Degenerate perturbation theory question

In summary, Griffiths discusses degenerate perturbation theory and the equations for determining the "good" unperturbed states. He shows that these states are orthogonal and that the first order correction to the energy is given by W_{bb}. The equations involve 3 unknowns, but by eliminating \alpha and \beta, Griffiths is able to solve for E_1 in terms of Ws. The question asks for clarification on what it means for the states to be "determined up to normalization" and for help in solving the equations.
  • #1
gulsen
217
0
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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  • #2
gulsen said:
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")
 
  • #3
nrqed said:
: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)!
 

What is degenerate perturbation theory?

Degenerate perturbation theory is a mathematical tool used in quantum mechanics to calculate the energy levels and wavefunctions of a system when there are multiple states with the same energy.

When is degenerate perturbation theory used?

Degenerate perturbation theory is typically used when the energy levels of a system are closely spaced or degenerate, meaning they have the same energy. It is also used when the perturbation (external influence) on the system is small.

What is the difference between non-degenerate and degenerate perturbation theory?

Non-degenerate perturbation theory is used when the energy levels of a system are well separated, while degenerate perturbation theory is used when the energy levels are closely spaced or degenerate. Non-degenerate perturbation theory is simpler and more accurate, but it cannot be used when the energy levels are degenerate.

How is degenerate perturbation theory applied?

Degenerate perturbation theory involves diagonalizing the perturbation matrix, which represents the effects of the external influence on the system, in order to obtain the new energy levels and wavefunctions of the system. This allows for a more accurate calculation of the energy levels and wavefunctions compared to non-degenerate perturbation theory.

What are the limitations of degenerate perturbation theory?

Degenerate perturbation theory assumes that the perturbation on the system is small, and it may not accurately predict the behavior of a system when the perturbation is large. It also does not take into account interactions between different energy levels, which can affect the accuracy of the results.

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