I need to find the probability of measuring certain energies with the given linear combination wavefunction I have a group of wavefunctions given by $\psi_{n}(x)=\left\{\begin{array}{ll}{\sqrt{2 / a} \sin (n \pi x / a),} & {0 \leq x \leq a} \\ {0,} & {\text { otherwise }}\end{array}\right.$
I also have $\Psi(x)=\frac{1}{\sqrt{2}} \psi_{1}(x)+\frac{1}{\sqrt{3}} \psi_{2}(x)+\frac{1}{\sqrt{6}} \psi_{3}(x)$
Now I need to find the values of the energies that can be obtained and with what probabilities. I found this answer so I understand that the probability of each energy is given by $P\left(E_{0}\right)=\left|\int_{-\infty}^{\infty} \Phi^{*} \psi \mathrm{d} x\right|^{2}$ where $\Phi$ is an energy eigenstate but I am unsure if each of the group of wavefunctions that I have are eigenstates. Don't the eigenstates need to be solutions to the time independent SE for the same value of E? If they are not energy eigenstates how would I solve this without being given a value of E?
 A: You already have the wavefunction at $t \ = \ 0$ as:
$$\Psi(x)=\frac{1}{\sqrt{2}} \psi_{1}(x)+\frac{1}{\sqrt{3}} \psi_{2}(x)+\frac{1}{\sqrt{6}} \psi_{3}(x)$$
Since this is a linear combination of stationary states, the probabilities of measuring different energies don't change with time, even when I introduce time-dependence, and they are equal to the squared coefficients of each of the position-space energy eigenstates in the expansion of $\Psi(x, \ 0)$. Thus, we will measure $E_1$ with probability $1/2$, $E_2$ with probability $1/3$, and $E_3$ with probability $1/6$.
Now, let's look at this in general. In order to find the probability of measuring any of the allowed energies $E_n$ corresponding to the time-independent wavefunction $\psi_n$ in the following expansion:
$$\Psi(x, \ 0) \ = \ \displaystyle\sum_{n} c_n \psi_n(x)$$
we essentially have to "project" this wavefunction onto the stationary state that corresponds to the energy we are trying to find the probability of measuring. It follows that we take the inner product of $\psi_a$ and $\Psi$ to get the probability that we will measure $E_a$. Since all of the wavefunctions are orthogonal, the inner product between $\psi_m$ and $\psi_n$ with $n \ \neq \ m$ will just be $0$, thus this inner product will "pick out" the coefficient on the $\psi_a$ term, giving us the coefficient $c_a$, which we then modulus square to get the probability:
$$P(E_a) \ = \ \Big| \displaystyle\int \ \psi_a^* \ \Psi(x, \ t) \ \text{d} x \Big|^2 \ = \ \Big| c_a \ \displaystyle\int \psi_a^* \psi_a \ \text{d}x \Big|^2 \ = \ |c_a|^2$$
