Why gauge bosons, but no gauge fermions

[dextercioby]

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.

 

[Haelfix]

"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

there's 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.

 

[seratend]

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.

 

[gptejms]

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?

 

[vanesch]

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.

 

[gptejms]

ok,probably i should read these things before i ask questions--regret the fact that no QFT was taught to us in our graduate school.though i managed to pick up some random aspects of it from here and there,that certainly is not enough.

 

[joshuaw]

What half-integer interaction particle are you speaking of?

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

And I had responded that I thought that they had to be integer spin due to the fundamental interactions, to which Dextercioby had asked why they had to have integer spin(from post number 5). I obviously don't have much knowledge of field theory, and did not realize how deep the original question was.

I read the original post as meaning that a force could be mediated by a fermion, but I don't understand how this can be physicaly possible since all fundamental interactions are mediated through integer spin particles. I found an intro level textbook on the subject and am reading it. I am sorry if I caused more confusion.

Josh

 

[marlon]

I think Vanesch gave the best explanation here. This is how i remember this stuff. Suppose that you want some physical system to be invariant under certain operations. Then indeed, the 'pieces' that make up that system must all transform like irreducibe representations (IR) of the group that contains all operations under which the system must be invariant.

Now, what does this mean ? Well, let's look at QM and how the J-operator is connected to rotations. A QM-system is invariant under rotations if

1) the normalization of the wavefunction is preserved
2) the expectation value of any observable is preserved
3) if the Hamiltonian does NOT change under rotations.

In order to obey these commands, the wavefunctions (these are the 'parts' that make up the physical system, caracterized by the three above conditions) must transform in a certain way : w' = Uw...Where w is the original wavefunction and R denotes the rotation. U MUST BE UNITARY in order to obey the conditions (this is just the same as asking why time-evolution must be unitary).

In QM one can prove that if a wavefunction transforms like w' = Uw, then this U (which is a rotation) can be written in terms of the component of the L-operator along the rotational axis. But what does this component look like ? Well QM proves that we can write it in terms of its eigenvalues l just by calculating the expectation value of the L-operator in the appropriate base.

So what does this mean ? Answer : the IR representations are directly connected to the eigenvalues of the L-operator, which is also called the generator of the rotations. So l = 0, 1, 2,... all represent a different IR of the rotation-group.

Just as an addendum. If a system is invariant under rotations (which all are put in the symmetry group of that system), the parts which make up the system MUST transform as IR of that symmetry group. This is a very important rule that is quite logical, if you think about it. Let's take three operations out of the symmetry group and we call them A, B and C = A ° B...

Now we perform them on the system:
1) A(system) = system
2) B(system) = system
3) A°B(system) = A(B(system)) = A(system) = system
4) C(system) = system = A°B(system) = system

This means that if you perform either A°B or C, the result must be the same. This implies that the parts of the system must transform in the same way under either A°B and C. But this means that the parts actually obey the multiplication table of the group and by definition, this means that they must be representations of that group.

regards
marlon

 

[dextercioby]

What does this analogy (if that can be called 'analogy') have to do with the possibility that the Lie hamiltonian gauge algebra have fermionic generators (which is basically the issue at hand) (i.e.be a supercommutative/\mathbb{Z}_{2} graded algebra)?

Daniel.

 

[marlon]

What does this analogy (if that can be called 'analogy') have to do with the possibility that the Lie hamiltonian gauge algebra have fermionic generators (which is basically the issue at hand) (i.e.be a supercommutative/\mathbb{Z}_{2} graded algebra)?

Daniel.

Sometimes i really wonder if you actually understand yourself, the words that you write down.

marlon...

ps, dexter, why didn't i see you in the Kapeldreef in Leuven this week ?

 

[dextercioby]

I understand myself,don't worry. Actually it's an interesting subject that i'll be thinking about:is there any connection between various irreducible reps of the restricted hom.Lorentz group (which give the classical field theories that we quantize in QFT),meaning their spin/weight,and the gauge algebra (either Lagrangian or Hamiltonian)...?

Daniel.

P.S.Been in Leuven for >7 months.What's Kapeldreef?

 

[marlon]

P.S.Been in Leuven for >7 months.What's Kapeldreef?

You don't know what the Kapeldreef is ?

Don't you know where IMEC is ?

 

[dextercioby]

Nope,i think this discussion could take place via private messaging,don't u think...?

I haven't gotten aquainted with their names & acronyms.Good thing i know the peoples' names and remember their faces,too...

Daniel.

 

[Haelfix]

"Actually it's an interesting subject that i'll be thinking about:is there any connection between various irreducible reps of the restricted hom.Lorentz group (which give the classical field theories that we quantize in QFT),meaning their spin/weight,and the gauge algebra (either Lagrangian or Hamiltonian)...?"

Could you be a little more specific?

 

[dextercioby]

The idea in this thread was:are there any fermionic gauge field theories (classical)...?If there aren't,i want to know why.I'll seek,when i have time,if there is a connection between the weight of the representation (i shouldn't call it 'spin',i discuss it classically) and the the constraints,if any.I suspect that fermionic theories cannot be gauge (I-st class),but i need to be sure...

Daniel.

P.S.The fundamental symmetry is Poincaré...

 

[Haelfix]

Hmm, I suspect there is no such restriction in general, even for classical systems. I'm not entirely sure of that, but I have a reason in mind.

Anyway in the quantum world, its the Higgs interaction that messes things up, at least under the standard model, gauge fermion theories would mess up the chirality of all observed particle spectrums. I don't however think they are precluded a priori.

Even in SUSY, gauginos are somewhat problematic (especially in so called extended supersymmetry), as for instance they transform under the adjoint representation of the gauge group along with their boson partners. However since the group has to transform in a chiral representation, so to do the superpartners. Invariably this restricts the amount of possible solutions.

 

[Hans de Vries]

Don't you know where IMEC is ?

Nope,i think this discussion could take place via private messaging,don't u think...?
I haven't gotten aquainted with their names & acronyms.Good thing i know the peoples' names and remember their faces,too...
Daniel.

IMEC has become the world's number one laboratory for future generation
Semiconductor Research over the last 5 years or so. Extremely successful,
Consulted by all of the industry giants. Many, many projects sponsored by
the industry as a world wide pre-competitive cooperation. Much of the work
done goes into actual production world wide.

They are generally the first to get the latest $30 million or so lithography
steppers for next generation work.


Regards, Hans

 

[dextercioby]

Haelfix,i still think the problem resides classically.Poincaré symmetry:Since both gauge fields of spin 1 & 2 have classical correspondent (the electromagnetic field & the Pauli-Fierz/linearized gravity field),meaning that a choice of a Lagrangian action is not arbitrary if i want to describe a physical system,i'll stick to spin 3.It's described by a 3-rd rank tensor with plenty of index symmetries,a component being the free abelian 3-form...Now,in building a classical theory,i could manipulate the Poincaré invariants as to build a 2-nd class theory even with free abelian 3-forms,even though i knew that this theory is actually 1-st class/gauge...And that would be done without adding a mass term (Proca lagrangian (W+-Z0),massive YM).

I referred only to bosonic case.In the fermionic one,i'd say that i could play around with Weyl,Dirac,Rarita-Schwinger,5/2,... fields (which are typically II-nd class) as to get Lagrangian actions which would be I-st class,ergo gauge,but i don't know if they're physical or not.I suspect they aren't.*

I only spoke about free field theories.Interacting theories would pose some problems.We only know how to couple gauge theories to scalar & charged II-nd class theories.
We know how to couple 2 gauge theories (describing electromagnetism in a curved space),but i don't know if we could couple II-nd class theories...That's divagation.

I think Poincaré symmetry (global)<->gauge symmetry (local)<->II-nd class systems could be an interesting topic,even in the absence of manifest supersymmetry at classical level...

*So "why no gauge fermions in the SM?".I guess they're not physical.


Daniel.