Understanding derivation of Klein-Gordon equation from Dirac equation

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?

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

$$\gamma^\mu \gamma^\nu \partial^2_{\mu\nu}= \frac 12( \gamma^\mu \gamma^\nu \partial^2_{\mu\nu}+ \gamma^\nu \gamma^\mu \partial^2_{\nu\mu})$$ 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 $$\frac 12( \gamma^\mu \gamma^\nu \partial^2_{\mu\nu}+ \gamma^\nu \gamma^\mu \partial^2_{\mu\nu})= \frac 12( \gamma^\mu \gamma^\nu + \gamma^\nu \gamma^\mu) \partial^2_{\mu\nu}\\ =\eta^{\mu\nu}\partial^2_{\mu\nu}.$$
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.
• I get that step, but not why: $\gamma^{\mu}\gamma^{\nu}\partial_{\mu}\partial_{\nu} = \frac{1}{2} \{\gamma^{\mu},\gamma^{\nu}\}\partial_{\mu}\partial_{\nu}$ Commented Mar 15 at 13:55
• 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? Commented Mar 15 at 14:01
• 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. Commented Mar 15 at 14:08