Darwin and relativistic kinetic energy correction for hydrogen

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SUMMARY

The discussion focuses on combining the Darwin correction with the relativistic kinetic energy correction for hydrogen, specifically for the case of l=0. The fine structure formula, represented as ΔE_{fs} = - (E^{(0)2}_{n})/(2mc^{2})[4n/(j + 1/2) - 3], is confirmed to remain valid for s-states where j=1/2 and l=0. The Darwin Hamiltonian primarily affects s-states, and the relevant formula for these states is given as 2/(na^{3/2}) * (1/√(4π)). The inquiry seeks clarity on whether to combine the corrections and validate the equation for the specified quantum numbers.

PREREQUISITES
  • Understanding of quantum mechanics, specifically fine structure and quantum numbers.
  • Familiarity with the Darwin Hamiltonian and its implications for s-states.
  • Knowledge of relativistic kinetic energy corrections in quantum systems.
  • Basic proficiency in mathematical manipulation of quantum equations.
NEXT STEPS
  • Study the derivation of the Darwin Hamiltonian and its effects on s-states.
  • Explore relativistic corrections in quantum mechanics, focusing on hydrogen atom models.
  • Research the fine structure of hydrogen and its implications in atomic physics.
  • Review the application of quantum numbers in determining energy levels in quantum systems.
USEFUL FOR

This discussion is beneficial for physics students, quantum mechanics researchers, and educators focusing on atomic structure and relativistic effects in hydrogen. It provides insights into advanced topics in quantum theory and fine structure calculations.

kraigandrews
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Homework Statement


Combine the Darwin correction with the relativistic kinetic energy correction for l=0 to show that the fine structure formula:

\DeltaE_{fs}= - \frac{(E^{(0)2}_{n})}{2mc^{2}}[\frac{4n}{j + 1/2}-3]

remains valid for l=0

Homework Equations



From a previous problem the Darwin hamiltionian is shown to affected only under s-states
where for any s-state the formula \frac{2}{na^{3/2}}\frac{1}{\sqrt{4\pi}}

The Attempt at a Solution



so I know for s-states j=1/2 and l=0 (obviously). Overall, I am unsure as to what to show here, meaing:

do I combine the correction for the darwin energy to the correctionn for kinetic energy and then just show that for l=0, j=1/2 the equation still remains valid?

Thanks.
 
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