In a complex (non-hermitian) scalar QFT, is it correct that the creation/annihilation operators $a,a^\dagger$ (particle) and $b,b^\dagger$ (anti-particle) commute, i.e. $[a,b] = [a,b^\dagger] = [a^\dagger,b] = [a^\dagger,b^\dagger] = 0$?
More generally asked, do different creation/annihilation operators like $a,b$ always commute, or are there situation where one has to be careful?
If this requires more context, it's the complex scalar field from Zee's QFT in a Nutshell book on page 65:
$$
\varphi(\vec{x},t) = \int\frac{d^Dx}{\sqrt{(2\pi)^D2\omega_k}}\left[a(\vec{k})\mathrm{e}^{-i(\omega_kt-\vec{k}\cdot\vec{x})} + b^\dagger(\vec{k})\mathrm{e}^{i(\omega_kt-\vec{k}\cdot\vec{x})}\right]
$$
The physical meaning of the field $\varphi$ is that $a$ annihilates a particle while $b^\dagger$ creates an antiparticle.
Best Answer
If both operators are associated with fermions they'll anticommute instead, but otherwise yes.