Why do quantum effects of particles dominate when the thermal de Broglie wavelength becomes comparable to the inter-particle spacing? [duplicate]

Question

Why do quantum effects of particles dominate when the thermal de Broglie wavelength becomes comparable to the inter-particle spacing?

Answer 1

As was stated in the comments, 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.