Laplacian in spherical coordinates

  • #1
BRN
69
2

Homework Statement


Hello at all!

I have to calculate total energy for a nucleons system by equation:

##E_{tot}=\frac{1}{2}\sum_j(t_{jj}+\epsilon_j)##

with ##\epsilon_j## eigenvalues and:

##t_{jj}=\int \psi_j^*(\frac{\hbar^2}{2m}\triangledown^2)\psi_j dr##

My question is: if I'm in spherical coordinates, should I use Laplacian defined by:

##\triangledown^2=\frac{1}{r}\frac{\partial^2 }{\partial r^2}r-\frac{l^2}{r^2}##

or

##\triangledown^2=\frac{1}{r}\frac{\partial^2 }{\partial r^2}r## ?


Homework Equations




The Attempt at a Solution


At the moment I try to use

##\triangledown^2=\frac{1}{r}\frac{\partial^2 }{\partial r^2}r##

I'm not sure it's the right solution...

Could someone help me?

Thanks!
 

Answers and Replies

  • #2
Orodruin
Staff Emeritus
Science Advisor
Homework Helper
Insights Author
Gold Member
16,829
6,652
The Laplace operator in spherical coordinates is
$$
\nabla^2 = \frac{1}{r^2}\partial_r r^2 \partial_r + \frac{1}{r^2\sin(\theta)}\partial_\theta \sin(\theta) \partial_\theta + \frac{1}{r^2\sin^2(\theta)}\partial_\varphi^2.
$$
If it is acting on an eigenfunction of the angular momentum operator then the angular part can be replaced by ##-\ell(\ell+1)/r^2##. It is not clear from your post whether this is the case or not. Please give the full problem statement exactly as given as per forum guidelines.
 
  • #3
BRN
69
2
Thanks Orodruin for your answer.
I have not a exercise text to post because it is not an exercise, but part of a work.
I try to explain better what I need to do.
I have created an Schrodinger equation solver for nuclear systems in spherical coordinates. In this case, spherical symmetry leads to an equation independent of angular coordinates and my Schrodinger equation depends only on radial coordinate:

##[-\frac{\hbar^2}{2m}\triangledown^2+\frac{\hbar^2 l(l+1)}{2mr^2}+V(r)]\psi(r)=E\psi (r)##

Now, I need to calculate a total energy of the system by equations:

##E_{tot}=\frac{1}{2}\sum_j(t_{jj}+\epsilon_j)##

##t_{jj}=\int \psi_j^*(\frac{\hbar^2}{2m}\triangledown^2)\psi_j dr##

I'm not sure if I have to use Laplacian definition, in spherical coordinates, as:

##\triangledown^2=\frac{1}{r}\frac{\partial^2 }{\partial r^2}r-\frac{l^2}{r^2}##

or as:

##\triangledown^2=\frac{1}{r}\frac{\partial^2 }{\partial r^2}r##

I hope I explained myself better.

Thaks!
 

Related Threads on Laplacian in spherical coordinates

Replies
4
Views
6K
  • Last Post
Replies
1
Views
495
  • Last Post
Replies
7
Views
974
  • Last Post
Replies
6
Views
2K
  • Last Post
Replies
0
Views
3K
  • Last Post
Replies
2
Views
2K
  • Last Post
Replies
16
Views
234
Replies
7
Views
835
Replies
4
Views
4K
Top