The Dirac $\gamma$ matrices follow the Clifford algebra: \begin{equation*} \{ \gamma^\mu,\gamma^\nu\} = 2\eta^{\mu \nu} \end{equation*}

The traditional construction of the Dirac Lagrangian involve choosing a basis of the $\gamma$ matrices where $\gamma^\mu$ is either Hermitian or antihermitian depending on the metric, and proceeding to define the Dirac adjoint $\bar{\psi} = \psi^\dagger \gamma^0$. The textbooks then proceed to show that $\bar{\psi}\psi$ is a Lorentz scalar, etc. and construct the Dirac Lagrangian with the 'kinetic' term and the mass term.

Is it possible to construct the Dirac Lagrangian WITHOUT using the hermiticity of the $\gamma$ matrices? As far as I can see, the $S[\Lambda]$ matrix defined by the non-hermitian $\gamma$ matrices is a valid representation of the Lorentz group. So why do the textbooks resort to the hermiticity of the $\gamma$ matrices to construct the Lagrangian?

\begin{equation*} \mathcal{L} = i\bar{\psi}\displaystyle{\not}\partial\psi -m \bar{\psi}\psi \end{equation*}

If we construct the Dirac Lagrangian without assuming the hermtiticty of these matrices, how do we now define the Dirac adjoint?

  • $\begingroup$ Hermiticity is not a basis-dependent property of a matrix, no? $\endgroup$
    – Charlie
    Jul 30 at 19:31
  • $\begingroup$ @Charlie I guess I was rather cavalier in stating that we are using a 'basis' of the $\gamma$ matrices... a 'representation' would be a better word to use $\endgroup$
    – Harsha
    Jul 30 at 19:47

The answer is no. The gamma matrices are defined by the Clifford algebra $$\{\gamma^{\mu},\gamma^{\nu}\}=2\eta^{\mu\nu}$$

Let's say we are working in the signature with $\eta^{\mu\nu}=\text{diag}(-1,+1,+1,+1)$. We have $$(\gamma_0)^2=-1,\quad(\gamma_i)^2=1$$ These equations say that $\gamma_0$ has eigenvalues $\pm i$ (hence anti-Hermitian), and the $\gamma_i$'s have eigenvalues $\pm1$ (hence Hermitian).

I see what you are saying about the generators $\frac{1}{4}[\gamma^{\mu},\gamma^{\nu}]$ being the important object for defining the representation, and that there may be room for choosing $\gamma^{\mu}$ which leaves the commutator invariant. But the commutator works as a generator of the Lorentz group as a consequence of the Clifford algebra.


What I wrote above is incorrect. In fact, one can find solutions to the Clifford algebra which are not Hermitian. Here's a set of solutions I found of the $2d$ Clifford algebra


You can probably play with the $4d$ case to find examples but it escapes me at the moment. I think a better answer as to why we don't use gamma matrices like this is they result in generators which are neither Hermitian nor anti-Hermitian. Typically it is much harder to make a theory unitary if the representations we use of whatever symmetry group are not unitary representations. Now, the Lorentz representations we use are not unitary, but the generators are either Hermitian or anti-Hermitian (rotation and boost generators respectively), and there is a procedure colloquially known as Wick-Rotation which enables us to define a unitary theory in this case.

  • $\begingroup$ Why does $\gamma^i$ having an eigenvalue of +1 imply that it is Hermitian? All it says is that it has positive eigenvalues. i agree that a Hermitian matrix has real eigenvalues, but just because a matrix has real eigenvalues it doesn't need to be Hermitian. $\endgroup$
    – Harsha
    Jul 30 at 19:49
  • $\begingroup$ Yes you're right, that slipped my mind. Let me think for a bit and I will edit my answer. $\endgroup$
    – fewfew4
    Jul 30 at 20:19
  • $\begingroup$ @Harsha I made an edit to correct my answer. $\endgroup$
    – fewfew4
    Jul 30 at 21:00
  • $\begingroup$ Thanks for your comment. So I guess the final conclusion is that it is not necessary to have hermitian $\gamma$ matrices, but just that its too much work for zero benefit? $\endgroup$
    – Harsha
    Jul 30 at 21:32
  • 1
    $\begingroup$ @Harsha I think it is worse than 'too much work'. Representations of this sort, neither Hermitian nor anti-Hermitian generators, don't have an obvious way to make consistent. $\endgroup$
    – fewfew4
    Jul 31 at 3:10

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.