1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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?

  2. jcsd
  3. Mar 24, 2012 #2
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Similar Threads - Darwin relativistic kinetic Date
Time since hot big bang - relativistic species Feb 26, 2018
Scattering angle in relativistic kinematics Feb 23, 2018
Commutator in the Darwin Term Jun 25, 2017
The Darwin Term Feb 27, 2013
[QM] Darwin term is not hermitian Jun 24, 2010