# Pauli exclusion principle - does it apply to everything

1. Feb 20, 2005

### so-crates

The Pauli exclusion principle states that no two fermions can have the same quantum state. My question is, if you could conceivable solve the wave equation for every electron in a large body, perhaps even the entire universe, does this mean that no electrons in that body can have exactly the same energy ?

2. Feb 20, 2005

### dextercioby

What doesn energy have to do with quantum state...?Think atoms and so-called "orbitals"...The electrons which have 3 identical quantum #-s (n,l,m_{l}) must have the m_{s} different,as to satisfy Pauli's exclusion principle,but they have the same energy...

Daniel.

3. Feb 20, 2005

### Davorak

My understanding was that was only true to a very good approximation dextercioby. The fact that the atoms are separated by a good distance or/and have very large local potentials make the energy levels to be almost exactly the same. So much so that is not possible for use to measure the difference.

Take for the example two potential wells if they are separated by a good distance or high potential the energy levels can be considered separately. However if they are near each other or have a lower potential separating them the potential of both wells must be consider while coming up with quantum levels.

When the quantum wells are close to each other you can still have particles to a good degree localized to one well. However a particle localized in the other well can not have the same energy level.

So to my understanding the Pauli exclusion does imply that all electrons have different energies, however most of the time as in the case of atoms this energy difference is so small that it does not matter for practical concerns.

4. Feb 20, 2005

### dextercioby

What does that have to do with spin and the exclusion principle...?

Daniel.

5. Feb 20, 2005

### Davorak

Well if we consider just spin then the levels will always be split because of magnetic fields. Out side out generated by the movement or spin or another electron. For example for the Positronium or hydrogen atom there is always spin-orbit interaction or spin-spin interaction spliting the energy levels.

Edit:
My pervious post was explaining why different atoms would not have the exact same energy levels, which would then allow for all electrons to have different energy levels. I do not know however how relativistic quantum mechanics has to say about states that are in different light cones and I will not guess here.

Last edited: Feb 20, 2005
6. Feb 20, 2005

### so-crates

They would have different energies if placed in a magnetic field.

7. Feb 20, 2005

### Davorak

Even if not placed in a magnetic field there are two other sources of magnetic field, magnetic field created by other electrons spins and the orbit of the electron around a nucleus also provides a magnetic field. Both of these will split the Energy level in the hydrogen atom. Or in any other atom.

8. Feb 20, 2005

### Edgardo

The Pauli-exclusion principle forbids two electrons being in the same state.
But that doesn't mean they can't have the same energy. Two different states can have the same energy (degeneracy).

9. Feb 20, 2005

### ZapperZ

Staff Emeritus
Then you yourself have contradicted your own position... "IF PLACED IN A MAGNETIC FIELD". If not placed in a magnetic field, they will not have different energies.

Note that spin-orbit coupling is literally non-existent for s-orbitals. So in this case, even the orbital angular momentum cannot cause such splitting. Thus, if you have 2 electrons in the s-orbital, they have the same energy, and n,l, and s are good quantum numbers.

I suggest reading the more generalized description of the Exclusion principle as applied to the symmetry of the wavefunction for indistinguishable particles.

Zz.

10. Feb 20, 2005

### Davorak

Wont you still have spin-spin though Zapper.
The spins will be anti-aligned and repel each other just like to dipole magnetic would.

11. Feb 20, 2005

### dextercioby

Compute the Hamiltonian for this interaction spin-spin...

Daniel.

12. Feb 20, 2005

### Davorak

Well with out looking anything up it should be:
$$H_{e_{1}-e_{2}} \propto \mu_{e_{1}} \cdot \mu_{e_{2}}$$

Edit1:
At least this is true for the hyperfine or spin-spin for the hydrogen or positronium. You can get a nice singlet and triplet state. Is this what you are trying to get at? But you can not add a third particle with out then relaying on perturbation theory or numerical analysis.

Edit2:
So my understanding is that you can calculate degenerate energy levels, but once you start putting electrons into these degenerate levels they lose their degeneracy.

Last edited: Feb 20, 2005
13. Feb 20, 2005

### kirovman

My understanding of PEP suggests the range in which PEP applies is where there is overlap of the two wavefunctions, and only to quantised systems (ie the atom) as opposed to continuous systems (ie free electrons).
Obviously all the electrons in the universe could not be on the same atom, so hence not in the same state.

Even if the electrons on two identical, but separated atoms had exactly the same quantum values they would satisfy PEP since they are in a different state (ie they occupy different atoms).

So electrons on neighbouring atoms wouldn't interact via PEP unless their orbitals had some overlap.

It is also worth noting that the Pauli Repulsion term in Solid State Physics goes as $$R^{-12}$$so it is extremely short range compared to say electrostatic interactions.
And only Fermions are subject to PEP (not Bose Particles)

Hope this helps! I also hope my info was right.

14. Feb 20, 2005

### Davorak

This is a practical explanation, but the impression I got was that so-crates was asking for a technical theoretical question rather then a practical what do I need to consider when making a calculation.
Is it truly possible to have a continuous system? Would this not require an infinite universe and not a finite one.
If the universe as a whole is consider a potential well or a repeating system then even the quantum states in atoms are determined from the shape an potential of everything in the universe. At least if we ignore propagation time.

See my first post I think this only true is a very good degree Orbitals will always overlap to some degree even a very small one.

But there is no range where you can say the Pauli repulsion is not exactly zero. It can only be said that at this range the Pauli repulsion no longer matters because the effects are too small to care about. Right?

15. Feb 20, 2005

### ZapperZ

Staff Emeritus
But the "spin" IS thing that is resulting in the Fermi-Dirac statistics! Furthermore, the magnetic moment from each electron spin is neglibile, specially for an s-orbital. If this is so predominant, it would be silly to consider the concept of degeneracy, since obviously we would NEVER have such a thing.

Zz.

16. Feb 20, 2005

### ZapperZ

Staff Emeritus
But by saying this, you are being utterly unrealistic. At what point do you stop considering the gravitational effects from Alpha Centauri? I could repeat what you've just said, that there is no range where you can say the gravitational effect is not exactly zero. You will now see the absurdity of making such a statement.

When one can no longer detect its effects, or if such effects are so small that they no longer play a significant role in making an accurate description or prediction, it is zero!

Zz.

17. Feb 20, 2005

### Davorak

Of course it is negligible for all practical purposes I have not been disputing that. I have not be focuses on the practical measurable effects because I did not think that was what so-crates was asking for.
I do not know weather or not we would never have such a thing as degeneracy if the dipole-dipole interaction is taken into account, however I can not think of case were their would be degeneracy either.

I am not claiming that these considerations are important or testable with current levels of technology. However, the Pauli exclusion principle combined with dipole-dipole interaction does seem to suggest that no degeneracies exist for identical fermions.

I said the effects were to small to care about what more do you want. Just because they are too small to care about does not mean they do not exist though.

18. Feb 20, 2005

### Davorak

I apologize for my bad grammar. Below is what I meant to state.

“But there is no range where you can say the Pauli repulsion is exactly zero. It can only be said that at this range the Pauli repulsion no longer matters because the effects are too small to care about. Right?”

If I have made an incorrect assumption, or if I am misunderstanding something please do not keep quite. I do not claim to be an expert on Quantum Mechanics so I will not mind being corrected. I came to these boards for the reason that I want to become more fluent in physics. I will need to be if I go to physics grad school. I also do not wish to sound crankish in my statements and if my statements are crankish then best to tell me now so I do not continue to believe something false.

19. Feb 22, 2005

### kanato

So to answer the original question, if you have a very large body, such as an infinitely repeating crystal with discrete translational symmetry, you can use density functional theory to calculate the electronic density, and go from there to calculate dispersion relationships, the allowed energy values for various k vectors, $$\varepsilon_n(\vec{k})$$

You will find the energies are given by a band index n, and you can do spin-polarized calculations to treat the spins very carefully, and you will find that there are situations in which the bands do cross one another, which means that there are two quantum states with exactly the same energy, if the body is indeed infinite.