## Tuesday, February 3, 2015

### Penetrating the Barrier: Some Calculations

Yesterday I calculated the difference between the two ground states of the double well (the infinte well with a finite barrier in the middle.) The two states were of course the symmetric and the antisymmetric cases, with the antisymmetric having a slightly higher energy:

I picked a wavelength of 44 angstroms (7 angstroms = 1 radian) for the sine wave, and a decay length of 2 angstroms for the exponential region, so the ration of the two parameters was 7:2. And then I picked the dimensions of the box so that the waves would just fit inside. And that's pretty much all you need to do the calculation. The Wikipedia formula asks you to calculate the transmission coefficient in terms of E and V, the energies in the two zones respectively, but the formula they give is a bit redundant...you can re-express it just in terms of the ratio (7:2 in this case) of the two characteristic lengths:
You can see I've got r = 7/2, and ka = 4, so plugging in the numbers, we get a transmission coefficient of 1/2483, or close to 0.04%.   Is this the same as I got using the steady state solutions?

Yesterday I calculated that the wavelengths of the symmetric and antisymmetric modes were different by 0.4% (about one part in 250). But in quantum mechanics wavelength is momentum, and energy is frequency (momentum squared). So the frequencies are different by one part in 125.

Here is where it gets interesting. You start out with the electron all on one side of the barrier. How do you do that? By having the symmetric and antisymmetric modes in phase so they re-inforce on the left and cancel on the right. After 125 cycles, they will be back in phase again. But after 62.5 cycles, the relative phases will be reversed...so all the wave function will be on the right hand side.

Is this the same result as we got from the Wikipedia formula? It's hard to say, because the Wikipedia formula is expressed in terms of the Copenhagen interpretation as the "probability" of a particle getting through the barrier. Where is my "particle" in terms of standing waves?

Well, one way to interpret it would be to imagine the particle bouncing back and forth in the potential well. If it bounces once on each cycle, that means after hitting the wall 62 times, it gets completely through to the other side. That's like a 1.5% penetration on each cycle...more like 3% or 4%, actually, because there's the competing probability of it returning from whence it came. That's a lot more than the Wikipedia calculation. Have we done something wrong?

It's actually not quite that bad. Remember in quantum mechanics there's a discrepancy between the phase velocity and the group velocity. For electrons, the group velocity is twice the phase velocity. We've been treating the electron as though it travels with the phase velocity of the wave function...it's actually twice that, so where we thought it was hitting the wall 62.5 times, it was actually 125 times. So our nominal penetration ration goes down by half, to below one percent. That's a little better, but still a long way off. What gives?

There is a fascinating answer to this question, and it gives us a very deep insight into the whole subject of quantum resonance. We said that each time an electron strikes the barrier, it has a 1/2500 chance of getting through. But what does this mean in terms of the wave function? It means the amplitude of the transmitted function is one fiftieth. The power goes as the amplitude squared.

In the coupled well system, the transmitted wave is in phase each time it re-strikes. So the amplitude on the transmitted side goes as one fiftieth, two fiftieths, three fiftieths, etc. After only fifty strikes, it is 100 percent through!

Not exactly....because once the amplitude starts building up on the right hand side, there is the probability that it will come back through the other way. But early in the game, that probability is negligible. So while the amplitude is growing as 1/50, 2/50, 3/50....   the probability is growing as 1/2500, 4/2500, 9/2500... in other words, it is growing parabolically.

How close does this parabola fit to the sine wave which represents the oscillating probability? Pretty close, as it turns out. I'm not going to do it in detail, but if you look at the Taylor Expansion for the cosine function, the x-squared term projects to -1 at when x = 1.41 radians (the square root of two. On our sine wave, where 125 strikes (the electron striking the wall) makes is a half-cycle, that comes to 56 strikes.

It's pretty close.

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