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I have two questions about using the concept of indistinguishability to determine the partition function in statistical mechanics, like for instance when determining the partition function of an ideal gas.

1: Why do we assume that the particles in the gas are indistinguishable? In QM, a set of N particles are indistinguishable only if their combined wave function is either symmetric (bosons) or antisymmetric (fermions) under interchange of two particles. Why do we make this assumption for the combined wave function of the particles in the gas (whose single particle wavefunctions are given by the solutions to the particle in a 3D box problem, as usual)?

2: It's found that if the number of possible single particle states at low energy is much higher than the number of particles, then you can approximate the partition function by introducing a factor of 1/(N!) (where N is the number of particles) in order to account for the indistinguishability of the particles. This is because most states of the system will be such that all the particles are found in distinct energy levels. But if we assume that the particles are fermions, then surely the particles HAVE to be in different states (due to Pauli's exclusion principle), meaning that introducing the factor 1/(N!) is exact (and not just an approximation)?

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  • $\begingroup$ Hi, I tried to answer your first question. I think you should post a separate question for the second part, I'm not sure in what context you have seen this approximation and you should add more detail, maybe a reference where you've seen it :) $\endgroup$ Apr 21 at 12:03
  • $\begingroup$ @user2723984 I am talking about how the partition function is calculated for an ideal gas. In the classical model with distinguishable particles, the partition function of the system can be found by multiplying the single-particle partition functions together. In the case of indistinguishability, you introduce an extra factor of 1/(N!) to the classical partition function (the one which is given by the product of the single-particle partition functions). $\endgroup$ Apr 21 at 12:45
  • $\begingroup$ @user2723984 I'm using my stat mech books as reference, but if you go to the "partition functions of subsystems" on this Wikipeida article, then they talk about the same thing: en.wikipedia.org/wiki/… $\endgroup$ Apr 21 at 12:57
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1: Why do we assume that the particles in the gas are indistinguishable?

Because if we don't, then we find that the entropy of the system is non-extensive (see the Gibbs paradox), which leads to apparent violations of the second law of thermodynamics. The solution proposed by Josiah Gibbs is to treat the particles as indistinguishable by introducting an additional factor of $1/N!$ in the multiplicity function. This is one of the ways in which a fundamental quantum mechanical property is manifested in an ostensibly classical system.

2: [..]But if we assume that the particles are fermions, then surely the particles HAVE to be in different states (due to Pauli's exclusion principle), meaning that introducing the factor 1/(N!) is exact (and not just an approximation)?

No, it's still an approximation. Imagine that your system has three energy levels $E=\{0,\epsilon,2\epsilon\}$ and three particles with a total energy of $3\epsilon$.

  • For classical distinguishable particles, we could have each particle in a different energy level or all three particles in the second energy level. There are six ways to arrange the former and one way to arrange the latter, for a total multiplicity of 7. Dividing by $3!=6$ yields a corrected multiplicity of $7/6$.
  • For indistinguishable bosons, the aforementioned six possible arrangements of one particle per energy level all correspond to the same microstate. Therefore, the total multiplicity is $2$.
  • For indistinguishable fermions, we additionally have that the microstate with all of the particles in the second energy level is forbidden, which means that the total multiplicity is $1$.

As a side note, the probability that all three particles have the same energy is $1/7$ as computed classically, $1/2$ for indistinguishable bosons, and $0$ for indistinguishable fermions. This informs the rule of thumb that bosons are more likely to occupy the same state than a classical analysis would suggest, with the opposite rule holding for fermions.

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Why do we assume that the particles in the gas are indistinguishable? In QM, a set of N particles are indistinguishable only if their combined wave function is either symmetric (bosons) or antisymmetric (fermions) under interchange of two particles. Why do we make this assumption for the combined wave function of the particles in the gas (whose single particle wavefunctions are given by the solutions to the particle in a 3D box problem, as usual)?

I think you have this backwards. We get the result that particles are either fermions or bosons precisely because we assume they are indistinguishable. When you assume particles are indistinguishable in statistical mechanics, you get either fermionic or bosonic models.

Why we assume particles are indistinguishable? Because "indistinguishable" is just a way to say that we are considering all of the characteristics of the particles in our model, that is to say, there is nothing I can measure that I'm not already considering as a variable that would allow me to distinguish two particles.

Example: suppose I have two particles of the same mass and no charge. They have 2 unique properties that identify them: their position and their momentum. They are "indistinguishable" in the sense that if I take particle $1$ and change it so that its position and its momentum mach those of particle $2$, and vice versa, then effectively in the model particle $1$ has become particle $2$, and vice versa, and none of the physics has changed. The system will behave exactly in the same way as if I hadn't done anything. If the particles are identical in everything except position and momentum, how else are you going to label them if not "the particle that's here and slow and the particle that's there and fast"? Hence, it makes no sense to assign labels such as "particle $1$ and particle $2$", but all the labels we need are in the states.

Where the indistinguishabilty assumptions become interesting is that usually we do have these labels for particles: they are the labels of the corresponding Hilbert spaces forming the combined space $ \mathcal H_1\otimes \mathcal H_2$. The process of getting bosons and fermions from distinguishable particles is the process of selecting the portions of $\mathcal H_1\otimes \mathcal H_2$ that is invariant under the swapping of these labels, which is a mathematical way of saying "if I assign to particle $1$ all the properties of particle $2$ and vice versa, nothing changes".

This drastically changes if you artificially make the particles distinguishable, for example, you assign then a fixed position, e.g. "the particle on the left and the particle on the right", and you want to model the behaviour of some other variable, say their spins. Then the system is not invariant under the exchange of the particles, i.e. you want "up down" to be a physically distinct state from "down up", and hence you don't have fermions or bosons anymore. This is the case for example in lattice spin systems.

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  • $\begingroup$ I am not sure how this explains why the particles in the gas are indistinguishable. If indistinguishability defines the notions of bosons and fermions, then why not just assign the particles in the gas some other name/properties? In the classical model, the particles in the gas are distinguishable. You can distinguish them by their respective trajectories. To my understanding, particles are indistinguishable in QM only if their "trajectories" have mixed, such that the probability that particle 1 is at x_1 and 2 is at x_2 is equal to the probability that particle 1 is at x_2 and 2 is at x_1. $\endgroup$ Apr 21 at 12:36
  • $\begingroup$ And surely two fermions don't have to be indistinguishable. If you have an unentangled system of a pair of electrons, then you can distinguish the particles by their single-particle wavefunctions (because of the Pauli principle, these wavefunctions must be different, and hence you can distinguish the particles). $\endgroup$ Apr 21 at 12:38
  • $\begingroup$ Not quite, particles are indistinguishable in QM if swapping them shouldn't change the physics, i.e. $|x\rangle_1 \otimes |y\rangle_2$ should be the same as $|y\rangle_1\otimes |x\rangle_2$. The only thing that matters is "there is a particle at position $x$ and a particle at position $y$", not which of the two particles is in which position, because the particles being identical in every other way except position, their position is the only way to distinguish them. "there is a particle at position $x$ and a particle at position $y$" is translated mathematically in Fock space. $\endgroup$ Apr 21 at 12:42
  • $\begingroup$ From this comes the fact that a proper wave function will always look something like $|x\rangle|y\rangle+|y\rangle|x\rangle$ or similar, because these are the only superpositions allowed. Classical particles are also indistinguishable if they are only characterized by their trajectory, but this as far as I know doesn't have very profound implications. $\endgroup$ Apr 21 at 12:43
  • $\begingroup$ Example with classical particles: you have two balls: ball $1$ and ball $2$. They are perfectly identical except that one is in my right hand and the other one is in my left hand. Now I ask you to close your eyes, swap them a few times, and ask you: which one is ball 1 and which one is ball 2? You won't be able to tell, because there wasn't a ball 1 or a ball 2 in the first place, just one in the right hand and one in the left hand. The balls are just distinguished by the point they occupy in phase space, i.e. just their position. $\endgroup$ Apr 21 at 12:45

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