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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 $a$.

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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 $U(1)_V$.

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  • $\begingroup$ Ok, I got it. I can choose the transformation as in my Eq. (2) because it is not dependent on the particle I transform, but in the symmetry I want to prove. Is this the way to understand it? $\endgroup$ – Vicky Dec 4 '18 at 8:19

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