Canonical quantization of bosons During my studies on QFT a fundamental question occurred concerning the canonical quantization. In our course, we mentioned that: 
"The canonical quantization of a field with values  in the complex numbers can lead only commutation relations, as opposed to anticommutation relations."
How can I interpret this statement? Is there any justification? 
 A: The product of two complex numbers is commutative and that's why you can't have anticommutation relations. If you want to have anticommutative fields you need Grassmann numbers that are numbers whose product is defined to be anticommutative.
This is only a problem for classical fields though. Once you impose the canonical commutation/anticommutation relations you are replacing classical fields (i.e. functions of spacetime) with operators on a Hilbert space. When you are working with classical fields, before replacing them with operators, and you need classical fields to be anticommutative (for example because they are fermion fields) you need to use Grassmann numbers. This is mostly used when working with path integrals.
I hope I was clear enough.
A: I think this is actually quite simple. It is due to the fact that the complex numbers are a commutative algebra. Let's illustrate this with a very simple example. If $\phi (x)$ is a complex scalar field that acts on an eigenstate as
$$\phi (x) | \chi \rangle = c | \chi \rangle$$
$$\phi (y) | \chi \rangle = d | \chi \rangle$$
with $c,d \in \mathbb{C}$. Suppose that we believe that complex valued fields gave rise to anti-commutation relations, then we would have to have $cd + dc = 0$ if we wanted the anti-commutator of $\phi (x)$ and $\phi (y)$ to vanish (i.e. the eigenbasis is not spacetime dependent). That equality will not vanish so long as $c,d$ are complex so we use commutation relations instead. That equality would vanish if $c,d$ were Grassmann numbers.
