How can we visualize the negative sign in electron wave functions?

[mccoy1]

Okay, bear with my ignorance.
1. Excited state and ground sate

Isn't ground state the highest stationary state occupied by valence electrons, while excited state the next highest stationary state(s) where the valence electron(s) can jump to? If yes, then why do all books talk about the ground states being the lowest stationary states and then everything else after that are excited sates? For example, how on Earth can N=2(N =Quantum principle number) for Uranium be an excited state?

2. We can best describe electrons as wavelike, more than "point(dot)-like" in most cases. No question asked. However, how can we visualize that wave -like- thing as having a negative sign?

 

[alxm]

1. The ground state of an atom/molecule has all its electrons in the lowest possible energy state (the Pauli principle limiting that to two electrons per orbital).

Out of the electrons, the valence electrons are the ones with the highest energy, and thus those that require the least energy to move to an excited state. When talking about molecules this is called HOMO (Highest Occupied Molecular Orbital), if you're talking solid-state it's called the Valence Band.

Above them you have the lowest unoccupied state, which is the state that it's easiest to excite to. That's the LUMO (Lowest Unoccupied MO) or Conduction Band. And the difference in energy between the two is thus the smallest possible excitation (Band Gap).

For example, how on Earth can N=2(N =Quantum principle number) for Uranium be an excited state?

Says who?

2. Well I'm not sure about whether the wave or particle viewpoint is 'better'. But anyhow, visualizing the negative sign is easy. Consider a standing wave in water or whatever. The crests above the waterline have positive sign and the troughs below it have a negative sign. Energetically the sign doesn't matter - a crest has the same energy as a trough. And where a 'positive' wave and a 'negative' wave meet, they cancel out. That's how it works with classical waves, and that's pretty much how it works with QM.

 

[mccoy1]

Thanks for the reply,
1. Na atom: 1s2 2s2px2py2pz2 3s1
obviously this atom is in its ground state configuration.So what are its stationary state that would be regarded as its ground state? my ans:3s-orbital and below.And the excited states? my ans: above 3s-orbital.
However, in introduction to QM by Griffiths p.32, there're a generic wavefunction in which he considers n=1 as ground state and n=2,3 etc as excited state. If we appliy that reasoning to Na here, then the only ground state is n=1 which corresponds to s-orbital inn the first shell and the rest are excited states. This is wrong..or am I missing something?
2. Electrons are negatively charged, so I was actually wondering how to 'see'/visualize a wave in your mind as negatively charged.Same applies to protons.
I know two waves can be superimposed and the result can either be constructive or destructive interference, but i don't see how it explains the fact that electrons are negatively charged.
Cheers.

 

[alxm]

1 - Is it a single-particle wave function? I don't have a copy of Griffiths but I don't recall it getting into many-electron wave functions (e.g. Slater determinants)

2 - It doesn't explain the fact that electrons are negatively charged, or purport to. Charge is neither here nor there, regarding the sign of the wavefunction. The w.f. sign isn't physically significant, and can't be observed except for when you have interference. From the basic QM point of view, the charge is just a thing that gives rise to a coulomb potential term in the S.E.