# Heaviside Step Potential - Tunneling

1. Jun 6, 2013

### klawlor419

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. Lets 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.

2. Jun 12, 2013

### klawlor419

Anyone? Maybe this was just a dumb question.. feel free to call it like it is

3. Jun 12, 2013

### Avodyne

You can think of it as a ratio of fluxes. The flux is the probability per unit time to see the particle pass by you. The square of the wave function gives you a probability per unit length, so you have to multiply this by a velocity to get probability per unit time.

A more sophisticated version of this is to compute the probability current. See
http://en.wikipedia.org/wiki/Probability_current

4. Jun 12, 2013

### klawlor419

Cheers, thanks for the response. I see why it is a probability per unit time now. Why can't I just look at the ratio of the outgoing and incoming amplitudes? Because the energies are different on the step sides? Why does that matter? I'm still missing something.. I understand the probability current derivation but I don't understand why the standard ratio of terms doesn't work.