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This's a question from Griffiths, about degenerate pertrubation theory:

I'm stuck.

(note: this's not a homework!)

Let the two "good" unperturbed states be

[tex]\psi^0_\pm = \alpha_\pm \psi^0_a + \beta_\pm \psi^0_b[/tex]

where [tex]\alpha_\pm[/tex] and [tex]\beta_\pm[/tex] are determined (up to normalization) by Equation 6.21 (or Equation 6.23), with Equation 6.26 for [tex]E_\pm[/tex]. Show explicitly that

(a) [tex]\psi^0_\pm[/tex] are orthogonal ([tex]<\psi^0_+ | \psi^0_-> = 0[/tex])

(b) [tex]<\psi^0_+ | H' | \psi^0_-> = 0[/tex]

(c) [tex]<\psi^0_\pm+ | \psi^0_\pm> = E_\pm[/tex]

[tex]\alpha W_{aa} + \beta W_{ab} = \alpha E^1[/tex] (6.21)

[tex]\alpha W_{ba} + \beta W_{bb} = \beta E^1[/tex] (6.23)

For [tex]\alpha=0, \beta=1[/tex] for instance, eq. 6.23 doesn't tell anything at all![tex]E^1_\pm = \frac{1}{2} \left( W_{aa} + W_{bb} \pm \sqrt{(W_{aa} - W_{bb})^2 + 4|W_{ab}|^2 } \right)[/tex] (6.26)

What does it mean "determined up to normalization"?. Equations 6.21 and 6.23 involve 3 unknowns ([tex]\alpha, \beta, E^1[/tex]), and Griffiths solved them for [tex]E^1[/tex] in terms of Ws, eliminating [tex]\alpha, \beta[/tex]. Any ideas about solving the question?[tex]\beta W_{bb} = \beta E^1[/tex] (6.23)

I'm stuck.

(note: this's not a homework!)

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