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  • $\begingroup$ I think you lost a factor of 2 in expanding $\nabla^2(r\psi)$. Also $\int xdx=0$, since it is an asymmetric function (under some regularization caveats, which are implied when using plane waves.) $\mathbf{r}$ is a vector, whereas $r=|\mathbf{r}|$ is its magnitude (there might be some confusion between gradient $\nabla r$ and divergence $\nabla \cdot\mathbf{r}$) $\endgroup$
    – Roger V.
    Commented Jul 4 at 14:28
  • $\begingroup$ @RogerV. you're right there's a factor of two missing that I've now added back in, but that still doesn't change the overall outcome of the argument. Wouldn't $\int xdx=0$ only work for 1. Cartesian coordinates 2. Integral limits being $a,-a$? This wouldn't be something that's generally true, especially when we change to spherical coords where $r$ is always positive. All the $r$s here are $\textbf{r}$, and it's by a quirk that $\nabla\textbf{r}$ acts like the identity, it's not $\nabla\cdot\textbf{r}$. $\endgroup$
    – Redcrazyguy
    Commented Jul 5 at 15:20
  • $\begingroup$ Integral $\int \mathbf{r}f(r)d^3r$ is going to have the same value regardless of the system of coordinates, in which it is evaluated. what is important here is that this is an integral of a vector function (essentially, three different integrals - one for each component) - not identical with $\int rf(r)d^3r$, which is a single integral of a scalar function. I really think that the problem here is poor handling of vectors. $\endgroup$
    – Roger V.
    Commented Jul 5 at 16:32
  • $\begingroup$ @RogerV. the question has to do with actually evaluating the integral, I realise Mahan's notation isn't ideal but I've managed to understand what it is. However, I can't use the same method I used on particle currents on energy currents, hence the question. $\endgroup$
    – Redcrazyguy
    Commented Jul 6 at 18:09
  • $\begingroup$ It is not about notation, but about the errors (or perhaps inconsistent use of natotation) in your Q., where you without justification replace vector (in bold font) by its magnitude. Once you fix this, the integral the you claim as divergent will vanish due to parity, as per my first comment. Then a more serious discussion would be possible. $\endgroup$
    – Roger V.
    Commented Jul 7 at 12:54