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Thanks to very helpful and detailed answers for my previous question, I understand the functional derivative of the energy with respect to the density. In addition, this functional derivative is equal to zero as follows (this screenshot is a page of the paper: "A bird's-eye view of density-functional theory [PDF]").

enter image description here

Here, this equation means that the sum of the potential terms (external $v_\text{ext}$, Hartree $v_\text{H}$, and exchange-correlation $v_\text{xc}$ (e.g., LDA)) and the functional derivative of the kinetic energy with respect to the density, $δT_s[n]/δn(r)$, is zero. In other words, does this mean that we have zero at all positions as shown in the following figure? Is it right?

enter image description here

Here, the $δT_s[n]/δn(r)$ term remains in the equation. Is it possible to write this term with a clear expression like these potential terms? More precisely, the kinetic energy is a functional of $\psi$ in the Kohn-Sham DFT (i.e., $T_s[\{\psi_i(r)\}_{i=1}^N]$, where $\psi$ is the Kohn-Sham orbital), but can't we write down this derivative using another equation without $δ$?

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Yes, the sum of these potentials is zero everywhere with the true ground-state density. But the true ground-state density is generally only expressible when the basis set is complete (if you know the true ground-state density in advance, you can of course construct a finite basis set that reproduces it exactly, but if you don't know the exact density, you normally have to use a complete basis set). So in an actual calculation, the sum of the potentials will in general not be everywhere zero, especially far away from the molecule and/or near the nuclei, where the quality of the basis set is generally poor.

As for the functional derivative of the kinetic energy - of course you may write it as something like $v_{T_s}(\mathbf{r})$ (I'm not sure about what the most common notation in the literature is, though). The literature you cited spelled out the functional derivative explicitly, because in the Kohn-Sham approach the real-space representation of this functional derivative never appears in the actual calculations - it is transformed away after introducing the Kohn-Sham orbitals. In orbital-free DFT, however, one may be more willing to write this term in a form similar to $v_{T_s}(\mathbf{r})$, because in this case the latter form is indeed involved directly in the calculations.

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  • $\begingroup$ Its an interesting discussion, but since its already gone on a few comments and is somewhat tangential to the main post, I moved it to chat $\endgroup$
    – Tyberius
    Sep 3 at 16:01
  • $\begingroup$ Thank you for the answer. The functional derivative of kinetic energy and orbital-free DFT are also interesting topic and I will try to study! $\endgroup$
    – neco
    Sep 9 at 12:19
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Since we are considering the left side of your Eq. 69 as a function of $\mathbf{r}$, it must be the constant function of $\mathbf{r}$ that is 0 everywhere: $0(\mathbf{r})$. If it were any other function, such that the left side of the equation were to be non-zero anywhere, we would not write a 0 there. So it's true that your Eq. 69 is saying that the RHS is zero at all of the points in your diagram.

As wzkchem5 said, Eq. 69 will not in practice be true in your real-world computer calculations, and is likely to be further away from 0 at places where the basis set is not as good of an approximation of the true function its aiming to represent.

About the definition of $\frac{\delta T}{\delta n(\mathbf{r})}$: Continuing from my answer to your question "Functional derivative of energy with respect to density", this is how $T$ changes when $n(\mathbf{r})$ changes. If the change in $T$ with respect to $n(\mathbf{r})$ stays the same no matter what your position $\mathbf{r}$ is, then your functional derivative is a constant with respect to $\mathbf{r}$; let's call it $v_T$. Let's say that at each position $\mathbf{r}$, the way that $T$ changes with respect to $n(\mathbf{r})$ is slightly different, then the functional derivative will have some $r$ dependence, so we can call it $v_T(\mathbf{r})$. As wzkchem5 said, this may make sense in orbital-free DFT, but in Kohn-Sham DFT the introduction of the KS orbitals replaces the need for dealing directly with any such $v_T(\mathbf{r})$.

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  • $\begingroup$ Thank you for the answer. I found some recent studies that consider the functional derivative of the kinetic energy with machine learning, and I will also try to study them. pubs.acs.org/doi/abs/10.1021/acs.jctc.0c00580 $\endgroup$
    – neco
    Sep 9 at 12:21

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