In quantum field theory, the time ordering operator (TOO) appears in the formal expressions for the scattering amplitudes. It acts upon a product of operators that each depends on time, and returns the product of these operators sorted by time.
But, there is a problem with this definition. Since if the argument of the TOO is a Hilbert space operator (HSO), then the HSO itself "has no memory" of the time of which its terms were taken. In more mathematical words, if $T$ is the TOO and $A(t_1), B(t_2), C(t_3), D(t_4)$ are four different HSOs at times $t_1 < t_2 < t_3 < t_4$, and assume that $B(t_2)*A(t_1) = C(t_3)*D(t_4) = E$, and also that $D(t_4)*C(t_3) = F$ which is different from $E$. Then:
$$T\left\lbrace B(t_2)*A(t_1)\right\rbrace = B(t_2)*A(t_1) = E \Rightarrow T\left\lbrace E\right\rbrace = E $$
But on the other hand:
$$T\left\lbrace C(t_3)*D(t_4)\right\rbrace = D(t_4)*C(t_3) = F \Rightarrow T\left\lbrace E\right\rbrace = F $$
So there is a contradiction.
This can't be solved by making $T$ be a function not of a single HSO, but of $n$ HSOs and $n\!-\!1$ times. This is because then you won't be able to do any algebraic manipulations inside the argument of T (like performing the products of the HSOs) as is done in QFT, since each HSO would be in a different "slot" of the function $T$.
So my questions are: is this really a problem? is there a more rigorous definition of the TOO?