How do you get from the polarisation operator to particle current?

Question

I'm going through Mahan's "Many Particle Physics", and I'm a bit confused about his reasoning. He introduces the polarisation operator as $$\textbf{P}=\int\textbf{r}\rho(\textbf{r})d^3r$$ he then says "recall that the time derivative of the polarisation is the particle current" $$\frac{\partial}{\partial t}\textbf{P}=\int\textbf{r}\frac{\partial}{\partial t}\rho(\textbf{r},t)d^3r$$ which I'm guessing is because the position operator is time independent (measuring probability at a fixed point and all). He then states that this can be "proved easily" by using the equation of continuity, followed by an integration by parts \begin{align} \dot{\rho}(\textbf{r},t)&=-\nabla\cdot\textbf{j}(\textbf{r},t)\\ \frac{\partial}{\partial t}\textbf{P}&=-\int\textbf{r}\nabla\cdot\textbf{j}(\textbf{r},t)d^3r=\int\textbf{j}(\textbf{r})\cdot\nabla\textbf{r}d^3r\\ &=\int\textbf{j}(\textbf{r},t)d^3r \end{align} I can follow the second step (assuming that the $\textbf{r}$ is an abuse of notation and is not actually a vector), but I can't see at all how the $\nabla\textbf{r}$ drops out, or why the $\int\nabla\cdot(\textbf{r}\textbf{j})d^3r$ term vanishes from the integration by parts. This is the first time he's talked about polarisation, and everything I find online is either about light or EM fields in matter.

Could someone please explain what he has done here please? As well as making sure I've explained the earlier parts correctly. Thanks!

Answer 1

$\vec\nabla\vec r=\mathbb I$ pretty much by definition, and the integration by parts surface term vanishes because for every physical current going in whatever direction, the positions cancel in opposing pairs. I mean, imagine that this surface is a gigantic sphere, and you would have this cancellation.

Answer 2

The confusion arises from how Mahan's notation. Writing the polarisation operator w.r.t. the basis vectors $e_i$ $$\textbf{P}=\sum\left(\int x^i\rho(\textbf{r})d^3r\right)e_i$$ Since $\partial_t\rho$'s continuity is tacitly assumed, and $r$ is time independent, we can use Leibnitz' rule to see that $$\frac{\partial}{\partial t}\textbf{P}=\sum\left(\int x^i\dot{\rho}(\textbf{r})d^3r\right)e_i$$ where $\dot{\rho}$ denotes the time derivative of $\rho$ as usual. If we then apply the product rule for divergence to each coordinate, $$\nabla\cdot(x^i\textbf{j})=x^i(\nabla\cdot\textbf{j})+j\cdot(\nabla x^i)=x^i(\nabla\cdot\textbf{j})+\textbf{j}\cdot e_i$$ Assuming we don't lose any particles (i.e. $\textbf{j}\rightarrow0$ as $x^i\rightarrow\infty$) the LHS is zero over a large (or small) enough volume integral. Thus, by using the continuity equaiton $\dot{\rho}(\textbf{r},t)=-\nabla\cdot\textbf{j}$ we can substitute this back into the time derivative of $\textbf{P}$ to find $$\frac{\partial}{\partial t}\textbf{P}=\sum\left(\int \textbf{j}\cdot e_i d^3r\right)e_i$$ which, when converted back into Mahan's notation, gives $$\frac{\partial}{\partial t}\textbf{P}=\int\textbf{j}(\textbf{r},t)d^3r$$