# Dirac Equation and the Klein-Gordon Equation

I am trying to solve an exercise in Halzen and Martin's Quarks and Leptons book and got stuck on doing some math. The Dirac equation reads $$i \gamma^{\mu} \partial_{\mu} \psi - m\psi = 0.$$ Now, I want to act $$\gamma^{\nu}\partial_{\nu}$$ on both sides. What I have done was:

$$\gamma^{\nu}\partial_{\nu}$$ $$\left(i \gamma^{\mu} \partial_{\mu} \psi - m\psi \right)$$ = 0

Apply linearity and the product rule to get

$$\gamma^{\nu} \partial_{\nu} \left(i \gamma^{\mu} \partial_{\mu} \psi\right) - \gamma^{\nu} \partial_{\nu} (m\psi) = 0$$

$$i\gamma^{\nu} \gamma^{\mu} \partial_{\nu}\left(\partial_{\mu} \psi\right) + i\gamma^{\nu}\partial_{\nu}(\gamma^{\mu})\partial_{\mu}\psi - m\gamma^{\nu} \partial_{\nu} \psi = 0$$

I am stuck on simplifying the first two terms. As in the key, they have written it the following fashion:

I am puzzled as to how the quantity in red box was derived. I do understand the process after this though, where they made use of the anti-commutator relation $$\left\{ \gamma^{\mu}, \gamma^{\nu}\right\} = 2g^{\mu \nu}$$. Can someome walk me through how to get from the last line of my work to the red box?

It's simple if you think about how you change the summation variables.

$$\gamma^\nu \gamma^\mu\partial_\mu\partial_\nu=\frac{1}{2}(\gamma^\nu \gamma^\mu\partial_\mu\partial_\nu+\gamma^\nu \gamma^\mu\partial_\mu\partial_\nu)$$

This is straightforward I didn't do anything fancy here. Now what you can do is in the second term relabel the variables like this $$\mu \rightarrow \nu$$ and $$\nu\rightarrow \mu$$ and then use the fact that partial derivatives commute, so finally one gets:

$$\gamma^\nu \gamma^\mu\partial_\mu\partial_\nu=\frac{1}{2}(\gamma^\nu \gamma^\mu+\gamma^\mu \gamma^\nu)\partial_\mu\partial_\nu$$

• Thank you for your response. I haven't done any classes on tensors thats why I am having difficulty. Is there a rationale behind why swapping indices introduces a factor of 1/2? Commented Nov 10, 2023 at 18:43
• @Newbie Summed variables are called "dummy variables" and the symbol you use for them does not matter, since it is not really a free symbol. For example $\sum_{i=1}^3 x_i = x_1 + x_2 + x_3$, but I can also write this as $\sum_{j=1}^3 x_j = x_1 +x_2 + x_3$ or $\sum_{\alpha=1}^3x_\alpha = x_1+x_2+x_3$ or $\sum_{\xi=1}^3x_\xi = x_1+x_2+x_3$. The symbol $i$ or $j$ or whatever symbol you choose is a "dummy variable" since it is not really there, it is just there to be set equal to the index of each term in the sum.
– hft
Commented Nov 10, 2023 at 18:50
• Hmmm. That makes a lot of sense. And another question, does it mean that I can readily factor out $\gamma^{\mu}$ and treat it as a constant, since i am taking derivative with respect to $\nu$? i am referring to the line where I applied linearity. Commented Nov 10, 2023 at 18:54
• yes, the derivative passes through the $\gamma$ matrices because they don't depend on spacetime. Commented Nov 10, 2023 at 19:32
• Also, the half factor doesn't really have anything to do with the swapping of indices. It's like writing $1$ as $(1+1)/2$. Commented Nov 10, 2023 at 19:35