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After introducing some vectors to the SM (in an invariant way), I get the following type of terms after EWSB:

$$\mathcal{L} = -\frac{1}{4}(\partial_\mu V_\nu - \partial_\nu V_\mu)^2 + \frac{1}{2}(\partial_\mu h)^2 -\frac{1}{2}m_h^2h^2 - \frac{1}{2}m_V^2V^2 + ah\partial_\mu V^\mu,$$

where $V_\mu$ is a just a massive vector field, $h$ is the usual Higgs field, and $a$ is an unknown real constant.

I do not know how to deal with the term proportional to $a$ (the last term). The Feynman rule for that term gives me something which change the identity of either the vector field and the Higgs field.

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Are you sure about the form of the last term? I'll show you a way to treat (almost) such terms in higgs mechanism.

A good way to describe a massive vector field is to use the respective massless one and afterwards exploit the Higgs mechanism. The initial Lagrangian density will be of the form:

$$L=-\frac{1}{4}F^{\mu\nu}F_{\mu\nu}+(D_\mu\Phi)^\dagger(D^{\mu}\Phi)-V(\Phi,\Phi^\dagger)$$

with $$V(\Phi,\Phi^\dagger)=\mu^2(\Phi\Phi^\dagger)+\lambda(\Phi\Phi^\dagger)^2$$

where $$\Phi=\begin{pmatrix} h\\ \phi \end{pmatrix}$$ and $$F_{\mu\nu}=\partial_\mu A_{\nu}-\partial_\nu A_{\mu}$$

The $\mu^2$ here is not the mass term, since it must be negative so that the $\Phi_0=\begin{pmatrix} \upsilon /\sqrt{2}\\ 0 \end{pmatrix}$ point will be stable to make a perturbation around it.

It's easy to observe that this Lagrangian is invariant under the gauge tranformation:

$$A_\mu \rightarrow A'_\mu=A_\mu+\frac{1}{q}\partial_\mu \theta(x)$$

From the first two terms of $L$ you will get the kinetic terms of $h$ and $\phi$, the mass term of $A_{\mu}$ and your "weirdo" term $+$ interactions.

In fact the "weirdo" term should not exist. It tells you that a scalar field becomes a vector field.. out of nothing! Of course, such a process has never been observed experimentally. So, you must find a way to get rid of this. Recalling the gauge freedom that you've already had, you can choose:

$$\theta(x)=\frac{\phi(x)}{\upsilon} \Rightarrow \frac{1}{2}(\partial_{\mu}\phi)^2+\frac{1}{2}q^2\upsilon^2A_\mu^2=\frac{1}{2}q^2\upsilon^2 A'^{2}_{\mu} -q\upsilon A_{\mu} \partial^{\mu} \phi$$

You see that you get again your "weirdo" but this time with opposite sign, so that they will get canceled and you'll have one worry less!

Note: After this process you realise that $\phi$ is a massless degree of freedom, which vanishes after the gauge fixing. For more info check the Goldstone's theorem.

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