# Many body Schrodinger equation - DFT / Exchange-Correlation term question

In principle if you solve the many body Schrodinger equation you will get the whole physics and nature of the problem. With the hamiltonian, in the most general case (without Born-Oppenheimer approximation), taking into account the protons, electrons, electron-electron interaction, proton-proton interaction, and the proton-electron interaction.

In DFT you have the exchange and correlation term. I get why they have to be added and what they correct... the fermionic nature of the particles give rise to the exchange term, and the Coulomb nature of charged particles give rise to correlation term.

My question is, if in a hypothetical scenario you would be able to solve any system (regardless of its size) with the Schrodinger equation, the physics that the exchange-correlation term plays in DFT will inherently be in the solutions, right?

Another question if the answer to the above is yes. In which part of the whole DFT approximations does that term gets lost? Which approximation or postulate looses those inherent properties you would get by solving the many body Schrodinger equation.

I had read the density had the same information as the wave function but I'm thinking that's where the information gets lost.

Edit:

Went to talk to a teacher today to ask this but I am still a bit lost. I told him that given the definition of the exchange-correlation term (the remainder of the total energy with the rest of the known energies) and how DFT accounts for the terms in the hamiltonian it should be zero (since there is a term that accounts for electron-electron interaction, their kinetic energy, and the potential of the nuclei)... so I asked him where in the approximations does information get lost so that we need to add an extra term. I asked if we were to put the proper wave function in the exchange-correlation term (hypothetical exact solution to the many body Schrodinger equation) it would be zero, and he said yes. So, if given an exact solution it's zero, then where does the information get lost? Are the formulas of the kinetic energy and potential energy not exact?

To answer the first question, if you could solve Schrodinger's equation with Coulomb interaction for n electrons you would have the exact wave function. The exact wave function has all the physics for the system. Because we do not know how to solve the exact equation we either use an ansatz for the wave function (think of slater determinant wave function) or find a doctored Hamiltonian that is easier and captures the physics we think is important.

As long as one is interested in just the ground state properties of a n body quantum system then the Hohenberg-Kohn theorems tell you that implementing DFT exactly gives you the exact information for the true ground state i.e you are not losing any information.

So we have an energy functional $\epsilon[n]= T[n]+U[n]+U_{ee}[n]$ where U is the potential due to the ions, $U_{ee}$ is the coulomb interaction , T is the kinetic energy and $n$ is the ground state density. Subject to the contraint $\int d\vec{r} n(\vec{r}) = N$ we find the ground state energy that minimizes the energy functional. Hohenberg-Kohn showed that $n$ essentially determines the Hamiltonian. Now we can write the energy fuctional as $\epsilon[n] =\int d \vec{r} \left( n(\vec{r}) U + U_{universal}[n] \right)$ where $U_{universal}= T[n] + U_{ee}[n]$. Now $U_{universal}$ is essentially the same for all n body problems and finding it would solve all external potentials U.

Next step is to minimize this energy functional. Note that at no point so far have we made any approximations. The procedure is exact.

So what about the exchange correlation terms? Remember that in the Hartree Fock Approximation we have the functional $F[\psi]=<\psi| H | \psi>$. We then make a guess for the form of the wave function (which is the slater determinant wave function) and then do variational minimization and then arrive at the Hamiltonian which will include the exchange correlation term. In the context of DFT getting the Hartee Fock Hamiltonian should be got by simply writing down the density got from the slater determinant wave function and then doing a variation since $< \psi | H |\psi> = \epsilon[n]$ So the approximation comes in what we guess as the ground state density.

• Ok, if I am not loosing any information, why do I have to add the exchange correlation potential (which has no known formula) in DFT? Where in the approximations and postulates of DFT are we loosing the information that makes us add the exchange-correlation term that otherwise wouldn't have been needed. That's the second question basically. – M.O. Dec 12 '16 at 0:46
• Looking at the equations that you have, if you'd move the kinetic and both potential terms to the left of the equation, then that's the definition of the exchange-correlation term. So why is it not zero? Are the functionals of the kinetic energy and potential energy not exact? – M.O. Dec 13 '16 at 3:04
• @Mariel Although DFT is exact we have to make a guess of the form of the density state which we then use to minimize. The guess is our approximation and this is where we lose some information. I have removed a sentence in my answer that was not quite correct after I thought about it. I hope this is helping. – Amara Dec 13 '16 at 3:41
• @Amara your last comment is not correct. " we have to make a guess of the form of the density state which we then use to minimize. The guess is our approximation and this is where we lose some information." The inexactness (lost of information if you will) of DFT comes from the fact that we do not know the exact form of XC functional. – physicopath Feb 5 '18 at 9:47