Product of time ordered products (exponentials) I have a simple question regarding a property used in QFT. The operator $U_{21}$ defined by 
$$
U_{21} = T\left\{\exp\left(-i\int_{t_1}^{t_2} dt' V(t')\right)\right\}
\tag{1}$$
satisfies $$U_{32}U_{21}=U_{31}\tag{2}$$ (where the $T$ denotes time ordering). My question is: how do we carry out the product of the $U$'s? Can we say that $$T(A)T(B)=T(AB)~?\tag{3}$$ (This seems plausible to me.)
But my problem is that to show the above identity, one must deal with things like $e^{A+B}$, which isn't necessarily $e^Ae^B$. Does the non-commutative part of $e^{A+B}$ (the part that makes it differ from $e^Ae^B$) get killed by the time-ordering? 
 A: It's not generically true that $T(A)T(B)=T(AB)$ for any operators $A$ and $B$. However, it IS true if the operators in $A$ all occur at later times than the operators in $B$. This is trivially true, because then $T(A)T(B)$ is already time-ordered in total (if you expand $A$ and $B$ into power series and multiply them term-by-term, each term will be time-ordered).
In your case, $U_{32}$ only involves the operator $V(t')$ for times $t_2<t'<t_3$, while $U_{21}$ involves the operator $V(t')$ for times $t'<t_2$, so the product $U_{32}U_{21}$ is thus time ordered, and equal to $U_{31}$
A: No, eq. (3) is not true. Think what happens if the operator $B$ is at a later time than the operator $A$.


*

*The time-ordered exponentiated Hamiltonian is formally defined as
$$ U(t_2,t_1)~=~
T\exp\left[-\frac{i}{\hbar}\int_{t_1}^{t_2}\! dt~H(t)\right] $$
$$~=~\lim_{N\to\infty}
\exp\left[-\frac{i}{\hbar}H(t_2)\frac{t_2-t_1}{N}\right] \cdots\exp\left[-\frac{i}{\hbar}H(t_1)\frac{t_2-t_1}{N}\right] $$
for $t_1 <t_2$. 

*However, mathematically it is better characterized as the solution 
$$ \Psi(t_2) ~=~ U(t_2,t_1) \Psi(t_1)$$
to the time-dependent Schrödinger equation (TDSE), cf. this Phys.SE post. 

*The group-property
$$U(t_3,t_1)~=~U(t_3,t_2)U(t_2,t_1)$$ 
is manifestly satisfied.
