# Does the Time Evolution Operator Satisfy the Schrodinger Equation?

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, $$$$\mathcal{U}(t,0) = 1 + \sum_{n=1}^\infty \bigg ( \dfrac{-i}{\hbar} \bigg )^n \int_0^t dt_1 \int_0^{t_1} dt_2 \dots \int_0^t dt_{n-1} H(t_1) H(t_2) \dots H(t_n)$$$$ satisfies Schrodinger's equation $$$$i\hbar \dfrac{\partial}{\partial t} \mathcal{U}(t,0) = H\mathcal{U}(t,0). \label{SE}$$$$

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

$$$$\mathcal{U}(t,0) = 1 + \sum_{n=1}^\infty \bigg ( \dfrac{-i}{\hbar} \bigg )^n \int_0^t dt_1 \int_0^{t_1} dt_2 \dots \int_0^t dt_{n-1} H(t_1) H(t_2) \dots H(t_n) = T\big \{ \exp{\big ( \dfrac{-i}{\hbar} \int_0^t dt' H(t') \big )} \} \nonumber$$$$

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: $$$$\partial_x e^{f(x)} = f'(x) e^{f(x)}$$$$

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!

• 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. – uniquesolution Feb 5 at 19:40
• There is also the question of whether the series converges. – Keith McClary Feb 6 at 5:33
• Yes, it is correct, if you want someone to verify it. – onurcanbektas Feb 16 at 6:21

## 1 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.