Question on integration, in the Hartree-Fock theory for free electron gas.

[leoneri]

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:

\int_{k'<k_{F}}\frac{d\textbf{k}'}{(2\pi)^3}\frac{4\pi e^2}{\left|\textbf{k}-\textbf{k}'\right|^2}

I cannot figure out on how the solution becomes like this:

\frac{2 e^2}{\pi} k_{F} F\left(\frac{k}{k_{F}}\right)

where

F(x)=\frac{1}{2}+\frac{1-x^2}{4x}ln\left|\frac{1+x}{1-x}\right|

Well, up to this point, I know that I can change the coordinate system to spherical. So I can change

\int_{k'<k_{F}}d\textbf{k}'

to

\int^{k_{F}}_{0}}4\pi k'^2 dk'

but I am hopeless with this component.

\frac{1}{\left|\textbf{k}-\textbf{k}'\right|^2}

I learn that I can change the latter into

\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}}

But what should I do with the \textbf{k}.\textbf{k}' part? Because I want to turn all these vectors into scalars, so I can integrate them. Or is there other way to change \frac{1}{\left|\textbf{k}-\textbf{k}'\right|^2} to spherical cordinate system that make it to scalar?

Any help, advice, or suggestion will be very much appreciated. Thanks in advance.

 

[CompuChip]

Going one step back, you can change the integral to
<br /> \int_0^{2\pi} d\phi \int_{0}^\pi \sin\theta d\theta \int^{k_{F}}_{0}} (k&#039;)^2 dk&#039;<br />
Now let's choose the orientation of the axes such that theta denotes the angle between k and k'.
Carrying out the phi integral gives a factor of 2pi, and
k . k' = k k' cos(theta).

Thus,
<br /> \frac{1}{\left|\textbf{k}-\textbf{k}&#039;\right|^2}=\frac{1}{k \sqrt{1+k&#039;^2/k^2-2\textbf{k}.\textbf{k}&#039;/k^2}} = \frac{1}{\sqrt{k^2 + k&#039;^2 + k k&#039; \cos\theta}}<br />

Now let x = cos(theta) and try to carry out the integral :)

 

[leoneri]

Hi thanks a lot. I've done the integral, and I also found a mistake that I made. The procedure you told me is right, but my first inception of the definition is wrong, it should be like this:

<br /> \frac{1}{\left|\textbf{k}-\textbf{k}&#039;\right|}=\frac{1}{k \sqrt{1+k&#039;^2/k^2-2\textbf{k}.\textbf{k}&#039;/k^2}} = \frac{1}{\sqrt{k^2 + k&#039;^2 + k k&#039; \cos\theta}}<br />

so without the square on the \frac{1}{\left|\textbf{k}-\textbf{k}&#039;\right|} denominator.

Thanks again.

 

[CompuChip]

Whoops, sorry, I also overlooked that.
Of course |v|² = v . v and the norm |v| of v is the square root of that.