Learning DFT: Inhomogeneous Electron Gas (Hohenberg) Question

In summary, the conversation discusses the form of the kinetic energy and potential in Hohenberg's paper on the Inhomogeneous Electron Gas. The kinetic energy is expressed as an expectation value with a factor of 1/2, possibly coming from a ##\frac{\mathbf{p^2}}{2m}## term. The potential is also expressed as an expectation value over all space. The integral of the first term over all space is zero due to the wavefunction and its derivative approaching zero at infinity.
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yosty222
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TL;DR Summary
Where does the kinetic energy term of the hamiltonian come from? Is it an expectation value of the kinetic energy?
I'm reading through Hohenberg's seminal paper titled: "Inhomogeneous Electron Gas" that help lay the foundation for what we know of as Density Functional Theory (DFT) by proving the existence of a universal functional that exactly matches the ground-state energy of a system with a given interaction potential v(r). I'm working through this paper and I'm a bit confused on the exact form of the Hamiltonian he builds:
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The form of kinetic energy looks like an expectation value, but not quite as I'd expect the expectation value of kinetic energy to look like ## \int \psi^* \nabla^2 \psi ##. The factor of ##\frac{1}{2}## out front makes me think it's coming from a ##\frac{\mathbf{p^2}}{2m}## term (with m = 1 in these units), but why does a ##\nabla## operator get attached to each ##\psi^*## and ##\psi##?

Similarly, for the form of V (equation (3) above), is this expressing the contribution of the external potential ##v(\mathbf{r})## as an expectation value of the external potential over some volume d##\mathbf{r}## then integrating over all space?
 
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##\psi^*\nabla^2\psi=\nabla\cdot(\psi^*\nabla\psi)-\nabla\psi^*\cdot\nabla\psi##
The integral of ##\nabla\cdot(\psi^*\nabla\psi)## over all space will be zero, since it's equivalent to integrating over a surface at infinity, and the wavefunction and its derivative go to zero as ##r\to\infty##.
 
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