Trouble converting integral from cartesian to spherical coords.

In summary, the conversation discusses computing the expectation value of the Hamiltonian for a variational calculation of a hydrogen atom, specifically neglecting the potential term and using a trial wavefunction of e^{-\alpha r}. The focus of the question is on converting an integral from Cartesian coordinates to spherical coordinates, and how to account for the Laplacian operator in this conversion. The Laplacian in spherical coordinates includes additional terms, and the question is whether these terms can be neglected in the conversion. The conversation also includes an example calculation using the equations and integral involved.
  • #1
Zoil
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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 [tex]e^{-\alpha r}[/tex]. 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 [tex]r^2\sin\(\theta\)[/tex] when going from [tex]dxdydz[/tex] to [tex]dr d\theta d\varphi[/tex]. but what if the expression I am integrating has an operator (specifically, [tex]\nabla^2[/tex])? Do I need to account for [tex]r^2\sin\(\theta\)[/tex] when I am taking the derivatives?

Homework Equations



Trial Wavefunction:
[tex]\Psi = e^{-\alpha r}[/tex]
Laplacian in spherical coords:
[tex] \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}[/tex]

The Attempt at a Solution



Here's the equation:
[tex]\langle \psi | H | \psi \rangle [/tex]
and then when I plug in [tex]\psi and \nabla^2[/tex] and the integrals, I get:
[tex]
\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
[/tex]

Essentially my question boils down to this: Can I neglect the second two terms of [tex] \nabla^2 [/tex] 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.
 
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  • #2
Added LaTeX to make it easier to follow what I'm getting at.

Hope that helps.
 
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FAQ: Trouble converting integral from cartesian to spherical coords.

Why do we need to convert from cartesian to spherical coordinates?

Converting from cartesian to spherical coordinates is often necessary in scientific calculations, particularly in fields such as physics and astronomy. It allows for a more intuitive representation of certain systems and can simplify complex equations.

How do you convert from cartesian to spherical coordinates?

To convert from cartesian to spherical coordinates, you can use the following equations:
x = r * sin(θ) * cos(φ)
y = r * sin(θ) * sin(φ)
z = r * cos(θ)
Where r is the distance from the origin, θ is the angle from the positive z-axis, and φ is the angle from the positive x-axis in the xy-plane.

What are the advantages of using spherical coordinates?

One major advantage of using spherical coordinates is that they can simplify calculations for systems with spherical symmetry. They also allow for a more natural representation of objects such as planets or stars in astronomy.

Can you convert from spherical to cartesian coordinates?

Yes, it is possible to convert from spherical to cartesian coordinates using the following equations:
x = r * sin(θ) * cos(φ)
y = r * sin(θ) * sin(φ)
z = r * cos(θ)
Where r is the distance from the origin, θ is the angle from the positive z-axis, and φ is the angle from the positive x-axis in the xy-plane.

Are there any limitations to using spherical coordinates?

One limitation of using spherical coordinates is that they are not suitable for all systems. They work best for systems with spherical symmetry and can be more complex for non-spherical systems. Additionally, they may not be as intuitive for some individuals compared to cartesian coordinates.

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