Simple time-independent perturbation problem. QM

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SUMMARY

The discussion focuses on the application of time-independent perturbation theory in quantum mechanics, specifically addressing a delta-function potential within an infinite square well. The Hamiltonian is modified to {H^{'}} = αδ(x-a/2), where α is a constant. The first-order energy correction is calculated, revealing that the energies for even quantum numbers (n) remain unperturbed due to the odd symmetry of their wave functions, which results in zero overlap with the delta-function potential.

PREREQUISITES
  • Understanding of quantum mechanics principles, particularly perturbation theory.
  • Familiarity with infinite square well potential and its wave functions.
  • Knowledge of delta-function potentials and their implications in quantum systems.
  • Ability to perform calculations involving first-order corrections in quantum mechanics.
NEXT STEPS
  • Study the derivation of first-order energy corrections in time-independent perturbation theory.
  • Explore the properties of wave functions in quantum mechanics, focusing on symmetry and parity.
  • Investigate the implications of delta-function potentials in various quantum systems.
  • Learn about higher-order perturbation corrections and their significance in quantum mechanics.
USEFUL FOR

Students and professionals 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 energy corrections in quantum wells.

armis
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Homework Statement


"Suppose we put a delta-function in the center of the infinite square well:
{H^{'}} = \alpha\delta(x-a/2) where a is a constant. Find the first order correction to the allowed energies. Explain why the energies are not peturbed for even n"

Homework Equations



The Attempt at a Solution



Just to make sure. The energies for even n are not peturbed because the wave functions are odd with respect to the center of the well, thus equal to zero at those points where the delta-function potential is (if our square well is from 0 to a). Is that correct?
 
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Yes, that's correct!
 
thanks
 

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