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

Yoni V

## 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.

Korak Biswas

## 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)##.

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.

Yoni V