# Covariance of the Dirac equation

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?

Edit

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

$$iγ^ μ ∂_μ′ − m ψ ′ (x ′ ) = 0,$$

where

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

and

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

• where the Dirac spinor $\psi' = S \psi$, please look Equ. (3.42) in the book .. – S.S. Dec 12 '16 at 17:32
• I thought it's a famous book like Peskin, so it's here on this link : dropbox.com/s/v3h25xfaee14nab/… – S.S. Dec 12 '16 at 17:48
• 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 .. – S.S. Dec 12 '16 at 17:52
• No problem , i will add my question .. – 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.

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