Determine the corerctions to the eigenvalue in the first approximation and the correct functions in the zeroth approximation, for a doubly degenerate level.(adsbygoogle = window.adsbygoogle || []).push({});

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] ?

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# Correction to the Eigenvalue -- doubly degenerate case

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