Where do fermion properties come from?

  • Thread starter jostpuur
  • Start date
2,099
16

Main Question or Discussion Point

I'm still struggling to understand, that why precisely have I not understood what the Dirac's field is.

hmhmh... okey, my question is this: Where does the anticommutation relation [itex]\{a_j,a_k^{\dagger}\}=\delta_{jk}[/itex] come from?

Is there something better than: "because the commutation relation would give absurd results, hence the relation must be an anticommutation relation"?
 

Answers and Replies

strangerep
Science Advisor
3,006
808
Where does the anticommutation relation [itex]\{a_j,a_k^{\dagger}\}=\delta_{jk}[/itex] come from?

Is there something better than: "because the commutation relation would give absurd results, hence the relation must be an anticommutation relation"?
If you're looking for some deeper theoretical reason, then,... no,
that's about it.

Although,... I vaguely recall a different argument involving spin-1/2
Poincare unirreps and isotropy of space. It had something to do with
the former being multiplied by -1 under a [itex]2\pi[/itex] rotation,
combined with exchange of two particles at different points.
But I suppose this is still just "...would give absurd results...".

In the end, it all arises from Poincare/SR and positive-energy.
 
410
0
I'm still struggling to understand, that why precisely have I not understood what the Dirac's field is.

hmhmh... okey, my question is this: Where does the anticommutation relation [itex]\{a_j,a_k^{\dagger}\}=\delta_{jk}[/itex] come from?

Is there something better than: "because the commutation relation would give absurd results, hence the relation must be an anticommutation relation"?
The reason you mentioned, known as the spin-statistics theorem, is actually pretty good. Basically, mathematics gives you two options, and in order to describe nature, you have to pick one that works.

You can think of the Dirac field first as having a classical limit field [itex]\psi_\alpha[/itex] where [itex]\alpha[/itex] is a spinor index. These fields are Grassmann numbers, and obey [itex]\{\psi_\alpha, \psi_\beta\} = 0[/itex], classically. When you quantize these fields using your favourite method (e.g. path integral), the anticommutation relations you mention pop out naturally.

If you want to know where fermions "come from", well... that is a whole other ball of wax. In supersymmetry (which can be spontaneously or explicitly broken) they arise when spacetime has extra Grassmann dimensions (superspace). A (different) process, called Bosonization, they arise when you take a bosonic field theory and "change variables" and end up with a fermionic field theory. See for example, the Sine Gordon / Thirring duality.

Finally, if you think that an anticommutation relation is unnatural, remember that you can always turn an anticommutator into a commutator by multiplying by grassmann parameters. So if your Lie algebra has Grassmann generators, then taking your parameters of your group to be Grassmann as well, you end up with good old fashioned commutators.
 
Last edited:
932
0
I don't know if you know about representations of the Lorentz group. If you do, you will have noted that spin-1/2, 3/2, 5/2 etc. representations get a minus sign if we perform a pure rotation of [itex]2\pi[/itex] on them.

Furthermore, you will know from exchange symmetry that a particle can have either a + sign or a - sign upon exchange.

The spin statistics theorem essentially makes a connection between the two - if the particle obtains a minus sign under [itex]2\pi[/itex] rotations, then it obtains a minus sign upon exchange with an identical particle - we call these fermions. We call the other types bosons.
 
samalkhaiat
Science Advisor
Insights Author
1,636
831
... okey, my question is this: Where does the anticommutation relation [itex]\{a_j,a_k^{\dagger}\}=\delta_{jk}[/itex] come from?
1) Math: From the definition of the vacuum state;

[tex]a_{i} |0_{1},..,1_{j},0,..> =\delta_{ij} |0_{1},..,0_{j},..>[/tex]
[tex]a_{i}^{\dagger}|0_{1},..,0_{i},..> = |0_{1},..,1_{i},0_{i+1},..>[/tex]

you see that both brackets (commutator & anticommutator) satisfy

[tex]\left[ a_{i} , a_{j}^{\dagger} \right]_{\pm} |0> = \delta_{ij}|0>[/tex]

2) Physics: From the defining property of fermions (pauli principle);

[tex]|1_{i},1_{j}> = - |1_{j},1_{i} >[/tex]
[tex]|2_{i}> \equiv |1_{i},1_{i}> = 0[/tex]

it is easy to see that

[tex]\left[ a_{i}^{\dagger}, a_{j}^{\dagger} \right]_{+}|0,0> = 0[/tex]
[tex]\left( a_{i}^{\dagger}\right)^{2}|0> = 0[/tex]

So, it is physics (i.e., Pauli principle) that picks up anticommutators for fermions. And please don't ask me "why fermions follow the pauli principle?" because it is as silly as asking "why massive objects attract eachothers?". Like Newton's law, Pauli principle is a law of nature. However, it is not at all trivial to ask "what kind of classical field would, when quantized, obey the Pauli principle?" The answer is remakably beautiful and unique:
1) construct a Lagrangian from field functions taking values in the Grassman algebra.
2) apply the usual quantization rule:
"quantum bracket = i times Poisson bracket"
This way you will end up with QFT for fermions.
Please do not confuse the meaning of "classical field" with "macroscopic observability". Classical field (function over spacetime) does not necessarily need to describe some macroscopic object.

regards

sam
 

Related Threads for: Where do fermion properties come from?

  • Last Post
Replies
9
Views
2K
  • Last Post
Replies
2
Views
2K
  • Last Post
Replies
19
Views
3K
  • Last Post
2
Replies
31
Views
2K
Replies
2
Views
2K
  • Last Post
Replies
3
Views
1K
Replies
3
Views
662
Replies
36
Views
2K
Top