One-dimensional potential well problem

In summary, the particle is bound in a potential well and has different energies depending on whether it is inside or outside of the well. When you solve the differential equation, you should get a couple exponentials.
  • #1
Ruddiger27
14
0
My head's melting right now, because I've been stuck on this for the past 6 hours.
There's a particle of mass moving in a potential well where
V(x) = infinity at x<0
V(X)=0, 0<x<a
V(x)= Vo, x>a

Vo>0
E<Vo

I'm assuming that the wavefunction at x<0 is 0, since there's an infinite potential there. The energy inside the potential well is just the kinetic energy, =(Hk)^2/2m, where H=h/2pi, so the wavefunction should be of the form
psi= Aexp(-ikx )
Now outside the well, at x>a, the energy should be E= Vo-Ek because the particle is bound in the well. We then get psi=Bexp(-Tx), where T is k with (Vo-E) instead of E.
Am I wrong in assuming this? When I try to find the radius using
X^2 + y^2 = R^2, where x=k=a*sqrt(2mE)/H and y=ai*(sqrt(2m(Vo-E))/H
I get out a negative radius.
Please help me see what I've done wrong, I'm sure I've got the energy value on the finite potential side wrong, but I can't see how.
 
Physics news on Phys.org
  • #2
Haven't gone through your entire question to know if this is the only mistake, but I believe you made a mistake determining psi in the well.

When you solve the differential equation you should get a couple exponentials...something like

psi=A*exp(ikx)+B*exp(-ikx).

Remember these are complex exponentials so they don't blow up at infinity and you can't eliminate terms like with real exponentials.

Now your boundary conditions require that psi=0 for x</= 0, so plugging in 0 for x we get A+B=0, and hence A=-B, and psi is some sort of sine wave of the form
psi=A(exp(ikx)-exp(-ikx))

I haven't gone through the rest of your problem, and I'm not sure what radius you are referring to at the end.
 
  • #3
  • #4
The radius comes in when we look at the wavefunctions across x=a, where they should be continuous, so psi1=psi2 and dpsi1/dx=dpsi2/dx. I have to divide one by the other to get a solution in the form -y=xcot(x), then rearrange the circle equation to get y, sub it in and the intersection gives the solution.
 
  • #5
Ok thanks, you've really helped me out, the website was especially helpful.
 

1. What is a one-dimensional potential well?

A one-dimensional potential well is a theoretical model used in quantum mechanics to describe the behavior of a particle confined to a one-dimensional space. It consists of a region where the potential energy is lower than the surrounding areas, creating a "well" that the particle can be trapped in.

2. How is the potential well problem solved?

The potential well problem is typically solved using the Schrödinger equation, which describes the wave function of the particle in the potential well. The solution to this equation yields the allowed energy levels and corresponding wave functions for the particle.

3. What is the significance of the potential well problem?

The potential well problem is an important concept in quantum mechanics as it helps us understand the behavior of particles in confined spaces. It is also used in various applications, such as in the design of electronic devices and understanding the properties of materials.

4. What are the different types of potential wells?

There are three main types of potential wells: rectangular, triangular, and parabolic. Rectangular wells have a constant potential within the well and infinite potential outside. Triangular wells have a sloping potential within the well and infinite potential outside. Parabolic wells have a curved potential within the well and infinite potential outside.

5. What happens to the particle in a potential well?

The behavior of the particle in a potential well depends on its energy level. If the particle has enough energy, it can escape the well. But if its energy is lower than the potential well, it will be trapped within the well and its wave function will oscillate back and forth between the walls. This leads to the quantization of energy levels in the potential well.

Similar threads

  • Advanced Physics Homework Help
Replies
10
Views
419
Replies
16
Views
516
Replies
7
Views
2K
  • Advanced Physics Homework Help
Replies
14
Views
846
  • Advanced Physics Homework Help
Replies
7
Views
1K
  • Advanced Physics Homework Help
Replies
4
Views
2K
  • Advanced Physics Homework Help
Replies
3
Views
980
Replies
2
Views
1K
  • Advanced Physics Homework Help
Replies
5
Views
1K
  • Advanced Physics Homework Help
Replies
2
Views
806
Back
Top