Find Momentum Operators for 1s Electron in Hydrogen Atom

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Homework Help Overview

The discussion revolves around finding the momentum operators for a 1s electron in a hydrogen atom, specifically < px >, < p >, and < p2 >. The context involves quantum mechanics and the application of spherical coordinates in the wave equation.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to apply momentum operators but encounters difficulties with the wave equation's dependence on the radial coordinate r. Some participants suggest using spherical coordinates and the chain rule for derivatives. Others raise questions about the convergence of certain integrals related to the problem.

Discussion Status

The discussion is ongoing, with participants providing guidance on transforming derivatives and addressing the original poster's concerns. There is an exploration of different interpretations regarding the convergence of integrals, indicating a productive exchange of ideas.

Contextual Notes

Participants note the complexity of transforming Cartesian coordinates to spherical coordinates and the implications for computing derivatives. There are also references to specific mathematical expressions and integrals that may not converge, which are under discussion.

leenaa
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Find < px >,< p > and < p2 > for the 1s electron ofa hydrogen atom.

i am tried the solution but momentum operators Differential for x or y or z and the wave equation depends on the r !
 
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You need to use the transformation to spherical coordinates and the chain rule to compute the derivatives. Also, it's very likely that your textbook discusses the transformation of derivatives in Cartesian coordinates to derivatives in spherical coordinates, so you might find some useful formulas already derived for you.
 
Thank you very much

http://img104.herosh.com/2010/11/18/211847127.jpg"


But the show of the integral of Sec[\[Theta]] does not converge on {0,2 \[Pi]}
Is this true!
 
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You can't just determine [tex]\partial r/\partial x[/tex] by considering dx alone, since dr appears in the dy and dz equations as well. You can however compute

[tex]\frac{\partial r}{\partial x} = \frac{\partial }{\partial x} \sqrt{x^2 + y^2 + z^2} = \frac{x}{r}.[/tex]

Therefore

[tex]\frac{\partial}{\partial x} = \sin\theta \cos \phi \frac{\partial}{\partial r} + \text{angular derivs}.[/tex]
 
Thank you very much
Clearer idea
 

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