## 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.

### Quantum Tunneling: A Different KInd of Calculation

I did some interesting physics yesterday so I thought I should write it up. The question was about quantum tunneling. It came from my friend's 3rd year Physical Electronics course in Electrical Engineering. You had a MosFET with a silicon dioxide insulating layer between the gate and the channel. They gave you the the barrier potential in the insulator, and told you to assume an electron with a certain energy (less than the barrier potential) in the gate. What was the "probability" or rate of tunneling?

This is a pretty standard problem which you can look up on Wikipedia. You calculate the form of the free-electron solutions in all three regions and match up the boundary condition. It's a little messy but it works.

I thought I could do better. The problem is messy because the conditions are different on the left and right hand sides. So I came up with the idea of working with symmetries. Instead of taking a freely propagating wave from the left and following it through the barrier, I put the whole thing inside a bigger potential well and considered the steady state solutions. Of course there are two minimum-energy solutions: the symmetric and the anti-symmetric, separated by a very tiny energy difference.

I don't like to carry too many letter symbols so I picked numbers that would come out nicely. From the terms of the problem, it's easy to calculate the free-propagation constants in both regions. So I allowed myself to tweak the problem parameters so the propagation constants come out to nice integers. Assuming pi = 22/7, here is the problem as I set it up:

You see I gave the well a width of 20 (you can call it 20 Angstroms if you like or 20 millimeters, it won't matter in the end...it's all about the geometry.) Can you see why this fits the sine wave perfectly? It's because the wave penetrates into the barrier. I've assumed that the penetration length is 2, so 20+2=22 and everything fits. The anti-symmetric solution is the same basic idea:

You can see I set it up so that there is a wavelength of 44, which fits perfectly into the box. But actually that's a bit of a cheat. If I were matching up a sine wave to an exponential at the boundary, then I would indeed get a wavelength of exactly 44, because of the well-known property of an exponential whereby the projection of the slope intersects the x-axis at exactly the decay length (I'm also assuming the sine wave is in the small-angle regime where sin(x)=x):

I don't really care about the constant in front of the sine wave, but I just threw it in here to show that it's easy. I used the property of sine waves that the projection of the tangent from the zero-crossing reaches the altitude of the sine wave after one radian. The point is that if I'm matching my sine wave (period = 44) to an exponential (decay length = 2) then this is how they line up. Everything fits.

Except I'm not exactly matching the sine wave to an exponential. I'm matching it to cosh (in the symmetric case) and sinh (in the antisymmetric case. Cosh is just a little higher and a little less steep...so it matches the sine wave at a slightly different position. Actually...the only way to match it up is to make the sine wave just a little bit longer. Instead of a half-length of 22 angstroms, the sine wave stretches to around 22.04....close to 0.2%, if you like. Similarly, in the anti-symmetric solution, the sine wave has to be shortened by the same amount. So the anti-symmetric solution has a slightly higher frequency than the symmetric solution.

We'll see what the implications of this are when we come back.

Oh...how did I calculate the 0.04 angstroms? That's the beauty of this method...it's pure geometry, just looking at the ration of the function with its derivative on both sides of the boundary, and using the small-angle approximation tan(x) = x. It falls right out.