Eigenvalues and Eigenfunctions for a Delta Potential Well

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SUMMARY

The discussion focuses on finding the eigenvalues and eigenfunctions for a delta potential well defined by the potential V(x)=λδ(x) for -aa or x<-a. The stationary Schrödinger equation is employed, specifically the form ##\psi '' (x) = -\frac{2mE}{\hbar^2}\psi(x) +\frac{2m\lambda }{\hbar^2}\delta (x)\psi(x)##. The delta function introduces a boundary condition on the wavefunction's slope at ##x = 0##, which can be derived by integrating the Schrödinger equation across a small interval around the origin and taking the limit as the interval approaches zero.

PREREQUISITES
  • Understanding of the stationary Schrödinger equation
  • Familiarity with delta functions in quantum mechanics
  • Knowledge of boundary conditions in wavefunctions
  • Basic concepts of quantum mechanics, particularly eigenvalues and eigenfunctions
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  • Learn about boundary conditions and their effects on wavefunctions
  • Explore the integration techniques for solving differential equations
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Students and professionals in quantum mechanics, particularly those studying potential wells and eigenvalue problems, as well as educators seeking to explain the implications of delta potentials in quantum systems.

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Homework Statement



Consider this situation, V(x)=λδ(x) ,-a<x<a. V(x)=∞,x>a or x<-a.
How to find the eigenvalue and eigen wavefuntion of the Hamiltonian.

Homework Equations


i can only reder to stationary Schrödinger equation.

The Attempt at a Solution


when it is ouside the well(x>a or x<-a), the wave function is zero.
But when it is inside the well, i just do not know how to solve the Schrödinger function because i do not know how to deal with the dela potential.
 
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The Schrödinger equation may be written as

##\psi '' (x) = -\frac{2mE}{\hbar^2}\psi(x) +\frac{2m\lambda }{\hbar^2}\delta (x)\psi(x)##

The effect of the delta function is to impose a boundary condition on the slope of the wavefunction at ##x = 0##; i.e., a boundary condition on ##\psi'(x)## at ##x = 0##. You can discover the specific form of this condition by integrating both sides of the Schrödinger equation from ##x = -\epsilon## to ##x = \epsilon## and then letting ##\epsilon## approach zero.
 
thanks
 

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