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  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!

asked Feb 27 in Mathematics by MathematicalPhysicist (170 points) [ no revision ]

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 Answer

+ 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 Vladimir Kalitvianski (132 points) [ no revision ]

@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?

Yes, my e-mail: vladimir.kalitvianski@wanadoo.fr

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