# Why are all observable gauge theories not vector-like?

Why are all observable gauge theories not vector-like?

Will this imply that the electron and/or fermions do not have mass?

How is this issue resolved?

Background:

The Standard Model is a non-abelian gauge theory with the symmetry group U(1)×SU(2)×SU(3) and has a total of twelve gauge bosons: the photon, three weak bosons and eight gluons.

Massless fermions can either have their spin pointing along their direction of travel or opposed to it, these two types of massless fermions are called right-handed and left-handed. A fermion mass can turn a left-handed fermion into a right-handed fermion (this is technically called a Dirac fermion mass). The ordinary electron for example has both left-handed and right-handed pieces. For this to be allowed in a gauge theory the left-handed and right-handed fermions must have the same charge. If all the fermions in the gauge theory can be paired up in this fashion so that they are all allowed to have masses, then the gauge theory is called vector-like.

• I don't even know what that first question means... please explain. Commented Jul 1, 2012 at 3:25
• @ChrisGerig: I edited the question with some background. Commented Jul 1, 2012 at 3:30
• Comment to the question(v2): Note that not all gauge theories are of (Abelian or non-Abelian) Yang-Mills type, cf. physics.stackexchange.com/q/8686/2451 Commented Jul 1, 2012 at 16:01
• Are you asking (1) why every conceivable gauge theory is not vector-like? Or (2) why the standard model is not vector like? As for (1) I can certainly write down a vector-like gauge theory (QED). As for (2) talk to God. Commented Jul 1, 2012 at 20:30
• @djbunk: trying to ask both but answering one leads to the answer to the other. But specifically since you asked Why the standard model is not vector like and does the use of higgs mechanism help break this symmetry Commented Jul 1, 2012 at 20:32

There is no "vectorlike" gauge theory in the standard model, and this is a consequence of naturalness. This means that all particles in the standard model are naturally massless, and the mass only comes from Higgs mechanism. This is one of the great features of the standard model that is easy to break in any modification or extension.

The teminology "vectorlike" comes from the 1950s, when people didn't like 2-component spinors and thought that the world is fundamentally parity invariant. A "vectorlike" gauge field couples to a 4-spinor according to $\gamma^\mu A_\mu$, while a "pseudovectorlike gauge field" couples to a 4-spinor according to $\gamma^5\gamma^\mu A_\mu$. Both are parity invariant, but in the first case, A is a vector (meaning it changes sign under reflection), and in the second case it's a pseudovector.

But the gauge fields in nature are neither vectors or pseudovectors, they are parity-violating. They couple as "V-A" meaning $(1-\gamma_5)\gamma^\mu A_\mu$, which is a projection operator to one two component part of the 4 component Dirac spinor. This means that 4-component language is a little obfuscatory for this (although 4 component spinor notation is still useful, becuase Feynman trace identities are easier than Fierz identities, and the 4-component notation most easily generalizes to higher dimensions). The point is that there is no parity, and the gauge fields are neither "vectors" or "pseudovectors", they are parity violating vector fields which don't have a definite transformation under parity, because nature is chiral.

So I would drop the "vectorlike" terminology, and use the term "naturally mass allowing". A vectorlike gauge theory is "naturally mass allowing" because you can make the Fermion massive. This means that the left and right partners have the same charges, and this can be considered an accident.

The correct question is "why are all gauge theories in nature mass forbidding?" This is true of all the fields on the standard model--- none of the right handed and left handed fields in the standard model can pair up to form a mass, because they are different SU(2) multiplets and have different U(1) charge. Why are they all unpartnered and charged?

There is a simple reason for this--- any field which can get partnered will have an arbitrary mass term in the Lagrangian, and this term, without fine tuning, will end up generically being of order the Planck mass. So the only Fermions we see at low energies are those which are forbidden to have a mass, and therefore are chiral fermions without a partner to make a mass term with.

Further, all the fermions we see at low energies need to have a gauge charge, because without a charge of some sort, the Fermion can get a Majorana mass even without a partner, just by mixing with it's antiparticle. This is only forbidden if the particle is gauge charged in some way, so that the antiparticle has the opposite charge and the Majorana mixing is forbidden.

So all the fermions are chiral fermions with no partner to make a mass, so none of the low energy theories are vector-like.

The simplest right way to formulate gauge theories in a parity violating universe is in terms of 2-component spinors, each with an independent coupling to a collection of gauge fields. This procedure can lead to an inconsistency, if there is an anomaly in one of the gauged symmetries, so there are global constraints on the type of chiral fermions and the representations they can be in. If none of the Fermions have a partner, then the theory is natural, meaning "naturally massless" and the Fermions can only get a mass from a Higgs mechanism. The naturalness arguments say that the Higgs mechnism must be the source of mass of all Fermions in nature.

But if the Higgs is a fundamental scalar, the Higgs itself can have a mass, and the naturalness argument fails for the Higgs itself. So there is the question of why the Higgs has an unnaturally light mass. This is the hierarchy problem.

• en.wikipedia.org/wiki/Naturalness_(physics) Commented Jul 10, 2012 at 6:22
• @RonMaimon - I am a little confused why you say there would be a fermion mass hierarchy problem for a 'vector-like' gauge theory - doesn't chiral symmetry protect the mass term in the limit $m \rightarrow 0$? Commented Jul 10, 2012 at 12:28
• @DJBunk: If you impose the chiral symmetry by hand, you are right. But this is why I don't like saying "chiral symmetry protects against mass terms", because in general it's not chiral symmetry protecting the mass terms, it's that the mass terms are forbidden by gauge invariance. This is a better way to say it. But you can also protect against a mass term by imposing a global chiral symmetry on a fermion that could get a mass. Commented Jul 10, 2012 at 19:30
• The two procedures are different. In the standard model, there are no gauge invariant mass terms allowed. If you added a pair of fermions with a possible mass term, and imposed chiral symmetry, you can ask "why is the global chiral symmetry present"? And this is the same type of question as "why is the mass of the Fermion fine-tuned to zero?" But tuning a scalar mass to zero is worse, because even after you do it, the magic point is not protected by a symmetry, so radiative corrections ruin your tuning. Commented Jul 10, 2012 at 19:42
• @RonMaimon You answered here quora.com/… that "A coordinate transformation with compact support (meaning one which only changes the coordinates inside a bounded region) is not physical at all, it is just a change in philosophy about how you describe the physical situation." I'm currently trying to understand why this is the case physics.stackexchange.com/questions/432017/… and maybe you have an idea?
– jak
Commented Oct 2, 2018 at 9:09