Undergrad Do ensemble-based predictions truly describe real finite systems?

[syed]

TL;DR

Title

In both classical and quantum statistical mechanics, we often rely on the concept of ensembles, either as a collection of hypothetical copies of a system in different microstates (classical) or as a weighted mixture of quantum states represented by a density matrix (quantum), to predict equilibrium properties, entropy changes, and the apparent arrow of time.

However, real physical systems are always finite and in continuous contact with finite environments. If the actual universe consists of a single, finite system rather than an infinite ensemble, how can we be sure that the predictions derived from ensembles (essentially averages over many possible microstates )accurately capture the typical time evolution of that one real system, rather than just describing an abstract average that may not correspond to its actual trajectory?

Does this reliance on ensembles introduce a fundamental conceptual gap between the idealized theory and the behavior of actual, finite classical or quantum systems, and if so, what are the limitations of using ensembles to describe reality?

 

[Demystifier]

A canonical or grand-canonical ensemble is best understood as a small (but macroscopic) subsystem of a larger system. The larger system is usually assumed to be in a micro-canonical ensemble, meaning that its energy is well defined with almost perfect precision. Under these conditions the canonical ensemble of the small subsystem can be derived, with almost no additional assumptions. See https://arxiv.org/abs/cond-mat/0511091

 

[Lord Jestocost]

TL;DR Summary: Title

Does this reliance on ensembles introduce a fundamental conceptual gap between the idealized theory and the behavior of actual, finite classical or quantum systems, and if so, what are the limitations of using ensembles to describe reality?

As far as the modelling of the behavior of actual, finite systems is concerned, the current models based on statisticsl physics are very successful.

 

[Demystifier]

what are the limitations of using ensembles to describe reality?

There is a simple but general principle how to determine whether a statistical ensemble describes well a real single system. The statistical ensemble is just a conceptual model for the notion of probability distribution. From the probability distribution $p(x)$ (where $x$ is a microscopic state) you can compute not only the average value of an observable
$$\langle O\rangle \equiv \int dx\, p(x)O(x)$$
but also the fluctuation (i.e. standard deviation) $\Delta O$ given by
$$(\Delta O)^2 = \langle O^2\rangle - \langle O\rangle^2$$
In general, the average value $\langle O\rangle$ of the ensemble does not need to represent well the actual value $O(x)$ of a single system in the actual microstate $x$. However, if the fluctuation is small, in the sense that
$$\frac{\Delta O}{\langle O\rangle} \ll 1$$
then the average value represents well the actual value, i.e. we can be pretty certain that
$$\langle O\rangle \simeq O(x)$$
Typically this happens when the system contains a large number $N\gg 1$ of particles (or some other elementary constituents) and $O(x)$ is an observable that scales with $N$, so the approximate equality above is valid due to the law of large numbers. Typically $(\Delta O)^2$ then also scales with $N$, so
$$\frac{\Delta O}{\langle O\rangle} \sim \frac{\sqrt{N}}{N} = \frac{1}{\sqrt{N}} \ll 1$$

Let us illustrate it by a simple example. Suppose that you have a single system containing $N=10^6$ unbiased coins, each of which is in the state tail or head. Let us associate a value 0 with each tail and a value 1 with each head. A typical microstate is something like
$$x=(0,1,0,0,1,0,1,1,1,0, ...)$$
meaning that the first coin is in the state tail, second coin in the state head, etc. Let $O(x)$ be defined as the sum of all these 0's and 1's. Clearly, the average value of $O$ is
$$\langle O \rangle = 0.5 \cdot 10^6$$
The computation of $\Delta O$ is more complicated, but it is of the order of
$$ \Delta O \sim \sqrt{N}=10^3$$
Hence
$$\frac{\Delta O}{\langle O\rangle} \sim \frac{10^3}{0.5 \cdot 10^6} = 2\cdot 10^{-3} \ll 1$$
Thus we can say that the actual value of $O(x)$ is very close to the average value $0.5 \cdot 10^6$.

 

[syed]

There is a simple but general principle how to determine whether a statistical ensemble describes well a real single system. The statistical ensemble is just a conceptual model for the notion of probability distribution. From the probability distribution $p(x)$ (where $x$ is a microscopic state) you can compute not only the average value of an observable
$$\langle O\rangle \equiv \int dx\, p(x)O(x)$$
but also the fluctuation (i.e. standard deviation) $\Delta O$ given by
$$(\Delta O)^2 = \langle O^2\rangle - \langle O\rangle^2$$
In general, the average value $\langle O\rangle$ of the ensemble does not need to represent well the actual value $O(x)$ of a single system in the actual microstate $x$. However, if the fluctuation is small, in the sense that
$$\frac{\Delta O}{\langle O\rangle} \ll 1$$
then the average value represents well the actual value, i.e. we can be pretty certain that
$$\langle O\rangle \simeq O(x)$$
Typically this happens when the system contains a large number $N\gg 1$ of particles (or some other elementary constituents) and $O(x)$ is an observable that scales with $N$, so the approximate equality above is valid due to the law of large numbers. Typically $(\Delta O)^2$ then also scales with $N$, so
$$\frac{\Delta O}{\langle O\rangle} \sim \frac{\sqrt{N}}{N} = \frac{1}{\sqrt{N}} \ll 1$$

Let us illustrate it by a simple example. Suppose that you have a single system containing $N=10^6$ unbiased coins, each of which is in the state tail or head. Let us associate a value 0 with each tail and a value 1 with each head. A typical microstate is something like
$$x=(0,1,0,0,1,0,1,1,1,0, ...)$$
meaning that the first coin is in the state tail, second coin in the state head, etc. Let $O(x)$ be defined as the sum of all these 0's and 1's. Clearly, the average value of $O$ is
$$\langle O \rangle = 0.5 \cdot 10^6$$
The computation of $\Delta O$ is more complicated, but it is of the order of
$$ \Delta O \sim \sqrt{N}=10^3$$
Hence
$$\frac{\Delta O}{\langle O\rangle} \sim \frac{10^3}{0.5 \cdot 10^6} = 2\cdot 10^{-3} \ll 1$$
Thus we can say that the actual value of $O(x)$ is very close to the average value $0.5 \cdot 10^6$.

Thank you for the detailed explanation