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The average occupation nr. of an energy level in a bosonic/fermonic gas is:

$$\langle n_i \rangle=\frac{1}{e^{\beta(\epsilon_i-\mu)}\pm 1}.$$

If we make the assumption that we have a quantum gas of low density or that the temperature is high enough so that the average distance that the particle can cover without colliding with another is larger then the thermal wave length (de-Broglie wave length), then we gain the average occupation nr. of an energy level for an ideal classic gas, which is:

$$\langle n_i \rangle=e^{-\beta(\epsilon_i-\mu)}.$$

Initially we consider an ideal quantum gas and later on an ideal classic gas. The difference is that in an ideal quantum gas, the particles are indistinguishable in difference from an ideal classic gas.

It appears to me that once the temperature is big enough of the density is small enough, then we somehow do not care about the type of the particles from which the gas is made of.

What is the argument made, that once we make the above mentioned changes, the particles (whether bosons or fermions) from indistinguishable become distinguishable. I am assuming that this is the difference between quantum and classic gas.

And a follow up question for an ideal classic gas:

What is the characteristics/quality that helps us distinguish the particles?

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The difference is that in an ideal quantum gas, the particles are indistinguishable in difference from an ideal classic gas.

The particles which make up a classical ideal gas are indistinguishable too. This is the origin of the Gibbs paradox, and of the standard factor of $1/N!$ which is added to the calculation of the classical partition function.

What is the argument made, that once we make the above mentioned changes, the particles (whether bosons or fermions) from indistinguishable become distinguishable. I am assuming that this is the difference between quantum and classic gas.

And a follow up question for an ideal classic gas: What is the characteristics/quality that helps us distinguish the particles?

There is no such characteristic or quality. The difference between classical and quantum statistics is not a matter of distinguishability vs indistinguishability - it is a matter of how carefully we account for indistinguishability when we're counting the microstates.

If we have $N$ particles which all occupy different states, then there are $N!$ ways to permute them. If the particles are indistinguishable, then all of those $N!$ configurations should really be counted as one, which means that we've overcounted by a factor of $N!$ and should divide by that factor to fix the error.

However, if some of the particles are occupying the same state, then this factor is no longer correct. For example, imagine that we have three particles and three distinct states which each particle may inhabit. It's easy to see that there are $3!=6$ ways to put them all in a different state.

On the other hand, imagine that these three particles only occupy two states, e.g. that two particles occupy state $A$ and the other particle occupies state $B$. In this case, there are only three possible configurations of the system, and so dividing by $3!$ is no longer appropriate. If all of the particles occupy the same state, then there is only one configuration of the system and we haven't overcounted at all.

In other words, the assumption that $N!$ is the correct overcounting factor amounts to the assumption that the number of configurations in which any of the states are multiply-occupied is negligible - which is the same as saying that the average occupancy of the single-particle states is $\ll 1$. That is,

$$\langle n_i\rangle = \frac{1}{e^{\beta(\epsilon_i-\mu)}\pm 1} \ll 1 \implies e^{\beta(\epsilon_i-\mu)}\pm 1 \gg 1$$

But this is precisely the circumstance under which $e^{\beta(\epsilon_i-\mu)}\pm 1$ is well-approximated simply by $e^{\beta(\epsilon_i-\mu)}$, from which we recover the classical Boltzmann factor.

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    $\begingroup$ If I do not misunderstood your first paragraph, then I think it is wrong, see for example my comment with several papers here. Classical identical particles are not indistinguishable (in the same sense as in QM). The Gibbs paradox is really misinterpreted very often; several reasons the claim made in most text books in statistical mechanics that we can avoid this with saying that particles are indistinguishable is wrong. I upvoted, tho, because the rest of this answer is very good. $\endgroup$ Feb 24, 2023 at 21:49
  • $\begingroup$ See also my (perhaps not very enlightening) comments on the most upvote answer here, which however contain one or two more very interesting papers. $\endgroup$ Feb 24, 2023 at 21:54
  • $\begingroup$ So actually the difference between the two types of ideal gases, has to do with whether particles can be in different energy states, or an arbitrary amount of them can occupy the same one particle state (and how many, for distinguishing between bosonic and fermionic gases). And this is how we explain the transition from quantum id. gas to classic gas? An increase in temperature, means the particles have higher prob. to be in an excited state, which translates, the prob. of having 2 particles in the same state diminishes ? $\endgroup$
    – imbAF
    Feb 24, 2023 at 21:54
  • $\begingroup$ How do you recover the classical Boltzmann factor from $e^{\beta (\epsilon_i - \mu)}$? The sign in the exponent is wrong. It implies that decreasing temperature leads to increase of expectead average of the occupation number $n_i$, which in Boltzmannian distribution makes sense only for the ground state... $\endgroup$ Feb 25, 2023 at 18:10
  • $\begingroup$ @JánLalinský The factor of $e^{\beta(\epsilon_i-\mu)}$ is in the denominator of $\langle n_i\rangle$ $\endgroup$
    – J. Murray
    Feb 25, 2023 at 23:27

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