Microstates and macrostates

A microstate is a specific arrangement of the constituent parts of a system that produces an externally observable macrostate.

For example, consider a box containing two particles and only one quantized unit of energy (meaning it can't be divided up; it must go entirely to one of the two particles). There are two microstates, since assigning the unit of energy to particle A represents a different microstate than when the unit of energy is assigned to particle B. However, in both case, the macrostate is the one unit of energy in the system. Notice that in discussing the macrostate, the arrangement of the particles inside the box is unimportant, so there is only one macrostate.

Microstate consideration are most useful in calculating entropy and internal energy. In fact, for calculating entropy, the only required input is the number of microstates, since entropy is defined in terms of a constant and the natural log of the number of microstates.

$$S=k_B \ln{\Omega}$$

The problems dealing with the counting of microstates generally require combinatorics.

A box contains 10 balls that could be either red or blue. How many different microstates satisfy the macrostate of exactly 3 of the balls being red?

Solution: We have an arrangement of 10 balls and want exactly 3 to be red, which can be acheived $_{10}C_{3}$ different ways.

$_{10}C_{3} = \dfrac{10!}{3!(10-3)!} =$ 120 microstates

To get a handle on this, consider the system of a poker hand drawn from a standard deck of 52 cards. The exact state of the system can be found by answering the question "what cards are in the hand?" to which there could be any of $\binom{52}{5}$ possibilities such as $5\spadesuit,\text{K}\heartsuit, \text{J}\clubsuit, 2\diamondsuit, \text{3}\heartsuit$. This kind of description is called the microstate of the system because it specifies exactly the value of every card. Further, the chance of that exact hand being drawn is simply $1/\binom{52}{5}$ because every hand has the same chance of coming out from the deck, i.e. every microstate is equiprobable.

Thankfully, poker is not usually interested in the exact microstate description of a hand, but instead in whether or not it is of a common pattern like a "flush", or a "straight", or "four of a kind". This is called the macrostate description of the system, because it captures the essential information (a description of the pattern) without specifying the details (the exact cards).

Consider the task of determining winners in poker through the microstate description. For each set of hands, the tedious solution consists of consulting a lookup table of $\binom{52}{5} = 2,598,960$ entries, searching through the list for the exact match to each hand, and comparing their values.

In the macrostate description, focus instead on a vastly smaller set of possibilities such as "hands for which any four of the cards in the hand have the same number", or "hands for which the five cards be arranged to give a consecutive sequence", whose probabilities can be easily calculated with combinatorics. In poker, the only macrostates that matter (and their probabilities) are

$$\begin{array}{|c|c|} \hline \text{Pattern} & \text{Microstates} \\ \hline \text{royal flush} & 4 \\ \text{straight flush} & (13-4)\times 4 =36 \\ \text{four of a kind} & 13\times 24 = 624 \\ \text{full house} & \binom{4}{2}\times\binom{4}{3}\times 13\times 12 = 3,744 \\ \text{flush} & 4\times\binom{13}{5} - 4 - 36 = 5,108 \\ \text{straight} & 10\times 4^5 - 40 =10,200\\ \text{any three of a kind} & 13\times\binom{4}{3}\times\binom{12}{2}\times\binom{4}{1}^2 - 3,744 = 51,168 \\ \text{two pairs} & \binom{4}{2}^2\times 13\times 12\times 2\times 11= 123,552 \\ \text{any two of a kind} & \binom{4}{2}\times 13\times \binom{12}{3} \times 4^3= 1,098,240 \\ \hline \end{array}$$

Managing nine possibilties vs $2.5\times10^6$ (a $\sim$ 100,000-fold reduction) frees up quite a bit of thinking for other tasks.