This question is about series expansion of an unitary operator . Suppose we have an unitary operator $U(t,t_{o}) = exp(- i \int^{t}_{t_{0}}H_{I}(t') dt')$, where $H_{I}$ is the interacting Hamiltonian. Then, we expand $U(t,t_{0})$ into series
\begin{align} U(t,t_{0}) &= exp(- i \int^{t}_{t_{0}}H_{I}(t') dt')\\ &=1 - i\int^{t}_{t_{0}}H_{I}(t') dt' + \frac{(-i)^{2}}{2}( \int^{t}_{t_{0}}H_{I}(t') dt')^{2} + ...... \end{align}
Then, we try to differentiate this with respect to t, and quadratic term gives us
\begin{align} \frac{-1}{2} \big(\int^{t}_{t_{0}}H_{I}(t') dt' \big)H_{I}(t) -\frac{1}{2} H_{I}(t) \big(\int^{t}_{t_{0}}H_{I}(t') dt' \big) \end{align} Where $H_{I}(t')$ and $H_{I}(t)$ are not commute
My difficulty is that I do not know how to derive the above equation. I tried to compute the differentiation of quadratic term
\begin{align} \frac{d}{dt} \big( \frac{(-i)^{2}}{2}( \int^{t}_{t_{0}}H_{I}(t') dt')^{2} \big ) = (-i)^{2} \int^{t}_{t_{0}}H_{I}(t') dt' H(t) \end{align} Which is incorrect. Therefore, could someone point put the incorrect step in my derivation? Thank you
This problem is based on the quantum field theory lecture p.52
For an operator $A(t)$ depending on time we have $$ \frac{d}{dt} A(t)^2 = \frac{d}{dt} (A(t) A(t)) = \frac{d A(t)}{dt} A(t) + A(t) \frac{d A(t)}{dt} = A'(t) A(t) + A(t) A'(t)\,, $$ where $A'(t)$ does not necessarily commute with $A(t)$. Thus you can NOT use the differentiation rule $\frac{d}{dt} A(t)^2 = 2 A'(t) A(t)$ when dealing with non-commutative objects.
Your problem corresponds to the case $A(t) = \int_{t_0}^t H_I(t') dt' $.