(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

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##

2. Relevant equations

$$HΨ=EΨ$$

3. The attempt at a solution

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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# Homework Help: Finding the Wavefunction for Tunneling,with tunnel lenght L

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