# Quantum and non-equilibrium properties of the radiation, emitted by “thermal light sources”

Free photon gas
Let us consider different thermal light sources, such as the Sun, a incadescent lamp, or a fluorescent bulb. In elementary quantum mechanics and statistical physics one describes the emission of such light sources using the Planck's law, which is readily obtained as (up to a definition-dependent coefficient) $$B(\omega, T) \propto \sum_{\mathbf{k},\mu}\delta(\omega-\omega_{\mathbf{k},\mu})\hbar\omega_{\mathbf{k},\mu}\langle a^\dagger_{\mathbf{k},\mu}a_{\mathbf{k},\mu}\rangle,$$ ($$\mu$$ are the photon polarization states), where the averaging is performed using the free-field density matrix (i.e., assuming the non-interacting photon gas) $$\hat{\rho} =\frac{1}{Z}e^{-\beta H_{ph}},Z= Tr\left[e^{-\beta H_{ph}}\right], H_{ph}=\sum_{\mathbf{k},\mu} \hbar\omega_k a^\dagger_{\mathbf{k},\mu}a_{\mathbf{k},\mu}$$ From this it is readily follows that

• the shape of the spectral density depends only on temperature, but not on the nature of the emission process
• the light emitted by different thermal sources of the same temperature is indistinguishable, as, e.g., discussed in this thread.

Light-matter interactions The thermodynamic description above ignores the source of radiation, which we will call "matter", and which can be trivially included in the Hamiltonian: $$H = H_{matter} + H_{ph}$$ Using this Hamiltonian would not change the properties of the black body radiation discussed above, but it brings to the surface the important assumption used in thermodynamic description: neglecting the residual interactions, which are responsible for equilibrating the state of the photons and the matter. In particular, if some of these interactions are not small, we can no longer neglect them, but need to include them in the Hamiltonian. E.g., In case of two-level atoms interacting with a photon gas we would write something like: $$H = H_{atoms} + H_{int} + H_{ph},\\ H_{atoms} = \Delta\sum_i\sigma_z^{(i)},\\ H_{int}=\sum_{i,\mathbf{k},\mu}\sigma_x^{(i)}\left(\lambda_{\mathbf{k},\mu}a_{\mathbf{k},\mu} + h.c.\right)$$ In this case the eignestates of the Hamiltonian are however no more the photon eigenstates, but rather polaritons - joints state sof atom-photon system. If we were to consider a partition function of such a system in the basis of the photon eigenstates, it would have non-zero diagonal elements.

Equilibration time Note that the interaction in the example above is not sufficient for the establishment of the thermal equilibrium, as most of the exchange will happen between the atoms and the photon modes with frequencies close to $$\hbar\omega_{\mathbf{k},\mu}=\Delta$$. To have thermal equiliubration we still have to consider weaker interactions, higher order processes (such as Raman scattering), Doppler broadening, etc. This is to say that the equilibration between the matter and the photons may be very slow. In fact, it may be divergently slow - such as the equlibration between the alternative polarizations of a ferromagnet (somewhat similar to the phase transition in Dicke model).

It is easy to point examples where this is the case: gas tubes produce light of different colors, depending on the gas they are filled with, even of their temperature is the same. Similarly, we are able to discern the chemical composition of stars by the intensity of different spectral lines in their radiation.

Non-equilibrium Thermal light sources are never in equilibrium, as they lose energy via emitting radiation. Thus, describing the radiation emitted by a heated slab of metal as a black-body radiation is valid only approximately, as long as we believe that this loss of energy is negligeable for practical purposes. Moreover, the light sources such as incadescent bulbs or starts are in a steady rather than a thermal state - the energy is generated inside of them (via electric current or thermonuclear reactions) and then dissipated as radiation.

Taken together, all these mean that the light originating from different types of "thermal light sources" (stars, incadescent lamps, fluorescent bulbs) possesses different properties and in principle distinguishable. The question is then: how can we do it? More specifically, I am interested in detecting the entangled photons and the non-equilibrium state.

• The spectra of fluorescent lamps are not like a black body spectrum, although modern phosphors are rather better than the older ones. See en.wikipedia.org/wiki/… – PM 2Ring May 19 at 9:13
• @PM2Ring Good point. The same can be said about many other light sources - e.g., stars have certain chemical compositions, so their spectrum is not really black-body radiation, even though the motion of the emitting atoms is thermal. – Roger Vadim May 19 at 9:22
• I’ve removed a comment discussion that was going in a not-great direction. Be kind, people. – rob Jun 8 at 21:08

## 1 Answer

The question seems largely circular to me, the problem being that a thermal photon state only depends on temperature by construction. The density matrix of the thermal photon field is only dependent on temperature as a whole, so digging into quantum correlations (such as photon correlation functions or higher order interference as entanglement witnesses) is not going to help.

Given this observation, if you introduce an emission Hamiltonian that allows you to resolve details of the emission process via measuring entanglement in the photon field, one conclusion is simply that the Hamiltonian does not produce a thermal state.

At that point, you are not really asking about thermal states any more, but about general quantum emission processes. The latter question is easy to answer: we can distinguish a multitude of emission processes. Going into quantum correlations is usually not even necessary, often spectroscopy techniques are sufficient.

As a practical side note: quantum correlations of light can be highly non-trivial to detect. Multi-photon detectors are still rather bad and usually require extensive technical setups that will like exclude them from being used in combination with a telecope setup to look at something like a star. Interference experiments can be used practically, the most famous example I know of being the Hanbury-Brown-Twiss effect, which is indeed used for increased resolution in astronomy and relies to some extent on thermal light being bunched.

As a simple example which the OP is probably aware of: we can determine elements in the sun by resolving the spectral structure of the light we receive on earth. While sunlight is largely thermal, it has missing absorption lines at certain energies.

If you want to express this well-known feature in terms of the language used in the question: the sunlight is not thermal at the spectral resolution of the absorption lines. The density matrix differs from thermal light at the frequencies that are absorbed. So you can identitfy the elements because you have something other than thermal light.

• What really interests me is how the structure of the photon density matrix depends on the emission mechanism, not only its diagonal elements. You can base the explanation on Cohen-Tannouji, Glauber or any other quantum optics text. – Roger Vadim May 29 at 8:37
• @RogerVadim Which emission mechanism are you interested in then? For thermal states, the density matrix is by definition diagonal in the energy basis. Or in different words: thermal light is spectrally incoherent. – Wolpertinger May 29 at 8:48
• You have mentioned stars, incadescent lamps and fluorescent bulbs. All of these work rather differently though. My answer is intended to point out that a thermal state is by definition not dependent on these details and what you are looking for are deviations from thermal behavior. You are not looking for quantum properties of thermal states, which are unambiguous. – Wolpertinger May 29 at 8:51
• Thermal state is an ad-hoc assimption. Assuming the thermal state and then claiming that there's no coherence in this state is precisely what you called circular logic. On the other hand, the emission process is clearly coherent. The question is thus whether/when/how the photon field becomes thermal. Equations could be helpful. – Roger Vadim May 29 at 9:07
• @RogerVadim I may have misunderstood your question then. Are you asking for emission Hamiltonians which precisely produce a thermal state and how deviations from a thermal state can be used to characterize the emission Hamiltonian? – Wolpertinger May 29 at 11:01