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We consider the normalized wave function: $$\psi(x,t) = \sqrt{\frac{2}{3}}\psi_0(x)\exp\left(\frac{-iE_0t}{\hbar}\right) + \sqrt{\frac{1}{3}}\psi_1(x)\exp\left(\frac{-iE_1t}{\hbar}\right) $$

To compute the probability of an energy state $n$ we can simply square the eigenvalue, thus: $a_n^2$.

If that's the case am I correct concluding that the probability of the energy state $E_0$ occouring is $\left(\sqrt{\frac{2}{3}}\right)^2 = \frac{2}{3}$, should I compute the integral $ c_0 = \int\psi^*\psi_0 dx$ and then compute $c_0^²$, or do they both give the same result?

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    $\begingroup$ "Eigenvalue" is not the right term for $a_n$. The eigenvalues in the formula you gave are $E_0$ and $E_1$. $\endgroup$ – Chris Jan 19 '18 at 2:39
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They will give the same result. Since the Hamiltonian is a Hermitian operator, assuming $E_0\ne E_1$, it is guaranteed that $\psi_0(x)$ and $\psi_1(x)$ are orthogonal. That is to say: $$ \int dx~\psi_0(x)^\star\psi_1(x)=0$$

So if you compute the integral of $\psi^\star$ with any eigenstate $\psi_n$, all the other eigenstates will vanish and you will be left with $a_n$.

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Both will give the same results, but which one is most useful depends on the specific problem. This works because $$ \int dx \psi_a^*(x)\psi_b(x)=\delta_{ab} $$ so you can think of the $c_n$ as components and $\psi_n(x)$ as basis vector in the space of function.s

If you are given the components explicitly, in the form of an explicit sum $$ \Psi(x,t)=\sum_n c_n \psi_n(x) e^{-iE_nt/\hbar} $$ and you can read-off the $c_n$’s, then calculating the probabilities is easy.

If you are given some function $\Psi(x,0)$ - say for instance $A e^{-(x-a)^2/a_0^2}$, then you will need extra work to compute the $c_n$’s via $$ c_n= \int dx \psi^*_n(x) \Psi(x,0)\, . $$

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