From Probability Theory Vol. 1 Feller Section 2 Chapter 5:

Maxwell-Boltzaman distribution: consider $r$ indistinguishable balls and $n$ cells. Assuming that all $n^r$ possible placements are equally probable, the probability to obtain he given occupancy number $r_1,...,r_n$ equals $\frac{r!}{r_1!...r_n!}n^{-r}$.

However, today fermions are said to obey Fermi–Dirac statistics and bosons are said to obey Bose–Einstein statistics; no particles are said to obey Maxwell–Boltzmann statistics except at limits.

Question: Is there an intuitive reason that Maxwell–Boltzmann statistics aren't used to model particle behavior in general cases?

  • $\begingroup$ What exactly are you asking? If there is a reason for Maxwell-Boltzmann to not really apply or for a proof of the equation you wrote down? $\endgroup$ Commented Aug 6, 2018 at 5:46
  • $\begingroup$ What modern theory? $\endgroup$ Commented Aug 6, 2018 at 5:52
  • $\begingroup$ @GabrielGolfetti I am asking why the 'common-sense' Maxwell-Boltzmann distribution does not apply to any known particle $\endgroup$ Commented Aug 6, 2018 at 11:36

2 Answers 2


The thing is that Maxwell-Boltzmann does hold for any system. Quantum particles, however, are not only identical, but completely indistinguishable. For such particles, the distribution is characterized solely by the occupation number.

Consider a system of noninteracting identical particles. Suppose the single particles have a known Hamiltonian $H$ with eigenstates $|k\rangle$, such that


If we have a bunch of say, $N\gg1$ of these particles, we write the quantum state of the whole system in a different basis (the Fock basis), given by the number of particles in state $k, n_k$:


And this basis completely characterizes our quantum state. These are eigenstates of the many body Hamiltonian, with


Suppose then that our ensemble of particles has a state probability


Maxwell-Boltzmann statistics is generated by maximizing the Gibbs Entropy: $$ S=-\sum\limits_{n_1,n_2,...}p(n_1,n_2,...)\log p(n_1,n_2,...),$$

Where $\log$ is the natural logarithm. We can then do it subject to the constraints of normalization, total energy and total number of particles:


The maximization condition then corresponds to (using Lagrange multipliers, the calculation is a little involved)

$$p(n_1,n_2,...)=\prod\limits_k\frac{e^{-\beta n_k(E_k-\mu)}}{\sum\limits_{n_k'}e^{-\beta n_k'(E_k-\mu)}},$$

where $\beta$ and $\mu$ are parameters corresponding to the inverse temperature, $(k_BT)^{-1}$ and chemical potential. If we take the marginal distribution for a single occupation number, say, $n_j$, by summing over all other possible values for all the other occupation numbers, we get the probability distribution for the occupation number in question:

$$p_j(n)=\frac{e^{-\beta n(E_j-\mu)}}{\sum\limits_{n'}e^{-\beta n'(E_j-\mu)}}.$$

The step further taken by Bose-Einstein and Fermi-Dirac statistics is now computing the average, or expected occupation number of the quantum state $j$:

$$f_j=\langle n_j\rangle=\sum\limits_nnp_j(n).$$

A nice computational trick is then writing $f_j$ as follows:

$$f_j=\frac{\sum\limits_nne^{-\beta n(E_j-\mu)}}{\sum\limits_{n'}e^{-\beta n'(E_j-\mu)}}$$

$$f_j=\frac{1}{E_j-\mu}\frac{\sum\limits_nn(E_j-\mu)e^{-\beta n(E_j-\mu)}}{\sum\limits_{n'}e^{-\beta n'(E_j-\mu)}}$$ $$f_j=-\frac{1}{E_j-\mu}\frac{\partial}{\partial\beta}\log\left(\sum\limits_ne^{-\beta n(E_j-\mu)}\right).$$

As such, if we can compute that logarithm on the right, we can find $f_j$. Here, we need to make an assumption.


For bosons, $n$ can be any integer from $0$ to $\infty$. As such, we write

$$\sum\limits_{n=0}^\infty e^{-\beta n(E_j-\mu)}=\frac{1}{1-e^{-\beta(E_j-\mu)}},$$ a convergent geometric series. We now plug this into our expression for $f_j$ and get

$$f_j=\frac{1}{e^{\beta(E_j-\mu)}-1},$$ which is Bose-Einstein statistics.


For fermions, $n$ can be either 0 or 1. We then write the sum over $n$ as


Which then gives us


The occupation number of Fermi-Dirac statistics.


I just realized that I got ridiculously deep into the mathematics and forgot to anwer the question. The intuition behind all these calculations.

Thing is, in its essence, Maxwell-Boltzmann statistics is concerned with system states, and Bose-Einstein/Fermi-Dirac are concerned with particle states. It's easy to get both these concepts mixed up, but they are fundamentally different. While M-B gives us the probability for our system (all of our $N$ particles) to be in a specific configuration, B-E/F-D gives us the average number of particles that are in a specific state.

Since, by definition, these are averages, we have to average over all configurations of the system. And that is exactly what we did here.

It's also nice to mention that we can compute $\mu$ by solving the equation


  • $\begingroup$ > "Quantum particles, however, are not only indistinguishable, but completely identical." You got that backwards. Already in classical theory electrons are identical particles, but they are usually distinguishable. It is only in quantum theory that permuted states of identical particles are considered as indistinguishable. $\endgroup$ Commented Aug 6, 2018 at 21:02
  • $\begingroup$ @JánLalinský Oops, right. Fixed. $\endgroup$ Commented Aug 6, 2018 at 21:17

It does not work for quantum particles beacause Boltzmann statistic is written for classical particles that are distinguishable, in contrast, bosons and fermions that are quantum particles, are indistinguishable.

First of all, what means that the Boltzmann particles are distinguishable?
It means that we might identify each particle by continually observing their trajectories, beacause the motion induced by the Hamilton equation is deterministic.
In other words, imagine the you can mark a specific particle with a red target, then you will exactly predict the motion of this particle by solving the associated Hamilton equations and you can really observe and recognize this red particle with respect to each others.
Another way to say that two particles are distinguishable, given two particles of the same kind labelled by A and B, is to say that the configuration of particle A in state $a$ and particle B in state $b$ is different from the case in which particle B is in state $a$ and particle A is in state $b$.
This assumption yields non-physical results for the entropy, like you can see from the Gibbs Paradox problem (hope you know what i'm talking about).

Now, what is the practical meaning of the fact that in one case the particles are distinguishable and in the other case they are not?
From a mathematical point of view, the difference is given by the number of different microstates $W$ (hope you know what is it) that a system can occupy. The computation of the number of microstates of a system, let's say, of $N$ particles, is a combinatorial problem that gives to you different results if you prescribe that the particles are distinguishable or not. This explain why the different kind of statistics are different.
Obviously, the Boltzmann one is computed with the classical prescription on distinguishable particles and is not correct. Is experimentally proved that elementary particles (i.e electrons ecc..) are absolutely undistinguishable, and so the conclusion is that there are no real particles which have the characteristics required by Maxwell–Boltzmann statistics.

  • $\begingroup$ Umm Maxwell-Boltzmann is used for indistinguishable balls (particles) - for example, according to Maxwell-Boltzmann statistics, if we have 3 particles and 3 possible energy states (theoretically), there is a $\frac{1}{9}$ probability of the occupancy number 1,2.0 $\endgroup$ Commented Aug 6, 2018 at 11:38
  • $\begingroup$ probably i have not explained it in a very good way, now i edit. $\endgroup$
    – MRT
    Commented Aug 6, 2018 at 13:01
  • $\begingroup$ @Daniele1234, a simple example: you have $r$ dollars, which you can spend in $n$ different ways (food, entertainment, transportation, etc.). Each dollar has a unique identifier (series and number). The allocation of spending will obviously be described by Bose–Einstein statistics , Maxwell-Boltzmann has no sense here, so particles have no unique identifier... $\endgroup$ Commented Aug 7, 2018 at 10:01

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