# Pauli Exclusion Priciple

1. Mar 10, 2006

### scott_alexsk

I was wondering why electrons exsist in pairs in orbitals? Is it that since they both have opposite spins according to the Pauli Exclusion Priciple that they cancel each other out and create a more stable system?
Thanks,
-Scott

2. Mar 11, 2006

### inha

Well there fits more than 2 electrons on the shells beyond the K shell. It's not really about stability but just the Pauli principle and minimizing energy.

3. Mar 11, 2006

### scott_alexsk

What do you mean by 'minimizing energy?'
Thanks,
-Scott

4. Mar 12, 2006

### inha

The spin-orbit interaction for example. It's energetically favourable to fill the shells as described by Hunds rules. With some notable exceptions though.

5. Mar 12, 2006

### MiGUi

The Pauli Exclusion Principle says that two fermions can't be in the same individual state. If we only consider the external degrees of freedom (orbiting and so) then, for one level of energy you have only one state. This stands, at least, for hydrogenoid atoms.

If you consider spin degeneracy (spin degeneracy = 2s + 1, s = 1/2 so spin degeneracy = 2) then for every energy level there are two different individual states: the one which has the 3rd component of the spin "up" ($$\hbar/2$$) and the other which its 3rd component "down" ($$-\hbar/2$$).

So there is no violation.

That is not a question of stability. The wave function of a fermionic system has to be antysimmetric and this is the only way.

MiGUi

6. Mar 12, 2006

### scott_alexsk

Maybe I should rephrase my question. Why at all do electrons exsist in pairs in orbitals. Why would a single electron just exsist by itself, always, in its own orbital? Inha would you please explain the situation of the K shell. I am not familar with it.
-scott

7. Mar 12, 2006

### MiGUi

It can exists, but if you want to 'fill' the energy levels you need at least two.

8. Mar 12, 2006

### lueffy

Why fermion has anti-symmetric wave function? i've read that this is related to the matter of indistinguishability of particles, but if we can assign certain quantum numbers to an electron, we can say that they're distinguishable, can't we?
(Pardon me for jumping in the middle of this thread, but all i can say is that i'm interested in MiGUi's statement that fermion has an anti-symmetrical wave function, and i can't find out why...)

9. Mar 12, 2006

### Gokul43201

Staff Emeritus
They do not. They exist in pair only in orbitals of filled sub-shells.

Consider, for instance the electronic configuration of Mn - look at the picture here : http://www.webelements.com/webelements/elements/text/Mn/econ.html

Having no unpaired electrons is the exception, not the rule. And this follows from Hund's Rules, which essentially says that (among other things) it costs less energy to fill all the orbitals with one electron each than to pair them up (at a Coulomb energy cost).

Once you've filled all orbitals within a sub-shell with one electron, and you still have electrons to spare, the cheapest option then (barring a few rare cases) is to put these electrons into the above half-filled orbitals (rather than start filling a new sub-shell). When all the orbitals are filled with 2 electrons each, you have to move on to the next sub-shell. An orbital can not hold more than 2 electrons (Pauli Exclusion).

Note : There are some cases when Hund's Rules are violated.

10. Mar 13, 2006

### scott_alexsk

Yes apparently with Inha's K Shell the Pauli Exclusion Priciple is violated. Can you please describe this situation. Is it only a theoritical situation?
-scott

11. Mar 13, 2006

### inha

No the K-shell is just a name for the n=1 orbital. That one is always filled up first.

Hunds rules are violated in transition and rare earth metals. In transition metals for example the 3d shell does not fill up completely even though there are 4s electrons present. It's not only theoretical and actually is the reason for many of the interesting properities of transition and rare earth metals.

12. Mar 13, 2006

### MiGUi

Cause fermions are 'defined' with the antisymmetry condition for its wavefunction and the bosons with the symmetric one.

After, the spin-statistics theorem proofs that particles with an integer spin have to obey Bose-Einstein's statistic and particles with semi-integer spin have to obey Fermi-Dirac's one.

A wavefunction is symmetric if changing the position of two particles, the wavefunction remains the same. And it is antisymmetric if doing that, the wavefunction is the same but with the sign changed. If you link this with the indistinguishability of particles, you have the Pauli's Exclusion Principle.

Mathematically, indistinguishability of particles means this:

$$\Psi(\xi_1, \ldots, \xi_i, \ldots, \xi_j, \ldots, \xi_n) = \Psi(\xi_1, \ldots, \xi_j, \ldots, \xi_i, \ldots, \xi_n)$$

And symmetric and antisymmetric condition is:

$$\Psi_{boson} (\xi_1, \ldots, \xi_i, \ldots, \xi_j, \ldots, \xi_n) = \Psi_{boson} (\xi_1, \ldots, \xi_j, \ldots, \xi_i, \ldots, \xi_n)$$

$$\Psi_{fermion} (\xi_1, \ldots, \xi_i, \ldots, \xi_j, \ldots, \xi_n) = - \Psi_{fermion} (\xi_1, \ldots, \xi_j, \ldots, \xi_i, \ldots, \xi_n)$$

If you link the first condition with the third condition, you see that the wavefunction is equal to 0. This means that particle i and particle j can't be in the same individual state.

MiGUi