Can Schrodinger's Equation be Transformed into Spherical Polar Coordinates?

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Homework Help Overview

The discussion revolves around transforming Schrödinger's equation into spherical polar coordinates, a topic within quantum mechanics and mathematical physics.

Discussion Character

  • Exploratory, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the application of the chain rule for partial derivatives and the conversion of Cartesian coordinates to spherical coordinates. There is mention of the algebra involved in rewriting the derivatives and the importance of the laplacian operator in different coordinate systems.

Discussion Status

Some participants have provided guidance on the necessary steps to approach the transformation, while others have noted the algebraic complexity involved. Multiple interpretations of the process are being explored, but there is no explicit consensus on the best approach.

Contextual Notes

Participants reference the need to consult textbooks for specific forms of the equations and operators involved, indicating a reliance on external resources for complete understanding.

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how do you change the Schrödinger's equation into the spherical polar coordinates?
 
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Look up the chain rule for partial derivatives, and the equations that give you [itex]x, y, z[/itex] in terms of [itex]r, \theta, \phi[/itex] for spherical coordinates. Use these to re-write the derivatives [itex]\partial^2 \psi / \partial x^2[/itex] etc. into the derivatives [itex]\partial^2 \psi / \partial r^2[/itex] etc. There's a lot of algebra. The final result (which you should be able to see in your textbook) contains both first- and second-order derivatives.
 
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The coordinate-free form of the S.E. is:

[tex]i\hbar\frac{\partial \Psi(\vec r,t)}{\partial t}=-\frac{\hbar^2}{2m}\nabla^2 \Psi(\vec r,t)+V\Psi(\vec r,t)[/tex]

You can (should) look up the laplacian operator [itex]\nabla^2[/itex] in various coordinate systems in your textbook. There's not much physics to be learned by deriving it.
 
thank you very much! :)
 

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