QM: Infinite Square Well -a/2 to a/2

In summary, the student is trying to solve for the constants A and B in terms of k, but is having trouble. He is unsure how to prove his results mathematically.
  • #1
5
0
I have read similar threads about this problem but I wasn't able to make progress using them.

Homework Statement



Consider an infinite square-well potential of width a, but with the coordinate system shifted so that the infinite potential barriers lie at x=[itex]\frac{-a}{2}[/itex] and x=[itex]\frac{a}{2}[/itex].

Solve the Schrodinger equation for this case to calculate the normalized wave function [itex]\psi[/itex]n(x) and the corresponding energies En

Homework Equations



Time Independent Schrodinger Equation (V(x)=0 within the well)
-\hbar2[itex]\frac{1}{2m}[/itex] ([itex]\frac{∂}{∂x}[/itex])2[itex]\psi[/itex]=E[itex]\psi[/itex]

The Attempt at a Solution



Skipping ahead to the general solution for [itex]\psi[/itex](x) , I get:

[itex]\psi[/itex](x) = Aeikx + Be-ikx, k = [itex]\frac{\sqrt{2mE}}{\hbar}[/itex]

Using the boundary conditions,

[itex]\psi[/itex]([itex]\frac{a}{2}[/itex])=Aeika/2 + Be-ika/2=0
[itex]\psi[/itex]([itex]\frac{-a}{2}[/itex])=Ae-ika/2 + Beika/2=0,

together with some substitution I am able to prove that,

k =[itex]\frac{\pi}{a}[/itex]n

En =([itex]\frac{n\pi\hbar}{a}[/itex])2[itex]\frac{1}{2m}[/itex]

which I believe to be correct for infinite square well.

I run into trouble trying to solve for the constants A and B. My approach was to try and normalize the wavefunction and then use substitution to solve for A or B, but I can't seem to get anywhere after a certain point.
I would expect to get cosines for n=1,3,5... and sines for n=2,4,6... based on what I know about the shape of the wavefunction, but I am unsure how to prove this mathemetically.
Any tips would be appreciated.
 
Last edited:
Physics news on Phys.org
  • #2
Try plugging in your allowed values of k into the boundary condition equations. You'll get a different relationship between A and B depending on whether n is odd or even.
 
  • #3
Ah, that's the ticket. It's probably easier without using exponential form, but I was able to prove that A=B for odd n (giving me a cosine) and A=-B for even n (giving me a sine). Thanks!
 

1. What is the "Infinite Square Well" in quantum mechanics?

The "Infinite Square Well" is a theoretical model used in quantum mechanics to describe a particle confined within a potential well with infinitely high walls. It is often used as a simplified example to demonstrate the principles of quantum mechanics.

2. What does the "a/2" represent in the Infinite Square Well?

The "a/2" represents half of the width of the well. In this model, the particle is confined within a well of length "a", with the center of the well located at "a/2".

3. How does the particle behave in the Infinite Square Well?

The particle in the Infinite Square Well behaves as a wave, with its energy levels determined by the well's width and the particle's mass. The allowed energy levels form a discrete set, and the probability of finding the particle at any point within the well is constant.

4. What is the significance of the potential energy in the Infinite Square Well?

In the Infinite Square Well model, the potential energy is infinite within the well and zero outside of it. This represents an idealized situation where the particle is completely confined within the well and cannot escape. It is used to demonstrate the effects of confinement on the particle's behavior.

5. How does the Infinite Square Well model relate to real-life systems?

While the Infinite Square Well is a simplified model, it can be used to describe some real-life systems, such as a confined electron in a semiconductor. However, most physical systems are not perfectly confined, and the Infinite Square Well is used mainly as a theoretical tool to understand the principles of quantum mechanics.

Suggested for: QM: Infinite Square Well -a/2 to a/2

Replies
12
Views
1K
Replies
7
Views
1K
Replies
2
Views
730
Replies
5
Views
2K
Replies
10
Views
84
Replies
3
Views
800
Replies
2
Views
2K
Replies
39
Views
9K
Back
Top