Time-dependence in QM When you have a system governed by some Hamiltonian H that starts in some state $\lvert a \rangle$, the way to calculate its time-dependence is to decompose into a linear combination of eigenstates of H, $\lvert a \rangle = \sum_{n}\psi_{n}$ and tack on the time dependence $e^{-iE_{n}t/\hbar}$, such that: 
$$\lvert a(t) \rangle = \sum_{n}\psi_{n}e^{-iE_{n}t/\hbar}$$
I would like to prove this. To do I say suppose that at time $t=0$ we have:
$$\Psi(x,0) = \sum_{n}\psi_{n}(x)$$ 
The Schrödinger equation says:
$$i\hbar \frac{\partial \Psi}{\partial t} = H\Psi \\ \Rightarrow
\Psi(x,t) = e^{-iHt/\hbar}\Psi(x,0)$$.
By now inserting the eigenstate decomposition of $\Psi(x,0)$ and using 
that 
$$e^{-iHt/\hbar}\psi_{n} = e^{-iE_{n}t/\hbar}\psi_{n}\quad\text{(since it is an eigenstate)}$$
we arrive at the result. Is this a way to prove it or did I assume something that I am not allowed to. I guess I assumed H is time-independent but are there other problems? 
 A: Everything in your proof is correct. I will fill in some details for you, in case you were wondering about them. 
Consider the operator form of the Schrodinger equation:  
\begin{align}
\hat{H} \lvert \Psi\rangle =i\hbar\frac{\partial}{\partial t}\lvert \Psi\rangle
\end{align}
where I have used the notation $\lvert \Psi \rangle$ to denote $\lvert \Psi(t=0) \rangle$. Now, it is clear that, if the hamiltonian is independant of time, the solution to this equation is that 
\begin{align}
\lvert \Psi(t)\rangle = e^{-\frac{i}{\hbar}\hat{H}t}\lvert \Psi\rangle
\end{align}
where the exponential operator is defined by the taylor series 
\begin{align}
e^{-\frac{i}{\hbar}\hat{H}t}\equiv\sum_{j=0}^{\infty}\frac{1}{j!}\left(\frac{-i\hat{H}t}{\hbar}\right)^j.
\end{align}
Now, we can see that if we insert resolution of unity given by
\begin{align}
\hat{\mathbb{1}}=\sum_{n=0}^{\infty}\lvert n\rangle \langle n \rvert
\end{align}
,where $\lvert n \rangle$ denotes an energy eigenstate, into the above equation for $\lvert \Psi(t) \rangle$, we obtain that 
\begin{align}
\lvert \Psi(t)\rangle &= e^{-\frac{i}{\hbar}\hat{H}t}\lvert \Psi\rangle\\
&=\sum_{n=0}^{\infty}e^{-\frac{i}{\hbar}\hat{H}t}\lvert n\rangle\langle n\rvert \Psi\rangle\\
&=\sum_{n=0}^{\infty}e^{-\frac{i}{\hbar}E_nt}\lvert n\rangle\langle n\rvert \Psi\rangle
\end{align}
The rigorous reason for why $e^{-\frac{i}{\hbar}\hat{H}t}\rightarrow e^{-\frac{i}{\hbar}E_n t}$ upon the left-action on an energy eigenstate is contained in the taylor series definition of $e^{-\frac{i}{\hbar}\hat{H}t}$. This reduces it to just acting on an energy eigenstate with $\hat{H}^j$, which simply returns $E_n^j$, for each term in the series. In this way, we obtain the familiar taylor expansion of $e^{-\frac{i}{\hbar}E_n t}$. Now, insert the unity resolved in the position basis
\begin{align}
\hat{\mathbb{1}}=\int\lvert x\rangle \langle x \rvert\  dx
\end{align}
so that 
\begin{align}
\lvert \Psi(t)\rangle&=\sum_{n=0}^{\infty}e^{-\frac{i}{\hbar}E_nt}\lvert n\rangle\langle n\rvert \Psi\rangle\\
&=\sum_{n=0}^{\infty}e^{-\frac{i}{\hbar}E_nt}\lvert n\rangle \int \langle n\rvert x\rangle \langle x\rvert \Psi\rangle \ dx
\end{align}
Now, we define the wave function $\Psi(x)$ to be the state $\lvert \Psi \rangle$ expanded in the position basis. That is, $\Psi(x)\equiv\langle x \lvert \Psi \rangle$. It follows that $\psi_n(x)=\langle x\rvert n \rangle$. Therefore, since we desire $\Psi(x,t)$ we take $\langle x \rvert$ to both sides of the last equation to obtain 
\begin{align}
\Psi(x,t)=\langle x\rvert \Psi \rangle&=\sum_{n=0}^{\infty}e^{-\frac{i}{\hbar}E_nt}\langle x \lvert n\rangle \int \langle n\rvert x\rangle \langle x\rvert \Psi\rangle \ dx\\
&=\sum_{n=0}^{\infty}e^{-\frac{i}{\hbar}E_nt}\psi_n(x)\int \psi_n^*(x)\Psi(x,0) \ dx.
\end{align}
We recognize that this is the familiar expansion of $\Psi(x,t)$ into energy eigenstates
\begin{align}
\Psi(x,t)=\sum_{n=0}^{\infty}c_n \psi_n(x) e^{-\frac{i}{\hbar}E_n t}
\end{align}
where $c_n$ is given by 
\begin{align}
c_n=\int \psi_n^*(x)\Psi(x,0) \ dx.
\end{align}
A: This is a valid way to prove it, but you will notice that indeed if $\hat H = \hat H(t)$ then the integral is not $\hat H ~ t$ and in fact  it is even possible that $\hat H(t + \delta t)$ does not commute with $\hat H(t)$ and therefore you need to be very careful with what this so-called "propagator" $$\hat U(T) = \exp\left[\frac{1}{i\hbar} \int_0^T dt~\hat H(t) \right]$$ even means. If you take $t_n = n T/N$ so $\delta t = T/N$ to approximate this time interval $t:0\to T$ then you want the time-ordered product,$$\hat U(T) = \lim_{N\to\infty} \exp\left[\frac{\delta t}{i\hbar} H(t_{N-1})\right]\exp\left[\frac{\delta t}{i\hbar} H(t_{N-2})\right]\dots\exp\left[\frac{\delta t}{i\hbar} H(t_{0})\right],$$and then once you've defined the propagator that way, you can always state that $$|\Psi(T)\rangle = \hat U(T) ~|\Psi(0)\rangle.$$So yes, you can certainly define this operator $\hat U$ and if the Hamiltonian is time-independent then it does indeed have the limit you've found, $e^{-i~\hat H~t/\hbar}$, and that can indeed be evaluated on $|\psi_n\rangle$ by replacing the $\hat H$ in the exponent by $E_n$. This last fact is just the infinite-dimensional analogue of the statement that the exponential of a diagonal matrix is, $$\exp[\operatorname{diag}(a_1, a_2, \dots)] = \operatorname{diag}(e^{a_1}, e^{a_2}, \dots).$$
A: Also remember how you solved Schrodinger's equation. You began with an ansatz for the wave function namely that it was separable so $\Psi(x,t)=\psi(x)\phi(t) $. You then broke the PDE into two ODEs for time part you got $ \phi(t) = e^{-iEt} $. Therefore the solution is of the form $ \Psi(x,t)= \psi(x)e^{-iHt} $. So $ \sum_{n} \psi_n e^{-iE_nt} $ follows linear structure of Schrodinger's equation(linear superposition of solutions).
