# Differential equation for evolution of probability density in Quantum Mechanics?

I have come up with this differential equation for the evolution of $\vert \Psi \vert^2$, the probability density in quantum mechanics.

Is there a name for this equation? Is the logic sound?

So I start from the conservation of the total probability:

$$\frac{d}{dt} \int _{all space} \vert \Psi \vert^2 d^3\textbf{r} = 0$$

so that $$\int _{all space} \vert \Psi \vert^2 d^3\textbf{r} = {\rm const}$$

This means that $$\int _{all space} \vert \Psi (\textbf{r},t) \vert^2 d^3\textbf{r} = \int _{all space} \vert \Psi (\textbf{r}+d\textbf{r},t+dt) \vert^2 d^3\textbf{r}={\rm same\,const}$$

So since the integration is over the same volume I can equate the intergrands?

This gives us:

$$\vert \Psi (\textbf{r},t) \vert^2 = \vert \Psi (\textbf{r}+d\textbf{r},t+dt) \vert^2$$

or $$\Phi (\textbf{r},t) = \Phi (\textbf{r}+d\textbf{r},t+dt)$$ if we call $\vert \Psi (\textbf{r},t) \vert^2 = \Phi (\textbf{r},t)$. A Taylor expansion to first order gives us:

$$\Phi (\textbf{r}+d\textbf{r},t+dt) =\Phi (\textbf{r},t) + \nabla\Phi\cdot d\textbf{r}+\frac{\partial \Phi}{\partial t} dt$$

Plugging this into the previous equation gives us:

$$\boxed{ \nabla\vert \Psi \vert^2\cdot \frac{d\textbf{r}}{dt} = -\frac{\partial \vert \Psi \vert^2}{\partial t} }$$

What is $\frac{d\textbf{r}}{dt}$ in the context of quantum mechanics? Is this equation correct? What is the physical meaning of this equation?

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No, you can't just equate the integrands because you integrate over the whole volume. Under the integral, it doesn't matter which of the two you take, but the functions are definitely not the same. –  Martin Mar 30 '14 at 20:55
To expand further: if you're thinking of an integral as an area under a curve, you should surely be able to think of two curves which have the same area under them, but which are not the same at every point! By equating the integrands, you're asserting that the functions are the same at every point. –  gj255 Mar 30 '14 at 20:59
OK, sorry, very stupid mistake :( Should I delete the post? –  SuperCiocia Mar 30 '14 at 21:04
@Harold Related topic: en.wikipedia.org/wiki/Probability_current –  Danu Mar 30 '14 at 21:49
@Harold no, you shouldn't delete the post. I mean, you are allowed to, if you really want to, but I'd request that you don't because this may be useful to anyone else who has the same misconception in the future. Instead, I'd encourage you to post an answer explaining your misconception. –  David Z Mar 30 '14 at 21:58

$$\frac{\partial \rho}{\partial t} + \nabla \cdot \textbf{J} = 0$$, where $\rho = \vert \Psi \vert^2$ and $\mathbf{J}=\frac{\hbar}{m}\text{Im}\left[\psi^*\nabla\psi\right]$ is the probability flux.