I'm learning about how the Dirac equation came about.

In pg. 20 of the Lectures on Quantum Field Theory book by Ashok Das, pg.20, the author starts with the energy momentum relation

$$p^\mu p_\mu=m^2$$ in matrix form and then defines the matrix square root of $p_\mu p^\mu $ as $$\overline{p}=\gamma^\mu p_\mu$$ where $\overline{p}^2=p_\mu p^\mu$. It was also stated that $\gamma^\mu$ are four linearly independent matrices.

He then wrote the following relationship for the $\gamma^\mu$ matrices in eqn. 1.78: $$\gamma^\mu\gamma^\nu={1\over 2}(\gamma^\mu \gamma^\nu+\gamma^\nu\gamma^\mu).$$

How did he arrive at this relationship? In general, linearly independent matrices don't commute, so $\gamma^\mu \gamma^\nu \neq \gamma^\nu \gamma^\mu$.

I'm think this relationship holds for Dirac matrices, but the author uses this relationship to later on derive the Dirac matrices. Isn't that a circular argument?


I assume that you've made this conclusion from the derivations in equation 1.78 in the cited book, which proceed as follows

$$\gamma^\mu \gamma^\nu p_\mu p_\nu = p^2 \mathbf{1} \Rightarrow \\ \Rightarrow \frac{1}{2}(\gamma^\mu \gamma^\nu + \gamma^\nu \gamma^\mu) p_\mu p_\nu = p^2 \mathbf{1}$$

This does not imply that $\gamma^\mu \gamma^\nu = \frac{1}{2}(\gamma^\mu \gamma^\nu + \gamma^\nu \gamma^\mu)$ (which, as you've already noticed, would in turn imply $\gamma^\mu \gamma^\nu = \gamma^\nu \gamma^\mu$, which is false).

Instead, what the author does is take the identity in the first row with indices swapped $\nu \leftrightarrow \mu$, which reads

$$\gamma^\nu \gamma^\mu p_\nu p_\mu = p^2 \mathbf{1}$$

and adds it to the original identity:

$$\gamma^\mu \gamma^\nu p_\mu p_\nu + \gamma^\nu \gamma^\mu p_\nu p_\mu = 2p^2 \mathbf{1}$$

Now, use the fact that $p_\mu$ and $p_\nu$ commute:

$$(\gamma^\mu \gamma^\nu + \gamma^\nu \gamma^\mu) p_\mu p_\nu = 2p^2 \mathbf{1}$$ $$\frac{1}{2}(\gamma^\mu \gamma^\nu + \gamma^\nu \gamma^\mu) p_\mu p_\nu = p^2 \mathbf{1}$$

  • $\begingroup$ Thank a lot! I'll accept this answer when I can. $\endgroup$
    – TaeNyFan
    Dec 16 '20 at 7:47
  • $\begingroup$ @TaeNyFan Glad to be of help! Good luck on your QFT path. $\endgroup$
    – lisyarus
    Dec 16 '20 at 8:04

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