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:

Halzen Derivation

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?


1 Answer 1


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:

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

  • $\begingroup$ 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? $\endgroup$
    – Newbie
    Commented Nov 10, 2023 at 18:43
  • $\begingroup$ @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. $\endgroup$
    – hft
    Commented Nov 10, 2023 at 18:50
  • $\begingroup$ 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. $\endgroup$
    – Newbie
    Commented Nov 10, 2023 at 18:54
  • $\begingroup$ yes, the derivative passes through the $\gamma$ matrices because they don't depend on spacetime. $\endgroup$ Commented Nov 10, 2023 at 19:32
  • $\begingroup$ Also, the half factor doesn't really have anything to do with the swapping of indices. It's like writing $1$ as $(1+1)/2$. $\endgroup$ Commented Nov 10, 2023 at 19:35

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