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In reading "Density functional theory of atoms and molecules" by Parr and Yang, I was not sure what is meant by this sentence when the Virial theorem was introduced.

Suppose I have a Hamiltonian $\hat{H}$ for that describes some molecule.

"The kinetic energy component $\hat{T} = \sum_i\frac{1}{2}\nabla_i^2$ is degree -2 in particle coordinates". I understand that the Laplacian takes the partial twice for each spatial coordinate $x,y,z$, and then takes the sum, but I'm struggling to see why that means it's degree -2? Unless I'm completely misinterpreting what is meant by "degree -2 in particle coordinates".

It is much clearer what degree in particle coordinates mean when the potential energy component is described. "The potential energy component $\hat{V} = \hat{V}_{nn}+\hat{V}_{ne}+\hat{V}_{ee}=\sum_{\alpha<\beta}\frac{Z_\alpha Z_\beta}{r_{\alpha\beta}}-\sum_{\alpha,i}\frac{Z_\alpha}{r_{\alpha i}} + \sum_{i<j}\frac{1}{r_{ij}}$ is clearly of degree -1 in particle coordinates since I can clearly identify the 1/r dependence for all potential terms.

So, can someone clarify what this degree of kinetic energy term being -2 means? Thanks!

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I imagine that what the book is referring to are the powers associated with the dimension of length L of the different terms. As you correctly say, the dimension of $1/r$ is L$^{-1}$, which I imagine is what "degree $-1$" refers to. Similarly, the dimension of $\nabla^2$ is L$^{-2}$, which is probably what "degree $-2$" refers to. To see that the length dimension of $\nabla^2$ is L$^{-2}$, you can simply consider what the term looks like in Cartesian coordinates:

$$ \nabla^2=\frac{\partial^2}{\partial x^2}+\frac{\partial^2}{\partial y^2}+\frac{\partial^2}{\partial z^2}. $$

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  • $\begingroup$ Thanks ProfM! As a side note, are you aware of any starter resources to get started with dft, outside of the textbook I mentioned? e.g. lecture videos, demos etc. such as MIT open courseware? $\endgroup$ – Houndbobsaw Jun 30 at 14:39
  • $\begingroup$ Major codes have tutorials on their websites, which is always a good way of learning "by doing" $\endgroup$ – ProfM Jun 30 at 14:43

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