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?
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?
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