The virial theorem and a delta function potential

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 ?

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!

• I'm not sure, but if you want a derivative of delta-function, you need to integrate your virial theorem over the space also. Then using integration by part the right hand side will be equal $-D V_0$. Oct 19 '13 at 21:40
• @swish For the quantum-mechanical expectation-value I should indeed integrate over the place with some wavefunction. Perhaps this might work :).
– Nick
Oct 19 '13 at 21:54
• I'm a little rusty, but interesting question. For me, some of the constants don't make sense. Problem is I don't think the derivative of the delta function is defined, at least in the sense used by the gradient. For fun, I attempted a generalized second derivative by taking the derivative of two parameterized step functions (and saying the delta was similar to a limit on the parameterization), and I got zero. If the delta jumps and goes back down an equal amount at a single moment, zero doesn't sound crazy, though not particularly meaningful. Oct 20 '13 at 2:07
• @DerekE Yes I've been thinking about the comment of swish, and for deltafunctions I can use the fact that $x(d/dx)\delta(x) = -\delta(x)$ which is proven by partial integration. But in the classical context (where the expectation value is an integration over time), I can't make a substitution like that, so there I'm still stuck :(.
– Nick
Oct 20 '13 at 10:41
• That's neat, I like it. :) Though, I still don't see the proof, I'll need to play with it. With Euler-Lagrange, one needed an independent variable function to conclude an equivalence under the integral. I'll toy around with finding one here. Interesting result, though. I almost had something similar by looking at $\delta(x) = \int_{-\infty}^{\infty}e^{2\pi ix\xi} d\xi$, but things were off at the boundary when applying integration by parts. Oct 20 '13 at 13:49

2. If we focus on one pairwise interaction (out of the many pairwise interactions), then the attractive delta function potential $$\tag{A} V(r)~:=~ -A\delta^d(\vec{r}),\qquad A~>~0$$ is classically ill-defined, and needs to be regularized. One could hope that quantum mechanical smearing of the wave function renders the potential (A) well-defined. However, this is not possible for $d>2$: The potential (A) is quantum mechanically unbounded from below for $d>2$ (See e.g. this Phys.Se post. The limiting dimension $d=2$ case is only bounded from below for sufficiently weak attractive delta function potentials (A).)
• Yes of course the particles need to be confined (I'd probably should have mentioned that) since they otherwise would float off to infinity. The delta-potential should be seen in classical mechanics as the hard-sphere potential and in quantum mechanics as an approximation for the atomic potential trough the mechanism of low-energy s-wave scattering. So you're practically saying that $\langle V\rangle$ can only be calculatedin D = 2 ?