Quantum and non-equilibrium properties of the radiation, emitted by "thermal light sources"

Question

Free photon gas
Let us consider different thermal light sources, such as the Sun, an incandescent lamp, or a fluorescent bulb. In elementary quantum mechanics and statistical physics one describes the emission of such light sources using 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 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 eigenstates of the Hamiltonian are however no more the photon eigenstates, but rather polaritons - joints states of 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 equilibration 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 equilibration 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 if 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 negligible for practical purposes. Moreover, the light sources such as incandescent 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, incandescent 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.

Answer 1

This is a very interesting problem. The short answer is that the textbook expression for a thermal state is not what you would have from any physics device that can produce a thermal state in your lab. When we compute the energy in the textbook thermal state we find that it has an infinite amount of energy.

So, thermal states that are physically realizable need more information in their descriptions. In addition to the temporal information represented by the wavelength spectrum, one also needs to incorporate spatial information. A simple exercise to demonstrate this situation is to considered an entangled state produced in spontaneous parametric down-conversion. When one of the two beams is traced out, the other beam represents a thermal state, but this thermal state does not look like the textbook case.

To write down the most general form of a thermal state, one needs a formalism that is more versatile so that it can capture all the spatiotemporal degrees of freedom. I prefer to use a Wigner functional approach for this purpose. The Wigner functional for a general thermal state is then given by $$ W_{\Theta}[\alpha] = \mathcal{N}_0 \det\{\tau\} \exp\left(-2\alpha^*\diamond\tau\diamond\alpha\right) , $$ where $\mathcal{N}_0$ is a normalization constant, $\tau$ is a kernel that defines all the spatiotemporal properties of the thermal state, the $\alpha$'s are field variables of the functional phase space, and the $\diamond$-contractions imply integrations over shared wave vectors, so that $$ \alpha^*\diamond\tau\diamond\alpha = \int \alpha^*(\mathbf{k}_1) \tau(\mathbf{k}_1,\mathbf{k}_2) \alpha(\mathbf{k}_2) \frac{d^3 k_1}{(2\pi)^2} \frac{d^3 k_2}{(2\pi)^2} . $$

This representation of a thermal state now allows one to model the physical thermal state more accurately, by specifying its spatiotemporal properties in $\tau$. Hopefully this addresses your question. Please let me know if more information is required.

Answer 2

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.

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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.