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!

  • 3
    $\begingroup$ If I am not mistaken, the Feynman legacy is that if you can prove something in Physics by differentiating or integrating, then you may assume it is mathematically allowed. $\endgroup$ Commented Feb 5, 2019 at 19:40
  • $\begingroup$ There is also the question of whether the series converges. $\endgroup$ Commented Feb 6, 2019 at 5:33
  • $\begingroup$ Yes, it is correct, if you want someone to verify it. $\endgroup$
    – Our
    Commented Feb 16, 2019 at 6:21

1 Answer 1


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.


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