# Antisymmetry of fermions

1. Jul 9, 2013

### Sivasakthi

I have a doubt regarding the antisymmetry in the wave function of fermions.The antisymmetry is in the complete wave function or it is in the spin?

2. Jul 9, 2013

### CompuChip

The anti-symmetry is in the complete wave function, under interchange of particles.
I.e. if the single-particle wave functions are located at x1 and x2 (*), then the wave function for the whole system should satisfy
$\psi(x_1, x_2) = - \psi(x_2, x_1)$

(*) Note: more generally, you should have $\psi(\vec a_1, \vec a_2) = -\psi(\vec a_2, \vec a_1)$ where $\vec a_i$ are the quantum numbers characterizing the state of particle i. The spin is just one of these numbers, e.g. $a_i$ could be $(\vec r_i, \vec p_i, \vec J_i, \operatorname{spin}_i, \ldots)$.

3. Jul 13, 2013

### Sivasakthi

Hi,

Can we say that the antisymmetry in the total wave function is because of the antisymmetry in spin?The exchange of particles just deals with their spacial symmetry...so finally ends with Pauli's principle...am i correct?

4. Jul 13, 2013

### Bill_K

No, the exchange is a complete particle exchange - space, spin and all internal variables such as isospin, color...

5. Jul 13, 2013

Staff Emeritus
No, because sometimes, as CompuChip says, spin is just one possibility out of many.

6. Aug 12, 2013

### Sivasakthi

So does it mean that there exists many other possibilities than spin through which we may distinguish the identical particles?I thought the complete wave function includes only the spacial and spin parts.And as the particles get interchanged, their spacial wave functions remain symmetric and thus to make the total wave function anti symmetric their spins should be anti symmetric..that is what i had in my mind till now...

7. Aug 12, 2013

### The_Duck

Right. For example the $\Omega^-$ particle contains three strange quarks, all in the same spatial and spin state. But they are in an antisymmetric color state, so the total wave function is antisymmetric, as required.

8. Aug 12, 2013

### tom.stoer

Don't know whether this helps, but using creation and annihilation operators this becomes rather obvious. Suppose we use an operator $b_a^\dagger$ to create a particle in state a = {momentum, spin, isospin, ...}. Then creating two particles in the same state simply means

$|a,a\rangle = \left(b_a^\dagger\right)^2|0\rangle$

But for fermionic operators we have

$\left(b_a^\dagger\right)^2 = 0$

and therefore the state $|a,a\rangle$ is the null-vector.

9. Aug 13, 2013

### Sivasakthi

Again its confusing.That operator corresponds to which state?The state that is described here contains so many quantum states such as momentum spin etc..The state is a null vector represents that there cannot be any such possibility.So does it be valid for all the quantum states?

10. Aug 13, 2013

### tom.stoer

The state contains all quantum number including momentum, spin, ...; there is noithing special for quantum number spin, except for the fact that spin 1/2 demands Fermi-Dirac statistics which results in the properties of the creation operator.

And yes, you are right, the fact that the state with two identical fermions is a null vector means that thee cannot be such a state.

11. Aug 13, 2013

### CompuChip

You should also be careful in distinguishing the following:
* |0> is the "vacuum" state in which you can create particles
* The scalar 0 means that the operation is unphysical.

12. Sep 3, 2013

### Sivasakthi

Thanks a lot for your replies.I am having another doubt regarding the antisymmetry of two electrons.Electrons being fermions are antisymmetric.But does it have any sense if we are asked to create a symmetric state for them?

13. Sep 3, 2013

### tom.stoer

This exactly what the above mentioned construction does:
But

Last edited: Sep 3, 2013
14. Sep 3, 2013

### CompuChip

A more intuitive way of looking at it: electrons are fermions, so their states have to be antisymmetric. The only vector which is both antisymmetric and symmetric is the null vector (which is not, as I claimed before, a scalar).

15. Sep 3, 2013

### CompuChip

Since it was indirectly pointed out to me just now that my earlier post was incorrect I have updated it in the above quote (it is too long ago to be able to edit the original post). Thanks!

16. Sep 3, 2013

### Jolb

I think all the answers given above are sort of getting too technical. Here's an easy way to think of it.

Let's say we just have a two-particle position wavefunction ψ(x1,x2). [Just a position wavefunction, no spinor part nor any other hidden quantum numbers.] If we are dealing with distinguishable particles, we can write ψ(x1,x2)=ψ1(x12(x2). For distinguishable particles this makes sense because since we know particle 1 from particle 2, we can say definitely that particle 1 is in its state ψ1 and particle 2 is in its state ψ2.

For indistinguishable particles, however, we cannot use that form of the wave function because it is in principle impossible to say which particle is particle 1 and which particle is particle 2. So the probability distributions must be the same for the two cases:
1) particle 1 is in state ψ1 and particle 2 is in the state ψ2
2) particle 1 is in state ψ2 and particle 2 is in the state ψ1

Therefore we require the probability distribution |ψ|2 satisfies:
|ψ(x1,x2)|2=|ψ(x2,x1)|2

If you stare at this for long enough you can convince yourself that there are only two ways to satisfy the equation:
ψ(x1,x2)=ψ1(x12(x2)±ψ1(x22(x1).

We define bosons as those particles which obey ψ1(x12(x2)+ψ1(x22(x1) and we define fermions as those particles which obey ψ1(x12(x2)-ψ1(x22(x1).

Notice that none of this had anything to do with spin. So fermions don't necessarily have any spin until you go into relativistic quantum mechanics/anyonic systems. The fact that there is a connection between spin and bosons/fermions comes out of the spin statistics theorem, which requires much more advanced ideas.

Note: this is all stolen from Griffiths' "Introduction to Quantum Mechanics", chapter 5.

Last edited: Sep 3, 2013