# Pauli principle confusion

1. May 13, 2007

### jostpuur

Usual way of expressing pauli exclusion principle is, that the two identical fermions must be on different states. This seems a confusing statement, since the meaning of the term "different states" isn't very clear. Suppose I have two electrons, otherwise in the same states, but other one being spin up, and other one spin down. So the spin states are $$(1,0)$$ and $$(0,1)$$. When I rotate the system about the x-axis with 90 degrees, the spin states become $$(1,-i)/\sqrt{2}$$ and $$(-i,1)/\sqrt{2}$$. Now they are not in the "different states" in a naive sense, but this certainly cannot be a violation of the pauli exclusion principle. I can see that these vectors are still orthogonal. This was expected, since their orthogonality doesn't depend on the chosen representation. So wouldn't it be more correct to say, that the states of two identical fermions must be orthogonal? When I read and think at the same time, I see the expression "different states" meaning "orthogonal states". Is this correct?

The more fundamental fact is that the two particle wavefunction must satisfy the identity $$\psi(x_1,x_2)=-\psi(x_2,x_1)$$. When two fermions are occupying states, whose spatial representations are $$\psi_1(x)$$ and $$\psi_2(x)$$, according to all books, the two particle wavefunction must be $$\psi(x_1,x_2)=(1/\sqrt{2})(\psi_1(x_1)\psi_2(x_2)-\psi_2(x_1)\psi_1(x_2))$$, so that the antisymmetry property is satisfied. The texts then say, that now we see that $$\psi_1$$ and $$\psi_2$$ must be different, because if they were the same the wavefunction would vanish. But how do you see that they must be orthogonal? Okey, if these states were eigenstates of some usual variable, everything would be fine, but if $$\psi_1$$ and $$\psi_2$$ are not eigenstates of some usual variable (like energy), then what? If antisymmetry of $$\psi(x_1,x_2)$$ implies orthogonality of $$\psi_1$$ and $$\psi_2$$, it can probably be shown with some calculation, but I haven't seen such anywhere.

It doesn't fully make sense to think particles occupying different (or orthogonal) states, since in the antisymmetrical wavefunction none of the particles is in some specific state. Still, in statistical stuff it is usually assumed that fermions simply are on the different states. Is this some kind of statistical approximation? You know, even if it wasn't fully correct, it would give correct statistical results and so on?

2. May 13, 2007

### rewebster

aren't you viewing from this from a different reference frame?

3. May 17, 2007

### jostpuur

Should I now assume, that these questions do not even belong to the main stream physics?

4. May 24, 2007

### malawi_glenn

if psi_1 and psi_2 is the same :

psi(x_1,x_2) = (1/sqrt2)[psi_1(x_1)psi_1(x_2) - psi_1(x_2)psi_1(x_1)]

and

psi(x_1,x_2) = (1/sqrt2)[psi_2(x_1)psi_2(x_2) - psi_2(x_2)psi_2(x_1)]

?

Last edited: May 24, 2007
5. May 24, 2007

### jambaugh

Yes that is correct. "Completely different" is expressed by the zero transition probability which in turn is expressed as zero (square of the ) inner product.

In truth it is not the vectors per se but the one dimensional complex subspaces they span which correspond to quantum modes (a.k.a. "states").
Completely distinct modes are orthogonal. Less than orthogonal modes overlap and hence systems prepared initially in one mode have a non-zero probability of being detected in the other mode.
You needn't worry about them being eigen-modes of a specific observable they are necessarily eigen-modes of some observable. You can anti-symmetrize two non-orthogonal modes but the result will eliminate the overlap.

Example suppose you have modes: $$\psi_1$$ and $$\psi_2$$.
Now suppose that $$\langle \psi_1|\psi_2\rangle = c$$
We resolve $$\psi_2$$ as a superposition of parallel and perpendicular components:
$$\psi_2 = c\psi_1 + \psi_\perp$$
i.e.$$\psi_\perp = \psi_2 - c\psi_1$$
(I'm assuming $$\psi_1,\psi_2$$ have been normalized but you will note that $$\psi_\perp$$ is not!)

OK so far?
Now consider the anti-symmetric product:
$$\Psi =\psi_1\wedge\psi_2\equiv \psi_1\otimes\psi_2 - \psi_2 \otimes\psi_1$$
$$= \psi_1\otimes(c\psi_1+\psi_\perp) - (c\psi_1+\psi_\perp)\otimes\psi_1$$
$$= c\psi_1\otimes\psi_1 -c\psi_1\otimes\psi_1 + \psi_1\otimes\psi_\perp-\psi_\perp\otimes\psi_1$$
$$= \psi_1\wedge\psi_\perp$$

(Again note that since the two normalized vectors were not orthogonal then their anti-symetrized product will not be normalized!)

(Note you can similarly expand the first in terms of perp and parallel components to the second.)

Thus in essence if you combine two fermions the anti-symmetry dictates that the composite structure will be the same as for two orthogonal fermions. However you do need to be slightly careful about saying each is orthogonal to the other. Once you use the anti-symetrized composite it isn't quite proper to speak of individual particles as if they were distinguishable except by using values of a physical observable as a label.
Your are right as I noted above. But think of it as short hand for saying "they are equivalent to a composite of two in orthogonal modes".

(You'll notice I'm careful to say mode instead of state or to quote state. The Hilbert space vector refers to an equivalent mode (method) of preparation and not an objective state. It is important to keep that distinction in mind when parsing through some examples such as EPR experiments.)

Regards,
James Baugh

6. May 24, 2007

### turbo

Let us consider the states of the particles. Can the particles somehow sense that there is a similar particle with the same spin in close proximity and somehow manage to repel the other particle? That idea is not tenable. Let's instead consider that the PEP arises from the inability of the vacuum to support the existence of two similar same-spin particles concurrently. Christine Dantas has recently brought up the concept of concurrency WRT quantum physics and I think its a great area of inquiry. Here is part of a recent email.

7. May 24, 2007

### jambaugh

Some other musings.

You can think of the phenomenon of stimulated emission as a corresponding "inclusion principle" dual to the PEP.

Also while doing some deeply mathematical research with my thesis advisor, we pondered whether one could recast all physical forces as statistical in nature in the sense that the PEP effects a virtual repulsive force. Note that in neutron stars this repulsive force becomes very physical in effect.

Also note that when we look at forces from a quantum perspective they are all expressible as "interference effects" in the "sum over histories" formulation. So who is to say PEP "repulsion" isn't a real force?

Regards,
James Baugh

8. May 24, 2007

### jostpuur

jambaugh, I'm glad to see those equations conserning antisymmetric wavefunctions. In fact I had made similar remarks on my own, but I was afraid that I was coming up with "my own stuff" (once again), since the books and notes I've seen do not discuss this matter. Now when I've started to think about this more, I think this is starting to make sense, and I've hit a following, a rather mathematical, idea:

A two particle wave function $$\psi(x_1,x_2)$$ is not a unique way of representing a two fermion state, but instead we can define an equivalence relation $$\psi\sim\psi'$$ meaning that

$$\psi(x_1,x_2)-\psi(x_2,x_1) = \psi'(x_1,x_2)-\psi'(x_2,x_1)$$

and then it is the equivalence class $$[\psi]$$, represented by some wave function $$\psi$$, that is really the relevant physical object, and what defines the state.

How correct is this?

The equivalence relation was motivated by a remark, that when an arbitrary wave function is written in a form

$$\psi(x_1,x_2)=\frac{1}{2}(\psi(x_1,x_2)-\psi(x_2,x_1)) + \frac{1}{2}(\psi(x_1,x_2)+\psi(x_2,x_1)),$$

the symmetric part vanishes in the expression

$$\int dx_1\; dx_2\; \psi(x_1,x_2) a^\dagger_{x_1} a^\dagger_{x_2}|0\rangle$$

if creation operators satisfy the usual anticommutation relation.