Statistical Thermodynamics and Rate Theories/Microcanonical ensemble

A microcanonical ensemble is a statistical ensemble which is used to represent the possible states of a mechanical system in which the total energy of the system is exactly specified. A micro-canonical ensemble is assumed to be isolated, thus the system cannot exchange energy or particles with the environment. The system, which can be composed of a solid, liquid or a gas is completely isolated from its surroundings and has constant energy (E), number of particles (N), and volume (V). This ensemble composes of an assembly of “copies” of this isolated system; the energy of each copy is the same, thus E=U, considering one copy of this system is sufficient.

Combinatorial Mathematics
Even though the energy of each copy of the system in a microcanonical ensemble is the same, these copies can exist in different quantum states. For example, consider a system composed of 2 hydrogen molecules, with the total vibrational energy of 2hυ.

$$E_{vib} = \sum_{i} n_i h\nu = 2h\nu $$

$$ \sum_{i} n_i = 2 $$

The vibrational quantum numbers must add up to 2, however, there are, in fact, three ways of achieving this. Two possible configurations are where one molecule has n = 2, and the other has n = 0; and the third configuration is where both molecules have n = 1.

It can be noted that in the example above that the three configurations (also called microstates) form two macrostates (states with unique sets of quantum numbers) - {0,2} with a weight of 2 (because there are two ways of arriving at this configuration), and {1,1} with a weight of one. The weights of these states play an important role in Statistical Mechanics, so it is important to know a way to quickly determine them, rather than just counting them directly.

Let's consider the weights of configurations {1,2}, {1,2,3} and {1,2,3,4}. Counting the weights gives 2, 6 and 24 or 2!, 3! and 4! respectively for these three states where

$$n! = 1 \cdot 2 \cdot 3 \cdot ... \cdot n $$

represents factorial of the number n. The weights of the configurations are just permutations or ways a given set of the quantum numbers can be arranged. Using combinatorial mathematics, the weight of a certain vibrational state in the example above can be found to be

$$ W(n_1, n_2, n_3, ...) = \frac{N!}{n_1 \cdot n_2 \cdot n_3 \cdot ...} = \frac{N!}{\prod_i n_i!}$$

where N is the total number of molecules and ni is the number of molecules in the state i.

Generalizing, the weight of a system with occupancies $$a_1, a_2, a_3,...$$ in an ensemble containing $$\mathcal{A} $$ systems is

$$ W(a_1, a_2, a_3, ...) = \frac{\mathcal{A}!}{a_1 \cdot a_2 \cdot a_3 \cdot ...} = \frac{\mathcal{A}!}{\prod_i a_i!}$$

Example
If there are three molecules and four units of vibrational energy, what are the possible macrostates of this system and what is the weight of each macrostate?

A number of microstates in the same system can have the same composition. These systems can be weighted depending on the occupancies of the systems. The weight of a macrostate can be defined by the equation:

$$W(a_1, a_2, a_3, ..) = \frac{\mathcal{A}!}{\prod_i a_i!}$$

Where, $$\mathcal{A}!$$ represents the total number of systems in the ensemble and $$\prod_i a_i!$$ represents the product of the factorial of the occupancies of each level.

For three molecules with four units of vibrational energy, we have four unique ways to distribute the four units of vibrational energy between the three molecules, {0,0,4}, {0,1,3}, {1,1,2} and {0,2,2,}. So for this example, $$\mathcal{A}$$ = 4 and the weights of each variation of the system would then be:

$$W(0,0,4) = \frac{\mathcal{A}!}{\prod_i a_i!}

= \frac{4!}{4!0!0!} = 1$$

$$W(0,1,3) = \frac{\mathcal{A}!}{\prod_i a_i!}

= \frac{4!}{0!1!3!} = 4$$

$$W(1,1,2) = \frac{\mathcal{A}!}{\prod_i a_i!}

= \frac{4!}{1!1!2!} = 12$$

$$W(0,2,2) = \frac{\mathcal{A}!}{\prod_i a_i!}

= \frac{4!}{0! 2! 2!} = 6 $$