- #1
acme37
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Question: What happens when the energy of the state is exactly equal to the barrier it hits?
Say the particle is traveling to the right and hits the barrier. Before hitting, you have your typical solutions, say Aexp(ikx) + Bexp(-ikx). After hitting, the V(psi) and E(psi) terms cancel. Integrating then yields some Cx + D, which diverges at infinity, so C=0 and you have constant wave function, psi(x)=D
Here's where my question is. You can either ignore normalization and find A,B in terms of D, the compute things like transmission/reflection coefficients. When I did this I got A=B=D/2.
I computed R=(|B|/|A|)^2 = 1, and also T = (|D|/|A|)^2 = 4.
So I normalize these by hand and say R=1/5, T=4/5
On the other hand, a constant wave function is non-normalizable. So what does this mean? My friend thinks it is a non physical solution, and thus all the wave must be reflected. But the free particle is also initially non-normalizable until we express it as an integral over a continuous wave number. But that's not exactly the same thing here, and the non-normalizable argument seems strong.
Our problem was to find the SE for E=V, without normalizing, then find the reflection coefficient.
So basically, what happens when E=V for a barrier potential
Say the particle is traveling to the right and hits the barrier. Before hitting, you have your typical solutions, say Aexp(ikx) + Bexp(-ikx). After hitting, the V(psi) and E(psi) terms cancel. Integrating then yields some Cx + D, which diverges at infinity, so C=0 and you have constant wave function, psi(x)=D
Here's where my question is. You can either ignore normalization and find A,B in terms of D, the compute things like transmission/reflection coefficients. When I did this I got A=B=D/2.
I computed R=(|B|/|A|)^2 = 1, and also T = (|D|/|A|)^2 = 4.
So I normalize these by hand and say R=1/5, T=4/5
On the other hand, a constant wave function is non-normalizable. So what does this mean? My friend thinks it is a non physical solution, and thus all the wave must be reflected. But the free particle is also initially non-normalizable until we express it as an integral over a continuous wave number. But that's not exactly the same thing here, and the non-normalizable argument seems strong.
Our problem was to find the SE for E=V, without normalizing, then find the reflection coefficient.
So basically, what happens when E=V for a barrier potential