gulsen
- 215
- 0
This's a question from Griffiths, about degenerate pertrubation theory:
For \alpha=0, \beta=1 for instance, eq. 6.23 doesn't tell anything at all!
I'm stuck.
(note: this's not a homework!)
Let the two "good" unperturbed states be
\psi^0_\pm = \alpha_\pm \psi^0_a + \beta_\pm \psi^0_b
where \alpha_\pm and \beta_\pm are determined (up to normalization) by Equation 6.21 (or Equation 6.23), with Equation 6.26 for E_\pm. Show explicitly that
(a) \psi^0_\pm are orthogonal (<\psi^0_+ | \psi^0_-> = 0)
(b) <\psi^0_+ | H' | \psi^0_-> = 0
(c) <\psi^0_\pm+ | \psi^0_\pm> = E_\pm
\alpha W_{aa} + \beta W_{ab} = \alpha E^1 (6.21)
\alpha W_{ba} + \beta W_{bb} = \beta E^1 (6.23)
E^1_\pm = \frac{1}{2} \left( W_{aa} + W_{bb} \pm \sqrt{(W_{aa} - W_{bb})^2 + 4|W_{ab}|^2 } \right) (6.26)
For \alpha=0, \beta=1 for instance, eq. 6.23 doesn't tell anything at all!
What does it mean "determined up to normalization"?. Equations 6.21 and 6.23 involve 3 unknowns (\alpha, \beta, E^1), and Griffiths solved them for E^1 in terms of Ws, eliminating \alpha, \beta. Any ideas about solving the question?\beta W_{bb} = \beta E^1 (6.23)
I'm stuck.
(note: this's not a homework!)
Last edited: