- #1
klawlor419
- 117
- 0
I'm working on a basic potential step problem from Griffiths QM. Its problem 2.34 in the textbook. I am having trouble understanding the transmission coefficient in this problem. I realize that I can calculate the reflection coefficient and just do 1 minus the result but I am looking for something more physical.
I understand the probability current view of the issue. Which is that the probability current must be conserved. Due to the fact that there is a different amount of energy on both sides of the well, the way in which this is justified is when you have the extra coefficient in front of the transmission coefficient.
I am missing the physicality of this coefficient. At least in the way that I am trying to think of it. Which perhaps is just wrong. Griffiths mentioned something about considering the wave speeds. Basically the factor turned out to be a ratio of the velocities in the following way,
$$\frac{v_{II}}{v_{I}}=\sqrt{\frac{E-V_0}{E}}$$
So that the transmission coefficient is,
$$T=\frac{v_{II}}{v_{I}}\frac{{\lvert C \rvert}^2}{\lvert {A}\rvert^2}$$
I don't understand this result though. At least not entirely. (As a note - I am assuming that Region II is the region where the potential step is. And that Region I has no potential.)
It seems to make sense that as the speed of the particle in Region II gets really small the tunneling probability is small. It has less energy, and is less likely to reach infinity. Let's say that we could fix $$v_{II}$$ and make $$v_{I}$$ larger, I don't necessarily understand why the tunneling probability would go zero.
My thought is that if it has more energy on the left side of the step shouldn't there be a larger probability of transmission? If there is more energy on the zero potential side of the step when the particle mets the step it should have more energy, which makes me think that it should be more likely to tunnel through.
If someone could help me clear up my confusion it would be greatly appreciated. I think that I am over-thinking the factor. I can't find the logic for why it should be there unless I consider the probability current, and I would like another explanation.
I understand the probability current view of the issue. Which is that the probability current must be conserved. Due to the fact that there is a different amount of energy on both sides of the well, the way in which this is justified is when you have the extra coefficient in front of the transmission coefficient.
I am missing the physicality of this coefficient. At least in the way that I am trying to think of it. Which perhaps is just wrong. Griffiths mentioned something about considering the wave speeds. Basically the factor turned out to be a ratio of the velocities in the following way,
$$\frac{v_{II}}{v_{I}}=\sqrt{\frac{E-V_0}{E}}$$
So that the transmission coefficient is,
$$T=\frac{v_{II}}{v_{I}}\frac{{\lvert C \rvert}^2}{\lvert {A}\rvert^2}$$
I don't understand this result though. At least not entirely. (As a note - I am assuming that Region II is the region where the potential step is. And that Region I has no potential.)
It seems to make sense that as the speed of the particle in Region II gets really small the tunneling probability is small. It has less energy, and is less likely to reach infinity. Let's say that we could fix $$v_{II}$$ and make $$v_{I}$$ larger, I don't necessarily understand why the tunneling probability would go zero.
My thought is that if it has more energy on the left side of the step shouldn't there be a larger probability of transmission? If there is more energy on the zero potential side of the step when the particle mets the step it should have more energy, which makes me think that it should be more likely to tunnel through.
If someone could help me clear up my confusion it would be greatly appreciated. I think that I am over-thinking the factor. I can't find the logic for why it should be there unless I consider the probability current, and I would like another explanation.