A Big Problem in Landau's Statistical Mechanics Book

This question regards: Landau; Course of Theoretical Physics, Volume 5, Statistical Physics, Part 1, Third edition, revised and enlarged.

In Chapter IV, section 37 the objective is to derive the Boltzmann Distribution 1 ; so we want to derive an expression for the mean occupation number, of a state $$k$$, for a perfect gas that has a very low mean occupation number: $$\langle n_k \rangle << 1$$ In other words: we are dealing with a rarefied gas.

The resoning reported on the book begins as follows:

For the Gibbs distribution has been derived for bodies which are relatively small, but at the same time macroscopic, parts of large closed systems. The macroscopic nature of these bodies made it possible to regard them as quasi-closed, i.e. to neglect to some extent their interaction with other parts of the system. In the case under consideration the separate molecules of the gas are quasi-closed, although they are certainly not macroscopic bodies. Applying the Gibbs distribution formula to the gas molecules, we can say that the probability that a molecule is in the kth state is proportional to $$e^{\mathcal{E}_k/k_BT}$$, and therefore so is the mean number $$\langle n_k \rangle$$ of molecules in that state, i.e. $$\langle n_k \rangle = a \exp{\left[\frac{\mathcal{E}_k}{k_BT}\right]}\tag{37.2}$$

Here we encounter the first problem: Gibbs distribution gives us a probability, but a probability for what? In Chapter III, section 28, where Landau is deriving the form of the Gibbs distribution, it is stated that:

Our object is to find the probability $$p_i$$ of a state of the whole system such that the body concerned is in some definite quantum state (with energy $$E_n$$), i.e. a microscopically defined state.

So we can say that the probability $$p_i$$ given by the Gibbs distribution is the probability that a system has to be in a certain microstate. I mean: suppose we are dealing with a macroscopic system, this macroscopic system is composed by a lot of little sub-systems (for example: a gas can be the macroscopic system and the particles of the gas are all the little sub-systems that compose it), of course our macroscopic system has some macroscopic state (it has a volume, a temperature ecc.) but it also has a microscopic state (all the position and momenta of all the little particles that compose it); well: in this context Gibbs distribution gives us the probability that the macroscopic system has of being in one specific microscopic state of energy $$\mathcal{E}_i$$.2

We can now see that the reasoning in the first citation is not sound! Gibbs distribution needs a macroscopic system that can have a macroscopic and a microscopic state to have sense, what is Landau taking as his macroscopic system in the first citation? The perfect gas? Can't be! In fact the energy $$\mathcal{E}_i$$ is not the energy of the entire microstate of the gas, the energy of all the particles in a specific state, instead it's the energy of a specific state that a single particle can assume! So what is he doing? Is he taking a single particle to be the system to which Gibbs distribution applies? But, if so, this makes no sense at all! A single particle is not a system, it cannot have a macrostate ad a microstate, Gibbs distribution is simply not defined for a single particle. Or at least Gibbs distribution is not well defined in this case with the definition provided by Landau himself! What is going on here? Why is Landau stating that $$\langle n_k \rangle$$ must be proportional to $$\exp{\left[\frac{\mathcal{E}_k}{k_BT}\right]}$$?

But this is not the only problem. In the same Chapter IV, section 37 it is then stated that:

The constant coefficient in (37.2) can be expressed in terms of the thermodynamic quantities for the gas. To do this we shall give another derivation of the formula, based on the application of the Gibbs distribution to the assembly of all particles in the gas that are in a given quantum state. We are able to do this (even if the numbers $$n_k$$ are not small) since there is no direct force of interaction between these particles and the remainder (or between any of the particles in an ideal gas), and the quantum exchange effects occur only for particles in the same state. Putting $$E = n_k\mathcal{E}_k$$, $$N = n_k$$ and adding the suffix $$k$$ to $$\Omega$$ in the general formula for the Gibbs distribution for a variable number of particles (35.2), we find the probability distribution for various values of $$n_k$$ as $$p_{n_k}=\exp{\left[\frac{\Omega _k+n_k\mu-n_k\mathcal{E}_k}{k_BT}\right]}\tag{37.4}$$

I also have a couple of problems here, and I strongly suspect that all this incongruencies are somehow linked, that's why I reported them in the same question: the first problem here is that (37.4) simply is the grand canonical distribution, right? By definition of it this formula should apply to a macroscopic system that can have both a macroscopic state and a microscopic state, just like the canonical distribution a.k.a Gibbs distribution. But again: what is the system took into consideration by the grand canonical distribution? The entire system of the perfect gas? Only the particles in a state? Landau's book is not clear.. And also: $$\Omega$$ si of course the grand potential, and by definition the grand potential is a potential of a macroscopic state of a macroscopic object! We can talk about the grand potential of the entire gas, so why, in the name of Gauss, it has an index $$k$$ that counts on the microscopic states?? The grand potential's formula has entropy in it! For crying out loud: talking about the entropy of a microstate makes no sense! Entropy is a property of a macroscopic system and is defined by the number of microscopic states that the macroscopic system is compatible with! All this reasoning reported in Landau's book does not make sense to me, for the reasons I have exposed.. What is going on?

: For Landau Boltzmann Distribution is not a synonym of Gibbs distribution, contrary to what the main wikipedia article on the topic states.
: An important detail: for the Gibbs distribution to apply the macroscopic system must be a canonical system; a.k.a Canonical Ensemble.

• First of all, your Gibbs distributions are missing a few minus signs. Mar 6 '21 at 4:23
• The Gibbs (canonical) distribution applies to a system that exchanges energy with a reservoir (heat bath). The condition is that the system is much smaller than the reservoir. The system can consist of a single particle, or of a macroscopic number of particles, as long as the reservoir is even bigger. In an ideal gas, the most straightforward application is to take the system to be a single particle, and the reservoir is provided by all the other particles in the gas. Mar 6 '21 at 4:26
• A general comment: Landau books represent essentially how this man understood and explained the physics - they are a kind of Feynmann lectures (even though more dificult mathematically). This is to say that they can give one lots of insights and precious solutions to specific interesting problems, but they are not really good as the first (or the only) study books; since they often sacrifice building general and comprehensive picture or take unconventional paths for rather conventional problems. Mar 8 '21 at 14:53

Gibbs distribution needs a macroscopic system that can have a macroscopic and a microscopic state to have sense

Doesn't say that anywhere, at least as far as the excerpts you've shown. It sounds like the formula includes only properties (the energy) of the microstate. As far as I understand a macrostate is simply grouping many microstates together and calling it the same' state because they have the same or very similar energy and other relevant properties. If you still think a macrostate is necessary, why couldn't a single particle have a macrostate?: you can group each microstate into a macrostate of one microstate.

The insistence on macroscopic' seems to mean that the whole system does not have to be viewed as being jointly in a single quantum state, but like classical objects individual molecules can be regarded as $$n$$ separate, noninteracting particles.

There really are difficulties with having the thermodynamic potential $$\Omega_{k}$$ for the $$n$$ particles in quantum state $$k$$, if you try to go from the familiar thermodynamic definition of $$\Omega$$ as constructed from entropy etc. But, the purpose of the last paragraph you quote is precisely to find the value of the constant $$a = e^{(\Omega_{k}-n_k\mu)/k_bT}$$ and of $$\Omega_k$$. Essentially $$\Omega_k$$ is inserted into this equation as an unknown variable, labelling it $$\Omega$$ in analogy to how in equations 53.1-35.2 the similar normalization constant $$A$$ turned out to be $$A=e^{\Omega/k_BT}$$.

Considerations like those you mention are just informal, intuitive illustrations, not stringent derivations. They anyway apply only to the simplest situations - ideal gases.

Landau and Lifshitz (and with them everyone else) are indeed somewhat inconsistent. This begins already when they introduce in Volume III wave functions for parts of a system, and later ensembles through the density matrix. The problem is that there is no consistent way to assign to a system both a macrostate and a microstate. I gave a detailed discussion of their assumptions and assertions on pp.21-22 of my paper

We can now see that the reasoning in the first citation is not sound! Gibbs distribution needs a macroscopic system that can have a macroscopic and a microscopic state to have sense, what is Landau taking as his macroscopic system in the first citation? The perfect gas? Can't be! In fact the energy $$\mathcal{E}_i$$ is not the energy of the entire microstate of the gas, the energy of all the particles in a specific state, instead it's the energy of a specific state that a single particle can assume! So what is he doing? Is he taking a single particle to be the system to which Gibbs distribution applies? But, if so, this makes no sense at all! A single particle is not a system, it cannot have a macrostate ad a microstate, Gibbs distribution is simply not defined for a single particle.

Landau and Lifshitz are simply applying the Boltzmann distribution formula (which they unfortunately call the Gibbs distribution) to states of single molecule. They do this as extrapolation of their previous considerations, where the formula was applied only to macroscopic systems. That is OK. Physics is not axiomatic science, one can just take formula derived for one case and try to apply it to a slightly different case. Sometimes it works (molecules of ideal gas), sometimes it doesn't (electrons, classical EM field, or stars in a globular cluster).