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

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.