I am going to focus on the precise meaning of the density matrix and the probability distribution $w$ in Landau's notation, which could be called the
quantum statistical distribution.
Almost all texts and wikipedia mess this up.
If the usual axioms of quantum mechanics are accepted, every system is in a
pure state, given by a wavefunction. Sometimes this is not a very useful
thing to know. As a practical matter, a light beam, for example, may seem
unpolarised. To speak of an unpolarised light beam is an abstraction, since
in fact every photon in it must have a definite state of polarisation, but
it is usually impractical to gather the information about those polarisations
and so to treat the beam as "unpolarised" is a very useful abstraction,
which led von Neumann and Landau to introduce a new concept into Quantum
Mechanics, the density matrix.
The density matrix can be motivated in two different ways. The first way,
which is very concrete, uses the new notion of a "mixed state". We suppose
that we only know the probability with which we will find a photon in that
beam to be polarized up, for example, suppose it is 50 percent. Then the
same measurement of its polarisation has a 50 percent probability of yielding
the result "horizontally polarised" instead of vertically polarised.
We can model our incomplete knowledge of the state of each photon in this beam
as a probabilistic mixture 1/2 (vertical) + 1/2 (horizontal). It is important
to realise that these coefficients are probabilities, and this is not a superposition
of states, but a mixture of states. Each photon is thought of as being in
one or the other of these two states, $v_1$ and $v_2$, and not as being in a
superposition of the two states. It really ought not to be written with a
plus sign, so as to avoid confusion with superposition!!! Well, what is
the expectation of a measurement of the polarisation if we assign the value
1 to vertical and 0 to horizontal? The expectation is, by the rules of
classical probability, 1/2. But something like this is true for every
observable, so by the laws of Quantum Mechanics plus the laws of classical
probability applied to our incomplete knowledge, we obtain a well-defined
mapping from this mixture to expectation values of observables: if $Q$ is
an observable, its expected value for this mixture is $1/2 E(Q,v_1) + 1/2
E(Q,v_2)$ with the obvious notation. But something like this is true for every mixture,
no matter what the coefficients are. Von Neumann proved that for any given
mixture, there exists a two by two matrix A such that this mapping is
given by trace$(AQ)$. This matrix, called the density matrix, describes the only
features of the mixture that we can conveniently observe by experiment. Exercise: find the formula for the entries of $A$ in
terms of the probability weights of $v_1$ and $v_2$.
Exercise 2: suppose we instead consider the pure state $.707107 v_1 - .707107 v_2$.
So the only probability weight is a 1. What is the
matrix $A$? (Check your work: its square is equal to itself, but it is not diagonal.)
The other motivation is less algebraic at first. Based on physical reasoning,
one can axiomatise the properties you would expect of the mapping that takes an
observable Q to its expectation on a fixed mixture of states. Then one can
prove that every such mapping is given by that same trace formula, for some $A$,
and that the $A$ is unique, and that different mappings yield distinct $A$.
The point of view is, the only results of experiments that we can obtain with this mixture
are, for each observable, set up the appropriate measuring apparatus for that observable, and measure the expectation value of that observable on this mixed state.
So we invent a new concept that only depends on these values. We consider the map which takes each mixed state to the set of expectations of a "complete" set of observables.
It is a linear algebra theorem that every such map, satisfying certain physically reasonable axioms, is given as trace$(AQ)$ for some $A$.
As always in classical probability theory, a knowledge of all the expectations determines the individual probabilities of each possible value of a measurement, so we have a probability distribution on the possible results of measuring, simultaneously, a "complete" set of observables.
Suppose, to fix ideas, each of these observables has a discrete set of eigenvalues, so the simultaneous eigenstates are described by quantum numbers.
Then the possible values are vectors, indexed by the observables in this set, whose entries are the quantum numbers.
This is a discrete set, so we have a probability distribution on a discrete set.
But if the eigenvalues form a continous range, as with position $q$ and momentum $p$,
then we get a continuous probability distribution.
There is another, more profound, motivation for the notion of a density matrix, based
on the idea that a subsystem $H_1$ of a larger system $H$, with which it is in interaction, so it does not itself form a closed system, need not be in a pure state: its
state might be entangled with the rest of $H$. Landau analyses this situation and shows that the expectation of the measurement of an observable that ignores the rest of $H$ and only interacts with $H_1$ satisfies the same axioms and must be given by
"tracing out over the ignored variables". He does this without assuming anything about mixed states or a probabilistic mixture, in fact, warns against interpreting
the density matrix that results in this way as if it were a probabilistic mixture.
Where the mess-ups are. The only thing we observe in this situation are the expectations (or probabilities of the results of measurements). We do not observe what state the individual photon was in before the measurement. So we never really justify the statement that the probability that a photon in this beam is in the state $v_1$ is 50 percent. So we have a density matrix, we have a probability distribution $w$ on the results of measurements, but we do not have a probability distribution on quantum states. To see that we never get access to this last, note that different probabilistic mixtures yield the same density matrix. Let $u_1= .707107v_1 + i v_2/1.41414$ and let $u_2= .707107v_1 - i v_2/1.41414$. (Note, we assumed the $v$'s were orthonormal.) (Here $i$ can be either one of the two square roots of negative one, it makes no difference.) Calculate the density matrix of the probabilistic mixture of $u_1$ at 50 percent and the rest of the time, $u_2$. You get the same density matrix as before. But as probability distributions on states, these are incompatible. They yield the same observable probability distributions and expectation values of measurements, so they are (by assumption) impractical to distinguish without spending more money than we are prepared to spend.
Landau explicitly warns against thinking that the density matrix represents a probability distribution on states. He calls our initial motivation of the density matrix, where we considered what would happen if our incomplete information about the photons in the polarised beam could be described by that probabilistic mixture of states, "merely formal". And now if you think physically about where unpolarised light beams come from (the sun, a lamp...) you see that it is ridiculous to think that half the photons are polarised vertically. It is more reasonable to suppose there is a uniform, continuous distribution of polarisations, including circularly and elliptically polarised photons... but in physics we never need to work with such a continuous classical probability distribution on this space of states. The density matrix does not describe a classically mixed system of pure states, it is a new quantum concept that models the incompleteness of our knowledge in a new, different, and specifically quantum way.
(Mackey and others have worked out something involving quantum logic in this regard. I do not think physicists have found it useful.)
The other mess-up is to keep saying and writing that one is "counting the number of states in the assembly that satisfy the given conditions" when what is really doing is something that can be expressed accurately in two alternate ways: counting the dimension of the space of states in question (this is what Mackey does),
or, choosing an orthonormal basis for this space and then counting the number of
elements in the basis. This makes the analogy with the Liouville measure in the Classical Stat Mech case weaker and more roundabout, although it's still there in some sense
(which Mackey made precise). What makes the way people mostly write about this even
worse is they write as if the system had to be in one of those eigenstates. Even Landau
writes this way, as I point out in my other answer. A quantum state of a system is also called a "pure state". One of these mixtures is, although it ought not to be, called a "mixed state". To confuse the situation, the knowledge given us by the density matrix (which is even more incomplete than what would be given to us if we had one of those mixtures) is also called a "mixed state". Now less in use is the terminology that an eigenstate of (usually) the energy is called a "sharp state" (except in the context of, say, spin or polarisation: here, a sharp state is one of the eigenstates of the polarisation observable we are considering). It is one thing to assume the system is in a pure state, but it is unjustifiable to assume it has to be in a sharp state. Schroedinger criticises this
quite severely in his short, classic, book "Statistical Thermodynamics". He writes, pp. 5f. and pp. 89--90,
"To ascribe to every system always one of its sharp energy values is an indefensible attitude.
It was challenged in the beginning of Chapter II, yet it was adoped throughout this treatise as a customary convenient short-cut.
The Appendix added to the Second Edition contains the general proof, that a consistent procedure, based on very simple assumptions, always gives the same results. The thermodynamical functions depend on the quantum-mechanical level-scheme, not on the gratuitous allegation that these levels are the only allowed states."
"We are faced with an assembly of $N$ identical systems. We describe the nature of any one of them by enumerating all its possible states, which we label $1,2,3,4,\dots,l,\dots$.
In principle we have always in mind a quantum-mechanical system whereby the states are to be described by the eigenvalues of a complete set of commuting variables."
\dots
"We shall always regard the state of the assembly as determined by the indication that the system No. 1 is in state, say, $l_1$, No. 2 in $l_2$,\dots , No. $N$ in $l_N$.
We shall adhere to this, although the attitude is altogether wrong.
For, a quantum-mechanical system is not in this or that state to be described by
a complete set of commuting variables chosen once for all.
To adopt this view is to think along severely `classicl' lines.
With the set of states so chosen, the system can, at best, be relied upon as having
a certain probability amplitude, and so a certain probability, of being, on
inspection, found in state No. 1 or No. 2 or No. 3, etc.
"The determination of the statistical entropy of a thermodynamical system always
rests on counting the number of permutations that comply with certain restrictions, or, speaking in terms of physics,
counting the number of different microstates that do not differ for the macroscopic observer
because they all agree with regard to the macroscopic properties which alone can be observed by him\dots
" For performing this count we have, in this book, always adopted the customary method, which rests on the assumption that every system \dots always finds itself in a state of sharply defined energy \dots
"In the beginning of Chapter II it was mentioned that this assumption is irreconcilable with the very foundations of quantum mechanics."
