Help pls have an exam in 2 days

topsquark

Forum Staff
Apr 2008
2,936
611
On the dance floor, baby!
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
 
Jul 2019
2
0
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 ?
 

Attachments

topsquark

Forum Staff
Apr 2008
2,936
611
On the dance floor, baby!
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