Dirac exchange energy and X_alpha LDA methods

Question

I want to make a DFT code (by myself) that works in the RKS LDA H2 configuration. This post is a continuation of here.

Now by adding the Vosko, Wilk, Nusair correlation I have much better results :

But when compared to results of this article, the energy I got seems greater compared to the article. I am thinking that the exchange energy might be the problem. The results can be found here. The most important picture to look at being the exchange-correlation energy Exc_VWN.png. I ran H2_RHF_STO3G_becke.py.

The Dirac exchange energy is given by : $$E_x^{Dirac} = \int \rho \epsilon_x(\rho) = -\frac{3}{4}(\frac{3}{\pi})^{1/3} \int \rho^{4/3}$$

Hence the exchange potential is : $$ v_x(\rho) = -(\frac{3}{\pi}\rho)^{1/3} $$

But I am not sure if it is right. I have two resources that defines the exchange energy in the case where you have $\alpha$ and $\beta$ densities :

• Introduction to Computational Chemistry, Frank Jensen, third edition, p 247 :

$$ E_x^{LSDA} = -2^{1/3} \frac{3}{4}(\frac{3}{\pi})^{1/3} \int \rho_{\alpha}^{4/3} + \rho_{\beta}^{4/3} $$

which is not consistant with the case where you consider an unique electron density (for RKS and RHF methods). By continuing to read, I stumbled upon the $X_{alpha}$ LDA methods. so I have a question:

How is LSDA exchange energy is derived? And why is it not consistent with the RKS case where $\alpha$ and $\beta$ electrons have the same density also the LDA case?

Answer 1

How is LSDA exchange energy is derived?

LSDA exchange is computed from the homogeneous electron gas from the expression for exact exchange; see any standard textbook on electronic structure with density functional theory.

And why is it not consistent with the RKS case where α and β electrons have the same density also the LDA case?

The two expressions above coincide. Density functionals are often expressed in the literature in terms of spin-unpolarized densities, $\rho = \rho_\alpha + \rho_\beta = 2\rho_\alpha$, from which you can derive the spin-polarized expressions applying spin scaling relations.

If you substitute this in the first expression, you see

$$ -\frac{3}{4}\left(\frac{3}{\pi}\right)^{1/3}\rho^{4/3}=-\frac{3}{4}\left(\frac{3}{\pi}\right)^{1/3}\int(2\rho_{\alpha})^{4/3} $$

Now, taking out the factor $2$ from the exponent gives you $2^{4/3}$, from which you take out the cube root in front and split the factor 2 as a sum of the alpha and beta terms

$$ -2^{1/3} \frac{3}{4}\left(\frac{3}{\pi}\right)^{1/3}\int \left[ \rho_{\alpha}^{4/3} + \rho_{\beta}^{4/3}\right] $$

Note that exchange is same-spin only, so there is no $\rho_\alpha \rho_\beta$ term.

Answer 2

You can get the LDA energy in spin-polarized form (LSDA), or spin-unpolarized (LDA). For a closed-shell system this will give exactly the same energy.

In LSDA you work with rho-alpha (=0.5*rho) and rho-beta, which are elevated to the power 4/3. In LDA, you work with rho elevated to the power 4/3. The trick is in the prefactors, Cx and Ax:

2 * (0.5*rho)^(4/3) * Cx = (rho)^(4/3) * Ax

You will find that Cx (ca. 0.9305) is equal to 2^(1/3) times Ax (ca. 0,7386).