# Pauli's exclusion principle Intrpretation

1. Dec 9, 2011

### TMSxPhyFor

Hi

I spent my last couple hours reading books and browsing this forum and web to get answer on my question, obviously in vain:

I know that wave function for fermions should be antisymmetric, I know fermions spin should be half integer otherwise causality will broken (but I don't why yet), I know spinors can be represented by SU(2) and we can get from there the half integer and exclusion principle...

But, as I understood from some books and ArXiv papers that basically, and despite the enormous implications of this principle starting from the "space occupied" by atoms to neutron stars formation, there is no clear intuitive explanation why two fermions with the same spin can't be at the same energetic level, and why there will be "repulsive" relations between them despite of that there is no actual "force" between them (contrary to what postulates some time in crystal studies), I agree that even so arguments such as particles has "consciousness" are interesting, they still meta-physical, why fermions are so different of Bosons? are there such a particles which with wave functions not symmetric nor antisymmetric?

So is there any explanation in modern main-stream physics ? if not it will be really strange for me why such a very important principle didn't get enough attention?

Last edited: Dec 9, 2011
2. Dec 9, 2011

### DrewD

The exclusion principle is derived from quantum particles being indistinguishable (I think it may have been postulated first, but later derived). If you have a wave function that describes two particle (say electrons orbiting a nucleus—could be any bound state), switching the positions of the of the particles should not change the expectation values. Since the expectation value is obtained from the square of the wavefunction, both positive and negative solution work. That is an antisymmetric solution. Switching the particle gives the same function, but negative. The symmetric solution doesn't change sign. It sound like you know this already, but I want to be sure.
The symmetric solution is in the form
$\Psi(a,b)=(\phi_u(a)\phi_d(b)+\phi_u(b)\phi_d(a))$
Now the antisymmetric solution is in the form
$\Psi(a,b)=(\phi_u(a)\phi_d(b)-\phi_u(b)\phi_d(a))$
This could be describing two electrons, one spin up, one spin down, but that doesn't matter, just that the states are different. If they were the same (the $\phi_u=\phi_d$) the wave function would be 0. Therefore antisymmetric wavefunctions for two particles must have the particles in different states.
Particles that are described by the top equation have been named bosons while those that follow the bottom are called fermions. Simply put, if two particles with an antisymmetric wave function dropped into the same state, they would cease to exist and violate many conservation laws!

3. Dec 10, 2011

### TMSxPhyFor

Yes you were right, why the fermions will repulse each other so that even an enormous gravity of neutron star can't overcome it? haw this repulsion happens if there is no real force? electrical charge basically is a intrinsic property as much as the spin, and both has "plus-negative" charge-orientation of spin), but one of them result in a real force, and the spin result in "virtual" force, how I should understand such a "dual standards" in QM ?

Hope my question now became clearer.

Last edited: Dec 10, 2011
4. Dec 10, 2011

### DrewD

The conservation laws are the most important. If the wave goes to zero, energy would not be conserved and, depending on the particle, charge may not be conserved. Its not a dual standard. Quantum mechanic at the basic level deals with energy before forces.

5. Dec 10, 2011

### TMSxPhyFor

yea I'm totaly agree with you that energy/potentials are more fundamental, so why there is interaction between spins while there is no field or energy exchange? and what you mean by :

charge not be conserved? according to my knowledge it is always conserved, am I wrong?

6. Dec 10, 2011

### Bill_K

TMSxPhyFor, In the first place, there is no guarantee that every important principle has a clear, intuitive reason "why" it is so. the universe does not come with an obligation to be easy to understand!

You say the principle hasn't got enough attention, but the connection between spin and statistics was examined carefully back in the 1930s and 1940s, from many angles. For example one reason "why" is that it prevents states from occurring with negative energy. Also it is necessary to insure that the theory is invariant under time reversal. You'll find a discussion of spin-statistics in any book on quantum field theory.

You seem to feel it's a mysterious thing that two fermions can't occupy the same state. On the contrary, I find it amazing that two bosons can occupy the same state!

The exclusion principle does not represent a "force", it simply says that certain multiparticle states do not exist. As an analogy, think - why can't the electron in a hydrogen atom sit closer to the nucleus than the ground state? It is not because the nucleus is "repelling" it. It's because there simply is no state that's closer.

7. Dec 10, 2011

### tom.stoer

One should have a look at fermionic creation and annihilation operators; b/c of

$$(b^\dagger)^2 = 0$$

one cannot create a state with two identical particles; starting with the vacuum

$$|0\rangle$$

one can create one particle

$$b^\dagger |0\rangle = |1\rangle$$

but creating two particles does not work b/c

$$(b^\dagger)^2 |0\rangle = 0$$

$$(b^\dagger)^2 |0\rangle = b^\dagger|1\rangle = 0$$

The second equation says that if one tries to "add" a second identical particle to a one-particle-state |1> does not work.

8. Dec 10, 2011

### ytuab

Even if Pauli exclusion principle exerts the "repulsive force" between particle's spin, it would not be included in the four fundamental forces, I think.
(This is due to the "mathematical" form of QFT.)

Because the four fundamental forces need to use bosons such as photon, gluon, W-boson, and graviton.
In these cases, next Feynman diagrams (interaction energy term) are very important.

$$L_{int} = eA_{\nu} \bar{\psi} \gamma^{\mu} \psi, \quad g W_{\mu}^{+} \bar{u} \gamma^{\mu} d \cdots$$

We can NOT express Pauli exclusion principle in this form.
Because Pauli exclusion principle is already included in Dirac's wavefunction such as $\psi$.
( As tom.stoer says, creation and annihilation operators of $\psi$ express the "force" of Pauli exclusion princile. )

So what we call "boson" of Pauli exclusion principle does not exist.
( "Mathematical" reason of QFT is related, I think. The upper Feynman diagram is everything in QFT. )

Last edited: Dec 10, 2011
9. Dec 11, 2011

### tom.stoer

Unfortunately there is the notion of "exchange interaction" (or "exchange force") http://en.wikipedia.org/wiki/Exchange_interaction which in a qm context looks like an interaction term in the Hamiltonian. But looking at the QFT concepts (ytuab's post) there is no force, no interaction, no associated field; it's nothing else but a consequence of the algebra of the fermionic operators. Therefore the term "interaction" or "force" is misleading.

10. Dec 11, 2011

### TMSxPhyFor

now we getting close to my question, but unfortunately I didn't studied Feynman diagrams yet, can you please explain that in some other way?

another analogy, dose such a "repulsion" happens between any two particles that has some opposite intrinsic quantum number? (spin, isospin ....)

11. Dec 11, 2011

### tom.stoer

I think you don't need any Feynman diagrams.

An interaction between two fermions is no direct (point-like) interaction, but is mediated via a field A or W (or some others). That means an electron interacts with another electron e.g. via the photon field A.

The Pauli principle holds as soon as you introcude a free spinor field, even w/o any interaction, i.e. w/o any field A or W.

12. Dec 11, 2011

### TMSxPhyFor

But that is the problem it self! there is no interaction at first place, becuase...
from http://en.wikipedia.org/wiki/Exchange_interaction

Lets do the flowing mental experiment: if I will try to force an electron of 1/2 spin to be in the first orbit (in atom) that already has electron with the same spin, what will happens?There will be resistance to do that , and this resistance appears as:

1- one of the electrons will flip it's spin (will change intrinsic property)
2- one of the electrons will change it's energy level or orbit (will change extrinsic property)

what will really happen?

13. Dec 11, 2011

### tom.stoer

Yes, something like that, but that is not (solely) due to the Pauli principle but due to the additional el.-mag. force which "mixes" with the fermion effects.

14. Dec 11, 2011

### ytuab

tom.stoer, What TmsxphyFor says is due to Pauli exclusion principle, I think.
Due to the first electron's spin, the second electron changes its orbit (to the higher energy level).
It is very difficult to explain from the viewpoint of the energy.
(For example, the energy level of the 2s orbit is much higher than 1s orbit. )

And the magnetic force of the electron spin is not strong enough to fix their states.

15. Dec 11, 2011

### tom.stoer

No, the el.-mag force is required, too.

Write down the Hamiltonian of two free (non-interacting) fermions and try to describe something like that; it will not work.

16. Dec 11, 2011

### ytuab

Sorry. I don't understand what you mean well.
Pauli exclusion principle is related to wavefunction itself (= determinant ) not Hamiltonian.

From the viewpoint of the energy, for example, the fine structure (2P3/2 and 2P1/2) is actually observed.
If only the energy of the magnetic force determines the spin state, 2P3/2 state would NOT be observed.
( As a result, the fine structure is not observed.)

The fine structure is caused by the spin and orbital magnetic interaction of one electron ( about "Bohr magneton").
The two electrons are apart from each other, so the spin-spin interaction is weaker than the spin-orbital interaction from the viewpoint of the energy.
First, the fine structure (= 0.00005 eV ) itself is very small, so the spin-spin interaction is weaker than that.

So the Pauli exclusion principle can not be explained by the magnetic force (energy).

17. Dec 11, 2011

### TMSxPhyFor

Yes, maybe the example not very accurate, but think about it in abstract way, no electromagnetism, or we have neutrons instead of electrons.

18. Dec 12, 2011

### Zarqon

Bolding mine.

Could you perhaps expand a bit on this answer. I mean, you're a basically saying that the Pauli principle is caused by a theoretical construct (the algebra), but I'm thinking that it must be possible to go deeper and state what part of nature cause a formulation of the algebra to be such that it does not allow two fermions in the same state.

We are basically deriving our knowledge of physics from a set of basic rules, such as energy/momentum/spin conservation, and descriptions of the 4 forces. So, the question is: is the Pauli exclusion principle directly derived from a more fundamental postulate, or is it a new postulate in itself?

19. Dec 12, 2011

### tom.stoer

First what I am saying is that Pauli's exclusion principle is NOT a force or an interaction. Suppose you have an Hamiltonian for free, non-interacting fermions. Then already at that level the exclusion principle does apply. There is no "force" between two fermions generated by this Hamiltonian; it's not that one electron "pushes away" the other one. Pauli's exclusion principle guarantuees that two identical fermions will NEVER EXIST.

Now when you try to change the quantum state of one fermion slightly such that it becomes equal to the state of a second electron, then you are talking about a dynamical system where the dynamics (e.g. the el.-mag. force between two electrons) is affected by the spins. The force itself becomes spin dependent (which is interpretetd as an exchange force or something like that), but even w/o any such force the Pauli principle does apply.

Regarding a deeper reason for the Pauli exclusion principle you may want to have a look at the Spin-statistics theorem
http://en.wikipedia.org/wiki/Pauli_exclusion_principle
http://en.wikipedia.org/wiki/Spin-statistics_theorem

20. Dec 12, 2011

### TMSxPhyFor

Yes right! on big distances, that also what historically happen, they tried at first to explain Hund first rule by "magnetic" nature of spin-to-spin interaction , but later they found that this is not enough, and Pauli exclusion principle is the lord in this, and this why it's considered as the one who makes atoms "occupy volume", and this way I'm saying if there is any explanation to it!

yes I'm aware of that, and i want to understand the physical reasons of such operators algebra!