Is the N factorial in the Partition function for N indistinguishable particle an approximation?

Question

I suspect that the N factorial in the partition function for N indistinguishable particles $$ Z = \frac{ Z_0^N } {N!} $$ is an approximation. Please someone correct me if I am wrong and why or why not. Thanks.

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A simple case:

each particle has two states with energy 0 and $E$. The partition function for a single particle is $$ Z_0 = 1 + e^{- \beta E} . $$ If there are only two particles, there is the total partition function $$ Z = \frac{ Z_0^2 } {2}. $$ But regarding the whole system consisting of these two particles, we can also write $$ Z = 1 + e^{- \beta E} + e^{-2 \beta E} . $$ And it is certain that $$ \frac{ Z_0^2 } {2} \neq 1 + e^{- \beta E} + e^{-2 \beta E} $$

This post imported from StackExchange Physics at 2014-03-31 13:18 (UCT), posted by SE-user Chengjian

Comments

Yes, it is an approximation. As the number of particles increase, the generic state is one where all of them are in different energy levels. The symmetry factor from indistinguishability is $N!$ in this situation. So dividing by $N!$ takes care of the generic situation correctly. This is sufficient to explain Gibbs paradox, for instance.

This post imported from StackExchange Physics at 2014-03-31 13:18 (UCT), posted by SE-user suresh

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@suresh that should be an answer

This post imported from StackExchange Physics at 2014-03-31 13:18 (UCT), posted by SE-user David Z

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@suresh If the OP is referring to a system of non-interacting particles, then there is no approximation being made. The hamiltonian splits into a sum of one-particle hamiltonians, but the Hilbert space is the symmetric subspace of the full Hilbert space, and if one computes the partition function, then one can show that the partition function is precisely what the OP writes, so I'm unclear on what you're getting at.

This post imported from StackExchange Physics at 2014-03-31 13:18 (UCT), posted by SE-user joshphysics

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@joshphysics The approximation is that dividing by $N!$ gets the answer wrong for all other situations but that is neglible in the thermodynamic limit.

This post imported from StackExchange Physics at 2014-03-31 13:18 (UCT), posted by SE-user suresh

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@suresh You're right; it's not exact in general. My "proof" was in error. However, I don't think $N$ large is the right limit, I think it's the high temperature limit in which that result is approximately true provided the number of single-particle states is infinite (like for a particle in a box). Otherwise, you could have, for example, a system with $N$ single-particle energy states and $N$ identical fermions. No matter how large $N$ is, there is only one allowed state for the system, and the partition function is just a single term.

This post imported from StackExchange Physics at 2014-03-31 13:18 (UCT), posted by SE-user joshphysics

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Josh, take a look at the answer I have written as a picture. I agree with what you say for fermions but the comparison is with $Z$ (the exact partition function) and $z^N/N!$. The second one is "wrong" for bosons and fermions.

This post imported from StackExchange Physics at 2014-03-31 13:18 (UCT), posted by SE-user suresh

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Chengjian, you're forgetting that there is degeneracy in the combined system. Your expression should be $1+2e^{-\beta E}+e^{-2\beta E}$, since there are two distinct ways the total energy can be $E$. Then you have to divide by two because you don't know which particle is which, giving the correct (exact) answer, $Z_0/2$. (I will explain that properly in an answer if I get chance.)

This post imported from StackExchange Physics at 2014-03-31 13:18 (UCT), posted by SE-user Nathaniel

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@Nathaniel If the particles are identical bosons, then there is only one admissible state with energy $E$, namely the totally antisymmetric combination $\frac{1}{\sqrt{2}}(|0\rangle|1\rangle + |1\rangle|0\rangle)$. Are you ignoring quantum statistics when you make your statement? I feel like we're all talking about different systems here.

This post imported from StackExchange Physics at 2014-03-31 13:18 (UCT), posted by SE-user joshphysics

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@joshphysics yes, I was ignoring quantum statistics. In fairness this is because the question quotes the classical partition function for classically indistinguishable particles. (I.e. where they are actually distinct but it doesn't matter which one's which.) In that case what I said is right. In the quantum case it's not so much an approximation as the wrong formula.

This post imported from StackExchange Physics at 2014-03-31 13:18 (UCT), posted by SE-user Nathaniel

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@Physics Right ok. I think all of the controversy in this entire thread and all of the answers comes from inconsistencies with which system each answerer is considering. I think that since the question never explicitly uses the term "classical," some people assumed that the OP was asking about quantum stat mech, and in that case the formula is certainly not correct, but it seems that for certain systems in some limit, it gives a good approximation.

This post imported from StackExchange Physics at 2014-03-31 13:18 (UCT), posted by SE-user joshphysics

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It seems that I should claim that the particles are bosons, in order to make sense. @Nathaniel, thanks a lot for pointing this out.

This post imported from StackExchange Physics at 2014-03-31 13:18 (UCT), posted by SE-user Chengjian

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I missed you last comment. That particles are bosons makes your expression of $Z$ for two particles correct. In this case, there is no $N!$ coefficient. For classical (uncorrelated) particles, there is a $N!$. These are two different cases, neither is the approximation of the other and both are exact.

This post imported from StackExchange Physics at 2014-03-31 13:18 (UCT), posted by SE-user V. Rossetto

Answer 1

Actually, it's exact. The flaw is "regarding the whole system consisting of these two particles, we can also write" $Z = 1 + e^{- \beta E} + e^{-2 \beta E}.$ Assuming the two particles are distinguishable, we have $$Z=\sum_ig_ie^{-\beta E_i}=1+2e^{-\beta E}+e^{-2\beta E}=Z_0^2,$$ with the $2e^{-\beta E}$ since the state of energy $E$ is doubly-degenerate. The additional factor of 1/2 in $Z=Z_0/2!$ accounts for indistinguishability.

EDIT: This seems to be wrong. See other answers.

This post imported from StackExchange Physics at 2014-03-31 13:18 (UCT), posted by SE-user DumpsterDoofus

Comments

@joshphysics. I am merely using the same definition as Gibbs i.e. particles which states are independent.

This post imported from StackExchange Physics at 2014-03-31 13:18 (UCT), posted by SE-user V. Rossetto

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@suresh,joshphysics: What I wrote certainly might be wrong, although I did notice the use of $N!$ in pg 229 of "Solved Problems in Quantum and Statistical Mechanics" 2012 edition, which involves the use of $N$ two-level systems. Is this an error in the book, then?

This post imported from StackExchange Physics at 2014-03-31 13:18 (UCT), posted by SE-user DumpsterDoofus

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This answer is correct. +1.

This post imported from StackExchange Physics at 2014-03-31 13:18 (UCT), posted by SE-user Nathaniel

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@Nathaniel. I think the controversy comes from the fact that the two particle partition function is different for different types of particles. See my answer below.

This post imported from StackExchange Physics at 2014-03-31 13:18 (UCT), posted by SE-user V. Rossetto

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@V.Rossetto Yeah I'd like to add that we all seem to be referring to different systems, namely quantum v. classical etc.

This post imported from StackExchange Physics at 2014-03-31 13:18 (UCT), posted by SE-user joshphysics

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I wish I could downvote too :-(

This post imported from StackExchange Physics at 2014-03-31 13:18 (UCT), posted by SE-user Mark A

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@joshphysics. The OP's counterexample is wrong and this answer is exact for classical particles. The $1/N!$ accounts for permutations of $N$ particles, there is no approximation.

This post imported from StackExchange Physics at 2014-03-31 13:18 (UCT), posted by SE-user V. Rossetto

Answer 2

In the figure above, consider the different configurations that are possible with 3 particles and 5 energy levels. Dividing by 3! gets the symmetry factor correct only for configurations of type 1 but is wrong for configurations of type 2 and 3. You can see this by explicitly writing out $Z$ and comparing with $z^3/3!$. That is why the OP's statement that $z^3/3!$ is an approximation is correct. I can add details to this, if necessary. (Notation: The lower-case $z$ is the single particle partition function)

To add to Josh's remark above, even for fermions where terms of type I are only allowed, the expansion of $z^N/N!$ contains terms of type 2 and 3 which are not present in $Z$. Nevertheless, the dominant contribution at large $N$ (as well as number of energy levels) is from terms of type 1. Hence my statement that it is a fairly good approximation holds.

This post imported from StackExchange Physics at 2014-03-31 13:18 (UCT), posted by SE-user suresh

Comments

@V.Rossetto I don't understand your remark.

This post imported from StackExchange Physics at 2014-03-31 13:18 (UCT), posted by SE-user suresh

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@suresh. To compute the two-particle canonical partition function for fermions, you do not make an approximation, you should count the occupation of quantum levels. See my answer below.

This post imported from StackExchange Physics at 2014-03-31 13:18 (UCT), posted by SE-user V. Rossetto

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There are several good answers and hard to pick which one to accept. I pick yours since your graph really helps a lot in the demonstration of the effectivity of the approximation. Thank you.

This post imported from StackExchange Physics at 2014-03-31 13:18 (UCT), posted by SE-user Chengjian

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@Changjian. I count three answers giving a yes (suresh +2, gj255 +2/-1, Mark A +3) and two giving a no (Dumpster +8/-4, mine +3). Such votes are rather unusual and denote and strongly debated question. I think it would be very interesting if you could gave us what you have understood and the conclusions you draw from these five answers and the comments.

This post imported from StackExchange Physics at 2014-03-31 13:18 (UCT), posted by SE-user V. Rossetto

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@V.Rossetto, I see. I think your answer is very helpful. I do not think it is giving no because your illustration is based on a different background of this question.

This post imported from StackExchange Physics at 2014-03-31 13:18 (UCT), posted by SE-user Chengjian

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I like this answer but the appropriate limit in which this approximation is valid is not large $N$ but high temperature/energy. It's the temperature which determines the number of relevant energy levels in the diagrams above, and therefore ensures that diagrams of type 1 are dominant. Conversely, it's clear that if the temperature is low enough that particles are confined into the lowest few levels then increasing $N$ makes diagrams of type 1 less dominant, not more.

This post imported from StackExchange Physics at 2014-03-31 13:18 (UCT), posted by SE-user Mark A

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Your answer is correct for fermions, but wrong for classical particles or bosons. For quantum particles, the partition function is not computed this way.

This post imported from StackExchange Physics at 2014-03-31 13:18 (UCT), posted by SE-user V. Rossetto

Answer 3

It is indeed an approximation, in particular an approximation that is usually good at high enough temperatures.

For $N$ distinguishable, non-interacting particles the partition function is $Z(\mathrm{dist.}) = {Z_0}^N$, where $Z_0$ is the single-particle partition function. If the $N$ particles are all in different quantum states then there are $N!$ microstates of distinguishable particles which all correspond to the same microstate of indistinguishable particles. However, if the particles are not all in different states then the number of microstates of distinguishable particles corresponding to the same microstate of indistinguishable particles is less than $N!$.

So in general the relationship between the sum over microstates for distinguishable particles and for indistinguishable particles is quite complex. However, if the temperature is high enough then (for most systems) the number of accessible quantum states will be much larger than the number of particles, so it is unlikely more than one particle will occupy the same quantum state. In this case we can make the approximation that the partition function for distinguishable particles is just $N!$ times that for indistinguishable particles. I.e. $Z(\mathrm{indist.}) = {Z_0}^N/N!$ as you wrote.

The claim that at high enough temperatures the number of accessible quantum states will be much larger than the number of particles is not true for all systems, for example it is false for those with bounded energy spectrum (such as the ideal paramagnet or the example the OP gives). Also, how high the temperature needs to be depends upon the system. For an ideal gas at atmospheric densities it's a good approximation for temperatures much above 1K.

The above is true for quantum particles (Fermions or Bosons). In the classical limit the quantum energy-level separation is taken to zero and the spectrum becomes continuous. For a continuous spectrum the probability that two particles are in exactly the same state is zero, and so the approximation becomes exact in this limit. Note however, that there is no such thing as a classical two-level system, such as the example given by the OP. A classical system with two local minima can in some cases be treated as a two level system, but this is in itself an approximation.

This post imported from StackExchange Physics at 2014-03-31 13:18 (UCT), posted by SE-user Mark A

Answer 4

The factorial factor $1/N!$ is exact but does not apply to all statistics.

Consider a two level system and let us call $\xi_i$ the grand-canonical partition function for the energy level $i$ ($i=0$ or $1$) and $z=\mathrm e^{\beta\mu}$ the fugacity. Keep in mind that $\xi_i$ is the partition function for a given energy level.

Classical particles are independent, indistinguishable, $\xi_i$ is computed as a usual partition function for undistinguishable particles. Here we divide by the symmetry $p!$ because the particles are undistinguishable and therefore the order in which we could label them should not matter. $$\xi_i^{\text{classical}}=\sum_{p=0}^\infty \frac1{p!}z^p \mathrm e^{-p\beta E_i}=1+z\mathrm e^{-\beta E_i}+\frac{z^2}{2}\mathrm e^{-2\beta E_i}+\cdots=\exp\left(z\mathrm e^{-\beta E_i}\right).$$ For quantum particles, the sum runs over the number of particles which can occupy the energy level. There is no division by $p!$ because the quantum states already are (anti)-symmetric; for instance the quantum state where there is one fermion in each level is $\frac1{\sqrt2}\left(\left|01\right\rangle-\left|10\right\rangle\right)$. $$ \xi_i^{\text{fermions}}=\sum_{p=0}^1z^p\mathrm e^{-p\beta E_i}=1+z\mathrm e^{-\beta E_i}$$ $$ \xi_i^{\text{bosons}}=\sum_{p=0}^\infty z^p\mathrm e^{-p\beta E_i}=\frac1{1-z\mathrm e^{-\beta E_i}}$$ The grand-canonical partition functions are $$ \mathcal Z^{\text{classical}}=\xi_0^{\text{classical}}\xi_1^{\text{classical}}=\mathrm e^{z\left(1+\mathrm e^{-\beta E}\right)} =\mathrm e^{zZ_1}=1+zZ_1+\frac{z^2}2Z_1^2+\cdots$$ $$\mathcal Z^{\text{fermions}}=\xi_0^{\text{fermions}}\xi_1^{\text{fermions}}=(1+z)\left(1+z\mathrm e^{-\beta E}\right)=1+zZ_1+z^2\mathrm e^{-\beta E}$$ $$\mathcal Z^{\text{bosons}}=\xi_0^{\text{bosons}}\xi_1^{\text{bosons}}= \frac1{1-z}\frac1{1-z\mathrm e^{-\beta E}}=1+zZ_1-z^2\mathrm e^{-\beta E}+z^2Z_1^2+\cdots$$ with $Z_1=1+\mathrm e^{-\beta E}$ is the one-particle partition function.

Now looking at the coefficient $z^2$ in $\mathcal Z$ gives the two-particle canonical partition function $Z_2$. We have $$ Z_2^{\text{classical}}=\frac{1}{2}Z_1^2,\quad Z_2^{\text{fermions}}=\mathrm e^{-\beta E},\quad Z_2^{\text{bosons}}=1+\mathrm e^{-\beta E}+\mathrm e^{-2\beta E}.$$ In none of these case was it necessary to make an approximation. Undistinguishability in the quantum cases is for quantum states, not for particles.

Remark The difference between $Z_2^{\text{classical}}$ and $Z_2^{\text{bosons}}$ is evidence for the fact that the bosons and the classical particles have a difference. Indeed, classical particles have no correlations, which is expressed by the fact $Z_2\propto Z_1^2$, whereas bosons have correlations: compared to the uncorrelated classical particles, the relative weight of the states where the particles are both at the same energy level is larger: bosons prefer to be in the same state. (Fermions avoid being at the same energ level and classical particles don't care.)

This post imported from StackExchange Physics at 2014-03-31 13:18 (UCT), posted by SE-user V. Rossetto

Comments

I am confused by this... in all the definitions I've seen the canonical partition function is given as a sum over microstates with fixed particle number $N$, and so the fugacity is not a variable in the function but given as a derivative with respect to $N$. The $\xi$ functions here don't look like canonical partition functions to me. However, it is correct to say that the $1/N!$ relationship is good in the classical limit. I've added a note on my answer to make this clear.

This post imported from StackExchange Physics at 2014-03-31 13:18 (UCT), posted by SE-user Mark A

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Okay, I noticed that in the first paragraph you refer to $\xi_i$ as a single-level grand-cononical PT whereas in the second you call it a cononical PT, so I guess the latter is a typo?

This post imported from StackExchange Physics at 2014-03-31 13:18 (UCT), posted by SE-user Mark A

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@MarkA. Thanks Mark, there was a typo. I have edited my answer and added a few thoughts about the $N!$ which might be better understood as a symmetry factor.

This post imported from StackExchange Physics at 2014-03-31 13:18 (UCT), posted by SE-user V. Rossetto

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@V.Rossetto I should specify that I am considering two identical bosons with two energy levels which is not that physical. BTW, in your calculation of $\xi_i^{classical}$, why do we need to divide by N!?

This post imported from StackExchange Physics at 2014-03-31 13:18 (UCT), posted by SE-user Chengjian

Answer 5

It is indeed an approximation. As suresh says, the approximation holds only when each particle is in a different single-particle state, or more precisely, when the number of states in which each particle is in a different single-particle state vastly outweighs the number of states in which more than one particle is in the same SP state.

The single particle partition function is $$ Z_0 = 1 + e^{-\beta E}$$ The two distinguishable particle partition function is $$ Z_\mathrm{dist} = 1 + 2e^{-\beta E} + e^{-2\beta E} = Z_0^2$$ The two indistinguishable particle partition function is $$ Z_\mathrm{in} = 1 + e^{-\beta E} + e^{-2\beta E}$$

You can hopefully see that $Z_\mathrm{in}$ is not equal to $Z_0^2/2!$ from the above equations. The logic for the $N!$ division is as follows: for each overall state of the two distinguishable particle system there are $N!$ ways of permuting the particles, but for identical particles these are all indistinguishable. However, for any overall state of the two distinguishable particle system with two particles in the same single-particle state, we are already considering the possible permutations as indistinct. Even for distinguishable particles, "particle A in state X and particle B in state X" is considered to be the same state as "particle B in state X and particle A in state X", as should be fairly obvious if you think about it! So only for those overall states for which all particles are in different single-particle states is the full division by $N!$ appropriate.

For systems with enormous numbers of states --- such as systems with translation degrees of freedom --- this is a very good approximation, since the typical number of available single-particle states vastly outnumbers the typical number of particles. On the other hand, high temperature is not generally a limit in which this approximation holds. You should be able to see this from the above expressions for partition functions. Only for systems with an infinite number of single-particle states does the approximation hold in this limit.

This post imported from StackExchange Physics at 2014-03-31 13:18 (UCT), posted by SE-user gj255