How Do Parity and Potential Affect Quantum Well Solutions?

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SUMMARY

The discussion focuses on solving quantum well problems using Schrödinger's equation, specifically addressing the implications of parity and potential in finite and infinite wells. The participant confirms the structure of the potential well and explores the eigenfunctions' parity, concluding that the lowest energy state is of even parity, while alternating states exhibit even and odd parity. The participant also analyzes the behavior of wavefunctions in regions defined by potential barriers, particularly for cases where energy levels are below and above the potential.

PREREQUISITES
  • Understanding of Schrödinger's equation in quantum mechanics
  • Familiarity with concepts of potential wells and bound states
  • Knowledge of parity in quantum systems
  • Ability to solve differential equations related to wavefunctions
NEXT STEPS
  • Study the implications of parity in quantum mechanics, focusing on eigenfunctions
  • Learn about finite and infinite potential wells in quantum mechanics
  • Explore the mathematical techniques for solving Schrödinger's equation
  • Investigate the behavior of wavefunctions in different potential energy scenarios
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Students and professionals in physics, particularly those studying quantum mechanics, as well as educators looking to enhance their understanding of quantum well solutions and the role of parity in wavefunctions.

thatguy14
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Homework Statement


The question is attached as a picture. Note: if someone would prefer I type it out I can.


Homework Equations


Schrodingers equation


The Attempt at a Solution



PART A
I am pretty sure I got the well right. It looks like a finite well inside an infinite well. I have attached a crudely drawn one called "well.png".

PART B
So I know that because the potential well is even (symmetric about 0) we can choose the eigenfunctions to have some definite parity. Then since we are dealing with bound states where the energy levels are non-degenerate. Then can we say that the lowest state should always be of even parity and every alternating state should be even also? and the ones inbetween are odd? Does this apply to the entire well? I'm not sure I understand this

PART C
This is where I lose all confidence in what I know. So there are 2 cases, one where E < V0 and E > V0. For the E<V0 the solutions would look exactly like the finite well but with the added restriction that the probability of the particle being at b must be 0. For E> V0 I am not really sure... It must be bound right? But how does it behave with the drop in potential?

I'll leave it there for now. Any help would be greatly appreciated
 

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Can anyone help please? If I am being unclear please let me know
 
I did some work on c, hopefully someone can give me a reply to this to make sure I am on the right path.

so I said that the solution is similar to the finite potential well. In region A which is between -b and -a the wavefunction is of the form:

\psi = Aexp(kx) + Bexp(-kx) (I know that there are i's here to indicate its complex but I am skipping a few steps).

Then I used a boundary condition that at x = -b, \psi=0. This led to

\psi = 0 =Aexp(-k*b)+Bexp(k*b)
then I solved for B

Bexp(kb) = -Aexp(-kB)
B = -Aexp(-2kb)

Then I plugged this back into the wave function

\psi = Aexp(kx) - Aexp(-2kb)exp(-kx)
\psi = A(exp(kx)-exp(-k(2b+x)))

Is this the correct approach? I know I have lots more to do but I wanted to make sure this was a step in the right direction. This has the behavior that I want, i.e., that it goes to 0 at b
 

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