A Simplification of the Hartree-Fock method.

+ 1 like - 0 dislike
186 views

I am reading the following paper by Slater:

https://journals.aps.org/pr/pdf/10.1103/PhysRev.81.385

On page 5 they write above equation (12) the following:

"If we now average over all wave functions, we find that the properly weighted average of $F(\eta)$ is 3/4."

Now, $F(\eta)=1/2 + \frac{1-\eta^2}{4\eta}\ln((1+\eta)/(1-\eta))$.

I don't understand what does it mean to average over wave functions, I thought that they calculated: $\lim_{T\to \infty} \frac{1}{T}\int_0^T F(x)dx$, but I have given maple to calculate this limit (for the additive part without 1/2), and it didn't gave me 1/4.

So I don't understand which average of this function did they calculate?

ANyone knows?

Thanks!

I wonder how you computed the limit with T going to infinity. The imaginary part is null only between -1 and 1...What is F(n) ? a classical correlation function?

@igael $F(\eta)$ is a function that appears in the exchange potential energy, where the ratio $\eta=p/P$ is the ratio between the magnitude of the momentum of the electron to the maximum momentum corresponding to the top of the Fermi distribution.

+ 1 like - 0 dislike

If you want to average "something" over a wave function $\psi$, i.e., to find a mean value of "something", itis calculated as follows: $\langle \text{something}\rangle=\int \text{something} \cdot |\psi|^2 dx$

answered Feb 27 by (132 points)

@VladimirKalitvianski  Ok, so it should be $<F> = \int F(x) |\psi(x)|^2dx$, but how would I calculate this integral if $\psi(x)$ isn't known, I think it should be the solution to equation (7) of Hartree-Fock equations ?

I could not read this PhysRev article. Yes, $\psi$ must be the solution of HF equation. If it is unknown, the only thing you can do is to apply a normalization condition which holds for any solution: $\int|\psi(x)|^2 dx=1$. Maybe your $F(x)$ is so simple that the integral with it can be reduced to calculation of $|\psi(x)|^2$, I do not know.

@VladimirKalitvianski I can send you a copy of the article.

Is there a way to send you a personal message through this website?

 Please use answers only to (at least partly) answer questions. To comment, discuss, or ask for clarification, leave a comment instead. To mask links under text, please type your text, highlight it, and click the "link" button. You can then enter your link URL. Please consult the FAQ for as to how to format your post. This is the answer box; if you want to write a comment instead, please use the 'add comment' button. Live preview (may slow down editor)   Preview Your name to display (optional): Email me at this address if my answer is selected or commented on: Privacy: Your email address will only be used for sending these notifications. Anti-spam verification: If you are a human please identify the position of the character covered by the symbol $\varnothing$ in the following word:p$\hbar$ysicsOve$\varnothing$flowThen drag the red bullet below over the corresponding character of our banner. When you drop it there, the bullet changes to green (on slow internet connections after a few seconds). To avoid this verification in future, please log in or register.