Question

To which degree is density functional theory variational?

Answer 1

A little intro - The Hohenberg-Kohn theorems:
In wave mechanics, the electron density is given by the square of the wave function integrated over N-1 electron coordinates and the wave function is determined by solving the schrödinger equation. For a system of \(N_{\sf nuclei}\) nuclei and \(N_{\sf elec}\) electrons, the electronic Hamiltonian contains the following terms:
\[H_e=-\sum_{i=1}^{N_{\sf elec}} \frac{ 1 }{ 2 } \nabla ^2 _i-\sum_{i=1}^{N_{\sf elec}}\sum_{j>i}^{N_{\sf nuclei}}\frac{ Z_A }{ \left| R_A-r_i \right| }+\sum_{i=1}^{N_{\sf elec}}\sum_{j>i}^{N_{\sf elec}}\frac{ 1 }{ \left| r_i-r_j \right| }+\sum_{i=1}^{N_{\sf nuclei}}\sum_{j>i}^{N_{\sf nuclei}}\frac{ Z_A Z_B }{ \left| R_A-R_B \right| }\]
Within the Born-Oppenheimer approximation, the last term is a constant. It is seen that the Hamiltonian operator is uniquely determined by the number of electrons and potential created by the nuclei, e.i. the nuclei charges and positions. This means that the ground state wave function (and thereby the electron density) and ground state energy are also given uniquely by these quantities.
Assume now that two different external potentials (which may be from nuclei), \(V_{ext}\) and \(V'_{ext}\) result in the same electron density, \(\rho \). Two different potentials imply that the two Hamiltonian operators are different, \(H\) and \(H'\), and the corresponding lowest energy function are different, \(\Psi\) and \(\Psi'\). Taking \(\Psi'\) as an approximate wave function for the hamiltonian and using the variational principle yields:
\[\langle \Psi'|H\Psi' \rangle>E_0\]\[\langle \Psi ' |H| \Psi ' \rangle+ \langle \Psi'|H-H'| \Psi ' \rangle>E_0\]\[E_0^{'}+ \langle\Psi '|V_{ext}-V_{ext}^{'} | \Psi' \rangle>E_0\]\[E_0'+\int\limits_{}^{}\rho(r)(V_{ext}-V_{ext}')dr>E_0\]
Similarly, taking \(\Phi\) as an approximate wave function for \(H'\) gives:
\[E_0+\int\limits_{}^{}\rho(r)(V_{ext}-V_{ext}')dr>E_0'\]
Addition of these two inequalities gives:
\[E'_0+E_0>E'_0+E_0\]
Showing that the assumption was wrong. In other words for the ground state there is a one-to-one correspondence between the electron density and the nuclear potential, and thereby also with the Hamiltonian operator and the energy. In other words: the energy is a unique functional of the electron density.
Using the electron density as a parameter, there is a variational principle analogous to that in wave mechanics. Given an approximate electron density \(\rho '\) (assumed to be positive definite everywhere) that integrates to the number of electrons, the energy given by this density is an upper bound to the exact ground state energy, provided that the exact functional is used.

So the question is:
This unique, universal functional is unknown and therefore we create the functionals our self based on a series of requirements. Let now assume we use an approximate functional is DTF then variational if I only increase the basis sets and hold the functional constant?
@Michele_Laino

Answer 2

Similarly, taking \(\Psi\) as an approximate wave function for H′ gives:*

Answer 3

My very limited understanding of DFT is that the theory is exact in and of itself, it's just that you can't actually know what the functional is so you have to guess and that's what you're varying over, is your guesses since I believe (although this actually might not apply to DFT) that you are minimizing over your guesses since you can't 'under guess' the true energy. But I think the proof I'm thinking of doesn't apply to DFT but they use it anyways.

I believe I read that in Computational Organic Chemistry - Steven Bachrach

Answer 4

I need some time to give an answer to this question

Answer 5

@Empty That is the whole thing, if you approximate the functional (as the universal functional is unknown), then you should be able to hit below the exact value, but if you increase the basis wouldn't it then go closer to the exact value?

Answer 6

basis sets*

Answer 7

I got 3 profs that can't agree to it. 1 who claim it is, 1 who claim it isn't, and one who said only if you do a comparison study the correct way. :)

Answer 8

Haha well if your professors don't even agree, a simpleton such as myself will not be able to weigh in on this one that's for damn sure. xD

Answer 9

But this is the funny part and why I think people here can handle it: They all use the Hohenberg-Kohn theorem for their argumentation. :P

Answer 10

DFT is gonna be outdated in 10 years anyway then we don't need to worry about it anymore.

Wave function based method is indeed the way forward (also what all my results suggest). MP2 for the win!

Answer 11

In nuclear physics, the variational method, is an useful method, even if, it can be more complicated. An important application of such method, is to describe the nucleon-nucleon interaction. Nevertheless, we can write this brief notes, which I enclose as PDF file below:

Answer 12

oops...these* brief notes...