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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.

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I'm sure this is a duplicate, but after some searching I can't previous instances. Anyhow, you assume $\Psi(x,t) = \psi(x)\phi(t)$ i.e. the wavefunction is a product of space dependant and time dependant functions. Feed this back into the Schrodinger equation, use the fact that $\psi(x) = \Psi(x,0)$ is an eigenfunction and you get the equation for $\phi(t)$ – John Rennie Jul 1 '14 at 5:20
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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.

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Going from your (1) to your (2) is bogus as you suspect. Your (1) isn't even really a differential equation: the independent variable $t$ does not appear. (1) is just using the initial condition to find what the time derivative is at $t = 0$. Indeed to directly conclude that the solution is (2), we would need your stronger assumption.

One correct way to arrive at (2) is to assume that the solution takes the form $\Psi(x,t) = f(t) \Psi(x,0)$, which will give you $f(t) = \exp(Et/i\hbar).$ This clearly matches the initial condition.

You could also just make the stronger assumption. This amounts to making an educated guess about the solution to the differential equation. This is a perfectly valid way of solving differential equations, as long as you can match the initial conditions.

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If this isn't at a higher level than the original lectures (which I haven't seen, actually), the correct way to go from $\psi (x,0)$ to $\psi (x,t)$ in Quantum Mechanics, is to employ the time evolution operator $\exp(-i{\hat H}t/\hbar)$, where ${\hat H}$ is the Hamiltonian, on $\psi (x,0)$, i.e.

$$\psi(x,t) = {\hat T} \left(\psi (x,0)\right) = \exp(-i{\hat H}t/\hbar) \ \psi (x,0)$$

Now, the exponential can be expanded into the series $\sum_i (-i{\hat H}/\hbar)^n/n!$, i.e. an infinite series in powers of ${\hat H}$. That can be simplified by using your original time-independent Schrodinger equation, $${\hat H} \ \psi (x,0) = E \ \psi (x,0)$$, for each term. This would lead you another infinite series (exponential), but in terms of $E$ rather than ${\hat H}$ this time around. On simplifying you would reach the desired equation $$\psi (x,t) = \exp(-iEt/\hbar) \ \psi (x,0)$$

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