What state does the particle in a box occupy? My textbook derives the equations for the different energy states $E_n$ of the particle in a box. But my professor in class said this example was a good one because it spoke about the "superposition of states".
I'm not seeing how there can be any superposition of states. I know differential equations allow for a superposition of solutions, but not in this case, because each solution $\Psi$ needs of a specific value of $E$.
What am I getting wrong here? If there is a superposition of states, how does it happen in a mathematical way? How would you calculate the probability of it being in a certain state? How can you know its in state, e.g., $n=1$.
 A: Eigenstates aren't the only allowed physical states. It's a postulate of quantum mechanics that the most general quantum state can be written as a superposition of eigenstates of some operator (the Hamiltonian for instance). For instance $\Psi(x)=\sum_nc_n\psi_n(x)$ is a general quantum state for a particle in a box, where $\psi_n(x)$ are the energy eigenstates. The probability of measuring an energy $E_n$ is simply $|c_n|^2$. If a particle exists in an eigenstate of the Hamiltonian however, you will always measure the same energy.  
A: It actually is the very essence of the QM. In short, when we observe a superposed state, the probability of observing specific eigenvalue is the square of the norm of the corresponding eigenstate in the superposed state. And this is more like a postulate, rather than a mathematical derivation.
For example, particle in a box has discrete eigenvalues, bounded below. So, we can denote $n$-th eigenstate as $|n\rangle $, who satisfies $\hat{H}|n\rangle=E_n |n\rangle$. 
Let $e_n=E_n/E_1$ be normalized eigenvalue. It can be shown that for this case, $e_n=n^2$. 
Let $\hat{h}=\hat{H}/E_1$ be the normalized hamiltonian.
A particle at $|1\rangle$ would have $e_1=1$. State $|2\rangle$ would correspond to $e_2=4$. 
But particle does not necessarily have its energy one of these discrete states. They are just eigenstates, which is special set of solutions for given DE.
Consider a particle, who has normalized energy of $e=2$. What state should this particle correspond to? (these states do exist, despite they are not one of the eigenstate.)
One of the answer could be 
$$
\left|\psi\right\rangle = \left. \sqrt{\frac{2}{3}}\middle|1 \right\rangle + \left. \sqrt{\frac{1}{3}}\middle|2 \right\rangle
$$
Why? 
Expected energy value is calculated by $e=\langle \psi| \hat{h}|\psi\rangle$, and for this case,
$$
e= \left( \sqrt{\frac{2}{3}}\middle\langle 1 \middle| + \sqrt{\frac{1}{3}}\middle\langle 2 \middle | \right ) \hat{h} \left( \sqrt{\frac{2}{3}}\middle|1 \right\rangle + \left. \sqrt{\frac{1}{3}}\middle|2 \middle\rangle \right)\\
=\frac{2}{3}e_1 + \frac{1}{3}e_2 = \frac{2}{3}+ \frac{1}{3}\times 4\\
=2
$$
Then, what happens when we observe energy?
Even though we have a state that is not an eigenstate, we cannot measure the energy other than the eigenvalue of the hamiltonian, which is one of the postulates of the quantum mechanics. So, either $e=1$ or $e=4$ would be observed. 
The probability of finding any state in $n$-th eigenstate is the square of the coefficient of the state $|n\rangle$ for the given superposed state, which is also one of the postulates of the quantum mechanics. Thus, finding $e=1$ is $2/3$, while probability of finding $e=4$ is $1/3$.
Wait. Then what happens to the energy conservation? This is why the Quantum Mechanics is weird. In the process of observation, the energy does not necessarily conserved, along with other observables--mostly conserved variables, such as angular momentum. As long as the energy expectation value does not change over time without observation, the energy is considered to be conserved.
While writing this, I referred to Shankar's the Principles of Quantum Mechanics, chapter 4. Since you study QM this semester, I think you should fully understand this to understand QM itself.
