In https://en.wikipedia.org/wiki/Photon#The_photon_as_a_gauge_boson is stated "...The quanta of an Abelian gauge field must be massless, uncharged bosons, as long as the symmetry is not broken; hence, the photon is predicted to be massless, and to have zero electric charge and integer spin ..."

Could anybody explain to me the reason why the quanta (aka gauge bosons = from mathematical viewpoint the basis vectors of its complexified adjoint representation of the gauge group) of an Abelian gauge field have to be be massless, uncharged bosons?

I'm familar with concepts of represenation theory and the interplay between Lie groups and Lie algebras unfortunately only with pure mathematical background. Thus I principally interested in a formal and not heuristical explanantion.

But I don't know the concrete argument why the claim about Abelian gauge field above holds. Moreover, could anybody explain maybe which problem occurs in general in theoretical physics when one tries to develop a field theory and is faced to deal with with concept of a 'mass'. I saw often that this causes a lot of troubles but nowhere found a nice explanation why this is so.

  • $\begingroup$ An unbroken gauge theory lacks a mass term, so the quanta are massless. Being abelian, it has no non-trivial structure constants, and so there cannot be photon-photon couplings, that is, the photon is chargeless, meaning it does not couple to itself. The problems with mass pertain to chiral theories, not photon couplings. It is all language, but you appear to be conflating all kinds of separate issues imagining connections. $\endgroup$ – Cosmas Zachos Jan 22 '20 at 20:05
  • $\begingroup$ Put in a mass term for the gauge potential and try making a gauge transformation. There is no way to have gauge invariance if the mass term is present. $\endgroup$ – G. Smith Jan 22 '20 at 21:17
  • $\begingroup$ @CosmasZachos: let me try to paraphrase your explanation in a way how I understood it as a non-physicist: We consider a Lagrangian and a gauge group acts on it but preserving it's shape (that's why gauge). By "lacks a mass term" you mean that the considered Lagrangian hasn't a summand which can be assocciated with mass, right? Futhermore, symmetry breaking is a procedure where the Lagrangian has a symmetry by gauge group, but not the lowest-energy vacuum solution. That is we take "choice" for one of the possible vacuum solutions and fix it for all time. $\endgroup$ – katalaveino Jan 22 '20 at 21:27
  • $\begingroup$ This "choice" changes the Lagrangian, so the new Lagrangian might be not more have the symmetry of the old one. This trafo of the L might give us a new "mass term". I if understood it correctly this procedure is called "symmetry breaking", yes? And what you concretly mean by "unbroken gauge theory"? That you consider a Lagrangian before you have performed such "trafo" to new one applying symmetry breaking or just that lowest-energy vacuum solution is stable (thus the symmetry breaking is useless in this can)? Is this the point? Sorry, if I confuse here some concepts. $\endgroup$ – katalaveino Jan 22 '20 at 21:27
  • $\begingroup$ Yes, in a way. When you realize nothing is lacking, the whole world belongs to you. There are so many texts teaching you the language, that e-chatting the logic before that stage stands in the way. $\endgroup$ – Cosmas Zachos Jan 22 '20 at 22:39

Adapting Cosmas Zachos' comment into an answer.

We say that a field is "free" if the Lagrangian is quadratic in the field and its derivative, eg. for a field $\phi^a$ with some kind of index $a$, the Lagrangian looks like $$ \mathcal L=Q_{ab}^{\mu\nu}\partial_\mu\phi^a\partial_\nu\phi^b+R_{ab}\phi^a\phi^b, $$ where $Q$ and $R$ are field-independent coefficients (in special relativity, usually constant, but in GR it isn't).

The first term is called the kinetic term, the second term is the mass term. It is named as such because upon quantization of the free field, the field quanta will obey a relativistic dispersion relation $E^2=p^2c^2+m^2c^4$ where the mass $m$ is related to the $R_{ab}$ coefficient.

It is usually assumed that the Lagrangian for interacting fields is polynomial in the fields (but if not then I guess one still usually looks at a power series expansion), and further terms containing fields are interaction terms. Eg. $$\mathcal L=Q^{\mu\nu}_{ab}\partial_\mu\phi^a\partial_\nu\phi^b+R_{ab}\phi^a\phi^b+S_{abcd}\phi^a\phi^b\phi^c\phi^d$$ is a Lagrangian where the field $\phi$ has a quartic self-interaction, which is represented by the term involving $S$.

As far as I am aware, if a Lagrangian contains more than 2 instances of derivatives then it is anathema, so these interaction terms are usually field-dependent but not field derivative dependent.

A (vector-) gauge boson is mathematically a connection on a principal fibre bundle, so locally it is a Lie algebra valued $1$-form $A=A^a_\mu T_a\otimes\mathrm dx^\mu$, where $T_a$ are a set of generators for a Lie algebra.

On two overlapping trivialization patches, the local gauge fields transform as $$ A^\prime=\mathrm{Ad}_{g^{-1}}A+g^\ast\Xi, $$ where $g$ is a Lie group ($G$) valued function on the overlap, and $\Xi$ is the left-invariant Maurer-Cartan form on $G$. For matrix groups we can also write $$ A^\prime=g^{-1}Ag+g^{-1}\mathrm dg. $$

The expression $$ F=\mathrm dA+\frac{1}{2}[A\wedge A],\ F=F^a_{\mu\nu} T_a\otimes\mathrm dx^\mu\wedge\mathrm dx^\nu \\ F^a_{\mu\nu}=\partial_\mu A^a_\nu-\partial_\nu A_\mu^a+C^a_{bc}A^b_\mu A^c_\nu $$is the local form of the curvature of the connection, it transforms homogenously as $F^\prime=\mathrm{Ad}_{g^{-1}}F$.

We want Lagrangians to be gauge-invariant, and (aside from some special cases like Chern-Simons terms, which is only weakly gauge invariant) a Lagrangian of the form $$ \mathcal L=\kappa F^a_{\mu\nu}F_a^{\mu\nu} $$ is the only gauge invariant option. Here the greek indices are raised and lowered by the metric tensor, the latin Lie algebra indices are raised and lowered by the Cartan-Killing form or some other Ad-invariant inner product, and $\kappa$ is an irrelevant constant.

If we expand the Lagrangian, then we get schematically (!) $$\mathcal L \sim (\partial A)^2+C\partial A A^2+C^2A^4.$$ The first schematic term is the kinetic term, the second and third are interaction terms (the second term involves only two powers of $A$ but also a $\partial A$ so it is not a mass term).

From this we can deduce some things.

  • Masslessness: The gauge invariant Lagrangian has no mass term, so the gauge boson is massless. Any attempts to artificially add a mass term of the form say $\sim m^2A_\mu^a A_a^\mu$ ruins gauge invariance.

  • Charge: Without constructing/defining "charge" explicitly, we see that the Lagrangian for the gauge field has two interaction terms, both involving the structure constants $C^a_{bc}$. This means that a general gauge boson interacts with itself, so we say it has charge, because charged particles are the ones who "feel" the interaction (here charge is not necessarily electic, but the charge of the given interaction, so to speak). However, if $C^a_{bc}=0$, which means that the Lie group is Abelian (all elements commute), then both interaction terms vanish. So an Abelian gauge field doesn't self-interact, so we say that its quanta is not charged.

  • Spin: This was also mentioned by OP, but is rather tangentially related. In a given Lorentz frame, the spatial components of $A^a_{\mu}$ (eg. $A^a_x,A^a_y,A^a_z$) transform as a 3-vector under spatial rotations, which is a spin-1 degree of freedom. The time component $A^a_0$ doesn't transform under spatial rotations, so it is a spin-0 degree of freedom. Gauge invariance however makes the spin-0 degree of freedom pure gauge, so the field quanta are particles of spin 1.

  • $\begingroup$ This was exacly the kind of answer I was looking for, great! The only point which stays still unclear is how (spontaneous) symmetry breaking comes into play? Cosmas started his explanation with "An unbroken gauge theory lacks a mass term ...". Formally, what is a unbroken gauge theory and what is a broken one? My understanding of it is then we have to look at lowest-energy vacuum solution and then decide about "brokeness" by checking if this state shares the same gauge invariance as the Lagrangian or not. If it does, then it is a "stable", thus "unbroken", right? $\endgroup$ – katalaveino Jan 23 '20 at 21:21
  • $\begingroup$ On the other hand assume that the lowest-energy vacuum solution is variant. Is this exactly that what we call spontaneous symm breaking? If yes, then I have two questions: firstly, what have we to do with the old Lagrangian if we "choose" one of the degenerated states of the vacuum states. Should we in this case also perform a certain transformation on the old Lagrangian which making it also not more gauge invariant? And secoundly: is there a precise formulation of symmetry breaking in diffgeometrical language? $\endgroup$ – katalaveino Jan 23 '20 at 21:22
  • $\begingroup$ @katalaveino I am not comfortable with this topic enough to give definite answers. But the old Lagrangian is the same but we express it with quantities that has been shifted so that the zero value of those quantities correspond to the vacuum state. Then from this shifting a mass term appears for the gauge field but nontheless the whole Lagrangian itself is gauge invariant. However then there appear also fields (Goldstone bosons) that are pure gauge, so they can be killed, but killing them will use up the gauge freedom of the broken subgroup... $\endgroup$ – Bence Racskó Jan 23 '20 at 21:29
  • $\begingroup$ @katalaveino ... so usually only the unbroken part is kept as gauge group to eliminate the goldstone modes. And yes, afaik reduction of structure group on principal bundles can be used to describe symmetry breaking. I think Gauge theory and variational principles by Bleecker and most of Sardanashvily's books treat this via differential geometry. $\endgroup$ – Bence Racskó Jan 23 '20 at 21:30
  • $\begingroup$ ok, so the choice of a concrete vacuum state corresponds in certain kind simply to a change of coordinates (e.g. a shift of $x$ by a constant or so) and thus in general the Lagrangian expressed in new coordinates possibly could obtain by such shift a 'mass', while the old one did not had one, right? One remark: what do you mean by fields that are "pure gauge"? $\endgroup$ – katalaveino Jan 23 '20 at 22:15

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