# $U(1)_V$ invariance

I'm working with an interaction Lagrangian of the form:

$${\cal L}_{int} = \bar{\psi}\Theta\chi \tag1$$

Where $$\Theta$$ contains other operators, coupling constants, etc. I'm trying to unveil if this kind of interaction has conserved fermion number, i.e., it's invariant under $$U(1)_V$$. Let's suppose that the free Lagrangian has that symmetry. I think that both fermion fields, $$\psi, \chi$$ transforms like:

$$\psi \rightarrow e^{-ia}\psi \Leftrightarrow \bar{\psi} \rightarrow e^{ia}\bar{\psi}, \quad \chi \rightarrow e^{-ia}\chi, \quad a \in \mathbb{C} . \tag2$$

If this is the way that they transform (to wit, same parameter $$a$$), then these fields has fermion number conservation. But this is the crucial point, I'm not sure if I can pick the same $$a$$.

Well, rather tautologically, the way a symmetry is defined determines what symmetry it is. For example, the transformation $$\psi \to e^{-i a} \psi, \quad \chi \to e^{-2 i a} \chi$$ could certainly be a $$U(1)$$ symmetry. However, since the $$\chi$$ field transforms twice as much, the corresponding conserved charge would count the number of $$\psi$$ particles plus twice the number of $$\chi$$ particles. So this would not count the total number of particles, but it could be the electric charge if $$\chi$$ has twice the charge of $$\psi$$.
If you want the symmetry whose corresponding conserved quantity is the number of $$\psi$$ particles plus the number of $$\chi$$ particles, then you have to choose $$\psi \to e^{-i a} \psi, \quad \chi \to e^{-i a} \chi.$$ This is what is usually called $$U(1)_V$$.