Solving Normalization with Psi(r)=Ne^(-ar)

[precondition]

Hi there
I'm stuck on one of the normalization question,
I'd be glad if you can help me out.
The questions is, psi(r)=Ne^(-ar) where N is a normalization constant and a is a known real parameter.
Solution reads, 1=int dr lpsi(r)l^2 = 4piN^2 int (from 0 to infinity) dr r^2 e^(-2ar)
What I don't get is how to get integral from 0 to infinity?(instead of -infinity to +infinity) with extra factor 4pi r^2?
Thanks for your help

P.S. the solution also says, use the following integral, n larger than or equal to 0: int(0 to infinity)dx x^n e^(-x)=n factorial
I don't get where you use above integral.

 

[Igor_S]

It seems you are stuck with math problem. Here is a little help:

In order to calculate normalization, you need to calculate volume integral from equation:

\int_{all space} |\Psi(\vec{r})|^2 d \vec{r} = 1.

Now the thing is, people use notation d\vec{r} to actually mean a little volume element (dV if you will). In spherical coordinates, small element of space (volume) can be written as:

dV = (r \sin{\theta}) (r d\phi) dr,

where r now represents distance from the origin (the length of vector r). To cover all space when using spherical coordinates, you need to take the limits for three variables (r from 0 to infinity, theta from 0 to pi, phi from zero to 2pi; these can be easily visualized if you draw a picture). Therefore, normalization integral is actually triple integral:

\int_{r=0}^{\infty} \int_{\theta = 0}^{\pi} \int_{\phi = 0}^{2\pi} |\Psi(r, \theta, \phi)|^2 r^2 \sin{\theta} d\phi d\theta dr.

If wavefunction depends only on variable r (\Psi(\vec{r}) = \Psi(r, \theta, \phi) = \Psi(r)), you can actually carry out integration over theta and phi. Try to do yourself from here