I'm looking for comments and references about an idea : gauging the Dirac representation of the Dirac matrices. What kind of field interaction would it give ?

Specifically, the Dirac equation is defined as this (free field, to begin with) : \begin{equation}\tag{1} \gamma^a \; \partial_a \Psi + i \, m \, \Psi = 0. \end{equation} By definition, the gamma matrices obeys the following relation : \begin{equation}\tag{2} \gamma^a \, \gamma^b + \gamma^b \, \gamma^a = 2 \, \eta^{ab}. \end{equation} Any set of 4 matrices which obeys this relation can be used in equation (1) above (usual Dirac representation, Weyl representation, Majorana representation, etc). All representations are related by an unitary transformation : \begin{align}\tag{3} \tilde{\gamma}^a &= U \, \gamma^a U^{\dagger}, \\[12pt] \tilde{\Psi} &= U \, \Psi. \tag{4} \end{align}

Now, suppose that the representation becomes a local symetry of the Dirac equation ; $U \Rightarrow U(x)$. We then need to change the partial derivative : \begin{equation}\tag{5} \partial_a \Rightarrow D_a \equiv \partial_a + i C_a(x), \end{equation} where $C_a(x)$ is a new gauge field.

I did not pursued further that idea by lack of time. But I would like to know if this idea has been explored by someone else (surrely it was already studied before !).

So what it gives ? What kind of interaction gauge field ? Is there any mathematical problem with this ?

EDIT : Just a few more comments :

The Lorentz group acting on the Dirac field is represented by $SL(4, \mathbb{C})$, and its elements aren't all unitary matrices : the rotations are represented by unitary matrices, but not the pure Lorentz transformations.

Gauging the Lorentz group gives gravitation (this is well known and is a part of classical General relativity). Then gauging the $\gamma$ representation will certainly interfere with the gravitation gauge field (veirbein and its spin connection), since some unitary matrices may represent some rotations (but not all unitary matrices !).

I don't think that the group of transformations that are changing the $\gamma$ representation is the same as the Lorentz group (i.e. $SL(4, \mathbb{C})$), but I may be wrong.

What is the full group that is defining the $\gamma$ representations ? Does it really need to be unitary, i.e. $SU(4)$ ? I suspect they are just similarity transformations, so any invertible 4 X 4 matrices may be good, not just unitary matrices.

In other words, is there a transformation of $SL(4, \mathbb{C})$ (from the Lorentz group) that may change the usual Dirac matrices to the Weyl matrices and to the Majorana matrices ?

  • 3
    $\begingroup$ This would be a gauge theory with gauge group the Lorentz group. The Lorentz group isn't compact which means technical complications so it might be that this hasn't been studied. Much work on gauge theory doesn't specify the gauge group, but I think it is the rule to assume that it is compact. $\endgroup$ Jul 3, 2016 at 23:08
  • $\begingroup$ The Rarita-Schwinger field has a fermionic gauge symmetry. Perhaps considering that would be helpful. See for instance: books.google.be/… $\endgroup$
    – Gerben
    Jul 4, 2016 at 0:48
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    $\begingroup$ I'm sorry, but I don't understand your question - gauge symmetries are defined on the level of the Lagrangian, and must be given by transformations of the fields. $\gamma\mapsto U\gamma U^\dagger$ is not a transformation of a field and hence not a prescription for a symmetry. In particular, you cannot gauge this symmetry because $\gamma^a$ does not depend on spacetime to begin with, so replacing $U$ by $U(x)$ does not make any sense. Also, the Lorentz group (or more precisely its universal cover) is $\mathrm{SL}(2,\mathbb{C})$, not $\mathrm{SL}(4,\mathbb{C})$. $\endgroup$
    – ACuriousMind
    Jul 5, 2016 at 12:52
  • $\begingroup$ @ACuriousMind, $SL(4, \mathbb{C})$ IS a representation of the Lorentz group, for the Dirac field. $SL(2, \mathbb{C})$ doesn't act of the Dirac field, which is a 4 components spinor. Also, the $\gamma$ representation is a global symetry of the Dirac lagrangian (changing representation means that you do change the field representation itself, not just the $\gamma$'s). It CAN be made a local symetry, exactly as for the global $U(1)$ phase symetry. So it does make sense. $\endgroup$
    – Cham
    Jul 5, 2016 at 14:51
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    $\begingroup$ You are using non-standard terminology. The Dirac representation is a group homomorphism $\mathrm{SL}(2,\mathbb{C})\to\mathrm{GL}(4,\mathbb{C})$, but $\mathrm{SL}(4,\mathbb{C})$ is not "a representation of the Lorentz group". The transformation $\gamma\mapsto U\gamma U^\dagger$ is not a "symmetry" of the Lagrangian in the formal sense because symmetries must be given by field transformations and $\gamma$ is not a field, it is a constant. It's something you can do to the Lagrangian, but it is not a symmetry, and it does not make sense to speak of "gauging" it. $\endgroup$
    – ACuriousMind
    Jul 5, 2016 at 15:03

2 Answers 2


Expanding on my comment, I think the Rarita Schwinger field (spin 3/2) has exactly the gauge symmetry you want: https://books.google.be/books?id=KFUhAwAAQBAJ&lpg=PA96&ots=vh0WtWM5rg&dq=rarita%20schwinger%20fermionic%20gauge%20symmetry&pg=PA95#v=onepage&q&f=false This gauge symmetry removes the spin 1/2 component of the field so only the spin 3/2 part is left. Now if you did the same gauging to the spin 1/2 field, you would gauge the entire spin 1/2 field away, the object would be made entirely out of nonphysical arbitrary gauge-stuff; I think.


Imposing local gauge symmetry on the Dirac equation produces the electromagnetic field interacting with it.



Before any down voting please see my comments below. The question was not about we whether Dirac's equation can be used to represent a spin 3/2 or higher fermion, though you could interpret it that way, and his example simply added a vector field as the gauge field. There is then no choice and it has to be electromagnetism. Weyl and Majorana fields are also consistent. See Peskin. Btw, the spin 3/2 Rarita Schwinger field I understand has problems, though I am not an expert on it.

If this is totally off the mark, just explain please.

  • 1
    $\begingroup$ Sorry I had to down-vote your answer. The EM gauge field is related to the $U(1)$ group, wich commutes with all the $\gamma$ matrices. It isn't related to the $\gamma$ representations at all. $\endgroup$
    – Cham
    Jul 4, 2016 at 0:42
  • $\begingroup$ Understand. But it seems to introduce that interaction in Diracs equation. See the reference. You still have U(1) from the EM equations, but this gauge symmetry creates the interaction term in Dirac $\endgroup$
    – Bob Bee
    Jul 4, 2016 at 0:47
  • $\begingroup$ Yes, the $U(1)$ local invariance introduces the EM field into the Dirac equation. This is very well known and is the best (i.e. simplest) example of how to introduce gauge fields into physics. But it isn't related to the question. $\endgroup$
    – Cham
    Jul 4, 2016 at 0:50
  • $\begingroup$ No? If you make your C to be A, what is the difference? $\endgroup$
    – Bob Bee
    Jul 4, 2016 at 0:54
  • $\begingroup$ Plus see also the Weyl equation and Majorana fermions derived from Dirac and Lorentz invariance in damtp.cam.ac.uk/user/tong/qft/four.pdf. It is well known also. $\endgroup$
    – Bob Bee
    Jul 4, 2016 at 1:21

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