What does it mean for a quantum particle to have energy $E_n$? And what is its general normalised state? In this particular case, I have found the energy to be quantised with energy levels $\frac{h^2n^2}{2m} >0 $ where $n$ is an integer.
Suppose a particle has energy $E=\frac{4h^2}{2m}$, then this clearly corresponds to $n=2$ or -2. 
Now, what is the general normalised state of a particle when it has this energy? 
(What does it mean for the particle to "have this energy"? That $H\psi=E\psi$, where $H$ is the Hamiltonian?)
Clearly a linear combination of the eigenstates corresponding to $n=2$ and $-2$ will give $H\psi=E\psi$. However, does the general state only involve these or is it a sum of all other eigenstates too? If the latter, how do we find the co-efficients?
 A: A state $\psi$ corresponds to an energy $E$ if:
$$H\psi = E\psi$$
Clearly, if there is a state $\psi = \sum_i c_i \psi_i$ where $H\psi_i = E_0\psi_i\ \forall i$, then
$$H\psi = \sum_i c_i H\psi_i = \sum_i c_i E_0\psi_i = E_0\psi$$
A linear combination of states with the same energy value again has the same energy value.
Now consider $\psi = c_1\psi_1 + c_2\psi_2$, with $H\psi_1 = E_1\psi_1$, $H\psi_2 = E_2\psi_2$, where $E_1 \ne E_2$, so that
$$H\psi = c_1H\psi_1 + c_2H\psi_2 = E_1c_1\psi_1 + E_2c_2\psi_2$$
In general, the right hand side reduces to (say) $E\psi$ only if $E_1 = E_2 = E$, or either of $c_1$ or $c_2$ equal zero. Thus, a linear combination of states with different energy values is not an energy eigenstate at all.
In general, if you know a state $\psi$ (say as a function $\psi(x)$ in the position representation), and you want to express it as $\psi = \sum_nc_n\psi_n$, where the $\psi_n$ are normalized energy eigenstates (more generally, normalized eigenstates of any Hermitian operator) then the $c_n$ can be found by making use of the following "orthonormality" relations:
$$\int dx\ \psi_m^*\psi_n = \delta_{mn}$$
where $\delta_{mn}$ is the Kronecker delta, which is $1$ when $m = n$ and zero otherwise.
Using this, we multiply $\psi$ by $\psi_m^*$ for some $m$, and integrate to get:
$$c_m = \int dx\ \psi_m^* \psi$$
With both $\psi_m$ and $\psi$ known, the integral (in principle) gives the coefficients of the linear combination.
