# Funny term in my Lagrangian

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.

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.