Laplacian in spherical coordinates

[BRN]

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!

 

[Orodruin]

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.

 

[BRN]

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!