- #1
bknock
- 7
- 1
I was curious what a delta-function potential barrier would do to a free (E>0) 1-dimensional particle, so I figured it out. I used V = b δ(x) for the potential energy. I got the expected result that full reflection will occur as |b|→∞ or as E→0. These results make sense intuitively when b>0. My intuition is mystified because I get the same result when b<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 call m, and define
k = √(2 m E / ħ²)
Since the wave function must be continuous at x=0, I get
ψ1 = A (ei k x+B e-i k x)
ψ2 = A (1+B) ei k x
where ψ1 is the incoming and reflected particle on the left, and ψ2 is the transmitted particle on the right. A is 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 for B, and I got
R = 1/(1+W) where W ≡ 2 ħ² E / (m b²)
Thanks in advance for any insight!
First of all, I give the particle mass, which I call m, and define
k = √(2 m E / ħ²)
Since the wave function must be continuous at x=0, I get
ψ1 = A (ei k x+B e-i k x)
ψ2 = A (1+B) ei k x
where ψ1 is the incoming and reflected particle on the left, and ψ2 is the transmitted particle on the right. A is 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 for B, and I got
R = 1/(1+W) where W ≡ 2 ħ² E / (m b²)
Thanks in advance for any insight!