# The Exchange-Correlation (XC)

1. Jul 28, 2007

### Modey3

Hello,

My understanding of the exchange correlation is basic. I know that the XC is the potential energy that keeps electrons in the same quantum coordinates {n,l,m and k (for periodic solids)} from having the same spin. Is there a analytical expression for the XC for say a 2-electron system such as He? In DFT calculations we approximate it by using the Local-Density-Approximation or Generalized-Gradient-Approximation. However, these expressions aren't correct and represent a major barrier in the calculation of the electronic ground-state. Thanks for any info.

Regards

Modey3

2. Jul 28, 2007

### meopemuk

There is no need for any special XC expression if you work in the rigorous many-electron approach. Then, all you need to know are standard Coulomb potentials between electrons and nuclei and representations of the many-electron wave function as a Slater determinants (this provides a description of exchange) or as a superposition of Slater determinants (this takes into account the correlation as well)

The idea of DFT is that one doesn't need the (very cumbersome) many-electron approach to calculate the ground-state energy. It is sufficient to know the electron density only. This doesn't present any difficulty in calculations of the electrons-electrons and electrons-nuclei Coulomb energy. However, general expressions for the exchange and correlation energies as functions of the electron density do not exist. There are various approximations, which work pretty well, but they are just approximations.

Eugene.

3. Jul 28, 2007

### Modey3

Eugene,

Alright, so in the Hartree-Fock approach exchange is taken into account by the Slater determinant of the electron-orbital basis functions. In the single-particle formulation of DFT, the basis functions are used to build up the electron density (analogous to the wave function in Schrödinger's Eqn.), however these basis functions (such as plane waves or atomic orbitals-written as a Bloch orbitals) aren't correlated meaning that there is nothing keeping two-electrons from having the same quantum coordinates (in the case of a simple cubic crystal with one atom-type the distinguishing quantum coordinate would be the wave-vector k in the ground-state) or equivalently the same spatial configuration. Therefore a XC term needs to be added in the Kohn-Sham equations to keep two electrons from having the same quantum coordinates.

Best Regards

Modey3

4. Jul 28, 2007

### meopemuk

Long time passed since I saw this stuff, so it is not fresh in my memory. In my undestanding, DFT does not deal with states of individual electrons, so I wouldn't say that "XC term needs to be added in the Kohn-Sham equations to keep two electrons from having the same quantum coordinates". DFT is all about total electron density. Whatever one-electron language is used in QFT (Kohn-Sham orbitals, n-electron determinants, exchange, correlation), it doesn't have precise physical meaning. These are some heuristic concepts and tricks whose only goal is to derive an approximation to the (yet unknown) functional which expresses the total energy of a many-electron system in terms of its total density.

Eugene.

5. Jul 29, 2007

### olgranpappy

Why do you think that? In the Kohn-Sham proceedure, for example, we solve for a set of "single-particle" wavefunctions $$\phi_i(\vec x)$$ and then we fill up the lowest N wavefunctions to find the density.
$$\sum_{i=1}^N\phi_i^2(\vec x)$$

We don't put more than one electron into a single state...

6. Jul 29, 2007

### olgranpappy

Anyways, your question about He is an interesting one. It is strange that Kohn's ideas work so well and there is a lot of room for improvement in DFT... Although, Kohn himself thinks that TDDFT is the place where one could make a name for oneself by coming up with improvements.

Also, atomic He is solved very well with the LDA, isn't it?

7. Jul 29, 2007

### meopemuk

That's right, Kohn-Sham determinants are constructed in very close analogy to Slater determinants of the Hartree-Fock method. but I think that Kohn-Sham determinants are not more than convenient parameterizations of the total electron density. In particular, it would be wrong to regard these determinants as representatives of many-electron wavefunctions.

Hartree-Fock determinants are approximations for the true many-electron wavefunction. One can make superpositions of such determinants and improve the approximation. With an infinite number of determinants one can obtain the precise many-electron wavefunction of the system.

I don't think the same is true for Kohn-Sham determinants. They surely look like Hartree-Fock ones, but there is no theory, which would allow one to get an approximation for the many-electron wavefunction, which converges to the exact result in the limit. If I remember correctly, some people attempted to interpret Kohn-Sham determinants as many-electron wavefunctions and calculate some physical properties (like transition probabilities) with them. Perhaps, one can even get some properties agreeing with experiment, by doing such things. But one should understand that this is an uncontrollable approximation, which is not supported by a rigorous theory.

Fundamentally, DFT is about total electron density in the ground state. Let's keep it that way.

Eugene.

8. Jul 29, 2007

### meopemuk

Ten years ago, when I was doing quantum chemistry, the absolute leader in improving DFT exchange-correlation functionals was Becke. His B3LYP and "half-and-half" functionals were competing with multi-configuration methods in accuracy and were much faster. I am not sure if there was much progress in this field lately.

Eugene.

9. Jul 30, 2007

### quetzalcoatl9

in practical terms: currently, the B3LYP hybrid method along with high level basis sets (i.e. the augCC T/Q zeta) is incredibly accurate, giving transferable results that are competitive with coupled cluster methods. in terms of accuracy/performance, it is a clear winner.

(nature has been kind in that the correlation functional can be so well parameterized from an electron gas. why would this be?)

i also think it should be pointed out that the Kohn-Sham equations are an EXACT rewrite of the schrodinger equation..the approximations are all lumped into the functional, which can be (and usually is) fit from high-level diffusion Monte Carlo simulations of a fermi gas (usually in the fixed-node approx). or fit from He data, in the case of LYP.

the Fock exchange energy is still tremendously accurate, so that energy contribution is still used in hybrid methods like B3LYP

but, as has been pointed out, there is no systematic method of improvement (in contrast to other quantum chemical methods).

10. Jul 30, 2007

### meopemuk

Kohn-Sham equations with exact functional (if such a functional is found) would allow us to obtain exact electron density and exact energy of the ground state. However, they would not produce exact many-electron wavefunctions. This is the fundamental limitation of DFT methods. So, there is no full equivalence between Kohn-Sham and the multi-particle Schroedinger equation.

Eugene.

11. Jul 30, 2007

### quetzalcoatl9

yes, that is true it is technically not an exact re-write since there is no wavefunction - only the density. it is more proper to say that it is an exact expression.

12. Jul 30, 2007

### meopemuk

I agree. Eugene.

13. Jul 30, 2007

### Modey3

It goes goes back to the major tenet of DFT: The electronic structure of the ground state is a unique energy-functional of the electron-density. Meaning that if you find the correct electron-density distribution that minimizes the Kohn-Sham energy functional that would be the correct ground-state electron-density. The electronic wavefunctions (the eigenvectors of the Hamiltonian matrix) are used to determine the electron density.

"the approximations are all lumped into the functional, which can be (and usually is) fit from high-level diffusion Monte Carlo simulations of a fermi gas (usually in the fixed-node approx). or fit from He data, in the case of LYP."

meopemuk is this an alternative way of determining the Kohn-Sham functional? From all the books I've been reading there are no "adjustable parameters" in the functional. Are talking about the XC-term?

Thanks

Modey3

14. Jul 30, 2007

### meopemuk

This phrase was written by quetzalcoatl9. So, we should ask him(her) for explanations. I don't know much about the theory used for fitting density functionals. You can read Becke's works in order to get a better idea.

You are right that the ultimate exact electron density functional must be given by some fundamental expression without any adjustable parameters, except, perhaps, electron's mass, charge, and the Planck constant. However, nobody knows how to obtain this functional. So, in practice, people try various semi-empirical formulas whose parameters are fitted to the properties of the electron gas, He atoms, and who knows what...

That's how I understand it.

Eugene.

15. Jul 30, 2007

### quetzalcoatl9

one way of finding an exchange-correlation functional is through quantum Monte Carlo simulations of an electron gas. i have seen these as path integral or diffusion MC and they are difficult in and of themselves (e.g. the "sign problem"). to be clear, i am talking about the term

$$V_{xc}$$

there are certainly "adjustable parameters" since there are many exchange-correlation functionals out there. this is an empirical fit, we do not have an exact equation for it. sometimes, as i mentioned before, the exchange portion of the energy is neglected in the functional and the Hartree-Fock exchange energy is used instead (since exchange is explicitly accounted for in the HF method via antisymmetry of the wavefunction). this is done with B3LYP (LYP functional with HF exchange energy).

16. Jul 31, 2007

### marlon

Though your description of the XC functional is correct, it is not accurate. The "exchange" part refers to the energy change of the system when the spatial coordinates of two electrons are interchanged. This more or less applies to what you described correctly. The "correlation" part refers to the notion that electrons are no longer described as point particles, but they are represented as electronic densities. This is a basic DFT element. Now, if the position of one point particle changes, the second point particle will change its position as well because they are interacting with each other. But using electronic densities changes this picture in the sense that changing the position of one density has not that direct influence onto the position of the second density. The accurate correlation that we have between two point particles is lost when we use densities. That is the point.

In DFT, the ground state energy is the basic quantity that is calculated and WILL BE ACCURATE ! The electronic properties (PDOS profiles, band gaps, bottom of conduction band)we calculate from that need to be interpreted with care. LDA, especially fit for metals, underestimates the magnitude of band gaps by 50 to 80 %. GGA gives better band gap values. Generally, DFT gives accurate valence band top-positions and Fermi Level positions. The bottom of the conduction band (hence the band gap) cannot be trusted. Different techniques can be used to acquire correct band gap values like the GW approximation.

So, tell us the nature of your calculations. What are you going to study with DFT. I assure you that using LDA or GGA will meet your goals as long as you know what to do. I mean, as long as you take into account the disadvantages of these functionals which can easily be bypassed if you compare one system to another !

marlon

17. Jul 31, 2007

### Modey3

Marlon,

I'm not worried about the XC-pseudopotentials that I'm using with VASP. The GGA has given good results in determining the lattice constant, elastic constants, and surface energy of transition metals. My purpose for posting this message is because I wanted to get a good description of the XC since one day when I defend my thesis I'm going to have explain to my advisers the nature of the XC. People shouldn't use any computational software like a black-box and I'm glad I've taken the challenge in learning this material. I'm looking for a deeper understanding.

"The "exchange" part refers to the energy change of the system when the spatial coordinates of two electrons are interchanged."

How is this true? When the spatial coordinates of two electrons (with same spin?) are interchanged the energy of the system should not change because electrons are indistinguishable particles. The idea of an electron having a specific spatial coordinate goes against quantum-theory. One can only predict the location of a electron within a spatial range and this prediction is given a probability.

Consider this thought experiment. For the He atom there are two electrons per atom. In the ground-state the 1s orbital is filled and they are spinning in opposite directions. The wave-functions will spatially be the same, but this is allowed since the electrons have opposite spin. The individual electron densities overlap so exchanging the spatial coordinates of two electrons doesn't change the electron density pattern for the individual electrons. If however, the electrons spins also "flipped" there should be a energy increase during the "flipping process" because at some point the spins will align. Perhaps this is the embodiment of the exchange energy.

Regards

Modey3

18. Jul 31, 2007

### olgranpappy

True. But, that has nothing to do with exchange energy.

No, not necessarily. Consider the full N-electron wavefunction
$$\Psi(x_1,...,x_N)\;.$$
Sure, this function picks up a minus sign if you switch any two of it's arguments. But, if I like I can still refer to the x_i as the positions of the electrons... it's vivid and useful language.

19. Jul 31, 2007

### olgranpappy

The exchange energy comes from approximating the ground state wave function as a slater determinant of "unperturbed" wavefunctions... but for He what should you choose as the "unperturbed" part? You might think that "obviously" I should choose the potential of the He nucleus as the unperturbed part, but I really could choose *any* single particle potential and then just write:
$$H=T_1+T_2+\frac{e^2}{|r_1-r_2|}-\frac{2e^2}{r_1}-\frac{2e^2}{r_2} =T_1+v_{\textrm{any}}(r_1)+T_2+v_{\textrm{any}}(r_2) +\frac{e^2}{|r_1-r_2|}-\frac{2e^2}{r_1}-v_{\textrm{any}}(r_1) -\frac{2e^2}{r_2}-v_{\textrm{any}}(r_2)$$
Then I could define:
$$H_0=T_1+v_{\textrm{any}}(r_1)+T_2+v_{\textrm{any}}(r_2)$$
and
$$H_1=+\frac{e^2}{|r_1-r_2|}-\frac{2e^2}{r_1}-v_{\textrm{any}}(r_1) -\frac{2e^2}{r_2}-v_{\textrm{any}}(r_2)\;.$$

Okay, so now I define my single particle wavefunctions as the eigenfunctions of H_0. Then I construct a slater determinant out of these single-particle functions and using these I can find some "exchange" energy... obviously this exchange energy is rather arbitrary because, as we just saw, the definition of single particle wavefunctions is arbitrary. I.e., there is some correction needed to get the correct ground state energy. The difference between the exchange and the true energy is just called the "correlation energy".

If you want to see explicitly what the exchange looks like for He then use the above procedure, but of course it will depend on which unperturbed functions you choose. Then, since He has been solved you can compare your slater determinant answer to the true answer to find the He correlation contribution.

Last edited: Jul 31, 2007
20. Jul 31, 2007

### marlon

The "spatial coordinates of electrons" refer to the wavefunctions, so the exchange part arises due to overlapping of wavefunctions. That is what the interchanging of spatial coordinates implies !

QM NEVER states that one cannot assign spatial coordinates to electrons. Actually you CAN ! The HUP just states that if you repeat the same measurement onto the same electron with, let's say, fixed position x, you will acquire a spread in it's momentum value ! The HUP manifests itself through the repition of experiments/measurements and has nothing to do with the system itself.

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