I would just like to confirm my solution to the following question. I'm a bit hesitant on my solution because of a specific step. I would just like confirmation if that step, which I will point out, is mathematically legal. The question I'm working on is:
Show that the time evolution operator, given by the Dyson series,
\begin{equation}
\mathcal{U}(t,0)
=
1 + \sum_{n=1}^\infty \bigg ( \dfrac{-i}{\hbar} \bigg )^n \int_0^t dt_1 \int_0^{t_1} dt_2 \dots \int_0^t dt_{n-1} H(t_1) H(t_2) \dots H(t_n)
\end{equation}
satisfies Schrodinger's equation
\begin{equation}
i\hbar \dfrac{\partial}{\partial t} \mathcal{U}(t,0) = H\mathcal{U}(t,0).
\label{SE}
\end{equation}
For this problem, I will evaluate the left-hand side of the Schrodinger equation to show that it is equivalent to the right-hand side. Firstly, we have that the Dyson series can be rewritten as
\begin{equation}
\mathcal{U}(t,0)
=
1 + \sum_{n=1}^\infty \bigg ( \dfrac{-i}{\hbar} \bigg )^n \int_0^t dt_1 \int_0^{t_1} dt_2 \dots \int_0^t dt_{n-1} H(t_1) H(t_2) \dots H(t_n)
=
T\big \{ \exp{\big ( \dfrac{-i}{\hbar} \int_0^t dt' H(t') \big )} \}
\nonumber
\end{equation}
The following steps is where my question is. The part I'm referring too is when I take the time-derivative with respect to $t'$.
Using this, we have that
\begin{align}
i\hbar \dfrac{\partial}{\partial t} \mathcal{U}(t,0)
&=
i\hbar \, \partial_{t'} \bigg [ T\big \{ \exp{\big ( \dfrac{-i}{\hbar} \int_0^t dt' H(t') \big )} \} \bigg ]
\nonumber
\\[.5em]
&=
i\hbar \bigg [ \partial_{t'} \big ( \dfrac{-i}{\hbar} \int_0^t dt' H(t') \big ) \bigg ] \bigg [ T\big \{ \exp{\big ( \dfrac{-i}{\hbar} \int_0^t dt' H(t') \big )} \} \bigg ]
\nonumber
\\[.5em]
&=
-i\hbar \dfrac{i}{\hbar} \partial_{t'}\int_0^t dt' H(t') \mathcal{U}(t,0)
\nonumber
\\[.5em]
&=
H \mathcal{U}(t,0)
\end{align}
Therefore, $i\hbar \dfrac{\partial}{\partial t} \mathcal{U}(t,0) = H\mathcal{U}(t,0)$ and $\mathcal{U}(t,0)$ satisfies the Schrodinger equation.
I'm essentially performing the following differential:
\begin{equation}
\partial_x e^{f(x)} = f'(x) e^{f(x)}
\end{equation}
Since the time-evolution operator is defined as the sum, I'm not sure how legal it is to take that derivative in the way that I did. If this is incorrect, I would appreciate an alternative method of working this problem.
Any guidance would be appreciated, thank you!
Best Answer
Yes, it is correct; you can check it by just expanding $e^{f(x)}$ in terms of the powers of $f(x)$ and then taking the derivative of that infinite sum.