Schrödinger's equation problem

Join the discussion
Ask a follow-up here, or get your own question answered by working scientists, mathematicians and engineers — people, not an autocomplete.
Real named experts · corrections over time · the nuance an AI answer skips
2 replies · 2K views
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: