Wave functions overlapping does not necessarily mean that they are at the same position. Consider plane waves. They would overlap if their wave vectors have approximately equal values, which would mean that their momenta are approximately the same. But, a plane wave is extended throughout the whole volume, so you cannot say the particles are nearby.
You should not ascribe too much weight to their 'condition'. It simply comes from the math. Namely, consider Fermi-Dirac or Bose-Einstein distribution:
<br />
\overline{n}_{i} = \frac{1}{\exp\left(\frac{\varepsilon_{i} - \mu}{T}\right) \pm 1}<br />
They would both reduce to the Maxwell-Boltzmann distribution:
<br />
\overline{n}_{i} = \exp\left(\frac{\mu - \varepsilon_{i}}{T}\right)<br />
when:
<br />
\exp\left(\frac{\varepsilon_{i} - \mu}{T}\right) \gg 1<br />
The left hand side is a monotonically increasing function of energy. So, it has a lower bound for the lowest possible energy of the particles. We may take this to be zero. Then, we still have to satisfy the condition:
<br />
\exp\left(-\frac{\mu}{T}\right) \gg 1 \Leftrightarrow \exp\left(\frac{\mu}{T}\right) \ll 1<br />
The chemical potential needs to be negative and equal several times the temperature by magnitude for this inequality to be satisfied (because of the exponential). The chemical potential is fixed essentially by the density of particles. Let us evaluate it for a free ideal gas under the Maxwell-Boltzmann distribution: Then, the plane waves are energy eigenstates:
<br />
N = \sum_{i}{\overline{n}_{i}} = \sum_{\mathbf{k}}{\exp\left(\frac{\mu - \varepsilon_{k}}{T}\right)}<br />
<br />
N = \exp\left(\frac{\mu}{T}\right) \, \sum_{\mathbf{k}}{\exp\left(-\frac{\hbar^{2} k^{2}}{2 m T}\right)}<br />
Going over from summation to integration with respect to \mathbf{k} according to the well-known rule:
<br />
\sum_{\mathbf{k}}{f_{\mathbf{k}}} = V \, \int{\frac{d^{3} k}{(2\pi)^{3}} f(\mathbf{k})}<br />
we get:
<br />
N = V \, \exp\left(\frac{\mu}{T}\right) \, \frac{4\pi}{(2\pi)^{3}} \, \int_{0}^{\infty}{k^{2} \, \exp\left(-\frac{\hbar^{2} k^{2}}{2 m T}\right) \, dk}<br />
The integral can be made dimensionless by introducing:
<br />
x \equiv \frac{\hbar^{2} k^{2}}{2 m T} \Rightarrow k = \frac{(2 m T x)^{\frac{1}{2}}}{\hbar}, \; dx = \frac{(2 m T)^{\frac{1}{2}} \, x^{-\frac{1}{2}}}{2 \hbar} \. dx<br />
<br />
N = V \, \exp\left(\frac{\mu}{T}\right) \, \frac{1}{2\pi^{2}} \, \frac{(2 m T)^{\frac{3}{2}}}{2 \hbar^{3}} \, \int_{0}^{\infty}{x^{\frac{1}{2}} \, e^{-x} \, dx}<br />
The value of the integral is \Gamma(\frac{3}{2}) = \pi^{1/2}/2[/tex] and we can identify n \equiv N/V. Then, we get for the chemical potential:<br />
<br />
<br />
\exp\left(\frac{\mu}{T}\right) = n \, \left(\frac{2 \pi \hbar^{2}}{m T}\right)^{\frac{3}{2}} \ll 1<br /><br />
<br />
We identify:<br />
<br />
<br />
\lambda_{T} = \left(\frac{2 \pi \hbar^{2}}{m T}\right)^{\frac{1}{2}} = \frac{h}{\sqrt{2 \pi m T}}<br /><br />
<br />
as the <i>thermal De Broglie wavelength</i> and r = n^{-1/3}[/tex] as the average interparticle distance. Then, the above condition can be written as:<br />
<br />
&lt;br /&gt;
\exp\left(\frac{\mu}{T}\right) = \left(\frac{\lambda_{T}}{r}\right)^{3} \ll 1 \Leftrightarrow \lambda_{T} \ll r&lt;br /&gt;<br />
<br />
, i.e. we can use classical statistics (the gas is non-degenerate) if the thermal De Broglie wavelength is much smaller than the average inter particle distance. I guess this is why they interpret it as their wavefunctions (whose wavelengths are of the order of \lambda_{T}) should not &#039;overlap&#039;.
