I'm reading the first chapters of Landau & Lifshitz's Statistical Physics and I don't understand the definition of the quantum microcanonical ensemble.

The microcanonical distribution for a quantum isolated system with energy $E_0$ is defined in the text as:

$$\text d w = \text{const}\times \delta (E-E_0) \cdot \prod _a \text d \varGamma_a.$$

where $\text d w$ stands for “probability” and $\text d \varGamma_a\!$ is the differential of the function $\varGamma_a(E_a)$, the number of quantum states of a subsystem with with $\text {energies}\leq E_a$ (which is treated as a smooth function).

I see two problems here: first of all, the distribution should be given in terms of a density operator $\rho$: I don't immediately see how to relate this density function to a matrix. However, forgetting for a second about the operator, what does $\text d w = ...$ mean? It is the probability density of what variables and how do these variables identify the state of the ensemble?

Finally, the text goes on asserting that, if the system is subdivided in $n$ small (quasi-)independent subsystems of the original system, then $\text d \Gamma$ can be written as the product $\text d \Gamma_1\cdots \text d \Gamma _n$. This fact is used later to derive the second law of thermodynamics: the joint probability density for the values $E_1 , \dots ,E_n$ of the subsystems energy is written as

$$\qquad\qquad\qquad\qquad\qquad \text d w = \text{constant} \times \delta (E-E_0)\cdot\prod_a \dfrac{\text d \varGamma_a}{\text d E_a} \text d E_a. \quad\qquad\qquad(7.15)$$

This one looks like the meaningful equation (to express the joint probability density of energies in the microcanonical ensemble). Is Landau's notation just a short-hand for the above one?


3 Answers 3


The first thing to point out is that there are two equivalent ways to describe a quantum statistical distribution: the density matrix, and a probability distribution on the results of measurements of a "complete" set of observables. (It is remarkable that one should think of givin such a probability distribution on the classical phase space of the quantum system, but let's not go into this here. It is remarkable since it seems to go against the whole principle of quantum mechanics to assign a probability to a point with simultaneously assigned values of both position and momentum. But it works somehow anyway.)
Landau explains this equivalence between $w$ as a probability distribution and $w$ as a density matrix. In his formula he wants to point out the analogy with classical mechanics and for that reason interprets $w$ as a probability distribution.

The microcanonical ensemble consists only of states with a given fixed energy $E_0$: any other value of the energy has a zero probability of resulting from a measurement. But of course there are other commuting observables, and they will take on possibly different values when measured, so $w$ is a distribution on those values. But it is, by definition, somehow "constant". Other statistical distributions on this same set of states would be possible, but would not be the microcanonical one.

The second edition of Landau--Lifschitz explicitly says that $dw$ is the probability of finding the system in one of the states that "belongs" to one of the energy levels infinitesimally close to $E_0$. He also says explicitly that $d\Gamma$ is the number of quantum states with energies in that same infinitesimal range around $E_0$. It is very hard to make rigorous sense out of this, although it has been done by other authors. That is why the density matrix is often preferred.

In fact the multiplication by the delta function means all states with a non-zero probability have the same energy. So $w$ on that ensemble is not a function of $E_0$, since $E_0$ is a parameter. But if you look at that formula, there are no other variables except $E$. So $w$ is constant on that ensemble, i.e. every state has the same probability. So we don't really need a formula at all: the constant $d\Gamma$ doesn't affect relative probabilities, and if you normalise this to make them genuine probabilities you just divide by $d\Gamma$.... Landau was a great physicist but he was not a very logical writer. Nor a very clear one. And throwing a junior co-author into the mix just makes things worse...

Varying the parameter and integrating $w$ over $E_0$ makes no real sense for the micro-canonical distribution.

The point is if you wish to calculate the entropy of the microcanonical distribution, as a function of the parameter $E_0$ (which is what Landau does in the very next section). Because the normalised density matrix has to have trace one, and be diagonal, and since every probability is the same, the value of the diagonal elements is just the reciprocal of the dimension of the space of states with that energy, which is the number or elements in a basis, which is what Landau really means by "the number of quantum states that 'belong' to an energy in that infinitesimal range." (Literally, of course, the number is uncountably infinite, not finite.)

I never recommend reading Landau unless you already understand the basic concepts. Landau is the kind of chess grandmaster who can be champion of the world but could not explain to you in writing how a knight moves.

  • 1
    $\begingroup$ But I did, just today, recommend reading Landau, and the pages next to the ones you are asking about, to an expert physicist with 40 years of experience in nuclear spectrocopy..... $\endgroup$ Nov 27, 2015 at 7:05
  • $\begingroup$ Thank you Joseph. About your very first paragraph, I'm not sure about what do you mean by "quantum statistical distribution": ofcourse specifying the probability of eigenstates of a single complete set of commuting operators doesn't determine the quantum state, since it cannot determine phases beetween the eigenkets. However, I (now!) understand that the idea of statistical QM is that these phases are somehow "lost" (at least in the energy basis), so I wonder if you are talking about a notion of "statistical distribution" more specific than "quantum state i.e. density matrix". $\endgroup$
    – pppqqq
    Nov 27, 2015 at 13:12
  • $\begingroup$ This requires an entire answer in itself. Briefly, it means the probability of obtaining x as the result of measuring the observable Q, where x ranges over all real numbers. But there is a confusion in your comment between mixture and superposition. $\endgroup$ Nov 27, 2015 at 15:29

I understand this formula in the following way:

$dw$ is a probability measure that aims at being general. It can be thought as being the probability measure for having a quantum system at energy $E_0$. The Dirac mass $\delta(E-E_0)$ specifies in any case the ensemble while $d\Gamma$ specifies the "natural" measure used to count the number of states.

From the way you phrased it (I don't have the book here to see for myself), it is implicitly assumed that there is no preference given to any quantum state when thinking of the function $\Gamma(E)$.

If that is true then you get that $d\Gamma = \frac{d\Gamma}{dE}dE = \Omega(E)dE$

where $\Omega(E)$ is the density of states. Replacing that in your formula you get that: \begin{equation} dw(E) = {\rm constant} \times \delta(E-E_0) \Omega(E)dE \end{equation} upon integration of the probability measure and imposing it to be normalized, you get:

\begin{equation} {\rm constant} = \frac{1}{\Omega(E_0)} \end{equation}so that: \begin{equation} dw = \frac{\delta(E-E_0)d\Gamma}{\Omega(E_0)} \end{equation}

Now, nothing prevents you from expressing $d\Gamma$ as being related to a more refined uniform measure on your Hilbert space of states.

To make an analogy, consider the measure $\mu$ of all the points inside a sphere of radius $R$ such that $\mu(R) = 4/3\pi R^3$ essentially "counts the number of points" which are at a distance less than $R$ from the center.

The notation $d\mu$ is a shorthand that can be used for anything ranging from $4\pi r^2 dr$ to $d\mu_{Lebesgue} = dxdydz$. The only constraint really is that whatever choice you pick, you have $\int d\mu = \mu(R)$.

More mathematically, I think this can be linked to the Radon-Nikodym theorem except that in the case of Landau, the notations used are ambiguous which is not the case is usual measure theory.

  • $\begingroup$ thank you for your answer. I agree that the symbol $\text d w$ is not referred to the proabability of values for a particular variable $A$ to which the probability density is referred. In fact the notation is used with this meaning in various parts of the book. However I'm not really convinced with the interpretation of $\text d \Gamma$ as a measure. $\Gamma(E)$ is precisely defined as the number of eigenstates with energy $E'\leq E$. The more I think about this, the more I'm convinced that this is just a cumbersome way to say that the probability of $N$ allowed states is $\propto N$. $\endgroup$
    – pppqqq
    Mar 12, 2015 at 20:14
  • $\begingroup$ if $d\Gamma$ is a mathematical object that ought to count stuff, what else (apart from a measure) could it be? Just to be sure that we are on the same page, a measure can be used to count the number of elements in a discrete set as well. $\endgroup$
    – gatsu
    Mar 12, 2015 at 23:25

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."


"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."

  • $\begingroup$ sorry, I should have stated my doubts more clearly. I'm not confused about the meaning of the density operator and, when talking about "quantum state" in my comment, I had in mind the general notion of "density operator" which the theorem you cite exhibits as the correct one of "quantum state". My point was much more simpler: you state that "a probability distribution on the results of measurements of a complete commuting set of variables" determines the statistical distribution, here understood as "quantum state" i.e. "density matrix". [...] $\endgroup$
    – pppqqq
    Nov 28, 2015 at 11:01
  • $\begingroup$ [...] But this is ofcourse false. Take for example a spin 1/2 system with basis $\lvert \pm \rangle $. Then { $S_z = \lvert + \rangle \langle + \rvert - \lvert - \rangle \langle - \rvert$ } is a complete set of observables for the system, and a probability distribution $p(\pm )=\frac{1}{2}$ for its two eigenstates cannot yield predictions about the results of measurements of, say, $S_x = \lvert + \rangle \langle - \rvert + \lvert - \rangle \langle + \rvert$, i.e. it cannot determine the quantum statistical distribution. $\endgroup$
    – pppqqq
    Nov 28, 2015 at 11:05
  • $\begingroup$ Yes, thanks. I have fixed the errors. Now, as to something else in your comment, there is not a universally observed agreement on terminology. Most precisely, a quantum state is a pure state. Not a mixed state. But people are sloppy about this. There are many physicists who now feel that one ought to revise the axioms of QM so that the basic axiom says that a system is in a mixed state, not necessarily a pure state. This is what you assert in your comment. But this is not the consensus. QIT people tend to agree with you, Quantum Optics people not so much. $\endgroup$ Nov 28, 2015 at 15:53
  • $\begingroup$ Perhaps this is because Quantum Optics technology has advanced to the stage where people really are trapping individual photons. But Quantum Computing technology has not advanced nearly so much. So they believe mixed states are something fundamental. Whereas Landau, who invented them, was clear that they were due to the incompleteness of our knowledge. $\endgroup$ Nov 28, 2015 at 15:55
  • $\begingroup$ Thanks for the attention Joseph. Totally agree with you that the word "state" in QM often requires some specification, expecially in statistical mechanics context. I've recently spent a consistent amount of time reading about decoherence, so it's quite natural to me to say "state" when I mean precisely "density operator". $\endgroup$
    – pppqqq
    Nov 28, 2015 at 16:50

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