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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
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accepted Finding Interatomic Spacing