7
$\begingroup$

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!

$\endgroup$
8
  • $\begingroup$ 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 $-D V_0$. $\endgroup$
    – swish
    Oct 19 '13 at 21:40
  • $\begingroup$ @swish For the quantum-mechanical expectation-value I should indeed integrate over the place with some wavefunction. Perhaps this might work :). $\endgroup$
    – Nick
    Oct 19 '13 at 21:54
  • $\begingroup$ 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. $\endgroup$
    – Derek E
    Oct 20 '13 at 2:07
  • $\begingroup$ @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 :(. $\endgroup$
    – Nick
    Oct 20 '13 at 10:41
  • $\begingroup$ 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. $\endgroup$
    – Derek E
    Oct 20 '13 at 13:49
0
$\begingroup$

Comments to the question (v8):

  1. The Virial theorem usually apply to periodic or bounded systems, but pairwise attractive delta function potentials would not constitute a bounded system unless the system is additionally confined in a box. (OP has in an update (v9) of the question introduced an additional potential to confine the particles.)

  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).)

$\endgroup$
2
  • $\begingroup$ 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 ? $\endgroup$
    – Nick
    Oct 20 '13 at 14:50
  • $\begingroup$ see also my edit to my question! $\endgroup$
    – Nick
    Oct 20 '13 at 14:53

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.