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The standard story goes as follows: gauge bosons cannot have a mass term because it would break gauge invariance in the lagrangian. This is clear, but why can't we just have massive vector bosons without any gauge transformations associated to them? My question arises from the standard model: spontaneous symmetry breaking gives you 3 massive bosons and a massless one, with a $U(1)$ residual gauge invariance from the subgroup of $SU(2) \times U(1)$ which leaves the $\varphi$ (vacuum) doublet invariant. My question is the following: why don't we just add massive vector fields with the desired masses and charges, without even defining gauge transformations for them?

On chapter 21 of Peskin&Schroeder they show how to quantize a spontaneously broken $U(1)$ theory in the $R_\xi$ gauge, and one interesting feature is that since the $\xi$ dependence drops out once you add both the (now massive) $A_\mu$ and the Goldstone boson $\varphi$ (with fake $\xi$ dependent mass) contributions, one can choose $\xi$ in different ways to prove different things, e.g. renormalizability, unitarity, and so on. Perhaps the theory wouldn't be renormalizable if we just added mass terms for the "non gauge" bosons, thereby making them fields with a Proca-like action.

user35319
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  • Depends on the model. Yang-Mills bosons are usually mass-less when the gauge symmetry is intact, and massive when the symmetry is broken. You can get topological mass in 2+1 dimensions and retain the symmetry. – puppetsock Jan 13 '20 at 19:14
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    Possible duplicate: https://physics.stackexchange.com/q/296975/2451 – Qmechanic Jan 13 '20 at 19:24
  • @Qmechanic but what if one does not interpret the W, Z boson as bosons associated to a gauge symmetry? In other words, what if one assumes that they are just massive vector fields coupled in a particular way to fermions and so on, without thinking of an underlying gauge invariance? – user35319 Jan 13 '20 at 19:38
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    @user35319 because if you add mass for these bosons mass explicitly (add mass terms into lagrangian of GWS model), you obtain non-renormalizable theory. Higgs mechanism preserves renormalizability. For theory with explicit mass term, you can obtain propagator with term $\sim -k_{\alpha}k_{\beta}/M^2$ and it does not have asymptote $\sim 1/k^2$ which means non-renormalizability of considered theory – Artem Alexandrov Jan 13 '20 at 20:01
  • @ArtemAlexandrov but that wouldn't work precisely because the unbroken GWS model has a nonabelian $SU(2) \times U(1)$ gauge invariance, which would be explicitly broken by mass terms, making it nonrenormalizable. I was wondering if it's possible to write down a renormalizable theory for the W, Z bosons which is not a gauge theory, but that has all the necessary vertices, and which doesn't come from a spontaneously broken nonabelian gauge theory such as the GWS model. – user35319 Jan 13 '20 at 20:07
  • @user35319 you mean a version of SM which represents all the experimental facts without "gauge nature", do not you? – Artem Alexandrov Jan 13 '20 at 20:14
  • @ArtemAlexandrov Exactly! A theory for massive W, Z bosons without a gauge symmetry breaking (or "hiding") – user35319 Jan 13 '20 at 20:22
  • No, you absolutely need gauge invariance, "hidden" or not, to ensure renormalizablity by constraining amps and constants. – Cosmas Zachos Jan 13 '20 at 22:33
  • @CosmasZachos why is that so? the theory of a single massive vector particle (which could describe the Z boson) is renormalizable even though the propagator is bad behaved as there's a term like $k_\mu k_\nu / M^2$ – user35319 Jan 14 '20 at 07:29
  • The full, nonabelian theory is not. Read up on main question. – Cosmas Zachos Jan 14 '20 at 11:30
  • @CosmasZachos i'm sorry if i can't understand correctly. To me the logic sounds kind of circular: mass terms for a nonabelian theory are forbidden by renormalizability, which is violated by mass terms because we assume that there exist some nonabelian gauge invariance? If we don't assume gauge invariance, the theory of a single massive boson is still renormalizable, isn't it? – user35319 Jan 14 '20 at 16:06
  • A nonabelian gauge theory loses renormalizability upon introduction of mass terms, unless there is Higgs hiding behind it. You may not cook up a renormalizable theory of massive Ws coupled to massive Z's, without Higgs hiding, independently of what you'd assume or not. – Cosmas Zachos Jan 14 '20 at 16:14

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