Free particle in one dimension

Click For Summary

Discussion Overview

The discussion revolves around a problem involving a free particle in one dimension, specifically analyzing the wave function and its implications for kinetic energy and momentum measurements. The scope includes theoretical reasoning and mathematical approaches related to quantum mechanics.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Technical explanation

Main Points Raised

  • The original poster (OP) presents a wave function and seeks help in determining the most probable kinetic energy and the values of momentum that cannot be measured.
  • Some participants question the validity of the wave function, noting that it is not continuous and therefore may not be a valid wave function.
  • Another participant suggests using a Fourier basis to express the wave function and indicates that the kinetic energy calculation may not yield zero when considering the edges of the wave function.
  • There is a suggestion to apply the definition of momentum in the context of the Fourier transform to find specific coefficients that may be zero.
  • One participant expresses suspicion that the problem may originate from a textbook, suggesting it could be a homework question.

Areas of Agreement / Disagreement

Participants express differing views on the validity of the wave function presented by the OP, with some agreeing that it is not continuous while others focus on the mathematical approach to the problem. The discussion remains unresolved regarding the correct interpretation and solution to the questions posed by the OP.

Contextual Notes

There are limitations regarding the continuity of the wave function and its implications for quantum mechanical validity. The discussion also highlights potential dependencies on definitions and mathematical interpretations that remain unresolved.

angass
Messages
1
Reaction score
0
Hi,
I´m not sure if my way of tackling a question, probably it's a trivial problem, but it's important for me to get it right so any help will be greeted.
The question is as follows:

Problem: consider a particle in a one-dimensional system. The wave function ψ(x) is as follows:
ψ(x)= 0 for (-∞<x<0),
ψ(x)= 1/√a for (0<x<a)
ψ(x)= 0 for (a<x<∞)

i) if the kinetic energy is measured what is the most probable value?
ii) which values of the momentum can never be found when measuring it?

My reasoning:
i) I use the operator for kinetic energy: K= -\frac{h^2}{2m}\frac{∂^2}{∂x^2}
which when applied: ∫ψ(x)*Kψ(x)dx gives me zero.
If this was right I assume the most probable value of the kinetic energy is zero.
ii) I have no clue whatsoever what the question means.

Thanks and forgive my bad english, regards from Spain!
 
Physics news on Phys.org
Hi angass, welcome to PF!

angass said:
The wave function ψ(x) is as follows:
ψ(x)= 0 for (-∞<x<0),
ψ(x)= 1/√a for (0<x<a)
ψ(x)= 0 for (a<x<∞)
Is that correct? Because this is not a continuous function, and therefore not a valid wave function.
 
This looks suspiciously like a griffiths or leibowitz homework problem...

You want to take express psi(x) in a Fourier basis (as a sum of e^i(kx-wt)). Doing the Fourier transform will be a simple integral from 0 to a with 1/a^1/2 as a constant.

i) You're forgetting the edges at 0 and a. The derivative is not zero there. Once you express this function as Fourier expansion, you'll see you don't get zero anymore.

ii) apply the definition of p = i hbar d/dx to the Fourier transform. The answer should drop out. I expect some of the coefficients for specific k terms in the Fourier transform will be zero. (the Fourier basis is an eigen value of the momentum and kinetic energy operator)
 
rigetFrog said:
This looks suspiciously like a griffiths or leibowitz homework problem...

Same here mate.

To the OP if it is we have a homework section.

But if it isn't - Google is your friend eg:
http://www.colorado.edu/physics/TZD/PageProofs1/TAYL07-203-247.I.pdf

Thanks
Bill
 
Last edited by a moderator:

Similar threads

  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 11 ·
Replies
11
Views
3K
  • · Replies 12 ·
Replies
12
Views
3K
  • · Replies 57 ·
2
Replies
57
Views
6K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 25 ·
Replies
25
Views
3K
  • · Replies 33 ·
2
Replies
33
Views
4K
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 8 ·
Replies
8
Views
2K