I was watching Allan Adams' lecture on energy eigenfunctions, and there's one part (around 43 minutes into the lecture) that confuses me.
Suppose we have the initial wave function $\Psi (x,0)$ such that $\hat{E}\,\Psi (x,0)=E \,\Psi (x,0)$ for some constant $E$. Then, plugging this into the Schrödinger equation, we'd get:
\begin{align}
i \hbar \frac{\partial}{\partial t} \Psi (x, 0) &= E \, \Psi (x,0) \\
\frac{\partial}{\partial t} \Psi (x, 0) &= \frac{E}{i \hbar} \, \Psi (x,0) \tag{1}\\
\therefore \Psi (x, t) &= \exp\left({-i \frac{E\,t}{\hbar}}\right) \Psi(x,0) \tag{2}
\end{align}
I'm a bit confused about how to go from $(1)$ to $(2)$.
Now if we make the additional assumption that $\hat{E}\,\Psi (x,t)=E \,\Psi (x,t)$ for all $t$, then the Schrödinger equation becomes:
\begin{align}
\frac{\partial}{\partial t} \Psi (x, t) &= \frac{E}{i \hbar} \, \Psi (x,t)
\end{align}
and I can solve this differential equation easily and get $(2)$. But from watching that part of the lecture, it seems we only need to assume a weaker statement – that the initial wave function is an energy eigenfunction. But then, it's not clear to me how I can get the solution $(2)$ from $(1)$. Am I missing something?
Update: Thanks for all the answers. After reading through the accompanying lecture note, we indeed need to assume that the energy operator is a constant over time.
Best Answer
By the simple form of the equation (1) you wrote down, Allan really meant $$ \frac{\partial}{\partial t} \Psi(x,t)|_{t=0} = \frac{E}{i\hbar}\Psi(x,0) $$ He just used the notation where $t=0$ is substituted from the beginning but he clearly did mean that $\Psi(x)$ is first considered as a general function of $t$, then differentiated, and then we substitute $t=0$.
This equation says that the time derivative of $\Psi(x,t)$ at $t=0$ is proportional to the same wave function. By itself, it does not imply that $\Psi(x,t)$ for an arbitrary later $t$ will be given by equation (2): if we only constrain the derivative at one moment $t=0$, the wave function may do whatever it wants at later (or earlier) moments $t$.
However, we may generalize (1) to any moment $t$ which is what you wrote down $$ \frac{\partial}{\partial t} \Psi(x,t) = \frac{E}{i\hbar}\Psi(x,t) $$ and this equation does imply (2). If the $t$-derivative of $\Psi(x,t)$ is proportional to the same $\Psi(x,t)$, then $\Psi(x,t)$ and $\Psi(x,t')$ are proportional to each other for each $t,t'$. That implies that $\Psi(x,t)$ must factorize to $\Psi(x)f(t)$ and the function $t$ is the simple complex exponential to solve the equation with the right coefficient.
Because you are more or less writing the same things, I find it plausible that you are not missing anything.