# Why aren't classical phase space distribution functions always delta functions?

The phase space distribution function (or phase space density) is supposed to be the probability density of finding a particle around a given phase space point. But, classically, through Hamilton's equations, the system's time evolution is completely determined once the initial conditions are specified. So for a 2D phase space, why isn't the distribution function always the same:

$$f(x,p,t)=\delta(x-x(t)) \ \delta(p-p(t))$$

I know that this thinking has to be wrong, and I am definitely confusing some things. I would like to ask for clarification.

## 2 Answers

You are right that if you know exactly the initial conditions of your system that is the exact location of your system's state in phase space then its evolution is entirely determined.

But that's where lies the issue; we don't know exactly the state of the system as described by a point in phase space.

Instead we may know some values of macro- or meso-scopic variables and try to infer from them compatible microstate. The reliability/accuracy of the inference we make is then captured by a (usually non-delta) distribution function in phase space.

• So, for example, we have some data that leads us to believe that the initial phase space point was located somewhere nearby $(x_0,p_0)$, say. And for that we construct a probability distribution $g(x,p)$, presumably centered around that point, such that $f(x,p,0)=g(x,p)$. And then $f(x,p,t)$ evolves according to the Liouville equation? – Spine Feast Oct 22 '14 at 15:26
• @DepeHb: Yes that's exactly that. – gatsu Oct 22 '14 at 17:24
• Also, it may be worth saying that the classical limit of a quantum state is a probability distribution on the phase space. So this is another context on which you obtain phase space distributions for a classical system (i.e. as an approximation of the quantum dynamics). The quantum states that classically correspond to delta distributions are the coherent states. – yuggib Oct 22 '14 at 18:55

Note that various density kernels (like the gaussian kernel used in classical phase-space) have delta functions as a limit. Physicaly, clasicaly delta (dirac) distributions are the correct ones. Mathematicaly more smooth distributions (e.g gaussians) may be used to calculate integrals and take limits (the limits should be same as having delta ditributions)

• @bobie, thankx, the history behind kinetic energy, vis viva (and the debate between newtonists, leibnizists), etc. is interesting and provides context for physics (even current ones) – Nikos M. Oct 24 '14 at 15:57