Formulation and probability of a wave-function I have got this problem where I have been given the following wave function: $$\Psi = 0\quad\text{if}~|x| > a\quad\text{and}\quad A(a^2-x^2)\quad \text{if} \quad |x|< a$$
Now the first question is "Determine the time-dependent wave-function. Your answer should be expressed as an inﬁnite sum." How do i express it as an infinite sum, can I just not multiply by $e^{-iEt/(\hbar)}$?
The second question is "What is the probability of ﬁnding the particle with energy $E_n = \hbar\omega(n + 1/2)$
at the time $t =2\pi/\omega$?" I don't really get where I shall start, if I express the time dependent wave function as $\psi (x,0)e^{-iEt/\hbar}$ I can substitute for $E$ and $t$, but then how do I find the probability, don't you need an interval to do that? 
EDIT: November 6th 2013
I think I have figured out what is meant to do now, this is how I have interrupted the problem with the help I have been given. Let's say for simplicity we want to find the probability of finding the particle with energy $E_1$ at time $t$. Firstly we decompse the wave function into a sum of eigen-functions, then we get that $$\Psi(x,0)=\sum_i^{\infty}c_i\psi_i(x)$$ then we find the time-dependent wave fucntion by doing the following: $$\Psi(x,t)=\sum_i^\infty c_i\psi_i(x)\cdot e^{-iE_nt/\hbar}$$. Since we know that $\vert c_i \vert^2 $ is the proability and $c_i$ is given by: $c_i=\int_{allspace} \psi(x)\psi_i(x)dx $ Now we are intrested in finding $c_1$ and thus we need to know $\psi_1(x)$ which by the formula given by WetSavannaAnimal aka Rod Vance gives the following: $$\psi_1(x)=\frac{1}{\sqrt2}(\frac{m\omega}{\pi\hbar})^\frac{1}{4}\cdot e^{-\frac{m\omega x^2}{2\hbar}}\sqrt{\frac{m\omega}{\hbar}}x^2$$ Substituting this into the formula of $c_i$ for the given wave fucntion and for space being between $-a$ to $a$ and pulling out the constants of the integral yields: $$c_1=A\sqrt{\frac{m\omega}{2\hbar}}(\frac{m\omega}{\pi\hbar})^\frac{1}{4}\int_{-a}^a(a^2-x^2)x^2e^{-\frac{m\omega x^2-2iE_1t}{2\hbar}}$$ after this we just find $\vert c_1 \vert ^2$ and if we want to find the probability for E_n we do the same just a bit more general with $\psi_i(x)$ but the priniple is the same. Is this the right way or at least one way of solving the problem or am I on a total wrong track?
 A: You cannot "just multiply" by the exponent as the function is not an eigenstate of the free Hamiltonian (I guess the free Hamiltonian (no potential) is assumed in the problem). So you should expand the initial function into a Fourier series (a sum of eigenstates of momentum) and only then multiply the terms in the expansion by the exponents with the relevant energies.
EDIT (11/3/2013): As the Hamiltonian turns out to be that of an oscillator, you should expand the function in a series in the eigenstates of the Hamiltonian. Please see http://www.gauge-institute.org/delta/HermiteDelta.pdf , formulas at the end of p.10 and at the beginning of page 11.
A: I'll help you answer your second question. 
But first, there are some difficulties with the problem you've been given. Firstly your question isn't fully specified: you need to have the Hamiltonian itself so you need at least the potential $V(x)$. The expression for $E_n$ you are given bespeaks either a quantum harmonic oscillator or an infinite well. Secondly, the there is a typo in your beginning quantum state. The two inequalities in the definition should read $|x|<a$ and $|x|>a$ (not $|x|<0$, which no $x\in\mathbb{R}$ fulfills!).
So I'll presume an infinite well as this ties in the with Fourier series method of akhmeteli's answer. 
As in akmetieli's answer, you expand the quantum state in energy eigenstates $\mathcal{N}^{-1} \cos\left(\left(n + \frac{1}{2}\right)\frac{\pi}{a}x\right);\,n=0,1,2,\cdots$ (when $|x|<a$, nought outside the interval) corresoponding to the energies $E_n$ (here $\mathcal{N}$ is the normalization to make $\int_{-a}^a|\psi|^2dx=1$. 
Note that this is a discrete series. The energy doesn't take on a continuous values, so the magnitude squared of your Fourier series (not integral as for continuously varying energy) weights are the probabilities, not probability densities, that the quantum state will be found in each energy.
So when you do your Fourier series, you should check that $\sum_n |w_n|^2 = 1$, where $w_n$ are the Fourier series weights. 
Now the probability $|w_n|^2$ does not vary with time. The phase of $w_n$ does, so the energy eigenstates interfere with one another to give a time varying wavefunction, but the probabilities to be in each energy eigenstate are constant. So you don't even need to know the time when you calculate your probability.
This should let you finish  your question.
Next stage: Since you are dealing with the quantum harmonic oscillator, your energy eigenstates are the following set:
$$\psi_n(x) = \frac{1}{\sqrt{2^n\,n!}} \cdot \left(\frac{m\omega}{\pi \hbar}\right)^{1/4} \cdot e^{
- \frac{m\omega x^2}{2 \hbar}} \cdot H_n\left(\sqrt{\frac{m\omega}{\hbar}} x \right), \qquad n = 0,1,2,\ldots.\qquad(1) $$
where $H_n(x)=(-1)^n e^{x^2}\frac{d^n}{dx^n}\left(e^{-x^2}\right)$ is the $n^{th}$ Hermite polynomial. 
These eigenfunctions are orthogonal in the sense that:
$$\left<\psi_n, \psi_m\right> \stackrel{def}{=} \int_{-\infty}^\infty \psi_n(x)^* \psi_m(x) {\rm d}x = \delta_{m,n}\qquad (2)$$
i.e. the "inner product" is nought for different discrete energy eigenfunctions and 1 for $n=m$, so each eigenstate is "normalized" or said to have unit "length" in the Hilbert space spanned by the energy eigenstates (don't worry if you don't understand all of this; these are the kinds of statements that'll become more and more wonted to you if you keep at your self study). The above orthogonality is the key to the kind of decomposition that Akhmeteli and I have spoken of: any initial quantum state $\psi(x)$ can be resolved into its energy eigenstates by assuming:
$$\psi(x) = \sum\limits_{m=0}^\infty w_m \,\psi_m(x)\qquad(3)$$
then multiplying both sides of (3) by the $m^{th}$ eigenstate in turn and integrating over the whole real interval, applying (2) to find (noting that we can integrate the series termwise):
$$w_m = \int_{-\infty}^\infty \psi(x)\, \psi(m)\, {\rm d}x \,\qquad(4)$$
Now it is the field of spectral theory that shows that our set of eigenfunctions (1) is complete, i.e. that a sum of the kind (3) can indeed represent (in the appropriate measure theoretic sense) any piecewise continuous $\psi(x)$ fulfilling $\int_{-\infty}^\infty |\psi(x)|^2 {\rm d}x < \infty$, which of course is true for valid quantum states since we must have $\int_{-\infty}^\infty |\psi(x)|^2 {\rm d}x =1$ (this is just a necessary condition for $|\psi(x)|^2$ to be a probability density in $x$).
You will ultimately learn that this "orthogonality" is a property of all eigenfunctions of any quantum observable, not only the Hamiltonian. This result comes to us from Sturm Liouville theory and is owing to the self-adjointhood of quantum observables (more generally it holds for any normal operator - one that commutes with its own adjoint).
Lastly, note that, since $\psi_n(x)$ is the $n^{th}$ energy eigenfunction, its full space and time variation is $\psi_n(x, t) = \psi_n(x) \exp(-i E_n t/\hbar)$. So once you've resolved your beginning quantum state into a superposition like (3), you can write down its general time dependence:
$$\psi(x, t) = \sum\limits_{m=0}^\infty \left(w_m \,\psi_m(x)\,e^{-i \frac{E_n}{\hbar} t}\right)\qquad(5)$$
This should let you get a bit further!
