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I recall from my undergraduate statistical mechanics and QM classes that Planck's Law may be derived fairly straight-forwardly by considering the density of states of EM radiation in thermal equilibrium with a particle inside a cubic box. Indeed the derivation on this wiki page is just the same and appears (IMO) to be fairly concise and accurate.

I wondered therefore, that if we invoke all the magic of QFT (such as the fields being promoted to operators and higher (than tree-level) order particle interactions taken into account,) can a similar derivation still be carried out? (Or indeed a dissimilar one?)

Associated to which, would such a derivation (if it is even possible) require consideration of QED alone or would the effects of the broken electro-weak symmetry have a significant effect on recreating the correct radiation spectrum?

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The derivation of the Planck distribution takes into account the quantization of the electromagnetic field: any particular wave-vector of the field can have quantized energies, which are interpreted as the number of photons times the energy of each one of them. At the time Planck derived his formula, this is all he needed. He had no formulation of a physical theory in which the electromagnetic field was quantized, but he showed that when you manage to get one, that quantization alone will be sufficient.

In quantum field theory, this is built-in as the starting point: you promote each field to an operator, and when you "second quantize"* you satisfy the requirement for the deriving the Planck spectrumm - the energy in each mode of the electromagnetic field is quantized. I believe second quantization of the Maxwell field (no need to add electrons to get QED) can be found in every text on the subject. After you do that, the derivation of the Planck spectrum is not all that different from Planck's.

This is only the starting point though. Once you have QFT you can add interactions and calculate corrections to the spectrum and any other thermodynamic quantity - due to QED, electroweak symmetry breaking, and what have you. Those will be normally small corrections**.

For the spectrum, the corrections comes effectively from adding interactions between the photons. The single photon stays massless (because of gauge invariance) and therefore its spectrum does not get modified. But the energy of two photons is no longer the sum of their individual energies. This should correct the measured spectrum in a small way.

(As a side comment: those calculations in thermal field theory are not that simple, as compared to usual loop integrals you get at zero temperature.)

  • Just for the record, I strongly dislike this term for other reasons, but here is seems appropriate.

** It is interesting whether or not such corrections have been measured - thermal effects of QED. I don't know the answer to that.

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Dear Moshe, don't you think that the masslessness of photons is exact to all orders in QED? So the Planck's black-body spectrum is unaffected by e.g. the box diagram with the electron loop, isn't it? Of course, the thermal ensemble also has to include some electrons and positrons if they may be created. But if you managed to have answered a question above, would you agree that the electroweak theory doesn't deform the photons' black body spectrum in any "quantitative" way? – Luboš Motl Feb 8 '11 at 18:22
Good point, I was thinking about more general thermal field theory calculations than just the spectrum. I'll add a clarification. – user566 Feb 8 '11 at 18:25
@Lubos: On second thought, maybe you can help me out here - the single particle spectrum of the electromagnetic field is not modified, but Planck's derivation needs also the multi-particle states, it is assuming that energies of multiple photons is additive. Doesn't that fact change when they interact through loops, like the one you describe? – user566 Feb 8 '11 at 19:52
@Lubos: So, I think I convinced myself that I am right...The photon of course stays massless, but the spectrum of multi-particle states gets modified by interactions. So, I'll edit the answer once again. – user566 Feb 9 '11 at 19:12
@Moshe: So it's clear that in the non-interacting limit, the distribution is Bose-Einstein in the single particle sector, and the multi-particle sector is a simple tensor product. In practise one might need some volume regularisation to get answers which were finite. With some interaction, the multi-particle sector is now Boltzmann, but cannot be written as a tensor product. Presumably it's sensible to ask "what is the overlap?", and conjecture that to leading order it would be a Bhahha scattering/$\alpha^4$ (?) correction from unity. Of course, there is a chance that it just doesn't converge. – genneth Feb 10 '11 at 11:25

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