Are fermions truly antisymmetric in their wave function?

[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?

 

[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 $(\vec r_i, \vec p_i, \vec J_i, \operatorname{spin}_i, \ldots)$.

 

[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 spatial symmetry...so finally ends with Pauli's principle...am i correct?

 

[Bill_K]

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 spatial symmetry...so finally ends with Pauli's principle...am i correct?

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

 

[Vanadium 50]

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

 

[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 spatial and spin parts.And as the particles get interchanged, their spatial 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...

 

[The_Duck]

So does it mean that there exists many other possibilities than spin through which we may distinguish the identical particles?

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.

 

[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.

 

[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?

 

[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.

 

[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.

 

[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?

 

[tom.stoer]

does it have any sense if we are asked to create a symmetric state for them?

This exactly what the above mentioned construction does:

... creating two particles in the same state [= creating a symmetric state] simply means

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

for fermionic operators we have

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

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

But

... the scalar 0 means that the operation is unphysical [i.e. that no such state does exist]

 

[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).

 

[CompuChip]

You should also be careful in distinguishing the following:
* |0> is the "vacuum" state in which you can create particles. It is a non-trivial state |n> with n=0 the label of the state.
* The null vector 0 (the analog of the origin in $\mathbb R^n$) holds no information whatsoever and means that the operation is unphysical.

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!

 

[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(x₁)ψ₂(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(x₁)ψ₂(x₂)±ψ₁(x₂)ψ₂(x₁).

We define bosons as those particles which obey ψ₁(x₁)ψ₂(x₂)+ψ₁(x₂)ψ₂(x₁) and we define fermions as those particles which obey ψ₁(x₁)ψ₂(x₂)-ψ₁(x₂)ψ₂(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.