Does the gradient of the probability density tell me in which direction the system is most likely moving?

Question

Suppose that a system is described by a probability density $\rho$ on the state space $S$. Of course we need to assume some additional structure on $S$ to define the gradient $\nabla\rho$. I am actually considering the simple case $S=\mathbb R$ such that $\nabla\rho$ is simply the derivative of $\rho$.

That being said, there seems to exist the following heuristic: If the system is prepared in the state $x\in S$, then the system is most likely moving in the direction of fastest increase of $\rho$, i.e. in the direction of $\nabla\rho(x)$ or $-\nabla\rho(x)$, depending on your convention. In particular, $\nabla\rho$ contains information about the most likely trajectories.

• Is there some sort of intuitive argument for this behaviour?
• Are there any examples where this method is applied? For example, what about the pathwise approach to metastability?
• Is my question related to probability currents discussed in QM?

Answer 1

The hypothesis is false. In some sense, in the case of a stationary probability density, things may go the other way around. Indeed, let's assume that we have a probability density $\rho(x)$ and that the values of $\rho$ at two neighbor points $x_1$ and $x_2$ are such that. $$ \rho(x_1)> \rho(x_2). $$ This condition implies that, on average, more representative points are in a region $dx$ around $x_1$ than around $x_2$. We can define a probability of transition from $x_1$ to $x_2$ as $\pi(x_1,x_2)$, and, symmetrically, from $x_2$ to $x_1$ as $\pi(x_2,x_1)$. Such transition probability contains the information about where a system is most likely moving. Now, a sufficient condition for a stationary $\rho$ is the so-called detailed balance, i.e., on average, the same number of systems go from $x_1$ to $x_2$ as from $x_2$ to $x_1$. But this implies $$ \rho(x_1)\pi(x_1,x_2) = \rho(x_2)\pi(x_2,x_1), $$ from where do we obtain $$ \frac{\pi(x_1,x_2)}{\pi(x_2,x_1)} = \frac{\rho(x_2)}{\rho(x_1)} $$ This is saying the opposite of the proposed hypothesis: the transition probability from a low to a higher probability region is higher than from high to low.

Notice that this is a counterexample. The detailed balance is not a necessary hypothesis.

Finally, the current probability in Quantum Mechanics cannot be expressed in terms of the gradient of the probability density.

Answer 2

Now that I am home, here is the general answer.

Your claim is, as far as I understood: Probability densities relax to a constant. Like Fick's law.

This means $\partial_t \rho \propto -\nabla \rho$.

My claim is: Let $S$ be unbounded. Then this is never true.

Proof: Suppose this is true. Then the steady state is given by $\nabla\rho=0$. This implies $\rho=|\psi(x)|^2=const$. Since $S$ is unbounded, the state is thus not in $L^2$, i.e. not normalizable. Thus it is unphysical.