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I am trying to answer this question: enter image description here

I have tried solving the Schrodinger equation using separation of variables.

However in the later time wave function i can not seem to get the exp(-3i...)

My current workings are:

enter image description here

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Expanding

\begin{align*} \Psi(x,0) &= \frac{1}{\sqrt{a}} \sin \frac{\pi x}{a}+ \frac{2}{\sqrt{a}} \sin \frac{\pi x}{a} \cos \frac{\pi x}{a} \\ &= \frac{1}{\sqrt{a}} \sin \frac{\pi x}{a}+ \frac{1}{\sqrt{a}} \sin \frac{2\pi x}{a} \\ \omega_{n} &= \frac{n^2 \pi^2 \hbar}{2ma^2} \\ \psi_{n} (x,t) &= \sqrt{\frac{2}{a}} e^{-\omega_{n} t} \sin \frac{n\pi x}{a} \\ \Psi(x,t) &= \frac{1}{\sqrt{a}} \exp \left( -\frac{n^2 \pi^2 \hbar}{2ma^2} \right) \sin \left( \frac{\pi x}{a} \right)+ \frac{1}{\sqrt{a}} \exp \left( -\frac{4n^2 \pi^2 \hbar}{2ma^2} \right) \sin \left( \frac{2\pi x}{a} \right) \\ &= \frac{\psi_{1}(x,t)}{\sqrt{2}}+\frac{\psi_{2}(x,t)}{\sqrt{2}} \end{align*}

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  • $\begingroup$ where does the exponential term come from in the psi n (x,t) $\endgroup$ – Robert Thompson Oct 6 '16 at 16:54
  • $\begingroup$ Note that $$i\hbar \psi_{t}=-\frac{\hbar^2}{2m} \psi_{xx}$$ $\psi_{n}(x,t)$ is an eigen state. $\endgroup$ – Ng Chung Tak Oct 6 '16 at 17:04
  • $\begingroup$ and why does the second exponential term in the solution have a 4/2 $\endgroup$ – Robert Thompson Oct 6 '16 at 17:05
  • $\begingroup$ Don't forget the factor $e^{-i\omega t}$ outside. $$e^{-i\omega t}(1+2e^{-3i\omega t} \ldots)=?$$ $\endgroup$ – Ng Chung Tak Oct 6 '16 at 17:07
  • $\begingroup$ I'm sorry but i really don't understand, why that happens and why the fact psi n is an eigen state means you then get psi n (x,t) has the exponential term in it $\endgroup$ – Robert Thompson Oct 6 '16 at 17:12

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