Start by initializing a particle in the left well and see if it can flow to the right well.

1. We would like to use the concepts of eigenstates to think about transport through a device, like a FET.

2. Assume that the 3 wells are a very simple model for a FET. If we can determine I1 and I2, we know the current flow through the devices

3. Start by initializing a particle in the left well and see if it can flow to the right well.

10. Obviously, it can flow.

11. Here is another example.

14. This one is clearly not flowing. Why?

15. Remember that in QM energy is related to frequency:

16. The 230 GHz indicates the rate at which the waveform in the left well is oscillating. The Schrödinger Equation is a linear equation and the above structure is a linear system. Therefore, if the input is 230 GHz, then the wave throughout the structure must remain 230 GHz.

17. We know that the 230 GHz corresponds to 0.94 meV. The middle well cannot oscillate at 230 GHz because it does not have an eigenstate at 0.94 meV. That is the reason the particle ca not flow: it doesn’t have an eigenstate in the middle well!

18. The previous waveform could go through the middle well because 3.75 meV corresponds to the ground state eigenfunction of the middle well.

19. We saw earlier that 15 eV was an eigenenergy of the 10 nm well, so we expect this one to go through.

22. Before too long, we can see the 2nd eigenfunction start to form in the middle well.

23. Eventually, it tunnels through.

24. Suppose I were to go back and store the values at the middle of the 10 nm well after I had initialized it in one of the eigenstates. What would I expect to see when I plotted the resulting data?

25. It would be an exponential with frequency determined by

26. I know that the Fourier transform of an exponential is a delta function in the frequency domain. In fact, each one of the eigenfunctions results in a spike at a certain frequency, which I can convert back to energy.

27. This leads me to believe that I will only get current flow if the particle is exactly at one of the eigenenergies of the middle well.

28. Of course, that represents the ideal situation where we had an infinite 10 nm well. In actuality, any given eigenfunctions is going to decay out.

31. I’d like to know how fast particles decay in and out of the channel. This will be the key to determining transport through the channel.

32. It is easier to measure the decay of the left well into the middle well.

40. The parameter gives the rate of decay: The parameter that is used in quantum mechanics is which is related to by has units of energy.

41. So at one of the eigenenergies, the time dependence is given by This is a causal function, i.e, positive time. Mathematically, we’d rather deal with the two-sided function

42. The Fourier transform of the function is We can convert this to a function of energy

44. We say that the eigenenergies have been “broadened.”

45. The broadening shows that even a waveform that is not exactly at an eigenenergy of the middle well has some probability of transmitting through.

47. Consider the total current flow through the channel:

48. A particle that starts in the left will only contribute to I1. We know the time characteristics of the particle are

49. If we could figure out how fast the particle flows out of the left well, that would tell us the current Due to normalization, integrates over space to 1. So I can say the current produced by initializing the left well to is

50. Of course, if we had initialized the right well, then The total current is The current flow depends on the following: 1. Availability of particles in the left or right well. 2. Availability of a corresponding eigenstate in the channel 3. The escape rate corresponding to the energy of the particle crossing the channel.