# Question about quantum numbers:

1. Jun 25, 2012

### VortexLattice

So, you got n,l,m_l,m_s. If you have two hydrogen atoms on opposite ends of the universe, it seems to me that the electron in each could have the set of quantum numbers (1,0,0,1/2).

Now, if you have a helium atom, because its two electrons are identical fermions, by the PEP they now "know" about each other, and can't have the same set of quantum numbers... So how close do two electrons have to get before this happens?

2. Jun 26, 2012

### jfy4

I have been wondering this same thing! Yet I still haven't gotten a straight answer from anyone.

3. Jun 26, 2012

### VortexLattice

My friend thinks it's that the "same quantum state" includes the spatial part of the function as well, which kind of makes sense. But I'm still not convinced.

4. Jun 26, 2012

### Bill_K

The quantum numbers n, l, ml, ms are origin-dependent, and so the states labeled by these numbers for atom 1 are not the same states as the ones labelled by the same numbers for atom2.

The antisymmetry is completely unimportant as long as the atoms are far enough apart that the wavefunctions barely overlap. Bring them closely together (an atomic diameter, say) and the fact that the total wavefunction must be antisymmetric starts to matter.

5. Jun 26, 2012

### jfy4

This has always bothered me though. How does Pauli "know" when the critical overlap is!? There is always finite overlap, however minuscule. How does more overlap work it's way into symmetrization?

6. Jun 26, 2012

### Bill_K

In place of the 3-D atoms, consider a simpler case: the double well, a pair of identical 1-D wells a distance R apart. This is a standard example covered in QM courses. When the wells are far apart the energy levels are essentially the same as for a single well, and the states are essentially independent of each other. Each well appears to be occupied as if the other did not exist. There's a twofold degeneracy.

But in reality there's a single wavefunction for both wells, and the true eigenstates are symmetric and antisymmetric. And each level that appeared to be twofold degenerate is slightly split: the symmetric function has one fewer node, and is therefore slightly lower in energy than the antisymmetric one. As the wells are moved closer together this difference becomes more and more noticeable. Very crudely, the increase in energy for the antisymmetric state can be thought of as a 'repulsion' caused by the Pauli principle.

Now if the particle occupying the wells is a fermion, the antisymmetric states are the only ones, and the wavefunction is always antisymmetric. For large R the effect of this (the energy level shift) is small and the wells appear to be independently occupied. But as R is decreased the Pauli effect becomes more and more noticeable.

7. Jun 26, 2012

### jfy4

So while it might seem pedantic, the wavefunction for a two fermion system, even when the fermions are far apart, must still be symmetrerized? If I had the technological power to build a machine, I could measure a difference in the quantum numbers of two fermions a meter apart because of spin-statistics and quantum number splitting?

8. Jun 26, 2012

### Bill_K

Yes in principle, there is only one electron field, and every electron in the universe is an excitation of that field and is antisymmetric with every other one.

9. Jun 26, 2012

### jfy4

Thank you

10. Jul 31, 2012

### VortexLattice

http://mlkshk.com/r/DBLT [Broken]

Last edited by a moderator: May 6, 2017