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

AI Thread Summary
Two wave functions with the same energy can only differ by a complex phase, as shown through the relationship derived from Schrödinger's equation, indicating that one wave function can be expressed as a complex multiple of the other. If the potential is real, it is possible to choose the wave function to be real as well, which simplifies the analysis. The discussion also highlights that the ground state wave function does not change sign, meaning it remains either entirely positive or negative. This property can be demonstrated by defining the ground state wave function and examining its behavior under the Hamiltonian's constraints. Understanding these principles is essential for solving quantum mechanics problems related to wave functions and energy states.
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) Schrodinger'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) Schrodinger'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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