# Fundamental assumption of statistical mechanics

I am confused about the statement of the 'fundamental assumption of statistical mechanics,' as one lecture would put it.

For an isolated system in equilibrium, all accessible microstates are equally likely.

What do we mean when we say 'microstate'? I am familiar with a 'state' of a dynamical system. But do microstates mean the different measurable quantities of each particle in a population?

## 1 Answer

A microstate is just a particular microscopic configuration of the system, where the state of each particle is fixed.

For example, take a three-level system with four particles. Treating the particles as indistinguishable, one particular microstate corresponds to two particles in the lowest state, which has energy $-\epsilon$, one particle in the second state with energy $0$, and one in the highest with energy $\epsilon$. The total energy of this microstate is then

$$E=2(-\epsilon) + 1(0) + 1(\epsilon) = -\epsilon$$

Now there are other ways you can get this energy. You can have three particles in the second state and one in the lowest state, for example. This is a different microstate. The equal probability of microstates says that those two microstates, corresponding to $E=-\epsilon$, are equally probable. Neither is "preferred."

Note that this idea of equal probability only applies to microstates with the same energy (and volume and number of particles). Microstates with different energies are not generally equally probable.

Further, if the system can exchange energy with a reservoir, the relative probabilities of two different energies is proportional to the number of microstates with that energy. This leads to the canonical ensemble, which uses the Boltzmann distribution along with this degeneracy of microstates to indicate the relative probabilities of different energy states.