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

1. Jan 13, 2010

### 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.

2. Jan 14, 2010

### CompuChip

Going one step back, you can change the integral to
$$\int_0^{2\pi} d\phi \int_{0}^\pi \sin\theta d\theta \int^{k_{F}}_{0}} (k')^2 dk'$$
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,
$$\frac{1}{\left|\textbf{k}-\textbf{k}'\right|^2}=\frac{1}{k \sqrt{1+k'^2/k^2-2\textbf{k}.\textbf{k}'/k^2}} = \frac{1}{\sqrt{k^2 + k'^2 + k k' \cos\theta}}$$

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

3. Jan 20, 2010

### 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:

$$\frac{1}{\left|\textbf{k}-\textbf{k}'\right|}=\frac{1}{k \sqrt{1+k'^2/k^2-2\textbf{k}.\textbf{k}'/k^2}} = \frac{1}{\sqrt{k^2 + k'^2 + k k' \cos\theta}}$$

so without the square on the $$\frac{1}{\left|\textbf{k}-\textbf{k}'\right|}$$ denominator.

Thanks again.

4. Jan 20, 2010

### 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.

Share this great discussion with others via Reddit, Google+, Twitter, or Facebook