- #1
leoneri
- 19
- 0
Hi all,
I am trying to learn Hartree-Fock theory on free electron gas. But I am stumbled on one integration that I cannot seem to figure out. Here is the integral:
[tex]\int_{k'<k_{F}}\frac{d\textbf{k}'}{(2\pi)^3}\frac{4\pi e^2}{\left|\textbf{k}-\textbf{k}'\right|^2}[/tex]
I cannot figure out on how the solution becomes like this:
[tex]\frac{2 e^2}{\pi} k_{F} F\left(\frac{k}{k_{F}}\right)[/tex]
where
[tex]F(x)=\frac{1}{2}+\frac{1-x^2}{4x}ln\left|\frac{1+x}{1-x}\right|[/tex]
Well, up to this point, I know that I can change the coordinate system to spherical. So I can change
[tex]\int_{k'<k_{F}}d\textbf{k}'[/tex]
to
[tex]\int^{k_{F}}_{0}}4\pi k'^2 dk'[/tex]
but I am hopeless with this component.
[tex]\frac{1}{\left|\textbf{k}-\textbf{k}'\right|^2}[/tex]
I learn that I can change the latter into
[tex]\frac{1}{\left|\textbf{k}-\textbf{k}'\right|^2}=\frac{1}{k \sqrt{1+k'^2/k^2-2\textbf{k}.\textbf{k}'/k^2}}[/tex]
But what should I do with the [tex]\textbf{k}.\textbf{k}'[/tex] part? Because I want to turn all these vectors into scalars, so I can integrate them. Or is there other way to change [tex]\frac{1}{\left|\textbf{k}-\textbf{k}'\right|^2}[/tex] to spherical cordinate system that make it to scalar?
Any help, advice, or suggestion will be very much appreciated. Thanks in advance.
I am trying to learn Hartree-Fock theory on free electron gas. But I am stumbled on one integration that I cannot seem to figure out. Here is the integral:
[tex]\int_{k'<k_{F}}\frac{d\textbf{k}'}{(2\pi)^3}\frac{4\pi e^2}{\left|\textbf{k}-\textbf{k}'\right|^2}[/tex]
I cannot figure out on how the solution becomes like this:
[tex]\frac{2 e^2}{\pi} k_{F} F\left(\frac{k}{k_{F}}\right)[/tex]
where
[tex]F(x)=\frac{1}{2}+\frac{1-x^2}{4x}ln\left|\frac{1+x}{1-x}\right|[/tex]
Well, up to this point, I know that I can change the coordinate system to spherical. So I can change
[tex]\int_{k'<k_{F}}d\textbf{k}'[/tex]
to
[tex]\int^{k_{F}}_{0}}4\pi k'^2 dk'[/tex]
but I am hopeless with this component.
[tex]\frac{1}{\left|\textbf{k}-\textbf{k}'\right|^2}[/tex]
I learn that I can change the latter into
[tex]\frac{1}{\left|\textbf{k}-\textbf{k}'\right|^2}=\frac{1}{k \sqrt{1+k'^2/k^2-2\textbf{k}.\textbf{k}'/k^2}}[/tex]
But what should I do with the [tex]\textbf{k}.\textbf{k}'[/tex] part? Because I want to turn all these vectors into scalars, so I can integrate them. Or is there other way to change [tex]\frac{1}{\left|\textbf{k}-\textbf{k}'\right|^2}[/tex] to spherical cordinate system that make it to scalar?
Any help, advice, or suggestion will be very much appreciated. Thanks in advance.