What are the quantum mechanisms behind the emission and absorption of thermal radiation at and below room temperature? If the relevant quantum state transitions are molecular (stretching, flexing and spin changes) how come the thermal spectrum is continuous? What about substances (such as noble gases) which don't form molecules, how do they emit or absorb thermal radiation? Is there a semi-classical mechanism (with the EM field treated classically) and also a deeper explanation using the full apparatus of QFT?

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    $\begingroup$ The reason you are confused is that you think that there is no way that atoms will emit thermal continuous spectrum when their lines are discrete. But there are severe constraints on atomic transitions which come from the fact that any atom placed in an electromagnetic field will absorb and emit thermal radiation in such a way that it maintains the radiation in equilibrium. This is Einstein's A and B coefficient paper, it is how he deduced the laws of stimulated emission, from the paradox you are considering. $\endgroup$
    – Ron Maimon
    Commented Jul 23, 2012 at 19:42
  • $\begingroup$ Perhaps this will shed some light on the matter. The paper linked there has a full many-body treatment of the problem as well. $\endgroup$ Commented Jul 24, 2012 at 8:30

5 Answers 5


The thermal radiation associated with some object is typically described in terms of the "black-body" spectrum for a given temperature, given by the Planck formula. This formula is based on an idealization of an object that absorbs all frequencies of radiation equally, but it works fairly well provided that the object whose thermal spectrum you're interested in studying doesn't have any transitions with resonant frequencies in the range of interest. As the typical energy scale of atomic and molecular transitions is somewhere around an eV, while the characteristic energy scale for "room temperature" is in the neighborhood of 1/40 eV, this generally isn't all that bad an assumption-- if you look in the vicinity of the peak of the blackbody spectrum for an object at room temperature, you generally find that the spectrum looks very much like a black-body spectrum.

How does this arise from the interaction between light of whatever frequency and a gas of atoms or molecules having discrete internal states? The thing to remember is that internal states of atoms and molecules aren't the only degree of freedom available to the systems-- there's also the center-of-mass motion of the atoms themselves, or the collective motion of groups of atoms.

The central idea involved with thermal radiation is that if you take a gas of atoms and confine it to a region of space containing some radiation field with some characteristic temperature, the atoms and the radiation will eventually come to some equilibrium in which the kinetic energy distribution of the atoms and the frequency spectrum of the radiation will have the same characteristic temperature. (The internal state distribution of the atoms will also have the same temperature, but if you're talking about room-temperature systems, there's too little thermal energy to make much difference in the thermal state distribution, so we'll ignore that.) This will come about through interactions between the atoms and the light, and most of these interactions will be non-resonant in nature. In terms of microscopic quantum processes, you would think of these as being Raman scattering events, where some of the photon energy goes into changing the motional state of the atom-- if you have cold atoms and hot photons, you'll get more scattering events that increase the atom's kinetic energy than ones that decrease it, so the average atomic KE will increase, and the average photon energy will decrease. (Or, in more fully quantum terms, the population of atoms will be moved up to higher-energy quantum states within the box, while the population of higher-energy photon modes will decrease.)

For thermal radiation in the room temperature regime, of course, the transitions in question are so far off-resonance that a Raman scattering for any individual atom with any particular photon will be phenomenally unlikely. Atoms are plentiful, though, and photons are even cheaper, so the total number of interactions for the sample as a whole can be quite large, and can bring both the atomic gas and the thermal radiation bath to equilibrium in time.

I've never seen a full QFT treatment of the subject, but that doesn't mean much. The basic idea of the equilibration of atoms with thermal radiation comes from Einstein in 1917, and there was a really good Physics Today article (PDF) by Dan Kleppner a few years back, talking about just how much is in those papers.

  • $\begingroup$ Pretty good answer. Suppose, however, in an idealised model we had a collection of "hot" gas atoms in "cold space" (therefore with no ambient electromagnetic field). I assume the gas would cool down by emitting thermal radiation. However there are no photons to couple with via Raman scattering. So how does that work? We could even consider the simplest case of two atoms colliding at "room temperature" speeds in a space with no EM field. Surely this would be an elastic collision and no "cooling" would take place? $\endgroup$
    – Nigel Seel
    Commented Jan 17, 2011 at 10:28
  • $\begingroup$ There is no such thing as a region of space with no electromagnetic field. In the real universe, there is always the cosmic microwave background radiation for the atoms to interact with. We sometimes talk about interacting atoms as if there's no radiation around, but we implicitly mean that there's no resonant radiation to worry about. In reality, the CMB is always there, and a hot gas of atoms will eventually come to equilibrium with it. $\endgroup$
    – Chad Orzel
    Commented Jan 17, 2011 at 11:59
  • $\begingroup$ Just to be clear, the presence of the CMB is a contingent fact. Are you suggesting that QM itself intrinsically forbids a vacuum state with zero EM field? $\endgroup$
    – Nigel Seel
    Commented Jan 17, 2011 at 12:45
  • $\begingroup$ If you want to talk about a "pure" vacuum state in QM, without any background thermal radiation, you still need to include the zero-point energy of all the field modes, which gives you a non-zero energy even in vacuum. I'm not sure of the details of the equilibration process in this sort of QFT regime (it's not something I've ever seen discussed), but I'm confident that the end result will be a thermalized radiation field in equilibrium with the atomic gas. $\endgroup$
    – Chad Orzel
    Commented Jan 17, 2011 at 15:39
  • $\begingroup$ I have doubts on the correctness of this answer. Raman scattering requires an electron to move to a higher excited state and the photon to provide the energy for that. Low energy photons don't have the required energy to excite electrons in simple molecules and in fact gasses don't come to an equilibrium with their surrounding thermal radiation unless there are relevant state transitions available. See also my answer below. $\endgroup$
    – JanKanis
    Commented Aug 30, 2019 at 21:40

Addressing only the question of how a continuous spectrum arises from what appears to be a set of discrete spectra associated with individual modes of interaction with a solid pahse material...

You can get a hint one of the things that is going on here by considering the Mössbauer effect. Individual photons are not required to interact with individual atoms, they can (and therefore sometimes will) interact with larger units of matter. IN the case of the Mössbauer effect they exchange momentum with a large unit of crystalline lattice rather than just the nucleus they interact with. You use the adjective "coherent" to distinguish these interactions.

The result is that a otherwise neat cluster of allowed energies is smeared out by the recoil associated with many different masses.


Here's a classical (non-quantum-mechanical) answer:

As you probably know, light is an electromagnetic (EM) wave. EM waves are produced by oscillating electrical charges. This is how an antenna works: connect it to an oscillating voltage source that pushes electrons back and forth in the antenna, and a radio wave is emitted. This is what Maxwell's field equations are all about.

When you heat an object, its constituent atoms/molecules vibrate, which results in oscillating charges, which in turn emit EM radiation. The velocities and oscillation periods of the atoms vary in a continuous distribution, and the peak of that distribution increases with temperature. Hence the EM radiation is also distributed over a range of wavelengths (forming the classic black body curve); the peak wavelength changes (gets shorter) with increasing temperature; and the total power integrated over all wavelengths also increases with temperature.

The above describes how EM radiation is emitted. As for absorption, when an EM wave hits an object, it is either reflected or absorbed. The part that is absorbed turns into heat, which is manifested by increasing vibration of its atoms. Eventually, the object's temperature will rise until emissions exactly equals absorption, at which point it will be in thermal equilibrium with its surroundings.

This is a complicated subject, and I've glossed over a lot of details, but this is a good mental model of what's happening.


Nigel asks among other things if there is a semi-classical mechanism. This is an interesting point which has not been addressed so I am going to give it a try.

The classical equilibrium between the electromagnetic field and a harmonic oscillator occurs when the energy per mode of the field is equal to the energy per mode of the oscillator. This is true classically and it is also true quantum mechanically. The real problem is that in a classical system of balls connected by springs, every mode gets the same energy no matter its frequency. This is true from the equipartition theorem which is based on maximizing entropy, and it is also true from a strictly mechanical argument of tracking collisions. This is the source of the ultraviolet catastrophe.

One can evade the ultraviolet catstrophe in an ad hoc manner by restricting the mode energies of the higher frequency e-m field components. Planck's formula is usually explained this way. But a reading of Planck's paper shows that he really solved the problem by redefining the way entropy is calculated for an oscillator. Applying this entropy calculation directly to the electromagnetic field gives the desired results. But one could just as well apply the entropy calculation to the purely mechanical system. If one does, it turns out that the higher frequency oscillators are not excited to the extent one would expect from analyzing them as classical billiard balls connected by springs. The very high frequency oscillations are suppressed. If one then allows the oscillators to carry a small amount of charge, and calculates the resulting radiation according to the classical laws of antenna theory, it turns out you get the correct black-body spectrum.

The system breaks down if you allow two or three high-frequency oscillators with the properties of a classical billiard ball on a spring to be added to the system. They will quickly equilibrate with the rest of the gas in terms of x,y and z velocities; and once the do this, they will share this energy with their oscillatory mode as well. It is the laws of quantum mechanics that forbid the existence of such oscillators, thereby preserving the thermal equilibrium.

EDIT: I've posted a series of calculations starting here showing how you get the correct thermal radiation by looking at the moving charges in a quantum-mechanical system and calculating the radiation classically using the Larmor formula.


The premise of your question is wrong. Gases consisting of very simple molecules in fact don't emit a continuous thermal spectrum, but have only very specific emission/absorption bands. Black body radiation is an approximation that assumes that the material has/consists of oscillators of every frequency. It works very well since for most solid matter there are often so many modes of vibration available that the black body spectrum is a good description, but materials deviate from a pure black body spectrum if you start looking at the spectral details. For a quantum mechanic description of thermal radiation from solid matter you will probably have to look at phonons, which are the quantum version of crystal lattice vibrations.

For the case of gases consisting of simple molecules such as noble gasses, but also regular simple molecules, these are often better described by spectral lines than by a continuous thermal spectrum. Some examples:

  • The atmosphere is transparent to visible and infrared radiation, except for those IR bands specifically absorbed by some of the molecules in it (which taken together is quite a bit of the IR spectrum, but not all of it).
  • If you look at rocket exhausts, rocket engines burning hydrogen have a transparent exhaust, and there are videos of icicles forming on the engine bell of hydrogen rocket engines that prove that the extremely hot exhaust is also not emitting much thermal radiation.
  • There is also a certain problem in the theory of the formation of the very first stars, which consisted of only hydrogen and helium, in that the clouds of gas that formed these stars would need to cool down during their collapse in order to condense into stars, but that atomic hydrogen and helium don't provide the necessary vibrational modes to emit the thermal radiation and so are unable to cool below 104 K. The solution is in cooling from molecular hydrogen.*

I don't know much about Raman scattering which the accepted answer offers as explanation, but it appears to include an electron getting excited to a higher energy state, so that is only applicable if an appropriate excited state is available and the incoming photon has enough energy. So it doesn't apply to monoatomic gasses and thermal energy photons.

* I don't have a very good reference, but this Scientific American article describes it: "The First Stars in the Universe", R.B. Larson and V. Bromm in Scientific American of December 2001. A different description is on this page. These astrophysics college slides also have interesting information, but the accompanying college is missing.


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