I'm reading my course on quantum mechanics and I just noticed something strange (or at least new to me) that I don't really understand :
$$\psi(\vec{r},t) = \psi_0 e^{\frac{i}{\hbar}(\vec{p}.\vec{r} -Et)}$$ $$\frac{d\psi}{d\vec{r}} = \frac{i\vec{p}}{\hbar}\psi \hspace{5mm}, \hspace 5mm\frac{d\psi}{dt} = -\frac{iE}{\hbar}\psi$$
I'm confused. I've never seen this notation $\frac{d\psi}{d\vec{r}}$ before (with the vector at the bottom). Usually, when there are vectors, I see grad, div, etc...
So what happened here ? Everything looks like the operation has been carried out by treating $\vec{r}$ like a simple variable and $\vec{p}$ like a simple constant, I didn't know that was possible. Exactly what rules have been applied here ? When can I do that ?
The notation is just a very compact way of writing $$\frac{d\psi}{dr_k} = \frac{i p_k}{\hbar}\psi,~~~~~\text{for}~~~~~k=1,2,3$$ where $r_k,p_k$ is the $k$'th component of the vector $\vec{r}$ and $\vec{p}$ respectively. The derivatives in the equation above are the good old standard derivatives.