# Fermion current and gauge interactions in Lagrangian density

I believe the answer to this is quite simple, perhaps so simple that I cannot find it in any book.

Usually, if there is a gauge symmetry in the theory, we add an interaction term in the covariant derivative, like: $$\begin{equation}D_\mu = \partial_\mu +i \frac{g}{2} V^i _\mu t_i\end{equation}$$ where, in this general case, the t's are the generators of the group corresponding to the symmetry, the V's are the bosons corresponding to each generators, the g is the coupling of the force that corresponds to the symmetry and the '2' is a "convenient normalization factor".

What happens next is to write the Lagrangian term which is usually: $$\begin{equation} \bar \Psi \gamma^\mu D_\mu \Psi\end{equation}$$

I understand that the fermion current $$\bar \Psi \gamma^\mu \Psi$$ comes from the dirac equation (I think? ).

My problem is that sometimes, usually when discussing BSM theories, 'they' start writing $$\Psi^T$$ instead of the $$\bar \Psi$$. For instance, Eq.(3.34), p.31 in this thesis.

I really don't understand when we should choose to write $$\Psi ^T \gamma^\mu \Psi$$ over $$\bar \Psi \gamma^\mu \Psi$$.

• Minor comment to the post (v1): Please consider to mention explicitly author, title, etc. of link, so it is possible to reconstruct link in case of link rot. Jun 7, 2020 at 11:12

The fermion current can be derived using Noether's theorem. Since the Dirac Lagrangian is invariant under the "global" action of the group $$U(1)$$ i.e $$\Psi\longrightarrow e^{iq\theta}\Psi$$, you get (by virtue of Noether's theorem) the conserved quantity

$$J^\mu= \dfrac{\delta \mathcal{L}}{\delta (\partial_\mu\Psi)}\delta \Psi+ \dfrac{\delta \mathcal{L}}{\delta (\partial_\mu\overline{\Psi})}\delta \overline{\Psi}=\text{constant}.$$

Then note that at first order in $$\theta$$ you get $$\delta \Psi=iq\theta\Psi$$ and the second term is zero because the Lagrangian does not depend on $$\partial_\mu\overline{\Psi}$$. So that you get $$J^\mu=q\overline{\Psi}\gamma^\mu\Psi$$, since the "theta parameter" can be reabsorbed in the constant.

For the second question I'm sorry but I don't know the answer since I'm not familiarized with BSM theories.

• Thanks for the answer, I always tend to forget the we need to consider Psi and \bar Psi as independent variables. Jun 9, 2020 at 9:23
• You are welcome C: Jun 9, 2020 at 16:40

Notice that there is a "$$C$$" in this equation, so the actual expression is $$\Psi^T C\ldots$$. Here $$C$$ is the charge conjugation matrix $$C\gamma^\mu C^{-1}= -(\gamma^\mu)^T.$$ The quantity $$\Psi^T C$$ is Van Nieuwenhuizen's "Majorana adjoint" and replaces $$\bar\Psi$$ when working with Majorana fermions. Since the section is about $${\rm SO}(10)$$ I assume that his "$${\bf 16}$$" is Majorana.

• I think you're right, as in most GUTs, the matter multiplet (here the 16) contains both particles and anti-particles, which means that there is Majorana nature to the 16. Thanks for the help :) Jun 9, 2020 at 9:21