Ground State Symmetry of Single Electron w/ Non-Interacting 2nd Electron

[positron]

for a symmetrical potential with one electron, i know that the wavefunctions are symmetric or antisymmetric. for the ground state why is the wavefunction symmetric?

Also, if you add a second electron that is non-interacting, (why) does it have the same wavefunction as the first electron?

 

[Physics Monkey]

I assume we are talking 1d here? So you've solved the 1d Schrödinger equation and have obtained the bound states of the system, right? You've put your first electron into the ground state, now what is the lowest energy single particle state available to the next electron? (Hint: the electron has spin 1/2)

 

[positron]

I assume we are talking 1d here? So you've solved the 1d Schrödinger equation and have obtained the bound states of the system, right? You've put your first electron into the ground state, now what is the lowest energy single particle state available to the next electron? (Hint: the electron has spin 1/2)

Yes, the problem is in 1-D. The potential is a general symmetric potential, so we don't know necessarily have an actual analytical solution to Schrödinger equation. So how do we know the ground state must be symmetric, and that if we put another electron in the potential, it has the same wave function?

 

[CarlB]

for a symmetrical potential with one electron, i know that the wavefunctions are symmetric or antisymmetric. for the ground state why is the wavefunction symmetric?

An intuitive explanation:

It's because the ground state has the smallest average energy. Energy in QM comes from two places, potential energy and kinetic energy. If you look at the form of the KE operator you will see that KE is smallest when your wave function slope is least (more flat). So the ground state only has one hump. That is, the more humps you got the more you have to go up and down and the higher the slopes.

Carl