Trouble converting integral from cartesian to spherical coords.

[Zoil]

Homework Statement

Specifically, this question is about computing the expectation value of the Hamiltonian of a variational calculation of a hydrogen atom *neglecting the potential term. I'm assuming the trial wavefunction e^{-\alpha r}. The question, however, is purely math based, not quantum mechanical.

I'm having trouble understanding how to properly convert an integral from Cartesian coordinates to spherical coordinates. I know that I must add a factor of r^2\sin\(\theta\) when going from dxdydz to dr d\theta d\varphi. but what if the expression I am integrating has an operator (specifically, \nabla^2)? Do I need to account for r^2\sin\(\theta\) when I am taking the derivatives?

Homework Equations

Trial Wavefunction:
\Psi = e^{-\alpha r}
Laplacian in spherical coords:
\nabla^2 = {1 \over r^2} {\partial \over \partial r} \left( r^2 {\partial \over \partial r} \right) + {1 \over r^2 \sin \varphi} {\partial \over \partial \varphi} \left( \sin \varphi {\partial \over \partial \varphi} \right) + {1 \over r^2 \sin^2 \varphi} {\partial^2 \over \partial \theta^2}

The Attempt at a Solution

Here's the equation:
\langle \psi | H | \psi \rangle
and then when I plug in \psi and \nabla^2 and the integrals, I get:
<br /> \frac{i\hbar}{2m} \iiint e^{-\alpha r} \left( \left({1 \over r^2} {\partial \over \partial r} \left( r^2 {\partial \over \partial r} \right) + {1 \over r^2 \sin \varphi} {\partial \over \partial \varphi} \left( \sin \varphi {\partial \over \partial \varphi} \right) + {1 \over r^2 \sin^2 \varphi} {\partial^2 \over \partial \theta^2}\right)+V(r)\right)e^{-\alpha r} r^2 \sin(\theta) \,dr\,d\theta\,d\varphi <br />

Essentially my question boils down to this: Can I neglect the second two terms of \nabla^2 that are nonzero when I convert to spherical coords?

P.S. Sorry if my LaTeX sucks, I just taught myself how to do it for this post.

 

[Zoil]

Added LaTeX to make it easier to follow what I'm getting at.

Hope that helps.