Degenerate perturbation theory question

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SUMMARY

This discussion centers on the application of degenerate perturbation theory as presented in Griffiths' textbook. The key equations involved are Equations 6.21, 6.23, and 6.26, which relate the coefficients \alpha, \beta, and the first-order energy correction E^1 to the matrix elements W_{aa}, W_{bb}, and W_{ab}. Participants clarify that the orthogonality of the states \psi^0_\pm and the evaluation of matrix elements can be derived by manipulating these equations, ultimately leading to a quadratic equation for E^1. The discussion highlights the importance of normalization in determining the coefficients and the method to isolate variables for solving the equations.

PREREQUISITES
  • Understanding of quantum mechanics concepts, particularly perturbation theory.
  • Familiarity with Griffiths' "Introduction to Quantum Mechanics" and its equations.
  • Knowledge of matrix mechanics and eigenvalue problems.
  • Ability to manipulate algebraic equations and solve for unknowns.
NEXT STEPS
  • Study Griffiths' "Introduction to Quantum Mechanics" focusing on Chapter 6 regarding perturbation theory.
  • Learn about eigenvalues and eigenvectors in the context of quantum mechanics.
  • Explore examples of degenerate perturbation theory to solidify understanding of the concepts.
  • Practice solving quadratic equations arising from perturbation theory scenarios.
USEFUL FOR

Students and researchers in quantum mechanics, particularly those studying perturbation theory and its applications in quantum systems. This discussion is beneficial for anyone seeking to deepen their understanding of degenerate states and energy corrections in quantum mechanics.

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

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!

\beta W_{bb} = \beta E^1 (6.23)
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?

I'm stuck.

(note: this's not a homework!)
 
Last edited:
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gulsen said:
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!
:confused: Of course it tells us something, it tells us that one of the first order correction to the energy is simply W_{bb}!
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?

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")
 
nrqed said:
:confused: Of course it tells us something, it tells us that one of the first order correction to the energy is simply W_{bb}!
Eh ^-^' I mean, it doesn't tell anything about \alpha

I was trying to ask how I would get \alpha and \beta. 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)!
 
Time reversal invariant Hamiltonians must satisfy ##[H,\Theta]=0## where ##\Theta## is time reversal operator. However, in some texts (for example see Many-body Quantum Theory in Condensed Matter Physics an introduction, HENRIK BRUUS and KARSTEN FLENSBERG, Corrected version: 14 January 2016, section 7.1.4) the time reversal invariant condition is introduced as ##H=H^*##. How these two conditions are identical?

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