# Why gauge bosons, but no gauge fermions

• Edgardo
In summary, the conversation discusses the question of why force carriers are always bosons and why there are no gauge fermions. The answer is related to gauge invariance and the requirement for gauge particles to be vector particles in a 4-dimensional space-time. The concept of "gauge fermions" in supersymmetry is also mentioned, as well as the role of group theory in understanding spin and Lorentz group. The conversation also touches on the contribution of gauge fermions in beta functions and the relationship between gauge fields and tensor fields.

#### 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 ?"

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

Daniel.

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

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

Daniel.

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

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.

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).

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

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, which 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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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.

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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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.

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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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.

put it in a language that's more comprehensible

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.

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.

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.

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.

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.

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.

what's local gauge invariance in gravity theories?

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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.

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.

dextercioby said:
Why would they have to carry integer spin...?

Daniel.
I was thinking about that all fundamental interactions have interaction particles that are integer spins(photon, W,Z,and graviton?).

One question I do have(pardon my ignorance on gauge theory please) is that if gauge theory is developed to better understand fundamental interactions, why do we consider adding a half integer spin interaction particle when we do not see evidence of them(as far as I know)?

Josh

joshuaw said:
I was thinking about that all fundamental interactions have interaction particles that are integer spins(photon, W,Z,and graviton?).

One question I do have(pardon my ignorance on gauge theory please) is that if gauge theory is developed to better understand fundamental interactions, why do we consider adding a half integer spin interaction particle when we do not see evidence of them(as far as I know)?

Josh

What half-integer interaction particle are you speaking of?

dextercioby,
If you replace the words 'local gauge invariance' by 'invariance under a set of local diffeomorphisms',you sound impressive but you are not explaining anything.

I know,but those transformations are called that way.Diffeomorphisms are typical to GR only.If u choose another I-st class theory,the Lagrangian infintiesimal gauge transformations will look differently...

Daniel.

I think none of my questions is one that can not be answered in a physically intuitive manner--though it may require a better understanding than the one at just a mathematical level.I repeat my questions with the hope that one of you answers them without resorting to terms like 'local diffeomorphism' and '(1/2,1/2) irreducible representations of restricted homogeneous Lorentz group'.

Why does a vector field have spin 1?Why does gravity which is a tensor field have spin 2 quanta?When is it supposed to be spin 3 or higher?Does it go with the rank of the tensor?
What's local gauge invariance in gravity theories---can anyone explain it in physical terms?

Do you know what a vector field is...?If u don't,there's no way you could understand the answer to the question:"Why does a vector field have spin 1?".

Daniel.

"can anyone explain it in physical terms?"

No not really gauge invariance is somewhat of a redundancy in description, one that so happens to be extraordinarily useful to calculate with.

This has to do with the geometry *within* the structure of our fundamental field equations

Theres a slightly more abstract, mathematical reasoning behind that, having to do with how fiber bundles behave (essentially they are an *abstract* enlargement of our spacetime manifold, and a generalization of the direct product). In this context, gauge fields (well not all Gauge fields but most of them) are described by the connection induced by the principal bundle onto the associated vector bundle. Local sections thereof are *choices* of how to fix the Gauge parameter.

Now Gravity is slightly different, some physicists (often who aren't careful) enlarge the meaning of what a gauge field is to include local diffeomorphisms. This is somewhat perilous territory, but it is doable.

gptejms said:
IWhy does a vector field have spin 1?Why does gravity which is a tensor field have spin 2 quanta?When is it supposed to be spin 3 or higher?Does it go with the rank of the tensor?

Well, when you say " a vector field, you say are speaking about a mathematical objects made of 3 scalar components that obey some geometric transformation (when you express them in different reference frames with spatial transformations) and not with physics.
In other words, if you analyse the properties of vectors, you know how they transform under rotations (just write the components of the vector in the new rotated frame), this is the symmetry: the object "vector" is the same even if expressed differently in another rotated reference frame.
What mathematics says (and to be simple) is that the rotations symmetries are sufficient to define a class of objects invariant under rotations: the spinors and their "rank" (=2s+1, s the spin).
A spinor of rank 1 is simply a scalar (spin 0), a spinor of rank 3 (spin 1) is a vector, etc ...

If you want to see a first introduction, in the QM context, you have the Messiah book, quantum mechanics, volume II, chapter XIII and XV. This introduction does not require much mathematical ground and cover most of the practical needs.

Seratend.

I think my questions insisting on a physical explanation have somehow conveyed a very wrong impression--whew(!) dexter asks me if I know what is a vector field--I felt like s* him!

Haelflix,I understand local gauge invariance.I read Moriyasu's 'Primer on gauge theory' a few years back.It's just that I wanted to know what it means in the context of gravity.

Seratend,so is the group for spin one O(3)?What about spin 2---local diffeomorphism?

gptejms said:
Seratend,so is the group for spin one O(3)?What about spin 2---local diffeomorphism?

Eh... The group for "spin" is SU(2), or SO(3) which is not very different from O(3). That same group (which corresponds to certain symmetries of space, called rotations) can be represented by different objects (meaning, different objects can obey the same group laws). For instance, a "number" (scalar) is a rather trivial representation of the group: to each group element corresponds the trivial transformation: number -> "same number".
But vectors can also be used as representing the group: the group element corresponding to a certain rotation in space then rotates the vector in a similar way. Besides scalars and vectors, there are other ways of building objects that can represent the group SU(2). And all these different types of representations can be numbered: this number is spin. Spin 0 is a scalar, and spin 1 is a vector representation. Spin 2 is s rank-2 tensor representation. Spin-1/2 is, well, a spinor representation

So when you say "spin", you automatically talk about a specific representation of the rotation group SU(2).

Why is this important ? Well, if you somehow assume that the system under study is invariant under this rotation group, then you can only use "building bricks" which are representations of that rotation group. Indeed, by hypothesis you can apply an element A of the rotation group to all elements, and your physical results must come out the same. You can apply an element B of the rotation group again. Or you can consider that you apply B o A = C of the rotation group. Clearly, whether you applied A and then B, or whether you applied C directly, you should somehow transform the building bricks of your theory in the same way if you are going to hope to get out identical results. But that means that those building bricks are a representation of the group.

cheers,
Patrick.