# Are all vector-bosons gauge-bosons?

All QFTs that I come across have vector fields appearing as gauge-bosons. Is there any problem with vector fields that are not gauge-bosons? I am not so concerned about the theory producing results that match observations at the LHC, I just want to write down a Lagrangian which does not belong to a guage theory, yet have it correspond to a local and unitary theory with a vector field. For example $$\mathcal{L} = A_\mu^a \square A^{a\mu} + gf^{abc}A^a_\mu A^b_\nu\partial^\nu A^{c\mu} \ .$$ I was unoriginal and decided to truncate some of the YM Lagrangian. Is there any mathematical obstruction to doing this?

My question appears to be different to this one because in it, guage invariance is assumed. Moreover, in the answer there is the statement

If one drops gauge invariance, there are lots of other possible Lagrangian densities, for example a mass term, products of it with the terms described, and even more.

I would like to verify that indeed it is possible to drop gauge invariance.

EDIT: In this paper the authors show that from considerations at four points, the three-point amplitudes must be dressed with totally antisymmetric coefficients $$f^{abc}$$ that obey a Jacobi identity. This, however, could be achieved by the Lagrangian that I presented, that does not have local gauge invariance.

• Interesting question, imho gauge invariance is a direct result of the requirement local invariance. So you cannot build up a local theory which is not gauge invariant also. But i might be wrong on this one. – Davide Morgante May 26 '20 at 15:01
• What is "local invaraince"? Surely you do not mean "local gauge invariance", since that is what I want to get rid of :) – maor May 26 '20 at 15:13
• Local invariance under some Lie group. You seem to ask whether we can construct a local $U(1)$ theory which is not gauge invariant. I rest my case that apart from the specific group you choose, a local invariance under that symmetry group implies the existence of gauge fields and a gauge invariant theory. But seem that somebody answered something opposite to my feeling, so I was wrong! – Davide Morgante May 26 '20 at 15:16

We are looking for a vector field $$A_{\mu}(x)$$ which has spin 1 particle excitations, and does NOT require gauge invariance to describe it. Let's figure this out systematically, although I won't go through the gory details (references are below). First of all, the vector field is in the $$(\frac{1}{2},\frac{1}{2})$$ representation of the Lorentz group, so translating this into what spins this field could possibly produce, it is spin $$0$$ and spin $$1$$. If we wish to kill the spin $$0$$ component of the field, which would be of the form $$A_{\mu}(x)=\partial_{\mu}\lambda(x)$$, we could

1) Require that our theory has a gauge invariance $$A_{\mu}\to A_{\mu}+\partial_{\mu}\lambda$$.

2) Require that the field $$A_{\mu}$$ satisfies the "Lorentz gauge" constraint $$\partial_{\mu}A^{\mu}=0$$ (although calling it a gauge in this context is misleading).

Or we could simply leave the spin $$0$$ excitation alone and let it propagate.

Let's now consider the effect of the particle's mass. Starting with a massless spin $$1$$ particle. It turns out, that on very general circumstances, it is impossible to construct a vector field with massless excitations which transforms under Lorentz transformations the following way

$$U(\Lambda)A_{\mu}U^{-1}(\Lambda)=\Lambda^{\nu}_{\mu}A_{\nu}$$

Where $$U(\Lambda)$$ is a unitary representation of the Lorentz group. To understand why this is, it has to do with how the vector field is represented in terms of creation and annihilation operators. In general we have $$A_{\mu}(x)=\int d^3p(2\pi)^{-3/2}(2p^0)^{-1/2}\sum_{\sigma=\pm 1}\Big(e_{\mu}(p,\sigma)e^{ip\cdot x}a(p,\sigma)+e^*_{\mu}(p,\sigma)e^{-ip\cdot x}a^{\dagger}(p,\sigma)\Big)$$

Where $$e_{\mu}(p,\sigma)$$ is the polarization vector, $$\sigma$$ is the helicity and $$a(p,\sigma)$$ is an annihilation operator of the massless spin $$1$$ particle. The $$U(\Lambda)$$ acts on the $$a(p,\sigma)$$ and they represent the true particle content of the field excitation. Somehow, we need a polarization vector $$e_{\mu}(p,\sigma)$$ which transforms appropriately to go from the massless spin $$1$$ particle representation the $$a(p,\sigma)$$ transform under, to the $$(\frac{1}{2},\frac{1}{2})$$ representation that the field $$A_{\mu}$$ should transform with respect to. This simply cannot be done.

The best that can be done is the following

$$U(\Lambda)A_{\mu}U^{-1}(\Lambda)=\Lambda^{\nu}_{\mu}A_{\nu}+\partial_{\mu}\Omega(x,\Lambda)$$

This is a combination of a Lorentz transformation and a GAUGE transformation. This is highly concerning, as this implies that the theory can no longer be unitary. In order to remedy this we must require gauge invariance! Thus any massless spin 1 particle must be described by a gauge invariant vector field. Similar arguments can be made for non-abelian vector fields.

Once we add a mass, these issues don't arise, and gauge invariance is not required. But constructing a unitary and renormalizeable theory is another story. One could imagine adding a simple mass term to the Yang-Mills action $$-m^2A_{\mu}A^{\mu}$$ to describe a massive spin $$1$$ particle, however the pesky minus sign in the Minkowski inner product creates an unstable vacuum. One can attempt to remedy this by adding a term $$-\lambda (A_{\mu}A^{\mu})^2$$, but then we run into the issue of renormalizability, as well as a spontaneous breakdown of Lorentz symmetry.

The fact of the matter is it is much "easier" to describe a massive spin $$1$$ particle using gauge invariance. This is done in the standard model, where the $$W$$ and $$Z$$ bosons are massive spin $$1$$ particles. These are described by an $$SU(2)\times U(1)$$ gauge theory which is spontaneously broken, leaving some of the modes massive. This can be done in a unitary and renormalizeable way.

Most of the details here can be found in Volume I of The Quantum Theory of Fields by Steven Weinberg, Sections 5.3 and 5.9.

• Thanks for the detailed answer, it is veryinteresting. I would like to ask how you got to "the best that can be done is the following $U(\Lambda)A_\mu U(\Lambda)=...$" ? And how is this issue resolved by adding masses? I think I am asking for the details of the transformations of $a(p,\sigma)$ and the polarization vectors. By transforming the RHS of eq. 2 I guess I may find what you said? – maor May 27 '20 at 8:49
• And I am curious about spontaneous breaking of the Lorentz group, how does that come about? Maybe a reference will do if it is a long explenation! – maor May 27 '20 at 8:51
• The details of "the best that can be done" is worked out in section 5.9 of Weinberg. If done correctly, you should arrive at the result if you apply the Lorentz transformation on eqn 2. The major difference between the massless and massive case is what group the spin of the particle transforms with respect to. For the massless case, only helicity is well defined, and so the group is SO(2). For the massive case it is possible to have the particle at rest, and so the full rotation group SO(3) characterizes it's spin. This accounts for the difference. – fewfew4 May 27 '20 at 15:57
• That Lorentz symmetry breaking comment was a tree level comment, simply because one of the components of the vector field has the opposite sign for a mass term. This sets up the canonical $\lambda \phi^4$ spontaneous symmetry breaking scenario, except now it's a Lorentz 4-vector which gets a non-zero VEV. – fewfew4 May 27 '20 at 16:01
• I understand that if we want to write something Lorentz invariant it seems like it needs to be gauge invariant too. So far this seems to be applicable to the free theory. But in any case I consider the question answered, thanks! – maor May 30 '20 at 18:14

According to Sredicki's QFT pg. 120:

Theories with spin-one fields are renormalizable for $$d = 4$$ if and only if the spin-one fields are associated with a gauge symmetry.

So I guess that means that you can have low-energy effective field theories of spin-1 bosons without gauge symmetry, but not UV-complete theories.

Is every vector boson a gauge boson? The answer is no. The Kalb-Ramond field in string theory is an example of an antisymmetric two-index tensor $$B_{\mu \nu}$$ ($$\mu , \nu$$ $$\in$$ $$\{1, ... ,d\}$$) whose components $$B_{\mu i}$$ (where $$i$$ are indices on a compactification space) behave as vector bosons after dimensional reduction and (strictly speaking) such vectors are not the gauge bosons of a gauge interaction. The resulting effective theory (obtained by taking the zero slope limit of string theory in a given compactification) containt vector bosons $$B_{\mu i}$$ whose interactions are local and unitary. Similar examples can be given with higher p-form fields.

• Dear Davide: No, it cannot be a gauge boson. Mathematically: $B_{\mu i}$ is not a connection on any principal bundle, is the component of a higher gauge field. Even if you restrict $B_{\mu \nu}$ to $B_{\mu i}$ you can not obtain nothing isomorphic to a gauge bundle because the deformation theories are not equal. – Ramiro Hum-Sah May 26 '20 at 15:49
• Physically: If you explore the behaviour of the effective Kalb-Ramond vector fields $B_{\mu i}$ at strong coupling, you should discover that they should start to behave as the components of a single multiplet is higher dimensions $B_{\mu \nu}$. Something not expected for gauge bosons. – Ramiro Hum-Sah May 26 '20 at 15:50