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In quantum optics, a perfect absorber of light is said to emit the "vacuum field". In practice, any beam dump will be at finite temperature, so it will emit blackbody radiation. How do these fields compare? Is there some critical frequency (for a given temperature) above which the vacuum field dominates?

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The vacuum state is the thermal state for $T=0K$. How to compare if a state is close enough to the vacuum state? By counting photons (for vacuum it is zero). The occupation for photons is given by Bose-Einstein distribution:

$$n = \frac{1}{\exp( E/(kT)) - 1},$$

where $E$ is the photon energy ($E = \hbar \omega = h \nu$) and $k$ is the Boltzmann constant. For room temperature ($T\approx 300K \Rightarrow kT \approx 0.025eV$) and visible light ($E\approx 2.5eV$) it gives $$n\approx 10^{-44}$$ that is, very very few (and in practice - the vacuum state).

See a gallery of Wigner functions (the so-called Winger function illustrates fluctuation of electrical field around 0; note that even fore $T=0$ there are some fluctuations (zero-point energy fluctuations), but for $T>0$ there are higher).

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The vacuum state is not at all like a thermal distribution. The "vacuum field" is just the zero-point fluctuations of the various modes of the electromagnetic field, with each mode containing half of the fundamental energy for that mode. A thermal distribution has the modes populated with a number of photons following a Planck distribution.

For the frequencies that most quantum optics experiments work with, this makes almost no difference. A black body at room temperature has the peak of its black-body spectrum at just under 10 microns, which is a photon energy of about 0.13 eV. Most quantum optics experiments are done at optical or near-infrared frequencies-- the fundamental mode of a YAG laser at 1.06 microns is a common wavelength used, and that's got a photon energy of 1.17 eV. The photon occupation number for that mode would be a factor of e^{-9} lower than the peak value (about four orders of magnitude), which is pretty insignificant. And a lot of quantum-optics experiments use visible light sources, with a frequency almost twice as high as the YAG light, with even less radiation in the thermal modes.

This is an issue for some experiments, particularly those in the microwave regime where the black-body contribution at room temperature is significant. People who do quantum optics experiments in the microwave regime generally do them at cryogenic temperatures (liquid nitrogen or lower) for just this reason, and that's enough to eliminate the problem.

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