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Old Jul 30th 2019, 09:23 AM   #1
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Help pls have an exam in 2 days

I don't know how to start solving this problem do i integrate the wave function and equal it to 1 or do i solve it without integration and equal it to 0 ?
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Old Jul 30th 2019, 04:22 PM   #2
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Originally Posted by medofx View Post
I don't know how to start solving this problem do i integrate the wave function and equal it to 1 or do i solve it without integration and equal it to 0 ?
I don't know what you are asking. Are you trying to show that $\displaystyle \Psi _n = \sqrt{ \dfrac{2}{a} } ~ sin \left ( \dfrac{ n \pi x }{a} \right )$

-Dan
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Old Jul 30th 2019, 05:08 PM   #3
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Sorry i uploaded it in german. In the question number a " second wave" Basically i want to know if i should integrate or just substitute it. how should i start ?
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Old Jul 30th 2019, 06:00 PM   #4
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Originally Posted by medofx View Post
Sorry i uploaded it in german. In the question number a " second wave" Basically i want to know if i should integrate or just substitute it. how should i start ?
Your problem is to see if the given state is an eigenfunction. If so then it has to be of the form
$\displaystyle \Psi = A ~ sin \left ( \dfrac{N \pi x}{a} \right )$
where N is an integer.

So let's start with the form:
$\displaystyle \Psi = \dfrac{1}{\sqrt{a}} \left ( sin \left ( \dfrac{ \pi x}{a} \right ) + sin \left ( \dfrac{ 2 \pi x}{a} \right ) \right )$

Start with this identity:
$\displaystyle sin( \theta + \phi ) + sin( \theta - \phi ) = 2 ~ sin( \phi ) ~ cos( \theta )$

Thus we know that
$\displaystyle \theta + \phi = \dfrac{ \pi x}{a}$
and
$\displaystyle \theta - \phi = \dfrac{2 \pi x}{a}$

The solution is
$\displaystyle \theta = \dfrac{ 3 \pi x}{2a}$

$\displaystyle \phi = - \dfrac{ \pi x}{a}$

Which gives us:
$\displaystyle \Psi = - \dfrac{2}{ \sqrt{a} } ~ sin \left ( \dfrac{ \pi x}{a} \right ) ~ cos \left ( \dfrac{3 \pi x}{2a} \right )$

Can this be simplified to the form for an energy eigenvalue? (I'm not saying that this works, it's just a thought: recall $\displaystyle sin( \alpha + \beta ) = sin( \alpha ) ~ cos( \beta ) + sin( \beta ) ~ cos( \alpha )$. It's a good place to start for trying to simplify the form.)

I leave the details to you. If you get stuck, please feel free to ask.

-Dan
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