This Wikipedia article states that at sufficiently low temperatures (or high pressure), quantum effects lead to a breakdown of Maxwell–Boltzmann statistics.

I know that the spacing of energy levels becomes larger at low temperatures, but doesn't the derivation of the MB distribution, as shown in the article, already consider energy levels to be discrete? Furthermore, it also takes the degeneracy of energy levels into account.

So why and what other quantum effects are there at low temperatures that still violate the formula?

  • $\begingroup$ Have you checked the approximations it is based on? $\endgroup$
    – Greg
    Mar 4, 2019 at 5:06
  • 2
    $\begingroup$ @JohnnyGui the wiki also has a section describing the limits of applicability and that it is related to the fact that actual particles are indistinguishable. $\endgroup$
    – Tyberius
    Mar 4, 2019 at 5:36
  • $\begingroup$ In a quantum system, such as an atom or molecule, the spacing of levels is a property of the system being studied and not of temperature. Rather it is that there is not enough energy at low temperatures to populate many of the levels, and just the lowest one at zero Kelvin. $\endgroup$
    – porphyrin
    Mar 4, 2019 at 8:16
  • $\begingroup$ @porphyrin Why is it a problem that lower energy levels get populated by many particles in the case of a low temperature? How does this violate the MB distribution? $\endgroup$
    – Phy
    Mar 4, 2019 at 17:59
  • $\begingroup$ @Tyberius I have indeed read that part. What I don't get though is why particles become all of a sudden indistinguishable in the case of low temperature and/or high density $\endgroup$
    – Phy
    Mar 4, 2019 at 17:59

2 Answers 2


Since the explanation was a little more complicated than I initially thought, I figured it would be worth it to combine my comments (and info from Physics SE) into an answer.

Quantum particles satisfy Fermi–Dirac or Bose–Einstein statistics depending on whether they are fermions or bosons. These distributions have the form $$\langle n_i\rangle=\frac{1}{\exp[(\epsilon_i-\mu)/k_bT]\pm1}$$ where the plus/minus is for fermions/bosons.

To consider the high temperature limit, we need to note that not only is there a direct temperature dependence, but $\mu$ is also dependent on temperature. Specifically, in the high temperature limit, $\mu<0$ and $|\mu|>k_bT$. Combining this information, we can make the (very accurate) approximation $$\exp[(\epsilon_i-\mu)/k_bT]\pm1\approx\exp[(\epsilon_i-\mu)/k_bT]$$ as the exponential function will be much larger than $1$. So at high temperatures $$\langle n_i\rangle\approx\frac{1}{\exp[(\epsilon_i-\mu)/k_bT]}$$ which matches the form of the Maxwell–Boltzmann distribution. We can see that at low temperatures these two distributions would not agree, as the additional $\pm1$ term in the denominator would become more significant as the exponential got smaller.

It's important to remember that the particles are always indistinguishable; all we have done in the high temperature limit is made an approximation that simplified the functional form. We should not take this coincidental agreement with the MB distribution (for which the particles are assumed to be distinguishable) to imply that the quantum particles have become distinguishable at high temperatures.

  • $\begingroup$ Thanks for your answer. So are you saying that the whole derivation of the MB Distribution by maximizing the number of permutations for the most probable macrostate, and thus considering the particles to be distinguishable, is coincidentally similar? $\endgroup$
    – Phy
    Mar 5, 2019 at 2:09
  • $\begingroup$ @JohnnyGui The derivations of the distributions from various ensembles are materially very similar as they make a lot of the same assumptions. I guess what I'm saying is that just because the approximation of FD/BE looks like MB doesn't mean it represents the same picture of reality (distinguishable particles) as MB. $\endgroup$
    – Tyberius
    Mar 5, 2019 at 2:29
  • $\begingroup$ Thanks a lot for clarifying. Part of my question is also about what quantum effects are present in the case of low temperature that makes the MB distribution inaccurate, perhaps you can help me with that as well? $\endgroup$
    – Phy
    Mar 5, 2019 at 2:45
  • $\begingroup$ @JohnnyGui I don't know if there are additional quantum effects, at least in these models. All of them ignore interactions between particles. As you mentioned, MB is quantum as well as it discretizes the energy levels. The big difference is really just that the other distribution deal with explicitly quantum particles, that is fermions and bosons, and how those behave. Beyond that, I don't know and you may need to go to the Physics SE for a deeper dive into this. $\endgroup$
    – Tyberius
    Mar 5, 2019 at 3:09

The Maxwell–Boltzmann distribution (MBD) is inextricably tied to ideal gas behavior, so where ideal gas behavior fails so will the MBD. It isn't just at low temperatures that the MBD fails, since the MBD will also fail at "very high" temperatures because the gases ionize. The gist is that to consider quantum effects you have to go to a more complicated model than the MBD since the MBD explicitly ignores quantum mechanics.

The situation is somewhat like comparing Newton's theory of gravity and Einstein's theory of relativity. So:

  • The MBD was known before quantum mechanics was figured out, so the classical derivation doesn't use quantum mechanics.
  • The MBD is a continuous function. The "real" distribution would have to be quantized because of the implications of the Plank constant, so it be a histogram rather than a continuous function.
  • The MBD doesn't ignores gravity thus doesn't predict that a massive enough gas cloud would collapse due to gravitational attraction.
  • Fermi–Dirac statistics and Bose–Einstein statistics are two quantum distributions that are opposites in the sense that Fermi–Dirac statistics applies to fermions and Bose–Einstein statistics applies to bosons. However the MBD ignores this bifurcation of particle types.
  • Fermi–Dirac statistics and Bose–Einstein statistics are discrete distributions, but in the limit (ie lots of particles, energies great enough for implications of the Plank constant not to apply) reduce to the MBD which is a "continuous" distribution. (Think of the binomial distribution approximation of the Gaussian distribution in statistics.)
  • A Bose-Einstein condensate, where Bose-Einstein statistics apply, isn't really a gas, but rather it is another state of matter.
  • The MBD requires particles to be distinguishable, and that many particles can have the same energy state.
  • Bose-Einstein Statistics requires particles to be indistinguishable, and that many particles can have the same energy state.
  • Fermi–Dirac statistics requires particles to be indistinguishable, and that no two particles can occupy the same state.
  • The MBD, as well as Fermi–Dirac statistics and Bose–Einstein statistics, ignore relativity in which the mass of a particle seems to increase as the velocity of the particle increases.
  • The MBD, as well as Bose–Einstein statistics, fail at very high temperatures because the gas starts to ionize.
  • $\begingroup$ Thanks for the clear explanation. I have seen another reason that could break down MBD, which is the fact that at low temperature, the energy is too low and therefore not many energy levels get populated. To which of your mentioned explanations does this reason belong? $\endgroup$
    – Phy
    Apr 24, 2019 at 10:41
  • $\begingroup$ When using at low temperatures Bose-Einstein Statistics has energy states that aren't "continuous" but rather the energy states have discrete energies like a histogram. Multiple particles can be in each energy level. So no more a few levels need be occupied. $\endgroup$
    – MaxW
    Apr 24, 2019 at 15:42
  • $\begingroup$ Are the steps between low energy levels larger in terms of absolute difference? Or is there another factor that causes this "discreteness" at low temperature? $\endgroup$
    – Phy
    Apr 24, 2019 at 15:49
  • $\begingroup$ Bose-Einstein Statistics use the Boltzmann constant, which in turn depends on the Plank constant. $\endgroup$
    – MaxW
    Apr 24, 2019 at 16:08

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