Infinite potential well- Delta potential inside

  • Thread starter noamriemer
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Hello again. Thank you guys. You have been great help...

I have another one:

Given a potential well- 2a is it's width, and in the middle - there is a delta potential:

[itex] V(x)= \frac {\hbar^2} {2m} \frac {\lambda} {a} \delta(x) [/itex]

I am looking for the odd solution to this problem.

I thought I should answer:
I know odd solutions vanish for x=0. Therefore, [itex] \psi'(0)=0 [/itex]
So I am looking at a "normal potential well": [itex] \varphi_n=\frac {1} {\sqrt a} sin(\frac {\pi nx} {2a} ) [/itex] for n=0,2,4,...

But in the solution- they got the same result, only for n=1,2,3,4,...
Why is that so? in the general solution for infinite well, the sin refers to the even n's...

Later on, I want to find the even states...
So I was trying to find a cosine based solution ... and I saw that again, in the solution, they were looking for a sine solution...
Why?

I think I am mixing things here, but as far as I understood- sine refers to the anti-symmetrical states, and to even n's, and the cosine - to symmetrical states and odd n's...
Thank you sooooo much!
Noam
 

Answers and Replies

  • #2
CompuChip
Science Advisor
Homework Helper
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Why don't you try starting from the Schrodinger equation... plug in the given potential, and try a solution of the form [itex] \psi(x) = A e^{\lambda x} [/itex].
 

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