Let A(t), $0\leq t\leq$ T be a random process taking on a value of N$\times$ N real matrices, consider the random matrices Q(t) that satisfy the equation
$$\partial_tQ = QA, Q(0) = \hat1$$
then the solution to this equation can be written in terms of the antichronological exponential:
$$Q(t)= \sum_{n}\frac{1}{n!}\int_{0}^{t}d\tau_1...d\tau_2A(\tau_{i_1})...A(\tau_{in}), \tau_{i_1}\leq...\leq\tau_{i_n}$$
or the alternative way:
$$Q(t) = \prod_{\tau=0}^t (1+A(\tau)d\tau)$$
I read about this in a journal paper but I don't know how these two solutions were obtained. Can anyone write more details about how the two solutions were obtained?
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1it looks like a [dyson series](https://en.wikipedia.org/wiki/Dyson_series) – Dabed Nov 22 '19 at 22:14
1 Answers
$\partial_t Q(t) = Q(t)A(t)$ implies $ Q(t+dt)Q(t) = Q(t) A(t) dt$, namely
$$
Q(t+dt) = Q(t)(1+A(t) dt)
$$
Now make two little steps in time, just to see the pattern...
$$
Q(t+dt_1+dt_2) = Q(t)(1+A(t) dt_1 )(1+A(t+dt_1) dt_2 )
$$
where the order of the matrices in the LHS is important because we are dealing with matrices that may not commute.
Therefore, for $t_N=dt_1+...+dt_N$, $t_i=dt_1+...+dt_{i1}$ for $i>0$ and $Q(0) =1$,
$$
Q(t) = \prod_{i=1..N} (1+A(t_i) dt_i )
$$
Imagine now that each little step has the same length given by $dt=t/N$ for a large number of steps $N$. For simplicity take a constant matrix $A$, so that
$$
Q(t) = \prod_{i=1..N} (1+A t/N ) = (1+A\, t/N)^N \rightarrow e^{A t}
$$
This is just the basic idea of why you get an exponential. If $A$ has a time dependence then the generalization of the exponential is the Tordered exponential (timeordered exponential or pathordered exponential, depending on the context).
You can "approximate" the factors $(1+A(t_i) dt_i )$ as
$$
1+A(t_i) dt_i \approx e^{A(t_i) dt_i}
$$
so that
$$
Q(t) = \prod_{i=1..N} (1+A(t_i) dt_i ) = \prod_{i=1..N} e^{A(t_i) dt_i}
$$
which is exactly the discrete definition of time ordered exponential since $t_1<t_2<t_3...$
This is only a sketch, but gives the basic idea. When expressed as a series and in the continuum limit it is possible to prove your integral formula for the chronological exponential, see https://en.wikipedia.org/wiki/Ordered_exponential
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