Infinite Well Solutions: How Can Different Techniques Yield Contrasting Results?

In summary, the conversation discusses the time-independent Schrodinger equation solutions for an infinite well and how they change when the well is moved from 0 to a to -a/2 to a/2. The equations for the new boundaries involve a halving of the period, which is not expected. The conversation then delves into the application of boundary conditions and the resulting solutions, which may differ for sine and cosine functions. The issue of dividing by A is also brought up as a potential problem.
  • #1
ehrenfest
2,020
1

Homework Statement


The time-independent Schrodinger equation solutions for an infinite well from 0 to a are of the form:

[tex] \psi_n(x) = \sqrt{2/a} \sin (n \pi x/ a) [/tex]

If you move the well over so it is now from -a/2 to a/2, then you can replace x with x-a/2 and get the new equations right?

If I try to actually apply these new boundary conditions to the sin function, I get something different, however:[tex] \psi_n(x) = \sqrt{2/a} \sin (2 \pi n x/a) [/tex]

How can these two techniques give different results?

Homework Equations


The Attempt at a Solution

 
Physics news on Phys.org
  • #2
Probably because you did it wrong. You managed to halve the period of a wavefunction by translating it. That's not supposed to happen, is it?
 
  • #3
I just applied the boundary conditions to sin(k_n * x) by plugging in (a/2, 0) or (-a/2, 0).

This yields k_n = 2 n pi/a?
 
  • #4
For the symmetric well, there are also solutions cos(n\pi x), with n odd.
 
Last edited:
  • #5
But I get A cos (k_n a/2) + B sin(k_n a/2) = 0 and A cos (k_n -a/2) + B sin (-k_n a/2) = 0 implies that B/A = tan (k_n a/2) = -B/A which implies that B is 0?
 
Last edited:
  • #6
Maybe the problem is that I am assuming that the k_n is the same for both sine and cosine?
 
  • #7
sin(n*pi*(x-a/2))=sin(n*pi*x-n*pi/2). Clearly this is +/- cos(n*pi*x).
 
  • #8
OK. The problem was that in post number 5 I divided by A which could have been 0.
 

What is a symmetric infinite well?

A symmetric infinite well is a theoretical model used in quantum mechanics to describe the potential energy of a particle confined within a certain region. It consists of two infinitely high potential barriers placed symmetrically on both sides of a finite well.

What is the significance of a symmetric infinite well?

The symmetric infinite well is an important concept in quantum mechanics as it serves as a simple model for understanding the behavior of particles in a bound system. It can also be used to study the properties of energy levels and wave functions of particles in confined environments.

How is the energy of a particle in a symmetric infinite well quantized?

The energy levels of a particle in a symmetric infinite well are quantized, meaning they can only take on certain discrete values. This is because the particle is confined within the potential barriers and can only exist in specific energy states determined by the size and shape of the well.

What is the wave function of a particle in a symmetric infinite well?

The wave function of a particle in a symmetric infinite well describes the probability of finding the particle at a certain position within the well. It is a complex-valued function that satisfies the Schrödinger equation and contains information about the particle's energy and momentum.

What are the applications of the symmetric infinite well model?

The symmetric infinite well model has various applications in physics, including understanding the behavior of electrons in atoms, analyzing the properties of molecules, and studying the behavior of particles in semiconductor devices. It is also used in engineering applications such as designing quantum well lasers and transistors.

Similar threads

Replies
16
Views
400
  • Advanced Physics Homework Help
Replies
14
Views
739
  • Advanced Physics Homework Help
Replies
19
Views
327
  • Advanced Physics Homework Help
Replies
7
Views
1K
  • Advanced Physics Homework Help
Replies
2
Views
1K
  • Advanced Physics Homework Help
Replies
9
Views
3K
  • Advanced Physics Homework Help
2
Replies
39
Views
9K
Replies
3
Views
2K
  • Advanced Physics Homework Help
Replies
3
Views
942
  • Advanced Physics Homework Help
Replies
11
Views
2K
Back
Top