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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?

Qmechanic
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imbAF
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1 Answers1

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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.

Albatross
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