Time derivative of path ordered monodromy matrix

Question

I am currently gettting familiar with integrably systems and came the following statement in my literature: $U=U(x,\lambda,t)$ some matrix (Lax component) we define $$T(\lambda,t) = \mathcal{P} \exp \int_0^{2\pi} \mathrm{d}x U\tag{1}$$
then $$\partial_t T(\lambda,t)=\int_0^{2\pi} \mathrm{d}x \mathcal{P} e^{\int_x^{2\pi} \mathrm{d}y U} (\partial_t U) \mathcal{P} e^{\int_0^{x} \mathrm{d}y U}.\tag{2}$$

$\mathcal{P} $ denotes path ordering.
I now spent quite some time on it but simply can't get it. It looks sensible to me and I could somehow derive something similar but I would like to have some rigrous derivation - so some help would be greatly appreciated!

Answer 1

For rigor, you might try to ask on Math.SE or MO.SE. In this answer, we will give a heuristic derivation via discretization. We will use a slightly different notation to connect to usual time-evolution in QM, but the idea is the same:

$$\begin{align} U_{\lambda}(t_f,t_i)~=~&T e^{-\frac{i}{\hbar}\int_{t_i}^{t_f}\! dt~ H_{\lambda}(t)}\cr ~=~&\lim_{N\to\infty} T\prod_{n=1}^N e^{h_{\lambda}(n)} \cr ~=~&\lim_{N\to\infty} e^{h_{\lambda}(N)}e^{h_{\lambda}(N-1)} \ldots e^{h_{\lambda}(2)}e^{h_{\lambda}(1)} , \cr h_{\lambda}(n) ~:=~&-\frac{i}{\hbar}\frac{\Delta t}{N}H_{\lambda}\left(t_i+ (n-\frac{1}{2})\frac{\Delta t}{N}\right), \cr \Delta t ~:=~&t_f-t_i. \tag{1}\end{align}$$

Here $\lambda$ is an external parameter. Therefore, if we are allowed to change the order of differentiation and limit, we formally get OP's formula $$\begin{align} &\frac{dU_{\lambda}(t_f,t_i)}{d\lambda}\cr ~=~&\lim_{N\to\infty}\sum_{n=1}^N e^{h_{\lambda}(N)}\ldots e^{h_{\lambda}(n+1)} \frac{de^{h_{\lambda}(n)}}{d\lambda} e^{h_{\lambda}(n-1)}\ldots e^{h_{\lambda}(1)}\cr ~=~&\lim_{N\to\infty}\sum_{n=1}^N e^{h_{\lambda}(N)}\ldots e^{h_{\lambda}(n+1)} \frac{dh_{\lambda}(n)}{d\lambda} e^{h_{\lambda}(n-1)}\ldots e^{h_{\lambda}(1)}\cr ~=~&-\frac{i}{\hbar}\int_{t_i}^{t_f}\! dt~ U_{\lambda}(t_f,t)\frac{dH_{\lambda}(t)}{d\lambda}U_{\lambda}(t,t_i).\tag{2}\end{align}$$

Answer 2

$T$ is the ordered product of many infinitesimal factors of $$ V(x,t) = \exp\{\epsilon U(x,t)\} \approx 1+ \epsilon U(x,t). $$ Apply Liebniz' product rule to differentiate. If you insist on a rigorous derivation then some properties of the operator $U(x,t)$ need to be adduced.