# In optics, how does the vacuum state compare to thermal radiation?

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?

-

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).

-