Derivation in section "Total energy" of DFT page is unclear

[RKBK]

I've been reading the DFT page, and I'm trying to understand the section on the Total energy. A short bit into that section, it is written that "If one is not careful about the potential associated with the eigenvalues, i.e., confusing V_{in} with V_{out}". However, where is "V_in" and "V_out" defined? It is a bit unclear if the section is trying to explain what would happen if the equations given there are used to evaluated the total energy on each iteration of the KS solution. But why would the energy after an iteration be evaluated using the previous external potential ("V_in"?)? The message of the section is a bit unclear to me.

Furthermore, the section seems to imply that the last equation of the section "Total energy" is somehow not preferable to a direct usage of the total energy expression (first equation of "total energy" section), despite the fact that several review works/text books on DFT indicate that the last expression is used in several DFT codes to calculate the total energy.

[certik]

Good point. I'll try to clarify that once I get to my computer. It's
actually very easy to program the total energy in an incorrect way and then
it will converge very slowly.

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On Nov 11, 2015 5:19 AM, "RKBK" notifications@github.com wrote:

I've been reading the DFT page, and I'm trying to understand the section
on the Total energy. A short bit into that section, it is written that "If
one is not careful about the potential associated with the eigenvalues,
i.e., confusing V_{in} with V_{out}". However, where is "V_in" and "V_out"
defined? It is a bit unclear if the section is trying to explain what would
happen if the equations given there are used to evaluated the total energy
on each iteration of the KS solution. But why would the energy after an
iteration be evaluated using the previous external potential ("V_in"?)? The
message of the section is a bit unclear to me.

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#58 (comment)
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[RKBK]

Thank you, I would certainly appreciate a clarification.

Furthermore, there is another thing in the derivation that is not obvious to me. In the equation following directly after
"If one is not careful about the potential associated with the eigenvalues, i.e., confusing V_{in} with V_{out}, one gets a slowly converging formula for the total energy. By expanding v_s using (12):"
the following equality is written:

\int (V_H + V_XC + v)n(r)d^3r = 2½ \int V_H n(r) d^3r + \int V_XC n(r) d^3r + \int v n(r) d^3r

How did the 2½ get into the RHS?

[certik]

So as mentioned in the section http://www.theoretical-physics.net/dev/quantum/dft.html#total-energy, the V_in is the potential that is used to solve the Schroedinger equation. It is defined by the line:

We use the whole input potential $V_{in}\equiv v_s$ and its associated
eigenvalues $\epsilon_i$ to calculate the kinetic energy $T_s[n]$, this follows
from the derivation of the expression for $T_s[n]$. Then we use the calculated
charge density to express $E_H[n]$, $E_{xc}[n]$ and $V[n]$.

The V_out is then the potential that you get if you take the charge density from V_in and use it to calculate the Hartree and XC terms in order to get a new potential for the next iteration. V_out is then put into some kind of a mixing scheme (non-linear iteration) in order to get a better guess for the next SCF iteration. For very simple systems, you can use V_out directly as V_in for the next SCF iteration.

@RKBK let me know if this clarifies the issue. I can then write this up in the section in a more clear way.

[RKBK]

Thank you, @certik , that does clarify the V_in vs V_out issue.

How about the question regarding the factor 2½, that I wrote about above?

[certik]

That 2½ really means 2 * 1/2 = 1. So it's just one. I should put parentheses there, then it will be clear. Good point.