Time-ordering of fermion operators

If $A$ and $B$ are fermion operators then the time ordering is defined as \begin{eqnarray} T(AB) = \left\{ \begin{array}{rl} AB, & \mbox{if $B$ precedes $A$}\\ -BA, & \mbox{if $A$ precedes $B$}\end{array}\right. \hspace{2in} (1) \end{eqnarray} On the other hand, the time-ordering operator that arises in the solution of $$i \partial_t U(t,t_0) = H_{_I}(t) U(t,t_0), \hspace{3.1in}$$ as $$U(t,t_0) = T \left[e^{-i \int_{t_0}^t H_{_I}(\tau) d\tau}\right] \hspace{3in} (2)$$ is the usual (bosonic) time-ordering operator even if $H_{_I}$ has fermion fields; we cannot use the definition of time-ordering shown in (1) in the derivation of (2) because introducing a negative sign when the order of fermion operators is flipped messes up the combinatorial counting and will not yield the exponential in (2). (In (2), $U(t,t_0)$ is the unitary evolution operator in the interaction picture and $H_{_I}$ the interaction Hamiltonian.)

So, isn't the definition of time ordering shown in (1) inconsistent with the definition used in (2)? What am I missing?

It's because always even number of Fermionic fields appear in the Hamiltonian. for example the Dirac Lagrangian for free electron: $\mathcal{L}=i\hbar\bar \psi(\gamma^\mu\partial_\mu-m)\psi$ has two $\psi$s. Invariance of the theory upon a global gauge transformation requires that each term in the Lagrangian have an even number of Fermionic fields. Since an even number of Fermionic Fields commute inside the time ordering operator, there isn't an extra minus sign.
• Good point, I was thinking of the operation of particle fields on vacuum, for example $|p> \approx \psi^\dagger|0>$. then a phase shift of $|p>$ state will results in gauge transformation on $\psi$ field. Maybe the relation is only for particle fields, I'm not sure anymore. – seyed May 3 '16 at 15:01