# Fermions and parity

1. Mar 20, 2009

### Niles

Hi all.

This isn't a homework question, but something I thought about. When looking at a system of 2 fermions, we have that:

$$\Psi(r_1,r_2)=-\Psi(r_2,r_1).$$

Now if we look at a 3 fermion system, then what is the demand for the waveequation? Does it have to be anti-symmetric when switching two of the particles or all three? And if it is all three, then in what order? I.e.:

$$\Psi(r_1,r_2,r_3)=-\Psi(r_2,r_1,r_3) \qquad \text{or}\qquad \Psi(r_1,r_2,r_3)=-\Psi(r_3,r_1,r_3).$$

I hope you can shed some light on this. Thanks.

2. Mar 20, 2009

### CompuChip

If you have three fermions, then it should be anti-symmetric under switching any two.
So
$$\Psi(r_1, r_2, r_3) = -\Psi(r_2, r_1, r_3) = -\Psi(r_1, r_3, r_2) = \Psi(r_3, r_1, r_2) = -\Psi(r_3, r_2, r_1) = \cdots$$

Any permutation of n fermions can be written in multiple ways as a series of transpositions (a transposition meaning: swapping only two of them), luckily group theory tells us that the parity of such a permutation (i.e. whether you need an odd or even number of transpositions) is always the same.

3. Mar 20, 2009

### Niles

Thanks for that. That helped!

4. Mar 20, 2009

### Niles

I am actually not quite sure I understand this part 100%. Should I understand from this that whenever we have n fermions, then interchanging two of them at a time will cause a minus-sign to appear?

5. Mar 20, 2009

### CompuChip

Yes, interchanging any two fermions will introduce one minus sign.

However, you may wonder about the following: if I start with the order (1, 2, 3) and want to get to (3, 1, 2) I may do this in several ways:
(1, 2, 3) --> (1, 3, 2) --> (3, 1, 2) [first swap the last two, then the first two]
(1, 2, 3) --> (2, 1, 3) --> (3, 1, 2) [first swap the first two, then the outermost]
(1, 2, 3) --> (3, 2, 1) --> (3, 1, 2) [first swap the outermost, then the last two]
In this case, you easily see that in both cases I need to do an interchange of two fermions twice, so I will get two minus signs and
$$\Psi(r_1, r_2, r_3) = + \Psi(r_3, r_1, r_2).$$

However, you may wonder if this always holds. For example, if I have 30 fermions and want to jumble them into some random order, will any way of doing it give me the same sign? Or can I do it one way with an odd number of interchanges and another way with an even number? Because if that is the case, then the whole thing we've been saying is useless (should the wave function get a minus or a plus, or should it be zero?)

The part you quoted and say you don't completely understand, was a brief justification that there is really no problem, because if you write any re-ordering of n fermions as a series of permutations of two of them at the time (so-called transpositions, in mathematics), then you will always get the same number of transpositions. In the example above, you always need an even number, whether you do (1, 2, 3) -> (1, 3, 2) -> (3, 1, 2) or (1, 2, 3) -> (1, 3, 2) -> (2, 3, 1) -> (2, 1, 3) -> (2, 3, 1) -> (2, 1, 3) -> (3, 1, 2).
If you have had this in a group theory course, this would make sense to you, if not, you'll just have to believe me that requiring the wave function to get a minus for each interchange of two fermions is consistent.

6. Mar 20, 2009

### Niles

Ahh, I see. It is a very nice explanation; thank you for taking the time to help me.

7. Mar 20, 2009

### clem

Your title is mislabeled, and your equation is not general. Fermions have internal properties, always including spin which is interchanged along with r, so the sign of interchanging only r can be plus or minus.

8. Mar 20, 2009

### alxm

$$r_x$$ should be interpreted as including spin as a coordinate, I suspect, since I'm guessing Nile's question was inspired by a https://www.physicsforums.com/showpost.php?p=2123369&postcount=2" I made the other day which defined r that way.

Last edited by a moderator: Apr 24, 2017