Quantum state - Probability of finding it in its initial state after a certain time. 
Consider a normed quantum state $|\Psi(0)\rangle$ and let $\Delta_\Psi E\equiv \langle (\hat{H}-\langle\hat{H}\rangle_\Psi)^2\rangle_\Psi^{1/2}$ be the energy uncertainty.
a. Compute the probability that the system is still in state $|\Psi(0)\rangle$ after a time $\delta t$. (Meaning $|\langle\Psi(\delta t)|\Psi(0)\rangle|^2$.
b. Determine the energy uncertainty of a one-dimensional harmonic oscillator in the state $|n\rangle$, whereas $\hat{H}=\hbar \omega(\hat{N}+1/2)$ and $\hat{N}|n\rangle=n|n\rangle$.
c. Let $|\Psi(0)\rangle =1/\sqrt{2}(i|1\rangle-|3\rangle)$. What's the energy uncertainty of that state at time t=0 and when is the probability of it being still in the initial state after $\delta t$ less than 0.9?

Hello everyone,
I've been doing exercises from my workbook and I'm kinda stuck at this one (and I tried looking for a exercises manual of that book, but apparently there is none).
As far as a. goes I have no idea how to start there. I'm guessing this is a different expression for the Born Rule, but I never really understood how to apply on a particular problem (but I get the theory of the rule a bit, but apparently not enough).
As for b: I just thought of taking the expressions for H and N and plug it into the term for the energy uncertainty (I will leave out the hats):
$\Delta_\Psi E=\langle(\hbar\omega(N+1/2)-\langle(\hbar\omega(N+1/2))^2\rangle_\Psi\rangle_\Psi^{1/2}=\langle(\hbar\omega(N+1/2)-\hbar\omega(n+1/2))^2\rangle_\Psi^{1/2}=\langle(\hbar\omega N-\hbar\omega n\rangle_\Psi^{1/2}=\langle\hbar^2 \omega^2N^2-2\hbar^2\omega^2Nn+\hbar^2\omega^2n^2\rangle_\Psi^{1/2}=0.$
Is that right?
And I'm also lost on c. there.
 A: $\newcommand{\ket}[1]{\lvert #1\rangle}$
$\newcommand{\bra}[1]{\langle #1\rvert}$
a. States evolve according to the Schroedinger equation, whose formal solution is given by exponentiating the Hamiltonian
$$ \Psi(t) = e^{-itH}\Psi(t=0) $$
Without further knowledge the best answer to the question is, that the "survival probability" is given by
$$ \lvert \bra{\Psi(0)}e^{-itH}\ket{\Psi(0)} \rvert^2 $$
If an eigenbasis $\{\ket{n}\}$of the Hamiltonian (this is a general feature and not constraint to the harmonic oscillator!) is known, one may expand states as linear combinations 
$$ \Psi(t) = \sum_k a_k(t) \ket{k} $$
Inserting into the above expression yields
$$ \bra{\Psi(0)}e^{-itH}\ket{\Psi(0)} = \sum_k\sum_m a_k^* \bra{k}e^{-itE_m}a_m\ket{m} = \sum_m e^{-itE_m} \lvert a_m\rvert^2$$
where the $a_m$ are the expansion coefficients of $\Psi(0)$ and $E_m$ are the eigenvalues of $H$.
b. You are correct. The system is in an eigenstate of the Hamiltonian so the uncertainty in energy must vanish. Make sure you understand this intuitively!
c. At least for $t=0$ you should be able to do it. The calculation is pretty much identical to part b. Use the results from a. for the seconds part. You will get an implicit equation for $t$.
I could be more explicit, but I urge you to try for yourself. It's the only way to learn properly. Get some clarity on the mathematical structure of the theory. Pick up a (different) textbook if you're not content with the current material. If you're still really stuck then, ask back.
