Why gauge bosons, but no gauge fermions

Edgardo

Hello all,

from Marlon's journal, I read the question "DO YOU KNOW WHY FORCE CARRIERS ARE ALWAYS BOSONS ??? WHY DON'T WE HAVE GAUGE FERMIONS ???"

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dextercioby

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Give me an example of a fermionic first class system,field theory,of course...

Daniel.

Meir Achuz

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Gold Member
Gauge invariance involves correcting problems caused in quantum mechanics by the appearance of the partial derivatives with respect ot x,y,z,t. These partials form a relativistic 4-vector. This requires 4-vector fields so that (in EM)
d/dx-->d/dx-ieA_x, etc. The particle excitations of vector fields have spin one.
So the requirement that gauge particles (the excitations of the gauge fields) must be vector particles follows from the fact that space-time is 4 dimensional.

joshuaw

I thought that because the exchange particles must be able to carry integer spin they must be bosons. (Spin flips and all).
Josh

dextercioby

Homework Helper
Why would they have to carry integer spin...?

Daniel.

marlon

Edgardo said:
Hello all,

from Marlon's journal, I read the question "DO YOU KNOW WHY FORCE CARRIERS ARE ALWAYS BOSONS ??? WHY DON'T WE HAVE GAUGE FERMIONS ???"

good question...

there are several ways to answer this : spin statistics being one of them

marlon

vanesch

Staff Emeritus
Gold Member
Meir Achuz said:
Gauge invariance involves correcting problems caused in quantum mechanics by the appearance of the partial derivatives with respect ot x,y,z,t. These partials form a relativistic 4-vector. This requires 4-vector fields so that (in EM)
d/dx-->d/dx-ieA_x, etc. The particle excitations of vector fields have spin one.
So the requirement that gauge particles (the excitations of the gauge fields) must be vector particles follows from the fact that space-time is 4 dimensional.
I like that answer :-) As gauge fields have to appear in the covariant derivative, they have to be vector fields. Nice.

cheers,
Patrick.

wangyi

In SuperSymmetry, maybe you will call the Fermions in the same doublet with the Bosons "Gauge Fermions", because they transform together with the gauge bosons, for
example wino or zino (they form chargino and neutrino).

marlon

Meir Achuz said:
Gauge invariance involves correcting problems caused in quantum mechanics by the appearance of the partial derivatives with respect ot x,y,z,t. These partials form a relativistic 4-vector. This requires 4-vector fields so that (in EM)
d/dx-->d/dx-ieA_x, etc. The particle excitations of vector fields have spin one.
So the requirement that gauge particles (the excitations of the gauge fields) must be vector particles follows from the fact that space-time is 4 dimensional.
This is indeed a great answer

marlon

BlackBaron

wangyi said:
In SuperSymmetry, maybe you will call the Fermions in the same doublet with the Bosons "Gauge Fermions", because they transform together with the gauge bosons, for
example wino or zino (they form chargino and neutrino).
"Gauge Fermions" in supersymmetry are generically called "gauginos".
As long as I know, in supersimetric models they contribute with the forces just like regular gauge bosons, but because of the exclusion principle their net effect is not very strong, wich also shows that if we only had "gauge fermions" and no "gauge bosons" the world wouldn't be as we know it at all.

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dextercioby

Homework Helper
Meir Achuz said:
Gauge invariance involves correcting problems caused in quantum mechanics by the appearance of the partial derivatives with respect ot x,y,z,t. These partials form a relativistic 4-vector. This requires 4-vector fields so that (in EM)
d/dx-->d/dx-ieA_x, etc. The particle excitations of vector fields have spin one.
So the requirement that gauge particles (the excitations of the gauge fields) must be vector particles follows from the fact that space-time is 4 dimensional.
Unless "vector particle" means something else than the quanta of a vector field,that conclusion is incorrect.

Daniel.

gptejms

I am not a QFTist and know only a bit of it,so let me ask-----does gravity have to necessarily be a gauge field?If so why?

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dextercioby

Homework Helper
Gravity is a gauge field.It's not a vector field (spin 1),but a tensor field (spin 2)...It's the idea behind post #11.

Daniel.

gptejms

Yeah ok,gravity is a gauge field(only potential differences matter).But some more stupid questions--why does a tensor field have to be spin 2 and vector field spin one?

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dextercioby

Homework Helper
Aaa,nice question.Group theory.Vectors are $\left(\frac{1}{2},\frac{1}{2}\right)$ irreducible reps of the restricted homogenous Lorentz group and therefore have total spin $\frac{1}{2}+\frac{1}{2}=1$.

Symmetric 2-nd rank tensors (the gravity field $h_{\mu\nu}$ which is the I-st order perturbation expansion of GR metric $g_{\mu\nu}$) are $(1,1)\oplus (0,0)$ irredcible reps of the restricted homogenous Lorentz group and have spin $1+1=2$.

Daniel.

gptejms

put it in a language that's more comprehensible

dextercioby

Homework Helper
I honestly can't."Spin" means group theory.Vectors & tensors mean group theory.That's all there is to it...All u need to understand is the 'coupling' between spin & Lorentz group...

Daniel.

Staff Emeritus
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gptejms said:
put it in a language that's more comprehensible
Quantum spin is a new concept. That means it can't really be explained in terms of older concepts. If you don't want to learn the group theory definitions, you can just accept the fact that the states of spin 1/2 only turn half as fast as the coordinates do when you perform a rotation.

Haelfix

Agreed with Dexter.

Although in Supersymmetry the fundamental axioms of Lie algebra are modified (into.. surprise Super Lie Algebra), which is why you can have 'gauge' fermions with the same standard model quantum numbers, and they do indeed participate. However for technical reasons they have to be Majorana fermions, and when you calculate the beta function you end up with a fraction of the contribution as the gauge boson (tho not too small, maybe an order of magnitude less). Of course super symmetry is badly broken, and depending on the mechanism choice, will reduce things further.

gptejms

If you don't want to learn the group theory definitions, you can just accept the fact that the states of spin 1/2 only turn half as fast as the coordinates do when you perform a rotation.
SU(2) you mean--i know that.please complete the argument.

Meir Achuz

Homework Helper
Gold Member
dc: What does your sentence "Unless 'vector particle' means something else than the quanta of a vector field,that conclusion is incorrect."
in post #11 mean? I thought I had written that VPs were the quanta of VFs.

re: post #13: Can you describe how a tensor field can be a gauge field?
You and I must be talking about different kinds of "gauge field".
Mine is related to LGI in QM and gauge invariance of a theory.

dextercioby

Homework Helper
In that case,concluding that gauge bosons must be vector bosons is incorrect...

I didn't assert that "a tensor field can be a gauge field".I said that (not literally) "gravity,which is symmetric 2-nd rank tensor field,is a gauge field".And there are many more 4D field theory examples ,actually counterexamples to your conclusion formulated in post #3 & quoted by me in post #11.

Daniel.

gptejms

what's local gauge invariance in gravity theories?

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dextercioby

Homework Helper
HE action is invariant under a set of local diffeomorphisms which can be shown to be the Lagrangian infinitesimal gauge transformations.Since the Hamiltonian formalism for GR is a real pain in the a,the analysis is simpler using Einstein's linearized theory (which uses the I-st order perturbations $h_{\mu\nu}$ mentioned earlier).

Daniel.

dextercioby

Homework Helper
gptejms said:
what's local gauge invariance in gravity theories?
Hmmmm,u changed your post. There's no difference to other theories wrt the definition of "local gauge invariance".Basically,it's a I-st class system which is called "classical gravity" and is based upon the HE action...

Daniel.

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