# The why of Pauli Exclusion

1. Mar 1, 2013

### anorlunda

I'm studying fermions with one of Leonard Susskind's video courses. The Pauli Exclusion principle seems odd and counterintuitive.

My question is this, if I learn enough quantum theory will I learn why Pauli Exclusion exists; not just how it works? Does the Pauli Exclusion principle pop out as a consequence of fundamental equations?

2. Mar 1, 2013

### Jorriss

Yes actually. It is a natural consequence of fermionic wavefunctions needing to be antisymmetric with respect to particle interchange. The connection between fermions and antisymmetric wavefunctions is a bit more advanced than I can speak about though.

3. Mar 3, 2013

### andrien

yes,you will learn it.the requirement is that if fermions will not obey pauli exclusion principle,then the used fermi dirac statistics will not hold for them.As a result there will not be any minimum energy state.For bosons,not using bose statistics will give noncommutativity of observables seperated by space like distance.

4. Mar 3, 2013

### tom.stoer

No, you will never learn why xyz is true ...

The Pauli principle "no two fermions can occupy identical quantum states" follows from the antisymmetry of fermions wave functions, which follows from the spin-statistics theorem, which follows from Lorentz invariance and some other assumptions (there are some exotic counteraxemples which become important when some assumptions do no longer apply!), which follow from ...

The are no ultimate justifications in physics.

5. Mar 3, 2013

### anorlunda

Uh oh, it sounds like "Nothing but turtles all the way down."

Thanks everyone.

6. Mar 3, 2013

### tom.stoer

in some sense this is the conclusion; if you agree with the assumptions of the spin-statistics theorem and believe in them as fundamental axioms, then you have your just justification; but if you start to question then it's the quest for the next turtle ...

7. Mar 3, 2013

### stevendaryl

Staff Emeritus
Well, less complicated than the spin-statistics theorem is to understand the fact that there should be symmetry under an exchange of particles. I don't remember the argument in full, but it went something like this...

If $\Psi(x_1, x_2)$ is the probability amplitude for particle 1 to be at position $x_1$ and particle 2 to be at position $x_2$, then, since particles of the same type are indistinguishable, we know that

$\Psi(x_1, x_2)$ and $\Psi(x_2, x_1)$

have the same physical content. That doesn't mean that they are equal, because wave functions have an arbitrary phase associated that doesn't make any observable difference. So switching two identical particles could possibly introduce a phase change. So we write:

$\Psi(x_1, x_2) = e^{i \alpha} \Psi(x_2, x_1)$

where $\alpha$ is a phase difference. Now, we can reason that if we switched the particles again, we have to get exactly back to where we started. So we reason that:

$\Psi(x_1, x_2) = e^{2 i \alpha} \Psi(x_1, x_2)$

which implies that

$e^{2 i \alpha} = 1$

So $\alpha = 0$ or $\pi$.

This is from memory, so there might be some subtle points that I've forgotten, but something along this lines is supposed to show that particles must either be Bosons (exchanging two identical ones makes no difference) or Fermions (exchanging two identical ones changes the sign of the wave function).

8. Mar 3, 2013

### tom.stoer

Yes, in principle this is right - provided that we believe in indistinguishable particles and some other subtleties ruling out anyons, ghosts and things like that

Last edited: Mar 3, 2013