# What goes wrong if we add a mass term for gauge bosons without the Higgs mechanism?

Question: Why can't we add a mass term for the gauge bosons of a non-abelian gauge theory?

In an abelian gauge theory one can freely add a mass and, while this breaks gauge invariance, as long as the coupling current is conserved everything works fine (i.e., the scalar modes decouple and the theory is renormalisable).

In non-abelian gauge theories, it is often stated that the only way to introduce a mass term is through the Higgs mechanism. If we added a mass term without introducing the Higgs field, but the coupling current is still conserved, at what point would the theory break down? It seems to me that the scalar modes decouple as well, at least to tree level. I failed to push the calculation to one loop order, so maybe the theory breaks down here. Is this the most immediate source of problems, or is there any simpler observable which fails to be gauge invariant?

One would often hear that if we break gauge invariance the theory is no longer renormalisable. I may be too naïve but it seems to me that a (gauge-fixed) massive gauge boson has a $\mathcal O(p^{-2})$ propagator and therefore (as long as the current in the vertices is conserved) the theory is (power counting) renormalisable. Or is it?

To keep things focused, let us imagine that we wanted to give gluons mass, while keeping self-interactions and the coupling to matter (and ghosts) unchanged. Could this work without a Higgs?

There are many posts about that are asking similar things. For example,

• Nothing goes wrong, just the theory becomes strongly coupled at $4\pi m_V/g$ where $m_V$ is the mass and $g$ the gauge coupling. It is simple to see this because the longitudinal polarization grows with energy. In fact, by a gauge redefinition one can make this apparent reintroducing the eaten Goldstone bosons. In practice, the Higgs mechanism is there in any case, it's the Higgs particle that could be missing. The theory has a cutoff which is at most $4\pi v$, where $v=m_V/g$ is the vev, as for any theory of Goldstone bosons that are derivatively coupled. The simplest UV completion adds h. – TwoBs Dec 6 '16 at 16:05
• @TwoBs thanks for the comment. What you describe is not really true for an abelian gauge theory: the longitudinal polarisations would grow, but in fact they cancel (because of current conservation) and therefore the S matrix elements don't grow with energy (massive QED is UV finite, and perturbatively unitary). But this seems to fail for non-abelian gauge theories. Why is a non-abelia gauge theory different than an abelian one? is it because, unlike QED, the longitudinal polarisations don't cancel (even if the current is conserved)? – AccidentalFourierTransform Dec 6 '16 at 16:20
• It's very simple to understand: for the abelian case the mass gives rise just a free kinetic term for the would be goldstone bosons, whereas for the non-abelian case it give rise to a non-trivial derivatively coupled interactions. The reason is because the coset structure of$U(1)\rightarrow 0$ is a circle which is one-dimensional and hence there is no non trivial Riemann curvature, whereas for non-abelian cosets, e.g. $SU(2)\rightarrow U(1)$, they have non trivial Riemann ( it's a sphere above). The longitudinal polarizations, that is the Goldstones, couple with coefficients given by Riemann. – TwoBs Dec 6 '16 at 23:05
• @TwoBs nice, thanks! (it would be nice of you to post an answer some day, whenever you have the time :-) ) – AccidentalFourierTransform Dec 6 '16 at 23:18
• I will try to post an actual answer, it's just hard to find the time to make one that is at once nice, short and accurate. – TwoBs Dec 6 '16 at 23:41

In electrodynamics, the situation is favorable. After separation of the gauge field into transverse and longitudinal components, the longitudinal part $k_\mu k_\nu/M^2$ which gives rise to the bad behavior in the propagator does not contribute to the $S$ matrix. This results from the noninteraction of longitudinal and transverse components and from the coupling of the field to a conserved current. In a nonabelian theory, none of these properties is satisfied. Longitudinal and transverse parts do interact, while the current to which the gauge field is coupled is not conserved. On the other hand, unexpected cancellations of divergences at the one-loop level make the theory look like renormalizable. This explains why it took some time to reach a consensus, namely, that the theory is not renormalizable. The way out of this unpleasant situation is to appeal to the mechanism of spontaneous symmetry breaking, to be explained in the next subsection.