I was curious what a delta-function potential barrier would do to a free ((adsbygoogle = window.adsbygoogle || []).push({}); E>0) 1-dimensional particle, so I figured it out. I usedV=bδ(x) for the potential energy. I got the expected result that full reflection will occur as |b|→∞ or asE→0. These results make sense intuitively whenb>0. My intuition is mystified because I get the same result whenb<0. Why should a potential well cause complete reflection? I was hoping someone could help my intuition grasp this result. Or maybe I did something wrong in my work, so I am showing my work below...

First of all, I give the particle mass, which I callm, and define

k= √(2mE/ħ²)

Since the wave function must be continuous atx=0, I get

ψ_{1}=A(e^{i k x}+Be^{-i k x})

ψ_{2}=A(1+B) e^{i k x}

whereψ_{1}is the incoming and reflected particle on the left, andψ_{2}is the transmitted particle on the right.Ais the normalization constant, but, since the particle is free, there is no need to solve for it. I then worked out that the fraction reflected is

R= |B|²

Using the rule about how the slopes of wave functions change across delta-function potentials, I solved forB, and I got

R= 1/(1+W) whereW≡ 2ħ²E/ (mb²)

Thanks in advance for any insight!

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# Bizarre results of 1-dimensional delta-function potential well

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