I seem to find a contradiction in the notion of probability density used by Landau and the notion of micro-canonical ensemble.

To see this, take an isolated classical system and we know experimentally that its energy lies between $E-\Delta$ and $E+\Delta$. So, we take a hypershell corresponding to these energies in phase space and say that at equilibrium, the probability density is constant in the whole shell. Now, we know that the system would be, in reality, at a fixed energy E' and the hypersurface corresponding to this energy would lie in the previous hypershell. Also, as the system is isolated, the representative point of the system would move only on this hypersurface. Now, take a point in the shell which lies outside the surface. Choose a small enough neighborhood of it that doesn't intersect the surface. Because, the probability distribution is constant, the probability of finding the system in this neighborhood is some non-zero positive number. But, as the system always remains on the surface, it never visits that neighborhood and hence the probability of finding it in that neighborhood is zero.

Am I doing something wrong?

  • $\begingroup$ Sorry, I personally found it very hard to find out what you're asking. First of all, it is totally unclear whether you want to talk about classical physics or quantum physics: the title and the word "shell" suggests that it is classical physics, the word "Hilbert space" makes it equally clear that you want to talk about quantum mechanics. Second, the rest of the text suggests that you don't want to allow distributions that are nonzero for states of differing values of energy: you would prefer if only states with a sharp, fixed energy were allowed. But I didn't understand why you think so. $\endgroup$ Sep 19, 2012 at 9:02
  • $\begingroup$ I am really sorry over there. I meant phase space and not Hilbert space. Guess, I wasn't paying much attention while typing. I will edit it. My question is in the domain of classical physics. $\endgroup$
    – user4235
    Sep 19, 2012 at 9:13
  • $\begingroup$ No prob! Still, the bulk of the question seems hard to deal with. The energy may be sharply known but it may be known with an uncertainty. Microcanical ensemble says that all states in the thick shell, interval, are equally likely. The energy is still conserved. If you're on a surface of a fixed E to start with, you will remain on the same surface. But if you don't know what the exact surface is, and in microcanonical ensemble, you don't know what it is for the initial state, you won't know the surface for the final state, either. $\endgroup$ Sep 19, 2012 at 9:48
  • $\begingroup$ @LubošMotl The energy is known with an uncertainty as mentioned in the question. My confusion is the apparent contradiction between the uniform probability distribution in the shell (as the system is in micro-canonical ensemble) and the meaning of this probability distribution as mentioned by Landau in his book (which I reviewed in the comment of the only answer to this question yet). Landau's definition implies that the probability distribution should be zero outside the "real" surface of energy $E'$ as I argue in the question above. $\endgroup$
    – user4235
    Sep 19, 2012 at 10:29

2 Answers 2


No, you're not doing anything wrong, this is all correct. As an analogy, imagine I roll a die and hide it under a cup. Since you don't know which side of the die is facing upward, you represent it with a probability distribution, with an equal probability assigned to each of the six spaces. This probability distribution doesn't change over time, in this case for the trivial reason that the die isn't moving.

You know that in reality, the die is sitting there with one particular side facing upward, and that it never "visits" any of the other sides. But unless I lift up the cup, you have no choice but to keep on thinking of it as being in a probability distribution, because you don't know which state is the true one.

With the microcanonical distribution it's the same. There is indeed one "true" energy $E'$ that doesn't change, and the system cannot visit states with any other value of $E$. But the assumption is that you don't have any way to measure the energy beyond a certain level of accuracy. So, in the analogy, the die remains hidden under the cup and you have to keep representing it with a probability distribution.

Although many text books fail to make this clear (because it was widely misunderstood for much of the 20th century), the probability distribution doesn't represent the set of states the system can visit, it just represents experimental uncertainty about which state the system is in. It is this uncertainty that remains invariant in equilibrium.

  • $\begingroup$ Hi, thanks for your answer. I understand what you are saying but the part that troubles me is the definition of this probability distribution attached to the system as defined by Landau and Lifschitz in their book. They define the probability distribution, $p$, such that if I take a neighborhood of volume $V$ in phase space, then $pV=\lim_{T \to +\infty}\frac{\Delta t}{T}$ where $\Delta T$ is the time system is found in that neighborhood if it is observed for time $T$. This definition implies that $p=0$ at any point outside the surface. $\endgroup$
    – user4235
    Sep 19, 2012 at 9:43
  • $\begingroup$ I haven't read Landau and Lifschitz. I'm aware that their book is famous and respected, but still I wouldn't be too surprised if they were just being inconsistent about this. The view I expressed in my answer is widely held now, but during much of the 20th century there was a strongly held opinion that the only reasonable definition of probability was the amount of time the system spends in a given state in the infinite limit (essentially the formula you quote). But that just turned out not to be a very useful way of thinking about it. $\endgroup$
    – N. Virgo
    Sep 19, 2012 at 10:47
  • $\begingroup$ Since from what you say they're talking about knowledge and measurement in saying that $E$ is known with uncertainty, it could even be that they're paying lip-service to the "standard" definition in terms of the infinite time limit, while actually doing calculations that refer to the uncertainty interpretation. That's a bit of a wild guess on my part though. $\endgroup$
    – N. Virgo
    Sep 19, 2012 at 10:54
  • $\begingroup$ Do you have some link where the problems with this infinite time limit is discussed? I have come across many different definitions of probability distribution and the successive establishment of formalism and the only way that appealed to me was Landau's way of doing things. In other definitions, the notion of ensemble doesn't make much sense to me. So, what is the currently accepted definition of probability distribution and the accepted meaning of ensemble? $\endgroup$
    – user4235
    Sep 21, 2012 at 15:44
  • $\begingroup$ The current sort-of-more-or-less-kind-of-accepted definition is the Bayesian one, where the probabilities represent an experimenter's imperfect knowledge of the system, and the "ensemble" arises because from the experimenter's point of view there are many possible states that the system might be in (although of course only one is the correct one). A very good (but long and at times a bit polemically worded) explanation of this can be found in Edwin Jaynes' Where do we Stand on Maximum Entropy?. $\endgroup$
    – N. Virgo
    Sep 21, 2012 at 16:43

In classical framework one defines an isolated system as that which is not interacting with any other system and thus whose energy is fixed.

Now the (naive) hypothesis would be that upon observation an isolated system will be found on its constant energy surface and its probability to be found in any of the states on its constant energy surface will be equal.

However as such this hypothesis is self contradictory because the very act of observation will make the system non-isolated and so in particular its energy may change by act of observation.

So one refines the hypothesis as:

Upon observation an (initially) isolated system (of energy $E$) will be found with equal probability in any of the states between energy $E-\Delta$ and $E+\Delta$ (Here $\Delta$ is a small number that takes into account the disturbance that your act of observation may produce.)

  • $\begingroup$ I don't think what you propose is correct. From whatever I know about classical statistical mechanics, it is the physical information about the system that constrains what region of phase space we are concerned with and no measurement is performed thereafter, that is, we analyze the problem statistically from that point onward. It is not that we are first concerned with the whole phase space and then some physical information about the system appears and we expect statistical probabilistic laws to still hold true given this new information. $\endgroup$
    – user4235
    Sep 21, 2012 at 15:49
  • $\begingroup$ In the end physics has to connect itself to experiments, and the point is that no experiment can ever observe a purely isolated system. $\endgroup$
    – user10001
    Sep 22, 2012 at 1:10

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