So the virial theorem tells us that:
$2\langle T\rangle = \langle \textbf{r}\cdot\nabla V\rangle$.
Now I was wondering what would happen if V has te form:
$V(\textbf{r}-\textbf{r}') = V_0\delta^{(D)}(\textbf{r}-\textbf{r}')$, where $\delta^{(D)}(\textbf{r}-\textbf{r}')$ is the delta-function in D dimensions. I'm not sure why, but I think that I should get that:
$\langle \textbf{r}\cdot\nabla V\rangle = \frac{1}{D}\langle V_0\rangle$ since the delta written out as a product of different components is:
$\delta^{(D)}(\textbf{r}-\textbf{r}') = \frac{1}{\sqrt{det(G)}}\prod\limits_{i=1}^D\delta(x_i-x_i')$, with $x_i$ the different components of the vector $\textbf{r}$, given in the base with metric G, where $\sqrt{det(G)}$ gives the D-dimensional volume-element in the basis $\textbf{e}_i$.
I don't know wether there is a more rigorous reasoning for this? Or wether this is even correct ?
Addendum: a different perspective:
Another way to look at it, is that if I rescale my vector $\mathbb{r}$ bij a factor $\lambda$, I get:
$\delta^{(D)}(\lambda\textbf{r}-\lambda\textbf{r}') = \frac{1}{\lambda^D}\delta^{(D)}(\textbf{r}-\textbf{r}')$. This makes me also think that i should get the above relation for the virial theorem. But still I'm not sure of my reasoning !
Extra demand on potential (necessary for finite system)
Next to my delta-potential, I also have an extra confining potential to keep the particles together. For simplicity I'll take an harmonic trap $V(r) = \frac{1}{2}m\omega^2r^2$ which keeps the particles together! So this is the other term of the potential, but this one I didn't consider in my question because that one posed no problem to my calculations!
