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

## 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.

Related Advanced Physics Homework Help News on Phys.org
blue_leaf77
Homework Helper

## 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 $c = exp (i \theta)$.

blue_leaf77
Homework Helper
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.

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.

blue_leaf77