We say that when the thermal de Broglie wavelength is much smaller than the interparticle distance, the gas can be considered to be a classical or Maxwell–Boltzmann gas. On the other hand, when the thermal de Broglie wavelength is on the order of or larger than the interparticle distance, quantum effects will dominate and the gas must be treated as a Fermi gas or a Bose gas, depending on the nature of the gas particles. $(V/N)^{1/3} < \lambda_{th} $. My question is why do quantum effects dominate?


This can be related to the uncertainty principle. The essentials are given on the thermal de Broglie wavelength Wikipedia page, but I'll try to give a little more detail.

We define a "typical" atomic momentum at temperature $T$ as $$ p_{\text{th}} = \sqrt{2\pi mk_BT} $$ where $m$ is the atomic mass, and $k_B$ is Boltzmann's constant. This is approximately equal to the root-mean-square value of any component of the momentum, say $p_x$, calculated from the classical statistical mechanics result (equipartition of energy) as $$ \langle p_x^2\rangle = m k_B T $$ In this kind of argument, numerical factors such as $\sqrt{2\pi}$ are not important.

From this, using de Broglie's relation, we can define a typical thermal wavelength of atoms $$ \lambda_{\text{th}} = \frac{h}{p_{\text{th}}} = \frac{h}{\sqrt{2\pi mk_BT}} $$ where $h$ is Planck's constant. This is the main quantity of interest here.

This also serves as an estimate of the uncertainty in the atom's position. We expect most atoms to have a momentum in the range $-p_{\text{th}}\ldots+p_{\text{th}}$, so we roughly set $\Delta p\approx p_{\text{th}}$ and the uncertainty principle gives us $$ \Delta x \approx \frac{\hbar/2}{\Delta p} \approx \lambda_{\text{th}} $$ where again we don't worry too much about numerical factors.

Then, if we had a one-dimensional system of $N$ atoms on a line of length $L$, so the typical separation between atoms is $d=L/N$, we would expect to be able to neglect quantum mechanical effects provided $\Delta x \ll d$, i.e. $\lambda_{\text{th}} \ll d$. The atoms could be treated as points, not significantly spread out by quantum mechanics. A similar argument in 3D leads to the condition $\lambda_{\text{th}}^3 \ll V/N$ or $N\lambda_{\text{th}}^3/V \ll 1$. So this is the "hand-waving" answer to your question.

It is possible to take this further, and this is done in statistical mechanical textbooks such as Statistical Mechanics and Applications in Condensed Matter by C Di Castro and R Raimondi. It's a bit of work (see their chapter 7), but the equation of state for an ideal gas obeying Fermi-Dirac or Bose-Einstein statistics can be expressed, in the regime where the classical description is just starting to break down, as an expansion $$ \frac{PV}{Nk_BT} = 1 \pm \frac{1}{2^{5/2}} \left(\frac{N\lambda_{\text{th}}^3}{V}\right) + \ldots $$ and the crucial ratio appears as the first-order correction term with a sign corresponding to whichever statistics applies, FD or BE.


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