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Correction to the Eigenvalue -- doubly degenerate case

  1. Oct 15, 2015 #1
    Determine the corerctions to the eigenvalue in the first approximation and the correct functions in the zeroth approximation, for a doubly degenerate level.

    The solution:
    Equation [tex]\left| V_{nn'}-E^{(1)}\delta_{nn'}\right|=0[/tex] has here the form

    [tex]\left|\begin{array}{ccc}V_{11}-E^{(1)}&V_{21}\\V_{12}&V_{22}-E^{(1)}\end{array}\right|=0[/tex]

    Solving, we find:

    [tex]E^{(1)}=\frac{1}{2}\left[ V_{11}+V_{22} \pm h\omega^{(1)}\right][/tex]

    Solving also equation [tex]\sum_{n'}(V-{nn'}-E^{(1)}\delta_{nn'})c_{n'}^{(0)}[/tex] with these values of [tex]E^{(1)}[/tex], we obtain for the coefficients in the correct normalized function in the zeroth approximation , [tex]\psi^{(0)}=c_{1}^{(0)}\psi_{1}^{(0)}+c_{2}^{(0)}\psi_{2)^{(0)}[/tex]


    [tex]c_{1}^{(0)}=\left\{ \frac{V _{12} }{2\left| V _{12} \right| } \left[ 1 \pm \frac{V _{11}-V _{22} }{h\omega ^{(1)} } \right] \right\} ^{ \frac{1}{2} }[/tex]

    [tex]c_{2}^{(0)}= \pm \left\{ \frac{V _{21} }{2\left| V _{12} \right| } \left[ 1 \mp \frac{V _{11}-V _{22} }{h\omega ^{(1)} } \right] \right\} ^{ \frac{1}{2} }[/tex]

    The first equation I know, the task is just to solve the problem [tex](V _{11}-E ^{(1)})(V _{22}-E ^{(1)})-V _{21}V _{12}=0 [/tex] ut why there is[tex] \left| V _{12} \right| ^{2}[/tex] factor in the final formula ? What it represents ? And the most important, how to derive the formulas for these coefficients [tex]c_{1}^{(0)}[/tex][tex] c_{2}^{(0)}[/tex] ?
     
  2. jcsd
  3. Oct 20, 2015 #2
    Thanks for the post! This is an automated courtesy bump. Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post?
     
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