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Let's say that we have a system such that

$$\Psi(x,0)=\frac{\sqrt3}{2}\phi_1(x)+\frac12\phi_2(x)$$

where both $\phi(x)$ are eigenfunctions of the Hamiltonian operator.

I want to find $\Psi (x,t)$ for all $t$ and I want to know if the current state of the system is stationary.

I did the following:

$$\hat{H}|\Psi\rangle=\frac{\sqrt3}{2}\hat{H}|\phi_1\rangle+\frac{1}{2}\hat{H}|\phi_2\rangle=\frac{\sqrt3}{2}E_1|\phi_1\rangle+\frac{1}{2}E_2|\phi_2\rangle$$

So the state would not be an eigenstate, ergo it would not be stationary. Is this correct?

For the time evolution of the whole system, would it be enough to just write the time-dependent exponentials to each of the terms? Namely,

$$\Psi(x,t)=\frac{\sqrt3}{2}\phi_1(x)e^{-\frac{-iE_1 t}\hbar}+\frac{1}{2}\phi_2(x)e^{-\frac{-iE_2 t}\hbar}$$

Or is that wrong?

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closed as unclear what you're asking by ACuriousMind, CuriousOne, user36790, Cosmas Zachos, Gert Jul 18 '16 at 2:07

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

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    $\begingroup$ It's a very straightforward linear algebra exercise to deduce whether the sum of two eigenvectors is an eigenvector. Just apply the Hamiltonian to that state and look whether it's an eigenvector or not. $\endgroup$ – ACuriousMind Jul 16 '16 at 22:21
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    $\begingroup$ If you know that these are eigenfunctions, then you know that they have an associated eigen-frequency. The time evolution is therefor trivial. If the two frequencies are not the same, then there is no one time dependent phase factor that makes the sum a stationary eigenstate and you will see beating. $\endgroup$ – CuriousOne Jul 16 '16 at 22:23
  • $\begingroup$ @ACuriousMind I've edited my question, could you please check it? $\endgroup$ – Tendero Jul 17 '16 at 17:33