Why is commutation bracket used instead of anti-commutation bracket on page 61 of Peskin QFT?

Question

Peskin&Schroeder was performing a trick where they used $$J_za^{s\dagger}_0|0\rangle=[J_z,a^{s\dagger}_0]|0\rangle\tag{p.61}$$ and claimed that the only non-zero term in this commutator would be terms having $a_0$ at the very right in $J_z$.

Now I am convinced that $J|0\rangle=0$ from a physical consideration. However I am not so sure about using commutator here, afterall you could have $$\{J_z,a^{s\dagger}_0\}|0\rangle$$ as well and then it makes sense that non-zero terms would involve $a_0$ in $J_z$. I find it hard to believe we should use commutator here given that we already knew the right way to quantise Dirac’s field is through anti-commutation and we already have $$\{a_{\vec p}^s,a^{t\dagger}_{\vec q}\}=(2\pi)^3\delta^3({\vec p}-{\vec q})\delta^{st}.\tag{3.97}$$

Answer 1

1. Long story short: One should use the super-commutator $$ [A,B]_{SC}~:=~ AB-(-1)^{|A|~|B|}BA. \tag{1}$$

2. It satisfies a non-commutative graded Leibniz rule $$ [AB,C]_{SC}~=~A[B,C]_{SC}+[A,C]_{SC}B(-1)^{|B|~|C|} \tag{2}$$ that we can use to reduce expressions with the help of fundamental CCRs and CARs.

3. Since $J_z$ is Grassmann-even and $a^{s\dagger}_0$ is Grassmann-odd, their super-commutator becomes a commutator.