Physical interpretation of $\frac{\partial \rho}{\partial t} + \frac{\partial j}{\partial x}=0$

Question

What is the physical interpretation of $\frac{\partial \rho}{\partial t} + \frac{\partial j}{\partial x}=0$? Here $\rho$ is the probability density and $j$ is the probability current of a particle of mass $m$ moving in one-dimension.

If the duplicate does not answer your question, please edit your question to be more specific what aspect of the usual interpretation of continuity equations you think is inapplicable to this specific case. — ACuriousMind, Mar 21 '20 at 08:28

score 1 · Accepted Answer · answered Mar 21 '20 at 08:14

Roughly speaking, the one-dimensional equation of continuity says that if the probability current going in is different from the probability current going out, then the probability density will change.

So, for instance, if there is more (of whatever) going in than going out, the probability will increase at that position.

Physical interpretation of $\frac{\partial \rho}{\partial t} + \frac{\partial j}{\partial x}=0$

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