# Why is the formula for entropy multiplicative when particles are said to lack haecceity?

The calculation for entropy counts each particle in a potential microstate as an individual. However, I have read in numerous places that physics dictates that we reject the haecceity (i.e. the essential thisness of identity) of quantum scale objects. That is, particles are interchangeable (e.g. Wheeler's example of thinking of just a single electron in multiple places across the universe at once rather than multiple electrons). This seems like a contradiction, what am I missing?

Common Entropy Example:

-There is a container of gas bisected by a wall.

-There is a small hole in the wall that allows the gas atoms to move from one side of the container to the other.

-There are 2,000 atoms of the gas in the container.

-The minimum entropy state is for all the gas to be on one side of container. There is only one way the gas can be arranged like this.

(I am aware that this is a simplification for the purposes of example)

So far so good... But then the common examples will continue:

-If there is one atom of gas on one side of the wall, and 1,999 atoms on the original side, there are now 2,000 possible configurations for this system (one for each atom).

-If there are two atoms on one side of the wall, and 1,998 on the other side, there are now 4,000,000 possible microstate configurations (2,000^2).

And so on. This explicitly counts each atom as having its own identity. It is saying the state where one atom moves to the empty side of the container is different from the state where a different atom moves to the other side of the container. But if atoms lack haecceity, then it does not matter which atom moves to which side.

For an example of the haecceity issue: https://plato.stanford.edu/entries/qt-idind/

In quantum statistical mechanics, however, we have two ‘standard’ forms: one for which there are three possible arrangements in the above situation (both particles in one box, both particles in the other, and one in each box), giving ‘Bose-Einstein’ statistics; and one for which there is only one arrangement (one particle in each box), giving ‘Fermi-Dirac’ statistics (which underpins the Pauli Exclusion Principle and all that entails). Setting aside the differences between these two kinds of quantum statistics, the important point for the present discussion is that in the quantum case, a permutation of the particles is not regarded as giving rise to a new arrangement. This result lies at the very heart of quantum physics; putting things slightly more formally, it is expressed by the so-called ‘Indistinguishability Postulate’:

If a particle permutation P is applied to any state function for an assembly of particles, then there is no way of distinguishing the resulting permuted state function from the original unpermuted one by means of any observation at any time. (The state function of quantum mechanics determines the probability of measurement results. Hence what the ; expresses is that a particle permutation does not lead to any difference in the probabilities for measurement outcomes.)

...conversely, it is argued, if such permutations are not counted in quantum statistics, it follows that quantum objects cannot be regarded as individuals in any of these senses (Post 1963). In other words, they are very different from most everyday objects in that they are ‘non-individuals’, in some sense.

I am not understanding how this is not a problem for the entropy formula. The SEP article says that the issue of particle haecceity is somewhat indeterminant. I can't seem to wrap my head around how the entropy formula can work the same way regardless of whether we treat the atoms as individuals. At the same time, based on other papers I read looking for an answer, I can see how particles fail Leibnitz'Law of the Identity of Indiscernibles. The number of potential microstates in Boltzmann's formula implies that one atom being in the place of another is a different microstate. Am I being led astray by the simplistic examples normally used for statistical mechanics? Any help would be appreciated.

• Are you referring to the Gibbs paradox? Commented Jul 1, 2022 at 21:00
• For those who, like me, don't know what haecceity is here's a link to Wikipedia's page which mentions it is connected with the philosophy of physics. Commented Jul 1, 2022 at 22:48
• I was not aware of the Gibbs Paradox, thank you. I think it's relevant here. The Gibbs Paradox is said to be "resolved" in that article due to the indistinguishability of particles. The makes sense; what doesn't make sense is how this doesn't cause issues for the entropy formula. Apparently, there is nonextensive entropy formula that is in use as well, but I'm at a total loss as to why one wouldn't always apply and not the other: en.wikipedia.org/wiki/Tsallis_entropy Commented Jul 1, 2022 at 22:57
• What is "haecceity"? Commented Jul 2, 2022 at 1:12
• @TimBrown What entropy formula are you referring to? Typically in classical stat. mech. we divide the naive number of microstates by $N!$ to compensate for the overcounting, but this only works well if the likelihood of any given state to be multiply-occupied is small (which in turn is valid at high temperatures). If we count more carefully, we obtain Fermi/Bose statistics instead. Commented Jul 2, 2022 at 2:08

If I understand you correctly, the fact that particles are indistinguishable just means that we need to be careful when computing the multiplicity function $$\Omega$$ which appears in Boltzmann's entropy equation $$S = k\ln(\Omega)$$.

In classical statistical mechanics, we typically compute $$\Omega$$ as though all of the particles were distinguishable and then divide by $$N!$$ to account for the overcounting. This works well if the expected occupation number of each single-particle state is extremely small, which occurs at high temperature; however, if the occupation numbers of the single particle states are of order $$1$$, then we are entering the regime of quantum statistical mechanics and need to be more careful. If we take indistinguishability into account more carefully, then we obtain Bose statistics. If we forbid the single particle states to have occupancy $$>1$$, then we obtain Fermi statistics.

As an example, imagine that we have two states in the system with energies $$\{0, \epsilon,2\epsilon\}$$ and three indistinguishable particles, and assume that the total energy in the system is $$3\epsilon$$.

1. Treating the particles as distinguishable, we could either have each particle in a different state ($$3!=6$$ ways) or all three particles in the second state ($$1$$ way) for a total multiplicity of $$\Omega=7$$ and entropy $$S=k\ln(7)$$.

2. The rough correction from classical statistical mechanics simply takes this number and divides by $$3!=6$$ to obtain $$\Omega_{classical} = 7/6$$.

3. If we're more careful, then we note that all 6 rearrangements corresponding to each particle being in a different state are in fact the same, because the particles are indistinguishable. Therefore, there are only two possible configurations (each particle in a different state, or each particle in the $$E=\epsilon$$ state) and so $$\Omega_{Bose} = 2$$.

4. If multiple occupancy is forbidden, then the particles may not all occupy the $$E=\epsilon$$ state, and so there is only one possible configuration, $$\Omega_{Fermi}=1$$.

I can't seem to wrap my head around how the entropy formula can work the same way regardless of whether we treat the atoms as individuals.

The difference is simply in how you count states to compute $$\Omega$$. Once you have your $$\Omega$$ - which will change depending on whether you treat the particles as distinguishable or indistinguishable - the Boltzmann formula works the same way, $$S=k\ln(\Omega)$$.

The number of potential microstates in Boltzmann's formula implies that one atom being in the place of another is a different microstate.

No it doesn't. This would be true for distinguishable particles, but for indistinguishable particles we need to count states more carefully as demonstrated in my toy example.

As a concrete example of classical statistics, you may be interested in the derivation of the Sackur-Tetrode equation for the entropy of an ideal gas, which explicitly adds the $$1/N!$$ overcounting factor - see the last section of Gibbs Paradox - Calculating the entropy of ideal gas, and making it extensive for an example.

• Thank you! That makes much more sense now that I have dug through the topics you referenced. The examples I had read through earlier were simplifications and my confusion was stemming from an incomplete picture of the issue it seems. Commented Jul 2, 2022 at 19:02