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Above is Tong's notes which shows how the Klein-Gordon equation is derived from Dirac equation. But I don't get why:

$\gamma^{\mu}\gamma^{\nu}\partial_{\mu}\partial_{\nu} = \frac{1}{2} \{\gamma^{\mu},\gamma^{\nu}\}\partial_{\mu}\partial_{\nu}$

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

But $\gamma^{\mu}\gamma^{\nu} \neq \gamma^{\nu}\gamma^{\mu}$ right?

$\gamma$ commutes with $\partial$, but $\gamma^{\mu}$ does not commute with $\gamma^{\nu}$ due to the relation:

$\{\gamma^{\mu},\gamma^{\nu}\} = 2\eta^{\mu \nu}1$

So how can the LHS be equal to the RHS?

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    $\begingroup$ $\partial_\mu \partial_\nu = \partial_\nu \partial_\mu$ $\endgroup$ Commented Mar 15 at 13:28
  • $\begingroup$ Yeah but $\gamma^{\mu}\gamma^{\nu} \neq \gamma^{\nu}\gamma^{\mu}$ $\endgroup$
    – Stallmp
    Commented Mar 15 at 13:38
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    $\begingroup$ You mustn't forget that you are contracting the gamma matrices over the derivative operators. $\endgroup$ Commented Mar 15 at 13:45

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$$ \gamma^\mu \gamma^\nu \partial^2_{\mu\nu}= \frac 12( \gamma^\mu \gamma^\nu \partial^2_{\mu\nu}+ \gamma^\nu \gamma^\mu \partial^2_{\nu\mu}) $$ where in the second term we have renamed the dummy indices $\mu\leftrightarrow \nu$. As $\partial^2_{\mu\nu}= \partial^2_{\nu\mu}$ we have $$ \frac 12( \gamma^\mu \gamma^\nu \partial^2_{\mu\nu}+ \gamma^\nu \gamma^\mu \partial^2_{\mu\nu})= \frac 12( \gamma^\mu \gamma^\nu + \gamma^\nu \gamma^\mu) \partial^2_{\mu\nu}\\ =\eta^{\mu\nu}\partial^2_{\mu\nu}. $$

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    $\begingroup$ Yes, exploit the fact that the order of differentiation can be changed. Good post. $\endgroup$ Commented Mar 15 at 14:39
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As you note in your post, $\{\gamma^\mu,\gamma^\nu\}=2\eta^{\mu\nu}1$. Thus, we do not require that the gamma commute, in fact we are making use of the fact that they are anti-commutative. The anti-commutative relation is all we need to make the argument in Tong's notes. So, we have that: $$\gamma^\mu\gamma^\nu\partial_\mu\partial_\nu=1\eta^{\mu\nu}\partial_\mu\partial_\nu=1\partial_\mu\partial^\mu.$$ Sticking this back in the Dirac equation gives: $$-(1\partial_\mu\partial^\mu+m^2)\psi=0$$ $$-(1\partial_\mu\partial^\nu\psi+m^2\psi)=-(\partial_\mu\partial^\nu\psi+m^2\psi)=0.$$ The last relation contains two independent equations, the Klein-Gordon equation for the particle and anti-particle respectively.

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  • $\begingroup$ I get that step, but not why: $\gamma^{\mu}\gamma^{\nu}\partial_{\mu}\partial_{\nu} = \frac{1}{2} \{\gamma^{\mu},\gamma^{\nu}\}\partial_{\mu}\partial_{\nu}$ $\endgroup$
    – Stallmp
    Commented Mar 15 at 13:55
  • $\begingroup$ Oh, you are saying you want to know why the gamma are anti-commutative? $\endgroup$ Commented Mar 15 at 14:00
  • $\begingroup$ I do know why the gamma matrices are anti-commutative, but $\gamma^{\mu}\gamma^{\nu}\partial_{\mu}\partial_{\nu} = \frac{1}{2} \{\gamma^{\mu},\gamma^{\nu}\}\partial_{\mu}\partial_{\nu}$ actually implies that they commute right? Because this equality seems to imply that $\gamma^{\mu} \gamma^{\nu} = \gamma^{\nu} \gamma^{\mu}$ or not? $\endgroup$
    – Stallmp
    Commented Mar 15 at 14:01
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    $\begingroup$ No, that relation implies that $\gamma^\mu\gamma^\nu\partial_\mu\partial_\nu=\gamma^\nu\gamma^\mu\partial_\mu\partial_\nu$, introducing the contraction over the derivatives is what changes things. $\endgroup$ Commented Mar 15 at 14:08
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    $\begingroup$ Well, if you try doing the brute force calculation, contract over the derivatives and use the fact that the order of differentiation doesn't matter, then you will see that the result spills out. If I get some time later I will try to sketch that out explicitly, I must go for now though. Hope that helps. $\endgroup$ Commented Mar 15 at 14:35

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