The canonical example of quantum mechanics in action is the harmonic oscillator, which is something like a mass on a spring. In classical mechanics, it wobbles back and forth periodically when it is given energy, if it's at a position $x$ , wobbling about $x_{0}$ and moving with velocty $v$ we can say its energy contains a potential term $V$ proportional to $(x - x_{0})^{2}$ , and a kinetic term $T$ proportional to $v^{2}$ , with an overall form:

$E = \frac{1}{2} k (x - x_{0})^{2} + \frac{1}{2} m v^{2}$

We could try and find a distribution over $x$ and $v$ , but continuous distributions tend not to "play well" with entropy. They're dependent on a choice of characteristic unit. Instead we'll go to the quantum world.

One of the major results of quantum mechanics is that systems like this can only exist in certain energy levels. In the harmonic oscillator these levels are equally-spaced, with a spacing proportional to the frequency associated with the classical oscillator. Since the levels are equally-spaced, we can think about the energy coming in discrete units called "phonons".

Our beliefs about the number of phonons $N$ in our system can be expressed as a probability distribution $P (N = n)$ over $n \in N$ :

This is progress: we've reduced an uncountably infinite set of states to a countable one, which is a factor of infinity! But if we do our normal trick and try to find the maximum entropy distribution, we'll still hit a problem: we get $P (N = n) = 0$ for all $n \in N$ .

The Trick: Distribution Families

Thinking back to our previous post, an answer presents itself: phonons are a form of energy, which is conserved. Since we're uncertain over $N$ , we'll place a restriction on $E (N)$ of our distribution. We can solve the specific case here, but it's actually more useful to solve the general case.

Maths, Lots of Maths, Skippable:

Consider a set of states of a system $s \in S$ . To each of these we assign a real numeric value written as $s \to x_{s} \in R$ . We also assign a probability $s \to p_{s} \in R^{+}$ constrained by the usual condition $\sum s \in S p_{s} = 1$ .

Next, define $E (X) = \sum s \in S p_{s} x_{s}$ and $H (S) = - \sum s \in S p_{s} ln p_{s}$ .

Imagine we perform a transformation to our distribution, such that the distribution is still valid and $E (X)$ remains the same. We will consider an arbitrary transformation over elements ${1, 2, 3}$ :

$p_{1} \to p_{1} + d p_{1}, p_{2} \to p_{2} + d p_{2}, p_{3} \to p_{3} + d p_{3}$

$d p_{1} + d p_{2}$ ...

Statistical Mechanics

Statistical Mechanics