Pauli Exclusion Principle, what does it say?

1. Apr 13, 2005

Inquisiter

I know I already posted this question, but it seems to have gotten "lost" among the other questions in the same thread. This is really confusing me now, so I'll ask it again.

The pauli exclusion principle says that no two fermions can be in the same quantum state. But if we have three eigenstates (as we do for an atom for n=2, l=1), can't we produce an infinite number of states (each one with the same energy) by superimposing the three eigenstates(each time with different coefficients)? So when we talk about quantum states as they relate to the Pauli principle, must these states be orthogonal or what?? What is the mathematics of it? I sort of know how we write out the states as a matrix and take the determinant to asymmetrize the wavefunction. But can the indivifual states be non orthogonal? Why or why not? I did a Google search, but didn't find much useful info. As far as I know, we can place only 6 electrons in n=2, l=1 state, not an infinite number of electrons. And a follow up queston: for one electron in an electric field of a point charge (the nucleus) we get a bunch of bound states as solutions to the Schr. eq. But when the Schr. equation is solved for an electron moving in a field produced by the nucleus AND other electrons (say, using the effective potential method), do we get basically the same solutions (just shifted in energy) as if there were no other electrons, but just the nucleus? I mean, we don't get any extra energy levels because of the other electrons which contribute to the total potential, right?

Last edited: Apr 13, 2005
2. Apr 13, 2005

Meir Achuz

If there are only 3 estates, no linear combinations can give more than 3 INDEPENDENT states. The key word is independent. So you could put no more than 6 electrons in.
It is like having three unknowns in algebra. You can write many equations for the 3 unknowns, but only 3 will be independet.
For your followup: The number of stats does not depend on what is producing the potential.

3. Apr 13, 2005

Inquisiter

That's what I was thinking. But it's usually not stated explicitly that the states must be independent. Don't the states have to be orthogonal, not just linearly independent though?

4. Apr 13, 2005

Meir Achuz

"Orthogonal" makes them easier to use, but "independent" is enough to expand in.
For instance, a Taylor expansion is an expansion in non-orthogonal functions.