# Why does a quantum gas lose its “quantum nature” in the limit $(\epsilon-\mu)/k_BT\gg1$?

Mathematically, the Fermi-Dirac (FD) distribution and Bose-Einstein (BE) distribution coincides with the Maxwell-Boltzmann (MB) distribution in the limit $(\epsilon-\mu)/k_BT\gg 1$. Therefore, in this limit the “quantum nature” of the particles i.e., the indistinguishability must be lost. What is physically happening inside the system in this limit and away from it?

Roughly I can understand that if the indistinguishability is lost then the counting of microstates becomes classical. The comparison of interparticle separation and thermal de Broglie wavelength reveals that the quantum mechanical nature of a quantum gas is lost at high temperature. However, this is apparently in contradiction with the limit $(\epsilon-\mu)/k_BT\gg 1$ at which a quantum gas becomes classical.e., the BE and FD distributions go over to MB distribution. This limit says for a quantum gas to behave classically, the temperature has to be low! But this is opposite to what the case usually is-a gas behaves quantum mechanically at low temperatures.

Firstly lets be clear about what the limit $\frac{\epsilon-\mu}{k_BT}\gg 1$ means. $\epsilon$ here is the single particle energy. In other words this is not looking at some gas where "the energy of the gas is much greater than its temperature" (what would that even mean?) we are taking a quantum gas at any temperature you like and looking at those particles with an energy much larger than $k_bT$, in other words those particle with an energy much larger than the average. This is a very different question to "what happens to quantum gasses at high temperatures?", where you do again recover the Maxwell-Boltzmann distribution.