum properties of the atoms are taken into account

So virtual energy states can be discrete? Does this mean that there's a quantum approximation on the selectivity of particular photon energies in Rayleigh scattering? For example, if virtual energy states can be discrete depending on the size of the atom (analogy to quantum dots), one could then explain that very small atoms only have virtual energy states that are equal to high energy (like blue) photons and thus mainly scatter these?