Time evolution of wave function in QM

Recently I've been studying quantum dynamics with Sakurai's modern quantum mechanics, but I am confused with why the time evolution operator is written as

$$U(t,t_0)=\exp\left[\frac{-iH(t-t_0)}{\hbar}\right]$$ for time-independent Hamiltonian, while $$U(t,t_0)=\exp\left[-\left(\frac{i}{\hbar}\right)\int_{t_0}^tdt'H(t')\right]$$ for time-dependent (commuting case). My thought is that, since we can Taylor expand the wave function at $t=t_0$

\begin{align}\psi(x,t)&=\sum_{n=0}^\infty \frac{1}{n!}\left(\left(\frac{\partial}{\partial t}\right)^n\psi(x,t)\bigg|_{t=t_0}\right) (t-t_0)^n\\ &= \sum_{n=0}^\infty \frac{1}{n!}\left(\left(\frac{-iH}{\hbar}\right)^n\psi(x,t)\right)(t-t_0)^n\\ &=e^{\frac{-iH(t-t_0)}{\hbar}}\psi(x,t) \end{align}

we only need to know the value of $H$ at $t=t_0$. If this is true, then $U(t,t_0)=\exp\left[\frac{-iH(t-t_0)}{\hbar}\right]$ should hold whether $H$ is time-dependent or not. What have I done wrong here?

• Where are you getting that the $n^\text{th}$ derivative is $(-i H / \hbar)^n$? – DanielSank Oct 27 '15 at 7:48
• Since $H\psi=i\hbar\frac{\partial}{\partial t}\psi$, we have $H=i\hbar\frac{\partial}{\partial t}$, then $\left(\frac{-iH}{\hbar}\right)^n=\left(\frac{\partial}{\partial t}\right)^n$. Correct me if I'm wrong, thanks. – Rick Pan Oct 27 '15 at 7:56
• @Rick Pan, the equation for derivative $\partial_t \psi =1/(i\hbar)H\psi$ holds only for $\psi$. It does not necessarily hold for its derivatives. – Ján Lalinský Oct 27 '15 at 8:00
• @JánLalinský Thanks for the comment. If we assume that the eigenstates of $H$ forms a complete basis, then we should be able to expand its derivatives with the complete basis. If this is the case, the relation should also hold for its derivatives right? – Rick Pan Oct 27 '15 at 8:24
• @RickPan, no, the relation does not hold because of the reason yuggib pointed out - the expansion coefficients are functions of time as well. Canonical momentum operator $p_x$ is always expressible as $-i\hbar\partial_x$, whatever the function $\psi$ may be, but the Hamiltonian is not always given by $i\hbar\partial_t$; it only holds for special functions - solution to time-dependent Schroedinger equation. – Ján Lalinský Oct 27 '15 at 18:51

The (omitted) starting hypothesis is that $$i\partial_t\psi(t)=H(t)\psi(t)\; .$$ If we iterate the derivation, we do not get simply $H(t)^2\psi(t)$, but rather (this is a simple application of the product rule, that actually works also in this case) $$(i\partial_t)^2\psi(t)=i\dot{H}(t)\psi(t)+H(t)^2\psi(t)\; .$$ As we can easily see, this is where the OP's argument goes wrong, since the derivative of $H(t)$ does not vanish in general for time dependent operators.
• Thanks for the answer, but I'm curious that if we expand $\partial_t \psi$ with eigenstates of $H$ (which is usually assumed to form a complete basis), don't we get $H^2\psi=(i\partial_t)^2 \psi$? – Rick Pan Oct 27 '15 at 9:36
• You could only use eigenstates for a fixed $t$, but they would not be eigenstates, in general, for another $t'\neq t$ (since $H(t')\neq H(t)$), and as well not eigenstates for $\dot{H}(t)$. Really, this is not the good way of thinking about that... ;-) – yuggib Oct 27 '15 at 9:40
• Thanks for the reply. It is true that eigenstates can change with respect to $t$, but I'm expanding the series at a fixed $t$, how is that not applicable? – Rick Pan Oct 27 '15 at 10:01
• Let $\psi_{n,t}$ be an eigenvector of $H(t)$; however in the expansion you have $\dot{H}(t)\psi_{n,t}$, and $\psi_{n,t}$ is not an eigenvector for it. – yuggib Oct 27 '15 at 10:04
• @RickPan Yes, for in general for a time dependent $H(t)$, the coefficients $c_k$ would depend on time as well as the functions $\psi_k$...believe me, this is not true/correct ;-) – yuggib Oct 27 '15 at 12:42