Valid Solutions for Harmonic Oscillator Potentials

Click For Summary

Homework Help Overview

The discussion revolves around the validity of a wavefunction solution for the time-independent Schrödinger equation in the context of harmonic oscillator potentials. The specific potential under consideration is V(x) = (1/2)*mω²x², and the wavefunction is given as ψ(x,0) = N exp[-α(x-a)²], where α = mω/(2ħ). Participants are tasked with demonstrating that this solution is valid only when a = 0.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss strategies for proving the validity of the solution, including the potential's expansion in a power series and the implications of differentiating the wavefunction. Questions are raised about what specific results to look for when substituting the wavefunction into the Schrödinger equation.

Discussion Status

There is ongoing exploration of different approaches to demonstrate the condition a = 0. Some participants suggest directly substituting the wavefunction into the Schrödinger equation and manipulating the resulting expressions, while others express uncertainty about the specific steps needed to reach a conclusion.

Contextual Notes

Participants mention the potential for a different form of the potential, V(x) = (1/2)*mω²(x-a)², and question whether the solution would hold for any value of 'a' in that case. There is also a shared context of completing similar homework assignments, indicating a collaborative atmosphere.

Mohandas
Messages
7
Reaction score
0
[tex]\psi(x,0) = N exp[-\alpha(x-a)^2][/tex] is a solution to the time-independent SE at time t = 0 for the potential

[tex]\ V(x) = (1/2)*m\omega^2x^2[/tex]

where N is a constant and [tex]\alpha = m\omega/(2\hbar)[/tex].

I'm asked to show that the solution is valid only if a = 0.

I'm a little at loss as to what strategy i should use to show this.

My initial thought was to somehow exploit the fact that the potential can be expanded in a power series about the origin, and that the solution (for a yet unknown reason) should also be expanded about the origin (and not about 'a'), but it kinda stopped right there..
Of course it might be as easy as just to differentiate (for a=0 and a=constant) the solution and plug it into SE and evaluate the result (in which case, what am i looking for?) ?

Btw, would the solution be valid for any 'a' if the potential was [tex]\ V(x) = (1/2)*m\omega^2(x-a)^2[/tex] ?
 
Last edited:
Physics news on Phys.org
Mohandas said:
[tex]\psi(x,0) = N exp[-\alpha(x-a)^2][/tex] is a solution to the time-independent SE at time t = 0 for the potential

[tex]\ V(x) = (1/2)*m\omega^2x^2[/tex]

where N is a constant and [tex]\alpha = m\omega/(2\hbar)[/tex].

I'm asked to show that the solution is valid only if a = 0.

I'm a little at loss as to what strategy i should use to show this.

My initial thought was to somehow exploit the fact that the potential can be expanded in a power series about the origin, and that the solution (for a yet unknown reason) should also be expanded about the origin (and not about 'a'), but it kinda stopped right there..
Of course it might be as easy as just to differentiate (for a=0 and a=constant) the solution and plug it into SE and evaluate the result (in which case, what am i looking for?) ?

Btw, would the solution be valid for any 'a' if the potential was [tex]\ V(x) = (1/2)*m\omega^2(x-a)^2[/tex] ?

The obvious thing to do is to simply plug this wavefunction in the time-ind Schroedinger equation and show that the equation is satisfied only if a=0
 
Mohandas said:
Of course it might be as easy as just to differentiate (for a=0 and a=constant) the solution and plug it into SE and evaluate the result

Yes, that's what "show that function f is a solution to the differential equation D" means.

(in which case, what am i looking for?) ?

After you plug the function and its derivatives into the differential equation, if the function is in fact a solution, you can cancel everything out (usually after some algebraic manipulation) and end up with 0 = 0.
 
Mohandas said:
[tex]\psi(x,0) = N exp[-\alpha(x-a)^2][/tex] is a solution to the time-independent SE at time t = 0 for the potential

[tex]\ V(x) = (1/2)*m\omega^2x^2[/tex]

where N is a constant and [tex]\alpha = m\omega/(2\hbar)[/tex].

I'm asked to show that the solution is valid only if a = 0.

I'm a little at loss as to what strategy i should use to show this.

My initial thought was to somehow exploit the fact that the potential can be expanded in a power series about the origin, and that the solution (for a yet unknown reason) should also be expanded about the origin (and not about 'a'), but it kinda stopped right there..
Of course it might be as easy as just to differentiate (for a=0 and a=constant) the solution and plug it into SE and evaluate the result (in which case, what am i looking for?) ?

Btw, would the solution be valid for any 'a' if the potential was [tex]\ V(x) = (1/2)*m\omega^2(x-a)^2[/tex] ?

Thats right, put it in the time independent schrödinger equation and do the gay maths.. Do you happen to do your home exam in the university in oslo? Me too, i got exactly the same task ^^
 

Similar threads

  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 2 ·
Replies
2
Views
2K
Replies
1
Views
2K
  • · Replies 16 ·
Replies
16
Views
3K
Replies
11
Views
2K
Replies
4
Views
2K
Replies
1
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K