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Darwin and relativistic kinetic energy correction for hydrogen

  1. Mar 22, 2012 #1
    1. The problem statement, all variables and given/known data
    Combine the Darwin correction with the relativistic kinetic energy correction for l=0 to show that the fine structure formula:

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

    remains valid for l=0

    2. Relevant equations

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

    3. 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.
     
  2. jcsd
  3. Mar 24, 2012 #2
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