Rayleigh–Ritz method - Yukawa coulomb potential

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SUMMARY

The discussion focuses on the application of the Rayleigh–Ritz method to a Yukawa coulomb potential, specifically analyzing the test function \(\phi(\alpha, r)=\sqrt{(\frac{\alpha^3}{\pi})}exp(-\alpha r)\) and the potential \(V(r)=-\frac{e^2}{r}exp(\frac{-r}{a})\). Participants confirm that the angular momentum operator \(\hat{L}^2 \phi\) yields a result of zero, as the test function does not depend on the angular variables \(\theta\) and \(\phi\). The discussion emphasizes the importance of recognizing the proportionality of the test function to \(Y_0^0\) in spherical harmonics.

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  • Understanding of quantum mechanics and Hamiltonians
  • Familiarity with spherical coordinates and angular momentum operators
  • Knowledge of the Rayleigh–Ritz variational method
  • Basic concepts of Yukawa potential in quantum physics
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  • Study the derivation of the angular momentum operator in spherical coordinates
  • Explore the properties of spherical harmonics, particularly \(Y_0^0\)
  • Investigate the Rayleigh–Ritz method for variational calculations
  • Learn about the Yukawa potential and its applications in quantum mechanics
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Students and researchers in quantum mechanics, particularly those studying angular momentum and variational methods in potential problems.

AwesomeTrains
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Hello everyone

Homework Statement


I have been given the testfunction \phi(\alpha, r)=\sqrt{(\frac{\alpha^3}{\pi})}exp(-\alpha r), and the potential V(r,\theta, \phi)=V(r)=-\frac{e^2}{r}exp(\frac{-r}{a})
Given that I have to write down the hamiltonian (in spherical coordinates I assume), and I have to calculate the angular momentum operator \hat{L}^2 \phi. (This is only a part of the whole problem. a) of a), b) and c) They should have used some other symbol for the testfunction than \phi, it's kinda confusing)

Homework Equations


Angular momentum operator in spherical coordinates.

The Attempt at a Solution


I guess the answer is 0, because \hat{L}^2 \phi contains derivations of \theta, \phi which the testfunction doesn't depend on. Is this true?
 
Last edited:
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AwesomeTrains said:
Hello everyone

I guess the answer is 0, because \hat{L}^2 \phi contains derivations of \theta, \phi which the testfunction doesn't depend on. Is this true?
Yes that's true. Another way to look at it is to realize that the test function is proportional to ##Y_0^0##.
 

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