# Postulate of a priori probability and harmonic oscillator

Physics Asked on January 5, 2021

According to the fundamental postulate of a priori probability in Statistical Mechanics:

An isolated system in equilibrium is equally likely to be in any of its accessible states.

But for a classical harmonic oscillator with constant energy, Aren’t the states around the extremum of position variable more likely than other states?

Aren't the states around the extremum of position variable more likely than other states?

I think your confusion is with microstate and macrostate. The fundamental postulate of equilibrium statistical mechanics says :

In a state of thermal equilibrium, all the accessible microstates of the system are equally probable.

Microscopic quantity refers to the individual degree of freedom as opposed to averages. Now that let us suppose that system has a fixed energy $$E$$. Then you can ask, Is the state in which all the energy possessed by one particle is equally probable to other states? The answer is Yes!.

Now suppose a system a simple probability example: A coin

Now If you have a coin, then possible microstates are $$H$$ and $$T$$. For two coins: $${HH,TT,HT,TH}$$. If you define a random variable $$X$$ which define as the number of heads then You can see this can take value $$0,1,2$$. The probability for this to take $$1$$ is greater ($$1/2$$) than the other ($$1/4$$). That's exactly what happens here. If you take a large number of coins then you get a probability distribution what is called a binomial distribution and Probability for $$X=N/2$$ will be higher than the other. Even though each microstate is equally probable.

The same thing happens in the case, You are talking about.

Answered by Young Kindaichi on January 5, 2021

There are a few finer details that one has to keep in mind here:

1. A single harmonic oscillator is really not a thermodynamic system, so the postulate doesn't apply to it.
2. All accessible states in case of a constant energy (i.e., when using a microcanonical ensemble) means all the possible states with the same energy. For a single oscillator there is only one such state. Of course, one cannot apply statistical description to a single state (statistics is the more precise, the more states we have, ideally of the order of Avogadro constant $$N_A=10^{23}$$) - hence point 1 above.
3. Finally, one could apply statistical description to an ensemble of oscillators. In fact, it is very instructive to consider a number of possible states with the same total energy for two oscillators, three oscillators, and so on. This exercise is a good entry point to understannding statistical physics, although most textbooks prefer to consider as an example ensembles of spin-1/2 particles.

Update

For a classical oscillator it is possible to consider that all the states on the constant energy ellipse in the phase space have the same energy. However, since the oscillator by itself does not form a thermodynamic system (it has only one degree of freedom), you will have to assume some averaging mechanism - e.g., that it is coupled to a bath (i.e., it is a part of a bigger thermodynamic system) or that you have some kind of time averaging or initial conditions randomization.

In this sense, the states near the points of the maximum amplitude are not more probable than other states with the same energy. However, the oscillator moves slower near these states, so, if you are more likely observe it near this points. This is a good example that averaging in time and statistical averaging are not always the same thing, i.e., for a single oscillator we cannot assume ergodicity, which is the essential assumption (or the other form) of the postulate in the original question.

"around the extremum position variable" is not a state. It is a description of behaviour within a state.

A pendulum may be oscillating with a small amplitude or a large amplitude. Statistical Mechanics makes a statement about the relative probability of these two cases. For either case the pendulum spends more time near the edges where it's moving slowly than in the centre which it moves past quickly. There's no conflict between these statements.

Answered by RogerJBarlow on January 5, 2021