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 Dec 18 awarded Famous Question Nov 5 awarded Popular Question Sep 13 awarded Popular Question Aug 27 awarded Popular Question Apr 10 awarded Popular Question Mar 21 awarded Popular Question Mar 17 awarded Notable Question Dec 22 awarded Popular Question Nov 27 awarded Popular Question Nov 10 awarded Notable Question Nov 3 awarded Popular Question Sep 26 awarded Popular Question Sep 24 awarded Autobiographer Aug 15 comment Determining the Wave Function From Initial Conditions Oh, I see. You are using terminology that I am unfamiliar with. I still do not see how this relates to the wave function being only a mixture of the first two stationary states. Aug 15 comment Determining the Wave Function From Initial Conditions Well, I did not define them, but the textbook I am using does, which is Griffith's Introduction To Quantum Mechanics. Here is what he says regarding stationary states: "Although the wave function itself does (obviously) depend on $t$, the probability density, $|\Psi(x,t)|^2 = \Psi^* \Psi = \psi e^{i Et/\hbar} \psi e^{-i Et/\hbar} = |\psi(x)|^2$ does not. Aug 15 comment Determining the Wave Function From Initial Conditions So, what other justification could be used? The answer key does not use this operator, nor is it spoken of in the chapter from which this problem comes from. Aug 15 comment Determining the Wave Function From Initial Conditions @ACuriousMind I do not know of this time evolution operator of which you speak. Aug 15 comment Determining the Wave Function From Initial Conditions Yes, exactly. As far as I understand, the most general solution of the infinite well is $\Psi (x,t) = \sum_{n=1}^{\infty} c_n \psi_n(x) \phi_n(t)$, where the coefficients are $c_n = \sqrt{\frac{2}{a}} \int_{0}^{a} \sin(\frac{n \pi}{a} x) \Psi(x,t)$. Why wouldn't I use these two equations to calculate the wavefunction for all future times? Aug 15 asked Determining the Wave Function From Initial Conditions Jul 29 accepted Finding Interatomic Spacing