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Say I have a potential well, between 0 and a. I also know how the wave function looks like for (t=0):

[itex] \psi(x,0)= \frac {2bx} {a} [/itex] for [itex] 0<x<\frac {a} {2} [/itex]

and

[itex] \psi(x,0)= 2b(1- \frac {x} {a} ) [/itex] for [itex] \frac {a} {2} <x<a [/itex]

Now, I wish to find the wave function of a general time, t.

I know I need to find symmetry. Therefore I move the x axis:

[itex] x\rightarrow x', x'\equiv x-\frac {a} {2} [/itex]

So now I have the same well, only symmetrical.

That means:

[itex] \varphi_n =\sqrt {\frac {2} {a} } sin(\frac {\pi lx'} {a}) \rightarrow l=2n

\varphi_n =\sqrt {\frac {2} {a} } cos(\frac {\pi lx'} {a}) \rightarrow l=2n+1 [/itex]

For the question before the axis transition, I know I have to look for a sin- solution.

But now, when I have this symmetry, I should look for a solution which vanishes at x=-a/2 instead of x=0

Meaning, the cosine must stay. But in this problem's solution, I see that the sine stays...

why is that so?

Thank you!

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# Potenitial well

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