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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 $.

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  • $\begingroup$ 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. $\endgroup$
    – Qmechanic
    Jun 7, 2020 at 11:12

2 Answers 2

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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.

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  • $\begingroup$ Thanks for the answer, I always tend to forget the we need to consider Psi and \bar Psi as independent variables. $\endgroup$
    – madcat
    Jun 9, 2020 at 9:23
  • $\begingroup$ You are welcome C: $\endgroup$
    – vin92
    Jun 9, 2020 at 16:40
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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.

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  • $\begingroup$ 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 :) $\endgroup$
    – madcat
    Jun 9, 2020 at 9:21

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