My textbook (Moore Physical Chemistry) defines a microcanonical ensemble by considering an isolated system containing $N$ molecules of gas with fixed volume $V$ and energy $E$. It states "we can construct an ensemble of $\mathcal{N}$ such systems, each member having the same $N$, $V$, $E$" and this is called a microcanonical ensemble. It then states two postulates for such an ensemble:

  1. The time average of any mechanical variable $M$ over a long time in the actual system is equal to the ensemble average of $M$ (the book notes that strictly speaking, this only holds in the limit $\mathcal{N}\to \infty$)

  2. The members are distributed with equal probability over the possible quantum states consistent with the specified values of $N$, $V$, $E$.

I have a few questions about this definition and the following postulates.

  1. It is not explicitly stated how many members $\mathcal{N}$ the ensemble must contain. The book states that "we can have any number of members of the ensemble to represent each state of the system (so long as it is the same number for each one)." I take this to mean that the ensemble must contain at least one member for each possible quantum state the system may attain (the word state seems to mean many different things (e.g. state of a system, state of an individual particle, microstate, macrostate, etc.) so I am not fully sure what it is referring to in the previous quote). However, I am not sure what the significance of having multiple members of the ensemble with the same quantum state as it seems redundant (in the sense that having multiple copies of the same quantum state in the ensemble will not contribute anything to the ensemble average because the states are exactly the same). Furthermore, to consider the limit $\mathcal{N}\to \infty$ required for the first posulate, it seems that the ensemble under consideration must contain infinitely many systems corresponding to each of the possible quantum states and again I don't see how the ensemble average of properties for this ensemble will be any different from the ensemble average of properties from the ensemble containing exactly one system for each possible quantum state.

  2. When considering microcanonical ensembles, is the word microstate synonymous with "one of the systems in the ensemble" or is there a difference.

  3. I am not fully sure how to rationalize the idea of a microcanonical ensemble within the realm of classical mechanics. In classical mechanics, the system can (in theory) occupy an uncountably infinite number states with the same $N$, $V$, $E$. The book describes: "in classical mechanics, we can imagine phase space to be divided into cells of volume $\delta \tau$; we can then imagine a microscopic state to be specified by giving the particular cell in phase space in which the system is situated." I was hoping someone could explain what exactly is meant by this phrase as I would imagine a microstate to involve a collection of points in phase space, representing the individual states of each of the particles, not an element of a discrete volume cell. Furthermore, it is not immediately clear why each of these cells must correspond to systems with the same $N$, $V$, $E$.

I apologize in advance for the length of my question.

  • $\begingroup$ Better split this into multiple independent questions. $\endgroup$ Nov 11, 2023 at 22:32
  • $\begingroup$ @JánLalinský As my answer should show, the three points raised in the question are interdependent. This is a clear case where a numbered list does not imply a non-focused question. $\endgroup$ Nov 11, 2023 at 23:17

1 Answer 1


The definition of a statistical ensemble, as in the citation, is usual in statistical mechanics. However, in a certain sense, it is not necessary. What matters is the definition of a probability measure on a suitable set of microscopic states of the physical system. The description of the ensemble as a set of an eventually infinite number of copies of the original system in different microscopic states is just a pictorial way to represent the probability measure in the frequentist interpretation of probabilities.

  1. The previous observation explains both the issues related to i) the number of microstates (a more probable microstate has to appear in more elements of the ensemble than a less probable) and ii) the limit $\cal N \rightarrow \infty$ is required by the frequentist definition of probability.

  2. Again, from previous observation, in a microcanonical ensemble, where each microstate is equiprobable, each state at a fixed energy value must appear the same number of times as any other in the ensemble. Even only once. In such a case, the ensemble may be realized by collecting all the microstates only once.

  3. In the case of classical mechanics, if a microstate would coincide with a mechanical state, i.e., a given set of values of the generalized coordinates and momenta, we would face the usual problem of every probability space with a continuous sample space: a point is always a set of zero measure, and consequently, zero probability. We must identify as events measurable set in the sample space. It is then convenient to use as a microstate the non-zero measure set obtained by a coarse-graining procedure where all the values of coordinates and momenta within a fixed phase space volume are assigned to the same microstate.

  • $\begingroup$ Thank you! When you say "each state at a fixed energy value must appear the same number of times as any other," do you mean each possible quantum state that has the same energy value $E$? Because as I understand it, in a microcanonical ensemble, $E$ is a constant for every member of the ensemble. Also, I understand your point about N -> infty being required for the frequentist definition. However, if we achieve this limit by adding more copies of the same quantum state to the ensemble. It feels like after each state is represented once, the average is fixed and won't change as N grows large. $\endgroup$ Nov 12, 2023 at 0:59
  • $\begingroup$ @JohnnySmith Yes, in the microcanonical ensemble, each microstate at energy $E$ must appear the same number of times as any other state with the same energy, and every state with a different energy doesn't appear. About the limit at large $\cal N$, it is mathematically required if you want to get frequencies represented by real numbers like $e^{-\beta E}$. $\endgroup$ Nov 12, 2023 at 8:10
  • $\begingroup$ Thank you!! If every state appears the same number of times as every other state, isn't the probability distribution automatically uniform (and not Boltzmann). Also, for a microcanonical ensemble, if all the states have the same energy, the weights e^(-beta *E) would all be the same. I assume these weights apply for canonical ensembles. However even for canonical ensembles, it doesn't make sense why simply adding replicas of already represented states will give convergence to real valued frequencies. Normally, in the frequentist interpretation, they talk about large, unique samples $\endgroup$ Nov 12, 2023 at 12:32
  • $\begingroup$ @JohnnySmith "in the frequentist interpretation, they talk about large, unique samples". Yes, but the justification of 1/2 probability of head or tail is that the same microstate ( head, for example) is repeated many times in the long run. The sample is unique, say one billion coin launches, but it is the same output about half of the time. $\endgroup$ Nov 12, 2023 at 12:41
  • $\begingroup$ That makes sense, but in that example, you don't have the restriction that for every ensemble, every possible quantum state is repeated the same number of times. This seems more akin to a scenario where I have ensembles of the form {H, T}, {H,H,T,T}, {H,H,H,T,T,T} where in constructing them, I have already guaranteed that the number of heads equals the number of tails to the proportion of heads is equal to tails in every sample and you don't see any sort of convergence (because as I understand, any two states at the same quantum state have exactly the same macroscopic parameters) $\endgroup$ Nov 12, 2023 at 12:46

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