Liouville equation with Dirac delta as probability density I would lke to know if the probability distribution given by
$$\rho(q,p,t)=\delta(q-q(t),p-p(t)) $$
with the initial condition $\rho(t=0)=\delta(q,p), $ where $q(t)$ and $p(t)$ are trajectories given by the Hamilton equations, is solution to the Liouville equation
$$\frac{\partial\rho}{\partial t}=-\left\{ \rho,H\right\} $$
 A: Yes, it is a solution. It is a special solution that goes by the name "Klimontovich distribution" (at least among plasma physicists). Specifically, you've written the Klimontovich distribution corresponding to a single particle. It is straightforward to verify that it solves the Liouville equation by explicitly inserting $\rho = \delta[q-q(t)]\delta[p-p(t)]$ into both sides of the equation.
On the right, you have
$$\begin{align}
\{\rho, H\} 
&= \frac{\partial \rho}{\partial q}\frac{\partial H}{\partial p} - \frac{\partial \rho}{\partial p}\frac{\partial H}{\partial q} \\
&= \delta'[q-q(t)]\delta[p - p(t)] \dot q(t) + \delta[q-q(t)]\delta'[p-p(t)]\dot p(t)
\end{align}$$
where $\delta'(x)$ is the derivative of the Dirac delta. On the left,
$$\begin{align}
\frac{\partial \rho}{\partial t} = -\dot q(t)\delta'[q-q(t)]\delta[p-p(t)] -\delta[q-q(t)]\delta'[p-p(t)] \dot p(t)
\end{align}$$
The two differ by only an overall minus sign. Observe that on the LHS, $\dot q(t)$ and $\dot p(t)$ arise from applying the chain rule to $\rho$, whereas on the RHS, they come from Hamilton's equations of motion.
That is the boring way to prove that the Klimontovich distribution solves the Liouville equation. The physically more interesting argument is to interpret $\rho$ as the density of a point particle. Any density which is conserved along the trajectory traced out by the equations of motion is a solution the Liouville equation. Since point particles under Hamiltonian dynamics cannot be created or destroyed, the Klimontovich distribution must be a solution.
