Dirac differential operator and anti-commutation relations

Question

I was studying Dirac's equation and my book says the following. Let $D = i gamma^{mu}partial_{mu}$, where $partial_{0} = partial/partial t$ and $partial_{mu} = partial / partial x_{mu}$ if $mu = 1,2,3$. Here, $gamma^{mu}$ are coefficients to be determined. Then:
$$D^{2} = frac{1}{2}{gamma^{mu},gamma^{nu}}partial_{mu}partial_{nu}$$
But why is that? I thought this would be (I'm expliciting the summations so I don't get lost):
$$D = isum_{mu=0}^{4}gamma^{mu}partial_{mu} Rightarrow D^{2} = -sum_{mu,nu=0}^{4}gamma^{mu}gamma^{nu}partial_{mu}partial_{nu}$$
Why is that wrong?

张嘉宝 · Answer

It is confusing, if we want to get that answer, we need $(-igamma^mupartial_mu)(igamma^nupartial_nu)$ since $gamma^mu$ are not Hermitian, we can't  use  $D^2=D^dagger D$.
if we use $D^2=DD$, we will get an extra minus sign.
Any way, I think the idea is show that Dirac equation implies K-G equation.
$D^2=D D=-gamma^mupartial_mugamma^nupartial_nu=-gamma^mugamma^nupartial_mupartial_nu=-gamma^nugamma^mupartial_nupartial_mu=-gamma^nugamma^mupartial_mupartial_nu$
$2D^2=-(gamma^mugamma^nu+gamma^nugamma^mu)partial_mupartial_nu=-{gamma^mu,gamma^nu}partial_mupartial_nu$
$D^2=-frac{1}{2}{gamma^mu,gamma^nu}partial_mupartial_nu$

J.G. · Answer

Since $partial_mupartial_nu=partial_nupartial_mu$, $[gamma^mu,,gamma^nu]partial_mupartial_nu=0$, so $gamma^mugamma^nupartial_mupartial_nu=tfrac12{gamma^mu,,gamma^nu}partial_mupartial_nu$.

Dirac differential operator and anti-commutation relations

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