# Scattering of Light

1. Jun 5, 2009

### Xian

I pretty much understand the idea of photon of some wavelength λ exciting an electron in atom above the ground state if the difference between the ground state and a higher energy level is exactly hc/λ. Not hard, but what happens to the electron and the atom when the photon energy is below the excitation energy? From what I've been reading in Optics by Hecht it seems that the photon will pass through the electron cloud and its electromagnetic nature will set the electron cloud into an oscillatory motion making it so the atom now behaves like an oscillating dipole (similar to those in radio towers). This oscillating dipole will then emit a photon of the same frequency as its oscillation (or the same frequency as the passing photon, or maybe coincidentally both, I'm not sure). What happens to the original photon however? Does it just continue to move on, gets weakened and moves on, or does it get absorbed by the cloud when it sets it into an oscillatory motion? I suppose conservation of energy would point to either the second or third option no?

Also, what happens when the photon is not carrying precisely the excitation energy, but more? Will the electron get promoted above ground level and then the atom absorbs the rest of the energy as just kinetic?

2. Jun 5, 2009

### fatra2

First of all, a photon does not weaken with time. Therefore, if the photon does not carry enough energy to excite the atom, then it will just carry on its way, like nothing happened.

If the photon carries more energy than what is needed for a one level excitement, the electron will be even more excited and "climb" the scale even higher. If the photon as more energy than what is "allowed" by the atom, the electron will simply leave the atom (process that is called ionizing an atom).

Cheers

3. Jun 5, 2009

### Xian

Hey thanks for the reply fatra, but I think I need to clarify myself a little.

In the first question when I said the photon weakens I meant that for instance the photon passes through the electron cloud, and as it does the oscillating electric component of the photon will cause the electron cloud to vibrate, so by conservation of energy the photons energy should be weaker since the atoms total energy has increased. This is what my textbook Optics by Hecht tries to illustrate. The only part that its not to clear on is what happens to the passing photon. I assume that it doesn't just fly on by like nothing happened, and so it either gets weakened by the transfer of energy (if thats possible) or gets completely absorbed by the electron cloud, though it does not excite the individual electrons, but instead is absorbed by the cloud in the sense that the cloud now oscillates. We could oscillate the cloud in a similar fashion by applying an external E-field. Also, even besides this, photons do weaken in certain situations such as the Cosmic Background Radiation of the universe. There the photons were once very energetic (since they were from the big bang) but then were weakened (though I'm not entirely sure what it was that weakened them, it would not just be distance alone but likely something from special or general relativity).

Also, for the second part, I meant to say that the photon has enough energy to excite the electron up one energy level, and has a just a little bit extra, not enough to make it climb higher, but enough so that there is a surplus of energy after that one electron gets bumped up. What happens if the photon has this much energy? My guess was that the whole photon gets absorbed, but the extra energy is absorbed as kinetic, though this doesn't feel right.

4. Jun 5, 2009

### Bob S

If you review the theory of Rayleigh scattering of sunlight in air, you will find that the theory is based on classical scattering of light waves (plane waves) off of uncharged atoms. Rayleigh scattering correctly predicts both the 1/lambda4 wavelength dependence (note:without any discrete transitions) and complete polarization (note: only with very smogless clear sky) of the scattered light at 90 degrees. The cross section (probability) for scattering is very roughly equal (within a factor of 32 pi/3) to the square of the classical electron radius (2.828 Fermi), therefore much smaller than atomic sizes (Bohr radius is 1372 x classical electron radius. So very roughly a visible photon "hitting" an atom has a very small chance of scattering.

5. Jun 5, 2009

### Xian

Though Rayleigh scattering is a useful to know and understand, I would like to understand as well as possible what goes on in the situation I mentioned above (which of course includes scattering). To really understand this though I will need to understand any quantum tricks that go on of course, and this is unavoidable to do the discrete nature of light and atoms. The classical picture of light as a wave is useful and more or less "accurate" when the number of photons is very large, however I'm looking for the best answer and that means looking at the single-photon-to-single-atom interaction. If possible I'd love to hear whatever you know in that area, and would greatly appreciate it.

6. Jun 7, 2009

### mn4j

Photons can and do transfer momentum to electrons and vice versa even when the energy of the photon is not sufficient to excite the electron. The photon will continue on it's way with the same energy but different momentum. The electron will be left with the same energy but different momentum as well.

7. Jun 21, 2009

### invisigo

I believe you are confusing yourself by mixing classical and quantum optics. Hecht's description uses classical optics, in which the atoms are dipoles and photons are EM waves that excite them. However, you cannot bring in the idea of "energy level" into this picture of light-matter interaction, because that is a quantum concept. In this description, yes, a low energy...and I really shouldn't even bring in the idea of energy, or even photon...let's say a low frequency EM wave has the potential to excite any atom (which is modeled as a dipole). The classical description is limited this way, because it can lead to incorrect deductions.

In the quantum description (or maybe even the semi-classical model), you can bring in the idea of energy levels. Now your whole model of physics changes. The atom is no longer a dipole, but an object with a bunch of energy levels. Throw away the dipole. I'm no expert (or even novice) in quantum electrodynamics, but I think your whole idea of electromagnetism changes too. But in any case, photons can be thought of as packets of energy, and if that energy does not match an atomic transition level, then it will pass through the atom.

8. Mar 24, 2011

I was under the impression that a photon is absorbed by the electron and excites it if the photon carries the minimum amount of energy to do so.
From what I've been taught, its an "all or nothing" process where the photon must have the minimum amount of energy or it simply won't affect the electron.

Also, when a photon of adequate energy excites an electron, I believe it is absorbed but is then re-emitted as soon as the electron falls back down to its original energy level. It is in this way that conservation of energy is preserved: a photon of equal energy is re-emitted. So, the photons being emitted in the atmosphere that are giving the sky its blue hue aren't actually the original photons, but the re-emitted ones.

This is what I can remember to the best of my abilities from class but I could be wrong. Someone please feel free to correct me if I am.

9. Mar 25, 2011

### Xian

Agreed, but my question is more about what happens below the threshold.

If the photon does not have enough energy to excite an electron beyond the ground state, my quantum intuition tells me that the photon would just pass through as if there was nothing there.

My classical (semi-classical?) intuition tells me however that since a photon is an electromagnetic oscillation that it should interact electromagnetically with the electrons (and protons) of the atom.

I know next to nothing about quantum electrodynamics but from what I get, an photon is described to be an excitation of the electromagnetic field... I'm not sure how to work with this interpretation (assuming it is a correct one).

Thoughts anybody?

10. Mar 26, 2011

### alxm

Right. Classically Rayleigh scattering is fairly intuitive in both pictures. As an electromagnetic wave, the atom is a polarizable 'medium', so you have diffraction. As particles, a photon simply scatters off the electron elastically. However your QM intuition is leading you wrong here; it's more classical than you might think. Consider what the atom's Hamiltonian looks like if you've got an external field - the electron has a coupling to the electrical field and so you have a response and polarization (Stark effect), even if the electron does not change states. So the classical wave picture more or less works here.

An alternate view (from QED) is that the photon is absorbed and that there's excitation to a virtual electronic level, followed by immediate re-emission. This is allowed because it's not actually real; the electron is never observed to be in that excited state. (It also means you don't have to worry much about selection rules) That as opposed to fluorescence, where the exited state is real.

11. Mar 26, 2011

### lightarrow

Anyway, if the photon does not have enough energy to excite an electron beyond the ground state, it's not correct to say that the photon "would just pass through as if there was nothing there": light *does* interact with the atom, just because it's scattered.

12. Mar 27, 2011

### johng23

My understanding of this process (From Boyd's "Nonlinear Optics") is that an electron could stay in a virtual level for a time $$\Delta$$t = h/$$\Delta$$E, where $$\Delta$$E is the energy difference between the virtual level and the nearest real level in the system. If that's a valid way to think of it, how can we say the re-emission is immediate? For example, with energy differences in fractions of an eV, the electron can stay in the virtual level for a few femtoseconds, appreciable compared to the duration of an optical cycle. So the re-emitted radiation would not be in phase.

13. Mar 28, 2011

### alxm

Because it is, the electron is not observed or observable in this virtual state, no measurable interactions occur or can occur between the absorption and re-emission.

By analogy to the ordinary uncertainty principle: You can measure the effects of an electron acting as if it were in two places at once, but we don't say it actually is in two places at once.

The system here is acting as if it'd been excited to a virtual level for that amount of time.

14. Mar 28, 2011

### johng23

I don't think I fully understand. If the re-emission is immediate, how is the system acting as if it were in the virtual level for the time $$\Delta$$t? What is the significance of this characteristic time?

15. Mar 28, 2011

### Antiphon

Johng23, you original question was about dielectric polarizability. This is not the same as electronic transitions, a much higher energy (and discrete) event.

You were asking how it is possible for a quantum to perturb a dipole and remain unaffected. The answer is that it does not remain unaffected. It is absorbed and reemitted with some probability by the dipole. Dielectric polarization is not a bound energy state with discrete states like an electronic orbit. A photon of nearly any energy can excite the dipole.

16. Mar 29, 2011

### lightarrow

Right. Using a metaphor, it's like hitting a ball which goes up a slope and then goes back because it doesn't find an equilibrium position at an higher altitude with respect to the ground.