the thermal de Broglie wavelength
/ duh-BROY /
How 'quantum' is a gas? Compare two lengths: how far apart the particles sit, and how big each particle's quantum wave-packet is at temperature T. The thermal de Broglie wavelength is that quantum size, roughly the de Broglie wavelength of a particle carrying a typical thermal momentum. When it is tiny compared with the spacing, the gas behaves classically; when the two become comparable, quantum statistics take over.
Precisely, lambda = h / sqrt(2 pi m k_B T), where h is Planck's constant, m the particle mass and T the temperature. Up to the 2 pi convention it is the de Broglie wavelength lambda = h/p for a momentum p of order sqrt(m k_B T) set by the thermal energy. It fixes the natural quantum volume lambda^3 per particle: the classical, Maxwell-Boltzmann regime is n lambda^3 << 1, dilute, hot and heavy, while quantum degeneracy sets in when n lambda^3 becomes of order one.
It is the single parameter that tells you when to use classical statistics rather than the Fermi-Dirac or Bose-Einstein distributions, and it appears throughout the ideal-gas partition function and the Sackur-Tetrode entropy, where the single-particle Z is V / lambda^3. Bose-Einstein condensation begins precisely when n lambda^3 reaches about 2.612. Honestly, lambda grows as the temperature falls, scaling like T^(-1/2), which is why cooling, and not compression alone, is the usual route into the quantum regime.
For helium atoms at room temperature lambda is about 0.05 nm, far smaller than the roughly 3 nm spacing in the gas, so it is classical; cool toward 1 K and lambda grows until n lambda^3 is of order one and quantum statistics, superfluidity and condensation, take hold.
The ratio of wave-packet size to spacing decides classical versus quantum behaviour.
Because lambda scales as T^(-1/2), lowering the temperature, not merely raising the density, is the practical way to reach the quantum-degenerate regime where n lambda^3 approaches one.