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Edit: I've tried to make the question a lot simpler.

The Laplacian of the function $1/r$ is $-4\pi\delta(r)$. If I take the trace of the Hessian matrix of the function $1/r$, I find that the trace is 0 even though I expect I should find $-4\pi\delta(r)$. I understand this could be happening because the Dirac delta function isn't a function, but a measure.

Now, my questions are as follows:

  1. In physics, one way of defining the delta function is by appealing to Poisson's equation as is done in Jackson's E&M in the link below. Is it possible to do something similar to get a matrix analogue of the Dirac delta function, such that the trace of this matrix analogue is $-4\pi\delta(r)$?
  2. If the above is not possible, should this be concerning? In other words, if I can't recover the density by taking the trace of a Hessian matrix of a potential, what is that telling me about this procedure?

Link, since I can't embed pictures yet

Sorry if this question isn't well-phrased. My background is not in math.

Miguel Martinez
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  • Isn't the trace of the Hessian by definition equal to the Laplacian? So if you understand how to find $\nabla^2 1/r=-4\pi\delta(\bf{r})$, you're done --- aren't you? – Carlo Beenakker Jun 23 '22 at 21:23
  • Please clarify your specific problem or provide additional details to highlight exactly what you need. As it's currently written, it's hard to tell exactly what you're asking. – Community Jun 23 '22 at 21:30
  • Yes, I agree that the trace of the Hessian is equal to the Laplacian by definition. However, what I am looking for is essentially the matrix version of the delta distribution, for which the trace is delta distribution – Miguel Martinez Jun 23 '22 at 21:34
  • See here for the second partial derivatives of $\frac{1}{r}$ in $\Bbb{R}^3$ in the distributional sense. The trace is indeed as expected since the principal value terms cancel out. – peek-a-boo Jun 23 '22 at 23:48
  • @peek-a-boo thank you so much! – Miguel Martinez Jun 24 '22 at 01:18
  • @peek-a-boo can you post that as an answer so I can accept it? Or this there a better procedure for this? – Miguel Martinez Jun 24 '22 at 01:21
  • I'm not sure, but if you want, you can write up your own answer and accept it. – peek-a-boo Jun 24 '22 at 01:29

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