Why does quantum mechanics become unnecessary at sufficiently high temperatures? In my statistical mechanics intro class, we are taught that at sufficiently high temperatures, the quantum treatment of things becomes unnecessary. Why is this? Can this be shown using certain equations? 
 A: Qualitatively, quantum effects can be ignored if the interchange properties of boson or fermions can be ignored, which is to say if the system is dilute.  When will a system be dilute?  A single particle occupies a volume of its thermal de Broglie wavelength cubed (in three dimensions).  For massless particles, the thermal de Broglie wavelength must go as $\lambda_{th}\sim1/T$ by dimensional analysis.  For massive particles, the thermal de Broglie wavelength goes as $\lambda_{th}\sim1/\sqrt{mT}.$  Thus for both massless and massive particles, as the temperature increases, the volume occupied by any one particle decreases and quantum effects become less and less important.
To flesh this out further, I'll essentially follow 8.4 and 8.5 of Baierlain's Thermal Physics book.  
The number distribution of particles is either Fermi or Bose, $$n(\varepsilon)\propto \frac{1}{e^{(\varepsilon-\mu)/T}\pm1},$$ where, for simplicity, I'm using units in which Boltzmann's constant $k_B=1$.
The "quantumness" of the distributions is in the $\pm1$, which is to say that the quantumness can be ignored when $e^{(\varepsilon-\mu)/T}\gg1$.  Then these two distributions can be well approximated by the classical number distribution of particles, the Maxwell distribution, $$n(\varepsilon)\propto e^{-(\varepsilon-\mu)/T}.$$
In the case that $e^{(\varepsilon-\mu)/T}\gg1$ we see that $n(\varepsilon)\ll1$, which is to say that the system has low occupation number.
When will a system have low occupation number?  Qualitatively, when the system is dilute. 
 Baierlein derives the following expression for the average energy of a semi-classical system, i.e. one in which quantum effects are only a small correction.  In three dimensions, $$\langle E \rangle = \frac{3}{2}NT\Big(1\pm\frac{1}{2^{5/2}}\frac{N \lambda_{th}^3}{(2s+1)V}\Big),$$ where $N$ is the number of particles, $s$ is their spin, $V$ is the volume of the system, and $\lambda_{th} = 1/\sqrt{2\pi mT}$.
One can see, then, that quantum effects can be ignored when $\frac{1}{2^{5/2}}\frac{N \lambda_{th}^3}{(2s+1)V}\ll1$, i.e. when $$T\gg\frac{1}{2\pi m}\Big( \frac{N}{2^{5/2}(2s+1)V} \Big)^{2/3}.$$ The denser the system---the large is $N/V$---the higher the temperature must be for one to ignore quantum effects.
