Wave Functions With Same Energies Are the Same (only differ by a complex phase)

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Homework Help Overview

The discussion revolves around the properties of wave functions for a particle described by the Schrödinger equation under a potential. The original poster seeks to demonstrate that two wave functions with the same energy differ only by a complex phase, explore conditions under which a wave function can be real, and investigate the behavior of the ground state wave function.

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants discuss the relationship between wave functions with the same energy and the implications of the Schrödinger equation. Some suggest examining the normalization of wave functions to establish a phase relationship. Others raise questions about the uniqueness of solutions in one-dimensional systems and the implications for degeneracy.

Discussion Status

Some participants have made progress on parts (a) and (b) of the problem, indicating an understanding of the principles involved. However, there remains uncertainty regarding part (c), with participants seeking clarification on how to express the ground state wave function in terms of Hamiltonian eigenvectors.

Contextual Notes

There is a mention of the constraints of the problem, particularly regarding the behavior of wave functions in one-dimensional systems and the implications of real potentials on the nature of the wave functions.

Yoni V
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Homework Statement


Assume a particle with a wave function ##\psi(x)## such that ##-\infty < x < \infty##, that move under some potential ##V(x)##.

Show that:
a) two wave functions with same energies can only differ by a complex phase;
b) if the potential is real, then you can choose the wave function to be real as well;
c) the wave function of the ground state (with real potential) doesn't change sign.

Homework Equations


a) Schrödinger's time independent equation.

The Attempt at a Solution


I'm stuck at (a). Need a push in the right direction for the very start.
I want to show that if two wave functions ## \psi_1, \psi_2## satisfy
$$ \psi_{1/2}''(x) + \frac{2m}{\hbar^2}\left(E-V(x)\right)\psi_{1/2}(x)=0$$
then I can find an equation that ties them in a phase relation.
But aside from writing this statement down, I don't know how to proceed. Thanks.
 
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Yoni V said:
a) two wave functions with same energies can only differ by a complex phase;
The answer to (a) can actually be found in numerous websites, e.g. in http://www.physicspages.com/2012/08/23/degenerate-solutions-dont-exist-in-one-dimension/. Try to understand the derivation, then ask if you are stuck on something.
 
Yoni V said:

Homework Statement


Assume a particle with a wave function ##\psi(x)## such that ##-\infty < x < \infty##, that move under some potential ##V(x)##.

Show that:
a) two wave functions with same energies can only differ by a complex phase;
b) if the potential is real, then you can choose the wave function to be real as well;
c) the wave function of the ground state (with real potential) doesn't change sign.

Homework Equations


a) Schrödinger's time independent equation.

The Attempt at a Solution


I'm stuck at (a). Need a push in the right direction for the very start.
I want to show that if two wave functions ## \psi_1, \psi_2## satisfy
$$ \psi_{1/2}''(x) + \frac{2m}{\hbar^2}\left(E-V(x)\right)\psi_{1/2}(x)=0$$
then I can find an equation that ties them in a phase relation.
But aside from writing this statement down, I don't know how to proceed. Thanks.

If the Hamiltonian be ##\hat H## and the two normalised energy eigen vectors be ##|\psi_1> ## and ##|\psi_2>##, then from Schroedinger's time independendent equation ##\hat H |\psi_1> = E|\psi_1>## and ##\hat H |\psi_2> = E|\psi_2>##. From these two equation you can conclude that ##|\psi_2> = c|\psi_1>##. Where ##c## is a complex number. Now as the wave vectors are normalised, you can write ##<\psi_2|\psi_2> = <\psi_1|c^* c|\psi_1> \Rightarrow <\psi_2|\psi_2> = c^* c <\psi_1|\psi_1> \Rightarrow c^*c =1##. Naturally ##c = exp (i \theta)##.
 
Korak Biswas said:
If the Hamiltonian be ##\hat H## and the two normalised energy eigen vectors be ##|\psi_1> ## and ##|\psi_2>##, then from Schroedinger's time independendent equation ##\hat H |\psi_1> = E|\psi_1>## and ##\hat H |\psi_2> = E|\psi_2>##. From these two equation you can conclude that ##|\psi_2> = c|\psi_1>##. Where ##c## is a complex number. Now as the wave vectors are normalised, you can write ##<\psi_2|\psi_2> = <\psi_1|c^* c|\psi_1> \Rightarrow <\psi_2|\psi_2> = c^* c <\psi_1|\psi_1> \Rightarrow c^*c =1##. Naturally ##c = exp (i \theta)##.
It does not explain why in 1D system like the one considered in the problem, there can be no degeneracy. Your method can be applied, for example, to a hydrogen atom. We know that for quantum numbers ##n>1##, the eigenstates are degenerate and those eigenstates are not merely related by a complex constant.
 
Thanks for your replies!
I managed both (a) and (b) and understood its underlying principles. I'm now left with (c).
We were suggested to define ##\phi = |\psi_0|## where ##\psi_0## is the wave function of the ground state, and then express ##\phi## in terms of the Hamiltonian eigenvectors and find its energy. Then, use (b) to finish. But how can I express ##\phi## in terms of the eigenvectors of the Hamiltonian when I don't know what it is?

Ok got it... sorry, I should have given it more thought before throwing the question in the air.
 
(c) asks you to prove that the ground state never crosses the x axis.
 

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