The time ordering operator is usually defined as $$\mathcal{T} \left\{A(\tau) B(\tau')\right\} := \begin{cases} A(\tau) B(\tau') & \text{if } \tau > \tau', \\ \pm B(\tau')A(\tau) & \text{if } \tau < \tau'. \end{cases}$$ The minus sign applies when $A$ and $B$ are fermion operators. My question is now: why appears there a minus sign? I think the answer lies somewhere in the argument, that one can write the operators in second quantization and use the canonical (anti) commutation relations for the fermionic creation and annihilation operators: $$\{\hat{c}_\nu,\hat{c}_{\mu}^\dagger\} = \delta_{\mu \nu}\\ \{\hat{c}_\nu,\hat{c}_{\mu}\} =0 \\ \{\hat{c}^\dagger_\nu,\hat{c}_{\mu}^\dagger\} = 0.$$ Can someone explain this more explicitly? I really don't see how this only changes the sign and does not introduce any constants coming from the $\delta_{\mu \nu}$ term which may appear.

Perhaps the easiest way to see that there should be a Grassmann sign factor $(-1)^{|A| |B|}$ in the definition of time ordering

$$\tag{1} {\cal T} \left\{ A(t_A) B(t_B)\right\} ~:=~ \theta(t_A-t_B) A(t_A) B(t_B) + (-1)^{|A| |B|} \theta(t_B-t_A) B(t_B) A(t_A), \qquad$$

is to go to the classical limit $\hbar\to 0$. Here $|A|$ denotes the Grassmann parity, which is $0~{\rm mod}~2$ if $A$ is a boson, and $1~{\rm mod}~2$ if $A$ is a fermion. Moreover, $\theta$ is the Heaviside step function. In the classical limit, all fields should supercommute, which mean that the supercommutator

$$\tag{2} [A,B]~:=~ AB - (-1)^{|A| |B|}BA~=~0 \qquad\leftarrow \text{classically}\qquad$$

should vanish. This follows from the correspondence principle between QM and classical mechanics: $$\begin{array}{ccc} \text{Operator} &\longleftrightarrow& \text{Symbol/Super-function}\cr A& \longleftrightarrow& a \cr B& \longleftrightarrow& b \cr [A,B]& \longleftrightarrow&i\hbar \{a,b\}_{PB}+ {\cal O}(\hbar^2)\cr \text{Supercommutator}&\longleftrightarrow& \text{Super-Poisson bracket.}\end{array} \tag{3}$$ In particular, time-ordering should not matter in the classical limit $\hbar\to 0$:

$$\tag{4} {\cal T} \left\{ A(t_A) B(t_B)\right\}~=~A(t_A) B(t_B)~=~(-1)^{|A| |B|}B(t_B)A(t_A).\qquad\leftarrow \text{classically}\qquad$$

But this will only be the case if we include the Grassmann sign factor $(-1)^{|A| |B|}$ in the definition (1).