Degenerate perturbation theory question

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

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)
[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)
For [tex]\alpha=0, \beta=1[/tex] for instance, eq. 6.23 doesn't tell anything at all!

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

I'm stuck.

(note: this's not a homework!)
 
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nrqed
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This's a question from Griffiths, about degenerate pertrubation theory:









For [tex]\alpha=0, \beta=1[/tex] for instance, eq. 6.23 doesn't tell anything at all!
:confused: Of course it tells us something, it tells us that one of the first order correction to the energy is simply [itex] W_{bb} [/itex]!!!
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?

I'm stuck.

(note: this's not a homework!)
It's not clear what is the "question" you have in mind. But if you simply isolate, say, beta from one of the two equations, plug in the other equation, you will see that alpha will drop out completely. Then you will have a quadratic equation for E_1 and you get the two roots Griffiths gives. That's all there is to it. What is the difficulty you are having?

(aside: it's even miore clean to realize that the two equations can be put in matrix form so that E_1 is really the eigenvalue of the "W matrix")
 
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:confused: Of course it tells us something, it tells us that one of the first order correction to the energy is simply [itex] W_{bb} [/itex]!!!
Eh ^-^' I mean, it doesn't tell anything about [itex]\alpha[/itex]

I was trying to ask how I would get [itex]\alpha[/itex] and [itex]\beta[/itex]. And secondly, how to solve the question (all seem "obvious", but how do I show them explicitly?).

Maybe I should add that I'm a complete newbie to the subject (perturbation)!
 

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