# Degree of kinetic energy component in Hamiltonian for any molecule

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 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!

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