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.

## 1 comment:

Mr. Green, do you think your methods can be generalized to other more compicated potentials? Or to quantum scattering theory?

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