# Electron gas and Pauli principle

1. Feb 23, 2012

### hokhani

According to Pauli principle, it is impossible for two electrons in the electron gas to have the same state. on the other hand, we say that each spatial state can be filled by two electrons with opposite spins.

But my question;
Suppose we have two electrons, one spin down and another spin up. In the relation below
Ψ(x1,x2)=1/√2(ψ1(x1)ψ2(x2)−ψ1(x2)ψ2(x1))
Each of x1s and x2s shows both position and spin, and wave functions' indices(1 or 2) show the particle states.

If we have the same wave functions' indices(say both 1 or both 2), then Ψ(x1,x2)=0
no matter what x1 or x2 (and hence the particles' spins) is. Therefore how can one say that two electrons can seat in the same spatial states?

Last edited: Feb 23, 2012
2. Feb 23, 2012

### questionpost

I guess a better way of putting it is that electrons can inhabit whatever state they want as long as their wave-functions don't combine to make 0. With atoms you don't really deal with space as much, you look at all the other factors and determine the quantum state that way. Obviously two electrons already can appear in the same space if you just look at an s orbital. I think with the way it works, any amount of electrons can occupy the same space at the same time so long as their quantum states are different, which is why when you look at models of atoms you see an s orbital being intersected by many p orbitals.

3. Feb 23, 2012

### Bill_K

No, the indices 1 or 2 designate which particle. Therefore it's impossible for them to be the same.

4. Feb 24, 2012

### hokhani

Thank you Mr.Bill_K

You really helped me.

5. Feb 25, 2012

### Ilmrak

I disagree.

To avoid confusion, let us rewrite the state of the two particle as

\psi_{a,b}(x,y;\sigma,\tau)= (\psi_a(x, \sigma)\psi_b(y,\tau) - \psi_a(y,\tau)\psi_b(x,\sigma))/\sqrt{2};

where $x, \sigma$ is the coordinate and spin of the first particle and $y, \tau$ coordinate and spin of the second particle.

Then "If we have the same wave functions' indices", i.e. a=b, the state of the sistem is:

\psi_{a,a}(x,y;\sigma,\tau)=0 \;\;\;\;\;\;\forall x,y;\sigma,\tau.

If a≠b but $\{x, \sigma\}=\{y, \tau\}$ then

\psi_{a,b}(x,x;\sigma,\sigma)=0 ,

that is the probability of finding two fermions in the same spot with the same spin is null regardless of their wave function.

I hope this helps,

Ilm

6. Feb 25, 2012

### hokhani

NO, you are wrong Ilm
Then if a=b, how can you place two fermions (with apposite spins) in the same state?
Please refer to my first question.

7. Feb 25, 2012

### Ilmrak

If a=b then, as you can easly check, it follows that:

\psi_{a,a}(x,y;\sigma,\tau)=0 \;\;\;\;\;\forall x,y;σ,τ.

This means that you have, for a system described by such a sate, zero probability to measure any value for any observable, i.e. this state doesn't actually exist ^^

This is infact exactly why we say it's impossible to place two identical fermions in the same state.

I hope we agree on this.

edit: If we still won't agree, I think this would be my fault. Maybe I'm assuming you give some symbols the same meaning I give, which could not be true. If this is the case please tell me so I'll try to define better symbols used.

Ilm

Last edited: Feb 25, 2012
8. Feb 25, 2012

### hokhani

As you know, the state of a system including one electron can be specified completely by expanding the State Ket in terms of space and spin so you needn't use the indices a and b to specify electrons' states. In fact Psi(wave function $\Psi$) itself explains the state. So you should only regard the indices 1 and 2 in $\Psi$s to specify your particles.

9. Feb 25, 2012

### Ilmrak

I agree with the premise, not with conclusion, so I'll try to express clearer what I mean.

i) A state is in general a ray $|\psi>$ in some Hilbert space $H$.

ii) Suppose the Hilbert space of a single fermion has a base of rays consisting of eigenvectors of coordinate and spin operators $|x;\sigma>$.

iii) The Hilbert space of two particle is the direct product of the two single particle spaces $H_1\otimes H_2$.

iv) If $|x;\sigma>$ span $H_1$ and $|y;\tau>$ span $H_2$, then $|x,\sigma;y,\tau>$ span $H_1\otimes H_2$.

Now let $|a>\; \in H_1$ be the single particle state of the first fermion and $|b>\; \in H_2$ be the single particle state of the second fermion.

Then the properly anti-symmetrized normalized state of the two identical fermion is

$|a,b> \equiv (|a>\otimes |b>-|b>\otimes |a>)/\sqrt{2}\;\; \in H_1 \otimes H_2$.

Defining

$\phi_a(x,\sigma)\equiv <x,\sigma|a>$,
$\phi_b(y,\tau)\equiv <y,\tau|a>$,
$\psi_{a,b}(x,\sigma;y,\tau)\equiv <x,\sigma;y,\tau|a,b>$,

we finally get my starting point equation:

\psi_{a,b}(x,\sigma;y,\tau)=(\phi_a(x,\sigma)\phi_b(y, \tau)- \phi_b(x,\sigma)\phi_a(y,\tau))/\sqrt{2}

It's then clear that we need both indices $a$ and $b$ to label the state of the system and indices $x,y;\sigma,\tau$ to label the basis we are using to evaluate the states.

Affirming the two identical fermions are in the same single particle state it's equivalent to say $a=b$, but then anti-symmetrized two particle state would be identical to zero in all points for all spins, so it doesn't actually exists.

I hope this (I thought close to useless to specify) definitions could bring some agreement in this topic ^^

Ilm

Last edited: Feb 25, 2012
10. Feb 27, 2012

### hokhani

You are making a little mistake.
Writing ab-ba, you are attributing the first wave function(first position in each term) to the first particle so to speak 12-12 and you don't let the particles to exchange their wave functions.
Do you agree?

11. Feb 28, 2012

### Ilmrak

Sorry but, again, I don't agree xD

The wave functions of the two single particle states is labelled by the indexes $a, b$ respectively for the fermion 1 and 2.

The indexes $\{x,\sigma;y,\tau\}$ only labels the base we choosed for the Hilbert space of the two fermions, $H_1\otimes H_2$.

If we want to permute the two fermions we only need to excange $a\leftrightarrow b$.

edit: Maybe I should remark that writing $|a> \otimes |b> \in H_1 \otimes H_2$ means that we are taking the tensor product of the state $|a> \in H_1$ with the state $|b> \in H_2$.

Ilm

Last edited: Feb 28, 2012
12. Feb 28, 2012

### hokhani

"The wave functions of the two single particle states is labelled by the indexes a,b respectively for the fermion 1 and 2."

If we supposed you are right, so a=b means that you are regarding two particles as one.Ok?

13. Feb 28, 2012

### Ilmrak

Not actually: the first particle would be in the state $|a> \in H_1$ and the second particle in the state $|a> \in H_2$.
So the two particles would be in the same state but still in two different spaces.

If you then try to describe the system of both (identical fermionic) particles you have to use an anti-symmetrized vector $|\psi> \in H_1 \otimes H_2$.
You find out (see post #11) that the state $|\psi> \in H_1 \otimes H_2$, corresponding to two identical fermions in the same single particle state $|a>$, is :

$|\psi>=0$.

Ilm

14. Feb 28, 2012

### hokhani

Let me explain my means more clearly
I think you should regard
H1*H2+H2*H1
while you are regarding
H1*H2+H1*H2
Do you agree?

15. Feb 28, 2012

### Ilmrak

Sorry, I think I'm not understanding, could you please rephrase your sentence?

Ilm

16. Feb 29, 2012

### hokhani

Ok, Sure, I explain my former reason again:

In your notation of the two-particle-wave function, say AB, you attributed the left position(A) just to the first particle and never let the second particle's wave function place in the left. So by the expression AB-BA, you are implying that A(1)B(2)-B(1)A(2)

However now i don't see any contradiction with your statement. I have got confused!
But
Do you agree with the trite expression that two particles with opposite spins can place in the same state? If yes, How can you explain it by your notation of the Pauli principle?
I can't explain that expression by your notation but The notation of Mr. Bill_K justifies it!

17. Mar 1, 2012

### Ilmrak

I agree that two identical fermions can have the same "orbital" wavefunction if they have opposite spins.

In "my" notation, if the two fermons have defined spins, the state of the first fermion would be something like

$|a>= |\alpha,\sigma>$ such that $<x,\sigma '|\alpha,\sigma>= \phi_\alpha (x)\delta_{\sigma,\sigma '}$, where $\sigma \in \{ -1/2;1/2\}$ labels the spin of the first fermion.

For the second fermion

$|b>= |\beta,\tau>$ such that $<y,\tau '|\beta,\tau>= \phi_\beta (y)\delta_{\tau,\tau '}$, where again $\sigma \in \{ -1/2;1/2\}$ labels the spin of the second fermion.

Now the wavefunction of the two fermions system is:

$\Psi_{\alpha,\sigma;\beta,\tau}(x,\sigma ';y,\tau ')\equiv\,<x,\sigma ';y,\tau '|{\large(} |\alpha,\sigma>\otimes|\beta,\tau>-|\beta,\tau>\otimes |\alpha,\sigma>{\large)}=\,\phi_\alpha (x)\delta_{\sigma,\sigma '} \phi_\beta (y)\delta_{\tau,\tau '}-\phi_\beta (x)\delta_{\tau,\sigma '} \phi_\alpha (y)\delta_{\sigma,\tau '}$.

Then if the "orbital" state of the two fermions is the same, i.e. $\alpha=\beta$,

$\Psi_{\alpha,\sigma;\beta,\tau}(x,\sigma ';y,\tau ')=\,\phi_\alpha (x)\phi_\alpha (y)\,{\large(}\delta_{\sigma,\sigma '} \delta_{\tau,\tau '}-\delta_{\tau,\sigma '} \delta_{\sigma,\tau '}{\large)}$

wich is non zero if and only if $\sigma \neq \tau$, i.e. the two fermions have opposite spins.

Vice versa if they have the same spin then, analoguosly, they have to be in different "orbital" states $\alpha \neq \beta$.

Ilm

18. Mar 15, 2012

### hokhani

Thank you very much Mr.Ilm.

Last edited: Mar 15, 2012
19. Mar 15, 2012

### Ilmrak

I'd be glad if I was of some help

Ilm