Not sure about shape of this quantum well

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SUMMARY

The discussion centers on understanding the shape of a quantum well described by a Dirac delta function potential, V(x) = -U δ(x), where U is a positive constant. Participants clarify the distinction between the potential function V(x) and the wavefunction ψ(x), emphasizing that V(x) resembles a negative Dirac delta function. The equation presented indicates that the slope of ψ(x) is discontinuous at x=0, which is critical for solving the Schrödinger equation in this context. Participants recommend consulting Griffiths' Quantum Mechanics, particularly section 2.5, for a clearer understanding of delta-function potentials.

PREREQUISITES
  • Understanding of quantum mechanics concepts, specifically potential functions and wavefunctions.
  • Familiarity with the Dirac delta function and its properties.
  • Knowledge of the Schrödinger equation and its applications in quantum mechanics.
  • Basic mathematical skills for manipulating differential equations.
NEXT STEPS
  • Study Griffiths' Quantum Mechanics, section 2.5, to grasp delta-function potentials.
  • Learn about the properties of the Dirac delta function and its graphical representation.
  • Research the implications of discontinuities in wavefunctions when solving the Schrödinger equation.
  • Explore examples of quantum wells and barriers to understand their solutions in quantum mechanics.
USEFUL FOR

Students of quantum mechanics, particularly those tackling introductory courses, as well as educators and tutors seeking to clarify concepts related to quantum wells and wavefunctions.

  • #31
Taking the solution with no time dependance: Cexp ikx + D exp -ikx
and then I substite into the equation: I will do left hand side then right hand side.

-(hbared)/2m d^2 phi over dx^2 = ( -k^2 Cexp ikx + k^2 Dexp - ikx) . -(hbared)/2m

= (right hand side) E. phi

E= left hand side divided by Phi

E=0

That surely is wrong?!

(dphi/dx = ik Cexp ikx - ik Dexp - ikx)

from my notes I notice i should get E= ((hbared . k)^2 / 2m)

Thanks.
 
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  • #32
Davio said:
Taking the solution with no time dependance: Cexp ikx + D exp -ikx
and then I substite into the equation: I will do left hand side then right hand side.

-(hbared)/2m d^2 phi over dx^2 = ( -k^2 Cexp ikx + k^2 Dexp - ikx) . -(hbared)/2m
On the RHS here, the expression in parentheses should be

( -k^2 Cexp ikx - k^2 Dexp - ikx)​


= (right hand side) E. phi

E= left hand side divided by Phi

E=0

That surely is wrong?!
Yes, it's wrong, E is not 0.

(dphi/dx = ik Cexp ikx - ik Dexp - ikx)

from my notes I notice i should get E= ((hbared . k)^2 / 2m)

Thanks.
Yes, that looks right. The important point is, this wavefunction has a positive energy. So it is not the solution you want, since you want a solution with a negative energy (E<0).
 
  • #33
Hiya, thank you for your help. I managed to meet up with some friends and we managed to do it. Your help was invaluable. Thanks :-D
 
  • #34
Great, glad it worked out.
 

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