Delta wall and infinite square well potentials ,and 2 other questions

[user3]

Consider the following potential function: V=αδ(x) for x=0 and V=∞ for x>a and x<-a , solve the shroedinger equation for the odd and even solutions.

solving the shroedinger equation I get

ψ(x)=Asin(kx) +Bcos(kx) for -a<x<0

and

ψ(x)=Asin(kx) +Bcos(kx) for 0<x<a


is it correct that I tried to solve the Shroedinger equation independently for both halves of the potential disregarding even or odd solution approaches (ψ(x)=ψ(-x) and ψ(x)= -ψ(-x))?

and also is it correct to assume that ψ(0)=0, since there's an infinite potential there?

final question (not related to this problem): consider the Delta potential well. In the book I am studying from, when the author was solving the Shroedinger equation he implicitly assumed that the energy of the particle before and after the well are the same . But isn't it careless , for the lack of a more polite word, to assume that? Isn't it possible that the particle loses or gains some energy as it crosses the well?

 

[DrDu]

and also is it correct to assume that ψ(0)=0, since there's an infinite potential there?

final question (not related to this problem): consider the Delta potential well. In the book I am studying from, when the author was solving the Shroedinger equation he implicitly assumed that the energy of the particle before and after the well are the same . But isn't it careless , for the lack of a more polite word, to assume that? Isn't it possible that the particle loses or gains some energy as it crosses the well?

No, as long as alpha is finite, psi(0) will in general be different from 0, at least for the even solution. Try to integrate the Schroedinger equation from -ε to +ε to obtain a relation between psi at x=0 and it's derivative.

To your second question: Where should the energy come from? A potential change already gives you the change of energy and the potential before and after the well are equal.

 

[user3]

I see. Thank You .