# 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 $(-i H / \hbar)^n$? Oct 27, 2015 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. Oct 27, 2015 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. Oct 27, 2015 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? Oct 27, 2015 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. Oct 27, 2015 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 $H^2\psi=(i\partial_t)^2 \psi$? Oct 27, 2015 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... ;-) Oct 27, 2015 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? Oct 27, 2015 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. Oct 27, 2015 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 ;-) Oct 27, 2015 at 12:42