Why does a quantum gas lose its "quantum nature" in the limit $(\epsilon-\mu)/k_BT\gg1$? - Physics Stack Exchange most recent 30 from physics.stackexchange.com 2019-08-22T22:51:27Z https://physics.stackexchange.com/feeds/question/350722 http://www.creativecommons.org/licenses/by-sa/3.0/rdf https://physics.stackexchange.com/q/350722 4 Why does a quantum gas lose its "quantum nature" in the limit $(\epsilon-\mu)/k_BT\gg1$? SRS https://physics.stackexchange.com/users/36793 2017-08-07T08:27:40Z 2018-01-26T13:14:51Z <p>Mathematically, the <em>Fermi-Dirac</em> (FD) distribution and <em>Bose-Einstein</em> (BE) distribution coincides with the <em>Maxwell-Boltzmann</em> (MB) distribution in the limit $(\epsilon-\mu)/k_BT\gg 1$. Therefore, in this limit the “quantum nature” of the particles i.e., the <em>indistinguishability</em> must be lost. What is <em>physically</em> happening inside the system <em>in this limit</em> and <em>away from</em> it? </p> <p>Roughly I can understand that if the indistinguishability is lost then the counting of microstates becomes classical. The comparison of <em>interparticle separation</em> and <em>thermal de Broglie wavelength</em> 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.</p> https://physics.stackexchange.com/questions/350722/-/382401#382401 1 Answer by By Symmetry for Why does a quantum gas lose its "quantum nature" in the limit $(\epsilon-\mu)/k_BT\gg1$? By Symmetry https://physics.stackexchange.com/users/48856 2018-01-26T13:14:51Z 2018-01-26T13:14:51Z <p>Firstly lets be clear about what the limit $\frac{\epsilon-\mu}{k_BT}\gg 1$ means. $\epsilon$ here is the <em>single particle energy</em>. 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 <em>much larger than the average</em>. This is a very different question to "what happens to quantum gasses at high temperatures?", where you do again recover the Maxwell-Boltzmann distribution. </p> <p>The next thing to look at is what is the difference between the Bose, Fermi and classical gases. A hand-wavy answer is that they disagree on the answer to the question "What happens if 2 particles try to occupy the same state?" This is essentially because if 2 particles are occupying different states you can distinguish those states, so the indistinguishably of the particles doesn't really matter. This means that if particles are only rarely trying to occupy the same state the distributions will generally give similar answers.</p> <p>So why do the distributions agree in the high energy limit? Well the vast majority of particles have an energy in the ballpark of the average energy, so if you look at states with an energy much higher than the average then there are very few particles competing for those states, so the question of trying to multiply occupy a state rarely arises, and all three distributions give essentially the same answer. </p>