I guess visible light is visible, because it has the right wavelength to be absorbed (or not) and emitted (or not) by many different molecules. Now visible light has a wavelength in the order of a few hundreds nanometers, while the typical size of a molecule is rather in the order of a few angstrom.

I guess the difference between the energy levels of the molecules will have the right magnitude that corresponds to the energy carried by a single photon. But it's still slightly strange, because a classical antenna would need a length of a quarter wavelength to be most effective at absorbing or emitting energy.

I also wonder what the effective size of the scattering cross-section of a molecule to absorb a photon of an appropriate wavelength will be. Is it more in the order of the size of the molecule, or more in the order of the geometric mean between the wavelength and the size of the molecule?

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    $\begingroup$ A scattering cross-section for an atom to absorb a photon strongly depends on the photon frequency. It shows a resonant behaviour for frequencies equalling atomic level distances (for allowed optical transitions) with the effect that they can be orders of magnitude larger than the atomic dimensions. $\endgroup$ – Urgje Jul 14 '14 at 10:27
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    $\begingroup$ Basically the same reason a microphone of a few millimetres can pick up bass frequencies of several meters wavelength. $\endgroup$ – leftaroundabout Jul 14 '14 at 12:34
  • $\begingroup$ @leftaroundabout: Not unless you have a microphone that is super-sensitive to low B-flat but rejects low A and low B. Which is what the atom does. $\endgroup$ – Marty Green Jul 14 '14 at 16:46

In quantum mechanics you calculate the charge density by taking the square of the wave function. If you do this for a hydrogen atom in a superposition of the ground state the first excited state (1s and 2p) you get an oscillating charge density. If you analyze this oscillating charge using Maxwell's equations, you get all the properties of the hydrogen atom: the absorption, the emission, the line-width...you name it. Everything the atom does in its normal interactions with light makes sense according to Maxwell's equations.

In quantum mechanics if you have a slab of material made of atoms, then at any given temperature there is a wave function made up of the superposition of the different thermal states of the slab. If you square the amplitude of this wave function you get a time-varying charge density full of oscillations. If you use Maxwell's equations and treat these oscillations as classical antennas, you will obtain the correct black-body spectrum for the solid slab. All the thermal properties of matter in its interaction with radiation (including the photo-electric effect) are consistent with Maxwell's equations.

It is true that the quantum oscillator is hundreds or thousands of times smaller than the quarter-wave dipole which is the most efficient classical absorber. But you can make a classical antenna shorter by adding an inductance. In an atom, the mass of the electron is the parameter which effectively increases the apparent inductance of the atomic antenna. The only difference with classical antennas is that you cannot normally get such a small size-to-wavelength ration because of the high resistance of copper.

The reason physicists don't talk about this is that they have a poor understanding of classical antenna theory. I explain how this works in more detail in a series of blog posts starting here:

  • $\begingroup$ I made an animation illustrating how an electron in a superposition state moves back and forth, just like you say in the first paragraph. See en.wikipedia.org/w/… $\endgroup$ – Steve Byrnes Jul 14 '14 at 14:22
  • $\begingroup$ You made that animation? That is excellent and really needed on the internet. I used to be able to find some pretty good three-dimensional animations of the hydrogen orbitals in superposition, but I can't find them anymore. So your animation, which clearly shows the basic idea, is really useful. $\endgroup$ – Marty Green Jul 14 '14 at 14:28
  • $\begingroup$ As an electrical engineer who works in the nitty-gritty details of classical EM and Maxwell's equations for practical problems, I find your claims fascinating! Do you know of any reference where they work out these ideas in detail, say with a single hydrogen atom? I'd love to see how the QM predicted charge density treated classically gives results consistent with QM. $\endgroup$ – rajb245 Jul 14 '14 at 15:35
  • $\begingroup$ I don't know anywhere you'll find the analysis except on my blog. I used to think that Jaynes and Scully "semi-classical" school represented my ideas, but it turns out they don't. Did you check out the link on Crystal Radio that I posted in my answer? I go into more detail on the comparision between the Copenhagen calculation and the semi-classical analysis in a series of articles starting with this one: marty-green.blogspot.ca/2012/02/… $\endgroup$ – Marty Green Jul 14 '14 at 16:30
  • $\begingroup$ About shortening classical antennas - typically, ceramic "chip" antennas can be used to reduce the quarter-wave antenna size due to their much higher dielectric constant. You can pack a copper antenna into a size one tenth of the quarter-wave (in vacuum). There are obviously other drawbacks with these though. $\endgroup$ – BjornW Jul 15 '14 at 13:25

I'm assuming you're talking about absorbtion lines. In that case it's a mistake to compare the size of the molecule to the wavelength of the photon. These absorption lines appear as a result of the electron energy levels of the molecule in question. These span a range of energies that, for most gases, will usually include visible light. Let's take a look at the Hydrogen atom as an example (a more realistic treatment would look at Hydrogen molecules, not atoms, but this doesn't substantially change the orders of magnitude involved) :

The transition from the second balmer line to the ninth represents an energy change equal to the energy of a violet photon (383 nm), whereas a transition from the second to the third balmer line corresponds to a red photon (656 nm).

As you rightfully point out, what matters is the energy levels that can be excited in your system. Comparison with molecular size are not useful, as the physics involved here is simply not the same as that of antennae.

  • $\begingroup$ I disagree. You can learn many interesting things by learning the classical theory of electrically-small antennas (e.g. en.wikipedia.org/wiki/Chu%E2%80%93Harrington_limit ), and applying those principles to atoms interacting with light. $\endgroup$ – Steve Byrnes Jul 14 '14 at 14:27
  • $\begingroup$ I'm not saying that the "molecule as an antennae" model is useless in understanding molecule/light interactions. I'm saying that, if "you're talking about absorption lines", then for that specific observational feature antennae physics is not relevant. I wasn't sure whether or not that's what he was asking for, so I pointed out that that's the context in which I'm working. $\endgroup$ – ticster Jul 14 '14 at 14:31
  • $\begingroup$ If you want to understand the cross-section or linewidth of an atomic absorption line, then it is not a mistake to compare the size of the molecule to the wavelength of the photon. $\endgroup$ – Steve Byrnes Jul 14 '14 at 15:37
  • $\begingroup$ While it could certainly explain some of the things you would see, it wouldn't be enough. Some of the features come from properties you could derive studying the individual atom (where you can't even speak of the "size of the molecule"). Sure, you could relate the size of the molecule to the energy levels of the individual atom, but it just seems like a roundabout way of doing things. $\endgroup$ – ticster Jul 14 '14 at 18:14

The absorption/radiation mechanism for antenna is purely classical. It depends on the current oscillation on the structure. For a molecule or atom, it is purely quantum. What only matters is energy level difference. There's no direct relation with the size of the particle.

However, the two mechanisms are not unrelated. They get connected for the case of nanostructures, e.g., nano antennas. There the size dependence is not exactly like that for a microwave antenna, but has important effects on absorption. On the other hand, typical plasmonic nano antennas can be understood in terms of collective excitations, which are coherent combinations of electron-hole transitions, so related to energy level spectrum of the many-body electron system.


It is also important to note here the difference between a single molecule or atom and a solid comprised thereof. A single atom is indeed not terribly likely to absorb a photon - a large bulk of atoms or molecules in a solid, however, is quite a different thing. The branch of physics that deals with this is called Solid State Physics. A relevant topic here is the electronic band structure of the solid.

When tightly packed together, the atomic or molecular orbitals of adjacent members of the solid can overlap and interact, generating a macroscopic electronic structure that extends throughout the solid. This, in part, is what gives rise to the optical absorption characteristics of solids.


Even classically the ability of an antenna to radiate or absorb is not related to its size but to it being "resonant" with the "ether" being its loading impedance. The size is related to the bandwidth over which this resonance can be achieved. For an antenna whose characteristic size is much less than a wavelength the relative bandwidth over which it can radiate reduces exponentially with that size. Think of a Hertzian dipole that is oscillating at one frequency, and there it radiates beautifully. Your cellphone antenna that is much smaller than 15cm (half wavelength at 2GHz) is made resonant with lossy matching circuits on the board over the band of interest, a lousy and inefficient radiator but a radiator, nevertheless. See, the classic results of J. L. Chu "Physical Limitations of Omni-Directional Antennas", JOURNAL OF APPLIED PHYSICS, vol. 19, December 1948, Figs: 5 & 6

  • $\begingroup$ This is not a bad answer, but it is unnecessary to talk about "lossy" components in understanding the classical antenna. We can imagine everything made of superconducting wires; and then, as user21748 correctly notes, it is not so much the absorption cross-section which goes down as you make the antenna smaller, but the bandwidth. And to be technically correct, it is not exponential but cubic: that is, if you make your ideal lossless antenna half as big, your Q factor (wL/R) and with it your bandwidth goes by a factor of 8. $\endgroup$ – Marty Green Jul 14 '14 at 19:17
  • $\begingroup$ Marty Green is correct about the "cubic" instead of "exponential" behavior of the Q vs. reciprocal size. $\endgroup$ – hyportnex Jul 16 '14 at 1:02
  • $\begingroup$ The reason to talk about loss because it is a mistake to think of an antenna without its matching circuit to its source or load; an antenna is not just its radiation pattern. If you wish the matching circuit is kind of boundary condition. $\endgroup$ – hyportnex Jul 16 '14 at 1:06
  • $\begingroup$ The lossyness of the matching circuit is crucial in understanding the problem with "wideband" matching of a small antenna:as Q increases so does the load current, and the loss is proportional to the square of the current. Superconducting wires don't help transmitting antennas because their superconductivity is limited by the maximum current; they may be useful in receive mode. Nobody has so far figured it out how to make a super-directive antenna radiate efficiently. $\endgroup$ – hyportnex Jul 16 '14 at 1:06

This is a fun question! Concerning the cross section for absorption of resonant light. I can speak to the case of Rubidium atoms and the 780 and 795 nm 5P doublet.
The cross section is greater than the atomic size, and also greater than the mean of the wavelength and atom size. In fact it's much nearer to the wavelength of the light! The experiment is fairly easy. You look at the transmission of resonant light through a cell with Rb as a function of temperature. (The vapor density of Rb is a strong function of temperature.)

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    $\begingroup$ This does not answer the general question how it is possible that molecules can absorb photons with wavelengths bigger than its size. $\endgroup$ – fibonatic Jul 15 '14 at 12:59
  • $\begingroup$ @fibonatic This general question is already answered more or less by Marty Green, but I need to check his answer thoroughly before accepting it, because it references his own blog instead of more "conventional" sources. $\endgroup$ – Thomas Klimpel Jul 15 '14 at 13:07
  • $\begingroup$ I'm not sure it's "proper" to ask how the absorption happens. We can use QM to calculate the absorption probability... but how it actually happens.. we just don't know. (Or at least I don't know.) I like to focus on things we can measure. $\endgroup$ – George Herold Jul 15 '14 at 14:19
  • $\begingroup$ The theoretical result whereby the absorption cross-section of a small classical antenna is on the order of the wavelength squared is a not-so-well-known classical result that nevertheless appears in such textbooks as Kraus (from the 1930's) and you can verify it on Wikipedia. It is normally derived by pure mathematics; however, on my website I give a physical picture of why it makes sense: see marty-green.blogspot.ca/2011/10/crystal-radio.html $\endgroup$ – Marty Green Jul 15 '14 at 16:37

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