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Finite Well

by 'AQF
Tags: finite
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'AQF
#1
Nov1-05, 12:27 PM
P: 33
I need to explain why, as the energy of a bound state in a finite potential well increases, the wave function extends more outside of the well. I need to do this from both a mathematical and a physical point of view. I think I know the mathematical explanation (see attached image). Can anyone help me with the physical explanation?
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elhinnaw
#2
Nov2-05, 03:51 AM
P: 13
I dont think you quite understand the physics or the math of the problem yet.

To answer your question quickly, the wave function inside the well is:
Asin(lx) + Bcos(lx)

outside the well is:
Fe^(-ka) where k: Sqrt[-2mE]/hbar (k is real because E<0)

Now when you INCREASE the energy, you are actually making the exponent "less negative." For example, if you original E made the wave function Fe^(-5) and then you increased the energy, your new wave function would look like Fe^(-1). Now you can easily check on your calculator that e^(-5) decays faster than e^(-1), hence the bigger energy will have a larger wave function outside the barrier, making it more probable.

You can also check this by explicitly calculating the transmission coefficient, which will yield the correct answer again (at least it did on this little napkin here).

Physically, things with higher energies have higher tunnelling coefficients. Imagine you are diving into a deep pool of sand (or punching thru a wall). The faster your body is moving right before you hit the sand (or the faster your fist is going right before you hit the wall) the more energy you will have and the deeper you will go into the sand (or the better your chances are to break a hole in the wall).


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