Gauge symmetry for spin 1/2 fields

[Lapidus]

Why is there no gauge symmetry for spin-1/2 fields?

Has gauge symmetry to be related to spin-1 fields/ particles?

thanks

 

[QuantumClue]

Who told you that? SU(2) has a gauge symmetry to spin 1/2 fields.

 

[RedX]

That does seem strange, that spin-1 fields seem to be all gauge bosons (transform in the adjoint representation), while spin-0 and spin-1/2 transform in fundamental representations.

But beyond this, the spin-1 fields, besides transforming in the adjoint representation, have an extra term that involves the derivative of the local group parameter. This term is missing from fields that transform in the fundamental representations. It's as if spin-1 fields are undergoing a local translation in addition to a local rotation.

I don't know the answers to these questions. Can you have a spin-1 field transforming in the fundamental representation, and have a covariant derivative for it that involves a fermion field as the gauge field?

 

[Lapidus]

Thanks for the answering so far.

I assumed that there is no gauge symmetry for spin-1/2 fields, since it is never mentioned in the textbooks. Gauge symmetry just always comes into play when spin-1 particles/fields are introduced.

 

[tom.stoer]

You have to distinguish between the fundametal gauge potentials A and the field strength F.

First of all a fermion field transforms in the fund. rep., i.e.

$$\psi \to U\psi$$

A gauge field strength transforms in the adjoint rep.

$$F \to UFU^\dagger$$

Only the gauge field A transforms as

$$A \to U(A-id)U^\dagger$$

which is strictly speaking not the adjoint rep. The adjoint rep. means that the elements are elements of an algebra; an algebra is - besides some other properties - a vector space. But you can't simply add gauge potentials to get a new one:

$$A^1 \to U(A^1-id)U^\dagger$$
$$A^2 \to U(A^2-id)U^\dagger$$

So

$$(A^1 + A^2) \to U(A^1-id)U^\dagger + U(A^2-id)U^\dagger = U((A^1+A^2)-2id)U^\dagger$$

But if you define

$$A=(A^1 + A^2)$$

you would get

$$A \to U(A-id)U^\dagger = U((A^1+A^2)-id)U^\dagger$$

You see the difference.

The geometrical reason is that A is not only a four-vector but a connection in the fibre bundle generated by the gauge group over spacetime, whereas fermions and F are just ordinary spacetime spinors, tensors and elements of a vector space the gauge group is acting on; you can add these vectors w/o violating the transformation law.

Btw. there are indeed covaraint derivatives both in the fund. rep. and in the adjoint rep.

 

[dextercioby]

Thanks for the answering so far.

I assumed that there is no gauge symmetry for spin-1/2 fields, since it is never mentioned in the textbooks.

Well, that's true. The massive spin 1/2 field doesn't possesses gauge invariance, just global U(1) invariance.

Gauge symmetry just always comes into play when spin-1 particles/fields are introduced.

Fields of spin 1, 3/2 and 2 have gauge invariance.

 

[tom.stoer]

Well, that's true. The massive spin 1/2 field doesn't possesses gauge invariance, just global U(1) invariance.

That's wrong!

The (massive) electron field in QED and the quark fields in QCD both have local gauge invariance, namly U(1) and SU(3), respectively. Only in chiral theories (like electro-weak theory) local gauge invariance of massive fermions is forbidden. But this problem is not solved via forbidding local gauge invariance but via demanding that the elementary fields are massless.

That means that all fermions of the standard modle have local gauge invariance U(1)*SU(2)*SU(3).

 

[dextercioby]

That's wrong!

No, it's true. The spin 1/2 field if massive is described by the Lagrangian density
$$\mathcal{L} = \frac{1}{2} \bar{\Psi}(x)\left(i\gamma^{\mu}\substack{\leftrightarrow \\ \partial_{\mu}}- m\right) \Psi(x)$$ (http://en.wikipedia.org/wiki/Dirac_field#Dirac_fields)

which possesses global U(1) symmetry (rigid).

If you wish to couple this (or a collection of this) to some gauge fields(s), then it will indeed <gauge> its global symmetries (=> QED or QCD as primary fundamental theories).

 

[tom.stoer]

What you are saying is trivial: if you use a Lagrangian w/o gauge symmetry then the Lagrangian has no gauge symmetry. You are using the wrong Lagrangian.

This has nothing to do whether the fermion field in your Lagrangian is massive or massless: your Lagrangian with massless fields (i.e. m=0) has still no local gauge symmetry. In any case you have to use the covariant derivative.

 

[dextercioby]

I'm using what's been known for 70 years as the Lagrangian for massive spin 1/2 fields in a flat 4D space-time. It can't be wrong.

As for covariant derivative, well, it enters the picture only if you want the collection of electrons/positrons described by the density I wrote to interact with something.

I've just addressed this post <I assumed that there is no gauge symmetry for spin-1/2 fields, since it is never mentioned in the textbooks. Gauge symmetry just always comes into play when spin-1 particles/fields are introduced.>

 

[tom.stoer]

You wrote:

The massive spin 1/2 field doesn't possesses gauge invariance, just global U(1) invariance.

I said that this has nothing to do with the mass of the fermion fields.

I've just addressed this post I assumed that there is no gauge symmetry for spin-1/2 fields, since it is never mentioned in the textbooks. Gauge symmetry just always comes into play when spin-1 particles/fields are introduced.

Yes, as I said, you have to introduce the covariant derivative.

What about the following:

The kinetic term of the spin 1/2 field doesn't possesses gauge invariance (just a global invariance) w/o introducing spin 1 gauge fields.

Could you agree?

 

[dextercioby]

What about the following:

The kinetic term of the spin 1/2 field doesn't possesses gauge invariance (just a global invariance) w/o introducing spin 1 gauge fields.

Could you agree?

Yes, the same can be said about a mass term.

 

[tom.stoer]

the same can be said about a mass term.

No, the mass term already has local gauge invariance, but of course this is useless w/o the kinetic term. Or let's say it the other way round: introducing local gauge invariance changes the kinetic term, but not the mass term.

 

[dextercioby]

No, the mass term already has local gauge invariance, but of course this is useless w/o the kinetic term.

I don't understand. Perhaps you mean the mass term for fermions as it appears from the electroweak interaction via the Higgs boson. But this is not what I meant. I was speaking only about QED.

Or let's say it the other way round: introducing local gauge invariance changes the kinetic term, but not the mass term.

The kinetic term is not changed by coupling to a single spin 1-gauge field. The lagrangian density altogether acquires the coupling term.

 

[tom.stoer]

The mass term in QED reads

$$m\bar{\psi}\psi$$

It is invariant w.r.t. a local gauge trf., that means you can transform it via

$$\psi \to U\psi$$

where U=U(x). But of course this is not reasonable w/o having introduced gauge symm. in the kinetic energy term. What I mean is that once you introduce gauge symm. via adding the gauge field this does not affect the mass term, neither in QED nor in QCD. In the electro-weak theory it's different because there is no mass term.

Regarding "the kinetic term is not changed by coupling to a single spin 1-gauge field." What I mean is that you replace the standard kinetic energy

$$\partial_\mu \to D_\mu$$

which creates the coupling term. Yes, stricty speaking you are right, the kinetic energy remains unchanged and an additional coupling term is introduced.