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This should be a very easy question. If you look at the bottom of "Identical Particles" in Wikipedia, you see Table 1, which gives you the two particle statistics for bosons, fermions and distinguishable particles. The problem is to extend this table for three, four and five particles, or give an equivalent formula. I think I know what the answer is for fermions and distinguishable particles, but what about bosons?

My first guess is it would always be evenly distributed, but the many body limit does not equal the classical result, so that can't be right. Or can it?

What is the formula, and the first few rows of the "Pascals triangle for bosons"?

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Dear Jim, it depends on how many different states - boxes - you have for those particles. If you still have 2 states, just like in the Wikipedia example, then for fermions, the probability is 0 everywhere - it's impossible to put more than 2 fermions to 2 states.

For distinguishable particles, each particle has 50% odds to be in state 0 and 50% odds to be in state 1. So for $N$ distinguishable particles, there is $(N {\rm\, \, choose\, \,} k)/2^N$ chance for $k$ of them to be in state 0 and the remaining $N-k$ of them to be in state 1. Note that the odds sum to one, if you sum over $k$ from $0$ to $N$.

For $N$ bosons, all separations to state 0 ($k$ particles there) and state 1 ($N-k$ particles there) are possible, and each of those separations only corresponds to one multi-particle state. So the odds are $1/(N+1)$ for each possibility - a direct generalization of the table that has $1/3$ in all columns for the $N=2$ bosons. So yes, this is probably what you mean by "equally distributed".

The Fermi-Dirac and Bose-Einstein distributions - and both of them, with the same chances - can only be reduced to the Maxwell-Boltzmann distribution if the number of particles is high (it's not really essential) but, more importantly, the number of states in which they can be found is even much higher. Whenever there is a significant chance that the particles want to share a state etc., the classical Maxwell-Boltzmann approximation is inapplicable. So the toy model with two states 0,1 clearly doesn't allow you to approximate the Bose-Einstein or Fermi-Dirac distrtibutions by the Maxwell-Boltzmann one. This is particularly clear for the fermions because the Fermi-Dirac probabilities can't even be calculated because there is no way to arrange more than two fermions to two states.

The Maxwell-Boltzmann distribution is applicable when the density of particles (number of particles divided by the number of states) is low. That's why the Maxwell-Boltzmann $\exp(-E/kT)$ is applicable as a replacement for high values of energy $E$ - because this is where a small number of particles will be found, so their arrangement won't be sensitive on their being bosons (collectivists who love siblings in the same state, more than the love between the distinguishable particles) or their being fermions (individualists who absolutely hate to share the state with others). For states at low $E$, that's where the concentration of particles per state is expected to be high, which is why you get the totally new quantum behaviors - such as the Bose-Einstein condensate in the ground state and/or the Fermi liquid - which obviously can't be described by the distinguishable statistics well.

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