The Dirac adjoint for Dirac spinors is defined as, $$ \bar{u} = u^{\dagger} \gamma^{0} \, . $$ However I have come across this, $$ \overline{\gamma^{\mu}} = \gamma^{\mu} \, , \tag{1} $$ (where $\gamma^{\mu}$ are the $4\times4$ gamma matrices). Naively applying the same rules as for the Dirac spinor clearly does not get us anywhere, $$ \overline{\gamma^{\mu}} = \gamma^{\mu \dagger} \gamma^{0} = \gamma^{0} \gamma^{\mu} \gamma^{0} \gamma^{0} = \gamma^{0} \gamma^{\mu} \neq \gamma^{\mu} \, . $$ So it seems that the Dirac adjoint for a matrix is defined differently, so in trying to figure this out I make the following reasoning, let $A$ be a $4 \times 4$ matrix and $u$ a Dirac spinor so that $Au$ is again a Dirac spinor. Taking the Dirac conjugate (which is defined) gives, $$ \overline{A u} = (A u)^{\dagger} \gamma^{0} = u^{\dagger} A^{\dagger} \gamma^{0} = u^{\dagger} \gamma^{0} \gamma^{0} A^{\dagger} \gamma^{0} = \bar{u} \;\underbrace{\gamma^{0} A^{\dagger} \gamma^{0}}_{ = \bar{A} ? } \, . $$ So my guess is that $\bar{A} = \gamma^{0} A^{\dagger} \gamma^{0}$. If this is the case it is straightforward to show that $\overline{\gamma^{\mu}} = \gamma^{\mu} $.

My question is the following, is the above statement correct? Is it so that the Dirac adjoint is actually only defined for Dirac spinors but it can be sort of extended to $4 \times 4$ matrices as above (allowing one to write $\overline{A u} = \bar{u} \bar{A}$)?

Link where I found eq. (1) (page 93, eq. 3.249)

Link where I found eq. (1) and the claim $ \overline{X} = \gamma^{0} X \gamma^{0} $ which appears to be missing a "$^{\dagger}$"? (page 9, eq. 5.54)

  • 1
    $\begingroup$ There are too many mistakes in the formulation of the question. It will be hard for the answerers to argue and win each one of them. The gamma matrices can't be universally real - if 3 spatial are real, then the timelike 1 is pure imaginary. Moreover, the bar for gamma matrices doens't mean Dirac conjugation but simple complex conjugation. For the complex conjugation, the formula for the complex conjugate matrix is completely different, too. Pretty much every statement meant above is seriously wrong and I gave up answering even though I had 1/3 of it completed already. $\endgroup$ Jan 22, 2014 at 11:31
  • 1
    $\begingroup$ If the bar means complex conjugation then indeed equation (1) would imply all the gamma matrices are real which is clearly not the case. So I thought Dirac conjugation was implied with the barred notation on the gamma matrix. I did find eq (1) in several (reliable looking) sources though (see links)... $\endgroup$ Jan 22, 2014 at 12:38
  • $\begingroup$ The equation 1 is completely wrong. The closest one you could have seen is $\gamma^{\mu\dagger} = \gamma_\mu$, note that it is dagger, Hermitian conjugation, and one index is upper and the other is lower. The complex conjugation of $\gamma$ matrices may be rewritten as a conjugation by $C$ which is a matrix that also includes $\gamma_2$ (a conventional one that has the other reality property than the other two spatial ones) aside from $\gamma_0$. $\endgroup$ Jan 22, 2014 at 14:32

1 Answer 1


In your first reference, page $58$, equation $(3.55)$, there is a personal definition by the author of what it calls "spinor adjoint of a matrix":

$\overline M \stackrel {def}{\equiv} \gamma_0 M^\dagger \gamma_0$

With this definition, as you noticed, you have obviously $\overline {\gamma^\mu}= \gamma^\mu$

The above definition of "spinor adjoint of a matrix" is compatible with the definition of the adjoint $(3.54)$:

$\overline \psi = \psi ^\dagger \gamma_0$

in the following way :

$\overline {M\psi} = (M\psi)^\dagger \gamma_0 = \psi^\dagger M^\dagger \gamma_0 = (\psi^\dagger \gamma_0 ) (\gamma_0 M^\dagger \gamma_0 ) = \overline {\psi}~ \overline {M}$

  • $\begingroup$ Wow, can't believe I missed that. Thank you for pointing it out. I backtraced the definition as my/your last equation. So it is indeed a personal definition by the author. $\endgroup$ Jan 23, 2014 at 7:53
  • 1
    $\begingroup$ I see - such a conjugation of a Dirac matrix isn't a terribly useful symbol then. ;-) $\endgroup$ Jan 24, 2014 at 6:49

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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