Schrodinger equation and boundary conditions

[BRN]

Hi at all,
I'm tring to solve Schrodinger equation in spherically symmetry with these bondary conditions:

$\lim_{r \rightarrow 0} u(r)\ltimes r^{l+1}$
$\lim_{r \rightarrow 0} u'(r)\ltimes (l+1)r^{l}$

For eigenvalues, the text I'm following says that I have to consider that the eigenfunctions are tending to zero at the extremes of integration, i.e. $r = 0$ and $r = 3 * r_{nucl}$

Why I need to consider an eigenfunction=0 in r=0? I would expect it to be maximum at that point...

Some idea?

Thanks.

 

[Ray Vickson]

Hi at all,
I'm tring to solve Schrodinger equation in spherically symmetry with these bondary conditions:

$\lim_{r \rightarrow 0} u(r)\ltimes r^{l+1}$
$\lim_{r \rightarrow 0} u'(r)\ltimes (l+1)r^{l}$

For eigenvalues, the text I'm following says that I have to consider that the eigenfunctions are tending to zero at the extremes of integration, i.e. $r = 0$ and $r = 3 * r_{nucl}$

Why I need to consider an eigenfunction=0 in r=0? I would expect it to be maximum at that point...

Some idea?

Thanks.

What is the potential $V(r)?$ In general, if you change the function $V$ you can/will change the nature of the solution $u(r)$.

 

[BRN]

Now, I'm using the harmonic oscillator potential:
$\frac{1}{2}m \omega^2r^2$

But these boundary conditions are used for Woods-Saxon potential too.