Electron in potential well equation

Click For Summary

Discussion Overview

The discussion centers around the Schrödinger equation as it applies to an electron confined in a one-dimensional potential well. Participants explore the formulation of the equation, the meaning of the potential function V(x), and the implications of different boundary conditions.

Discussion Character

  • Technical explanation
  • Debate/contested
  • Conceptual clarification

Main Points Raised

  • One participant presents the equation Hφ(x) =((p2op)/2m + V(x))φ(x) and expresses confusion about its origin and the meaning of V(x).
  • Another participant asserts that the correct form of the equation for a particle in a one-dimensional potential well is -hbar^2/2m.d^2/dx^2 = E φ(x), suggesting that the original equation contains a typo.
  • Some participants clarify that V(x) represents the potential energy, which is zero inside the well and may be infinite or a fixed value outside, depending on the type of potential well.
  • There is a discussion about the implications of V(x) being zero and how it relates to the Schrödinger equation, with some participants arguing that the equation does not explicitly state the particle is in a square well.
  • One participant notes that for a finite potential well, wavefunctions outside the well do not equal zero but decay exponentially, requiring continuity and differentiability at the boundaries.
  • Another participant acknowledges their previous misunderstanding and confirms their understanding of the time-independent Schrödinger equation.
  • There is a reiteration of the equation's form and its components, with one participant expressing gratitude for the clarification.

Areas of Agreement / Disagreement

Participants express differing views on the correct formulation of the Schrödinger equation for a one-dimensional potential well, with some asserting the original equation is incorrect while others defend its validity. The discussion remains unresolved regarding the precise implications of V(x) and the nature of the potential well.

Contextual Notes

Participants reference boundary conditions and the nature of potential wells (finite vs. infinite), but there are unresolved aspects regarding the mathematical treatment and assumptions underlying the discussion.

AmyJ
Messages
5
Reaction score
0
Hφ(x) =((p2op)/2m + V(x))φ(x) = (-hbar2/2m.d2/dx2+ V(x)φ(x) = E φ(x)


This equation is supposed to relate to an electron confined in a one-dimensional potential well. I'm really confused about where it comes from and I do not know what V(x) represents. If anyone could help explain this, I'd be very grateful :D
 
Physics news on Phys.org
umm no that's not the eqn for a particle in a 1d potential well

-hbar^2/2m.d^2/dx^2 = E φ(x) with the apropo boundary conditions is the eqn that describes a particle in a 1d well.
 
ice109 said:
umm no that's not the eqn for a particle in a 1d potential well

*scratch my head*

But of course it is. Ok, there is a closing bracket missing after the V(x), but I guess it is a typo.

This is the standard Schrödinger equation for a particle. I do not know, on which level your knowledge of physics is, but I suppose: Either you know, what a Schrödinger equation is or you need a crash course in quantum mechanics.

The V(x) is the potential you have, so in case of a potential well it will be 0 for x inside the potential well and will be infinite or have some fixed value (depending on whether your potential well has finite or infinite height) for x outside the potential well.
 
Cthugha said:
*scratch my head*

But of course it is. Ok, there is a closing bracket missing after the V(x), but I guess it is a typo.

This is the standard Schrödinger equation for a particle. I do not know, on which level your knowledge of physics is, but I suppose: Either you know, what a Schrödinger equation is or you need a crash course in quantum mechanics.

The V(x) is the potential you have, so in case of a potential well it will be 0 for x inside the potential well and will be infinite or have some fixed value (depending on whether your potential well has finite or infinite height) for x outside the potential well.

:rolleyes: and what does v(x) = 0 imply for the shrodinger eqn?

nowhere in the actual shrodinger differential eqn is the fact that the particle is in a square well expressed except that v(x) = 0. hence what i said stands.
 
ice109 said:
nowhere in the actual shrodinger differential eqn is the fact that the particle is in a square well expressed except that v(x) = 0. hence what i said stands.

V(x) is 0 for x>a and x<b if a and b denote the left and right border of the well. However V(x) has some constant value C or is infinite outside the potential well, which is the reason for the "short version" of just applying boundary conditions in the infinite case.

I agree, that this is not very different from using a Schrödinger equation without potential and just applying some boundary conditions in the case of the infinite potential well. In the potential well with finite height however, for the bound state you will find wavefunctions, which are not 0 outside the potential well, but fall off exponentially and the wave function must be continuuous and differentiable at the borders of the well. In this case the exact height of the potential outside the well is needed to determine the normalization constants of the piecewise defined wave function.
 
Thanks for the help :D. I see now that it is just the time-independent Schrödinger equation, and that the potential is zero inside the well.
 
ice109 said:
umm no that's not the eqn for a particle in a 1d potential well

-hbar^2/2m.d^2/dx^2 = E φ(x) with the apropo boundary conditions is the eqn that describes a particle in a 1d well.

That's the same equation that I had. The term -hbar^2/2m.d^2/dx^2 was written as the mometum operator squared, pop^2. I see where it comes from now, thanks :)
 
Cthugha said:
V(x) is 0 for x>a and x<b if a and b denote the left and right border of the well. However V(x) has some constant value C or is infinite outside the potential well, which is the reason for the "short version" of just applying boundary conditions in the infinite case.

I agree, that this is not very different from using a Schrödinger equation without potential and just applying some boundary conditions in the case of the infinite potential well. In the potential well with finite height however, for the bound state you will find wavefunctions, which are not 0 outside the potential well, but fall off exponentially and the wave function must be continuuous and differentiable at the borders of the well. In this case the exact height of the potential outside the well is needed to determine the normalization constants of the piecewise defined wave function.

now that is actually something i did not know till yesterday.
 

Similar threads

  • · Replies 15 ·
Replies
15
Views
3K
  • · Replies 5 ·
Replies
5
Views
1K
  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 4 ·
Replies
4
Views
4K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K