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Homework Statement
Let us suppose we have a particle with energy ##E## and ##E<U## and the potential defined as
##U(x)=0## for ##x<0## (I)
##U(x)=U## for ##0<x<L## (II)
##U(x)=U_0## for ##x>L## (III)
In this case ##E>U_0## and ##U>U_0##
Homework Equations
$$HΨ=EΨ$$
The Attempt at a Solution
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I find the general form of solution, which it is
##Ψ(x)=Ae^{βx}+Be^{-βx}## for ##β^2=2m(U(x)-E)/\hbar^2##
For region (I) I find that
##Ψ_1(x)=c_1sin(\frac {\sqrt{2mE}} {\hbar}x)+c_2sin(\frac {\sqrt{2mE}} {\hbar}x)##
For region (II)
##Ψ(x)=De^{\beta x}+Ee^{-\beta x}## for ##\beta=\frac {\sqrt{2m(U-E)}}{\hbar}##
Is this true ? Because in the site of the hyperphysics it says it should be,
##Ψ(x)=Ee^{-\beta x}## for ##\beta=\frac {\sqrt{2m(U-E)}}{\hbar}##
I am not sure how we can derive this mathematically ? Why the
##De^{\beta x}## term vanishes ?
http://hyperphysics.phy-astr.gsu.edu/hbase/quantum/barr.html
For the region (III)
I find that
##Ψ(x)=Ge^{iαx}## for ##α=\sqrt{2m(E-U_0)}/ {\hbar}##
So ##D=0## or not ? If so why its 0 ?
If ##U(x)=U##, ##x>L## then it was logical thing to say that ##D=0##, but ##U## is just for some distance.
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