In the Ashok Das book "Lectures on quantum field theory" , it's written that in page 76 : therefore, the matrix $ S~ \gamma^0~ S^\dagger ~ \gamma^0$ must be proportional to the identity matrix (this can be easily checked by taking a linear combination of the sixteen basis matrices in (2.100) and calculating the commutator with $\gamma^\mu$ ). As a result, we can denote

$$ S~ \gamma^0~ S^\dagger ~ \gamma^0 = b~ 1 $$

Do any one know how this can be made?


Where Dirac equation

$$iγ^ μ ∂ μ − m ψ(x) = 0 .$$

Under a Lorentz transformation $$x^ μ → x ^{′ μ} = Λ^μ_ν x^ ν ,$$

the transformed equation has the form

$$iγ^ μ ∂_μ′ − m ψ ′ (x ′ ) = 0,$$


$$\psi'(x') = \psi'(\Lambda x)= S(\Lambda) \psi(x) $$


$$\psi(x) = S^{-1}(\Lambda) \psi'(x') .$$

  • $\begingroup$ where the Dirac spinor $\psi' = S \psi $, please look Equ. (3.42) in the book .. $\endgroup$ – S.S. Dec 12 '16 at 17:32
  • $\begingroup$ I thought it's a famous book like Peskin, so it's here on this link : dropbox.com/s/v3h25xfaee14nab/… $\endgroup$ – S.S. Dec 12 '16 at 17:48
  • $\begingroup$ The covariance of Dirac equation is a whole section , so it's better to seen in the book, also Equ. (2.100) and (3.42) consist of many equations .. any way i think studying Dirac equation is common in different references, the transformation matrices and so .. $\endgroup$ – S.S. Dec 12 '16 at 17:52
  • $\begingroup$ No problem , i will add my question .. $\endgroup$ – S.S. Dec 12 '16 at 17:53

We need two lemmas:

$$B\mathrm e^A B^{-1}=\mathrm e^{BAB^{-1}}\tag{1}$$ (see here for the proof).

And the fact that the gamma matrices satisfy $$(\gamma^{\mu})^\dagger=\gamma^0\gamma^\mu\gamma^0\tag{2}$$ (the proof is easy; one just considers the cases $\mu=0$ and $\mu=i$ separately).

Here, $S$ is defined as $$ S=\exp\left[-\frac i2 \omega_{\mu\nu}\gamma^{[\mu}\gamma^{\nu]}\right] $$ where $\gamma^\mu$ are the Dirac gamma matrices.

Using $(1)$ together with $(\gamma^0)^2=1$, we can see that $$ \gamma^0 S\gamma^0\overset{(1)}=\exp\left[-\frac i2 \omega_{\mu\nu}\gamma^0\gamma^{[\mu}\gamma^{\nu]}\gamma^0\right] \tag{A} $$

Now, we use $(2)$ together with $(\gamma^0)^2=1$ to conclude that $$ \gamma^0\gamma^{[\mu}\gamma^{\nu]}\gamma^0=\gamma^0\gamma^{[\mu}\gamma^0\gamma^0\gamma^{\nu]}\gamma^0\overset{(2)}=-(\gamma^{[\mu}\gamma^{\nu]})^\dagger\tag{B} $$

Finally, using $(\mathrm A)$ and $(\mathrm B)$, we see that $$ \gamma^0 S\gamma^0=\exp\left[+\frac i2 \omega_{\mu\nu}(\gamma^{[\mu}\gamma^{\nu]})^\dagger\right]=S^\dagger $$ as we wanted to prove.

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  • $\begingroup$ Thanks. You mean here with the brackets $\gamma^{[\mu} \gamma^{\nu]}$ that they are anti commuting matrices ? $\endgroup$ – S.S. Dec 12 '16 at 20:21
  • $\begingroup$ Yes. More precisely, $\gamma^{[\mu}\gamma^{\nu]}\equiv\frac12 [\gamma^\mu,\gamma^\nu]$. $\endgroup$ – AccidentalFourierTransform Dec 12 '16 at 22:20

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