Derivative of Spherical Harmonic for negative m

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Homework Help Overview

The discussion revolves around evaluating the derivative of spherical harmonics with respect to the azimuthal angle and expressing it in terms of spherical harmonics, particularly focusing on the case of negative m values.

Discussion Character

  • Exploratory, Assumption checking

Approaches and Questions Raised

  • Participants discuss the derivative of spherical harmonics, with initial confusion regarding the correct angle of differentiation. There is an attempt to clarify the relationship between the derivative and the spherical harmonics for both positive and negative m values.

Discussion Status

Some participants have provided insights and partial calculations, while others express a desire for feedback on their approaches. There is acknowledgment of differing results, and the discussion remains open for further contributions and corrections.

Contextual Notes

Participants are navigating the complexities of expressing derivatives in terms of spherical harmonics, with specific attention to the implications of negative m values. There is mention of potential resources for further information.

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Hello!

Homework Statement


I want to evaluate the derivative of spherical harmonics with respect to the azimuthal angle and express it in terms of spherical harmonics.2. Homework Equations and 3. The Attempt at a Solution
I have calculated the derivative of the spherical harmonic with respect to the azimuthal angle.
I know how to express the derivative of the spherical harmonic in terms of spherical harmonics for positive m. But I don't know how to do this for negative m.

View attachment Spherical_Harmonics.pdfThanks for your help.
 
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Isn't the azimuthal angle the argument of the Exp[], in this case θ?
 
Yes you're right. I meant that I want the derivative wrt the polar angle.
 
Hope this is what you were looking for.

It isn't the final answer because you can still express Cot ø and Exp(iθ) in terms of the spherical harmonics but that part is not the worst. May even be able to find them in a table somewhere.
 

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Hello!

Yes! That is exactly what I was looking for.
Although I recalculated it and I get a slightly different result.
I will upload my calculations tomorrow or so. Right now I am too tired.

thanks
 

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