I am currently taking a class on Quantum Field Theory. The propagator was defined as:
$$K(x,t;x',t') = \langle x|\hat{T}e^{{\frac{-i}{\hbar}\int_{t}^{t'}dtH(t)}}|x\rangle$$
where, $\hat{T}$ is the time ordering operator. I fully well agree with this so far. However, we next considered '$n$' time slices (which is later considered to approach infinity) and equated:
$$\hat{T}e^{{\frac{-i}{\hbar}\int_{t}^{t'}dtH(t)}} = e^{{\frac{-i}{\hbar}\int_{t_{n-1}}^{t'}dtH(t)}} e^{{\frac{-i}{\hbar}\int_{t_{n-2}}^{t_{n-1}}dtH(t)}}... e^{{\frac{-i}{\hbar}\int_{t}^{t_1}dtH(t)}} \tag1$$
The argument given for this was, "Due to time ordering we now have all operators at later times to the left of earlier times". I do not fully agree with this because I believe in order to use the identity $e^{A+B} = e^Ae^B$, we need to have $[A,B]=0$. Here, our Hamiltonian does not necessarily commute with itself at different times.
However, texts I have looked up, seem to fully ignore time-dependent Hamiltonians and instead use time-independent Hamiltonians.
So my question is: How do I properly deal with the derivation in the case of a time-dependent Hamiltonian? Is eq(1) after all correct? What Am I missing?
