Schrödinger's equation problem

Click For Summary
SUMMARY

The discussion centers on solving the time-independent Schrödinger's equation for a unidimensional potential defined as V(x) = V(-x). Participants confirm that the solutions exhibit defined parity, specifically odd or even functions, provided the eigenvalues are non-degenerate. The equation in question is expressed as -\frac{\hbar ^2}{2m} \cdot \frac{\partial ^2 \Psi}{\partial x^2}+V(x) \Psi =E\Psi. Key insights include the use of the principle of superposition and the decomposition of functions into odd and even parts to facilitate finding solutions.

PREREQUISITES
  • Understanding of the time-independent Schrödinger's equation
  • Familiarity with ordinary differential equations (ODEs)
  • Knowledge of eigenvalues and their properties
  • Basic concepts of function parity (odd and even functions)
NEXT STEPS
  • Study the properties of eigenvalues in quantum mechanics
  • Learn about the principle of superposition in linear differential equations
  • Explore methods for solving ordinary differential equations, particularly non-constant coefficients
  • Review the concepts of odd and even functions in mathematical analysis
USEFUL FOR

Students and researchers in quantum mechanics, particularly those focusing on the Schrödinger equation and its applications in potential theory. This discussion is also beneficial for anyone studying differential equations and their solutions in physics.

fluidistic
Gold Member
Messages
3,934
Reaction score
286

Homework Statement


Show that for a unidimensional potential of the form V(x)=v(-x), the solutions to the time independent Schrödinger's equation have a defined parity as long as these solutions does not correspond to eigenvalues not degenerated.

Homework Equations



[tex]-\frac{\hbar ^2}{2m} \cdot \frac{\partial ^2 \Psi}{\partial x^2}+V(x) \Psi =E\Psi[/tex].

The Attempt at a Solution



My idea: replace [tex]V(x)[/tex] by [tex]V(-x)[/tex] in the equation I just gave and then find [tex]\Psi[/tex]. Then show that [tex]\Psi[/tex] is either odd or even.
I have no idea about the non degenerated eigenvalues of [tex]\Psi[/tex]...
I'm stuck on starting to solve the equation. I'm a bit familiar with differential equations but not with partial ones.
Is it an equation of the form [tex]c_1y''+gy=c_2y[/tex] where [tex]y=\Psi[/tex] and [tex]g=V(-x)[/tex]? I realize I've no idea why I even bothered changing V(x) for V(-x).
I don't really know what to do. I'd like a little push.
 
Physics news on Phys.org
fluidistic said:
Is it an equation of the form [tex]c_1y''+gy=c_2y[/tex] where [tex]y=\Psi[/tex] and [tex]g=V(-x)[/tex]?
Yep, that's correct. A partial differential equation when there is only one independent variable is no different from an ordinary differential equation.

As for a hint: I think you can use the fact that any function [itex]f(x)[/itex] can be expressed as the sum of an odd part and an even part,
[tex]f_\text{even}(x) = \frac{1}{2}[f(x) + f(-x)][/tex]
and
[tex]f_\text{odd}(x) = \frac{1}{2}[f(x) - f(-x)][/tex]
Use this along with the principle of superposition (any linear combination of two solutions to the Schrödinger equation with the same energy is also a solution).
 
Thank you diazona.
Oh now I remember from calculus I the property you mention in the hint.

If I'm not wrong, the ODE I must solve is [tex]\Psi ''+\Psi \left ( \frac{g-c_2}{c_1} \right ) =0[/tex].
All my knowledge on ODE comes from a self study of Boyce-Di Prima's book. I don't really remember how to solve such an equation since g depends on x and isn't constant. If I still remember well, I should propose a solution of a particular form. Problem is, I don't know how to "guess" the particular form of the solution.

And yes, I do know that if I get 2 linearly independent solutions of the ODE, then any linear combination is also a solution. :smile:

I'll check out if I can encounter Boyce-Di Prima's book.

Edit: Ok I found the book but I'm still stuck. It seems I can get a second solution and therefore the general form of the solutions if I already know one solution. But since I don't know it, I'm totally stuck.
I realize the solution would be a simple harmonic motion kind of function if g would be constant. But since it's not, I have no idea.
 
Last edited:

Similar threads

  • · Replies 6 ·
Replies
6
Views
2K
Replies
29
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
Replies
7
Views
3K
  • · Replies 10 ·
Replies
10
Views
3K
Replies
4
Views
2K
  • · Replies 8 ·
Replies
8
Views
5K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K