April 24: Thanks to some awesome work by
students in this class, I have some information on wave-functions and their parameters for the 2-square-well potential. This was obtained using the
TISE, boundary conditions associated with the TISE, and the normalization condition:
***You don't have to derive these parameters yourself. You can use the ones provided here! The most interesting part of the problem is how you use this information.
For the ground state:
\(\psi_1(x) = 0.593 cos [1.13(x-1.47 nm)]\), inside the well between x=0.5 nm and 2.5 nm.
\(\psi_1(x) = 0.139 cosh(b_1 x)\), between the two wells (between x=-0.5 nm and +0.5 nm).
\(\psi_1(x) = 0.238 e^{-b_1 (x-2.5 nm)}\), to the right (beyond x= 2.5 nm).
where \(b_1= 2.57 nm^{-1}\).
There are a lot of interesting things to notice about this. Since the individual wells are -0.3 eV deep and 2 nm wide, comparison to the single well ground state could be interesting and provide some key insights into how the quantum world works. For example, I notice that the k value for this double-well ground state is a little smaller than that for a single well. Why is that?
For the 1st excited state:
\(\psi_2(x) = 0.62 cos [1.17(x-1.48 nm)]\), inside the well between x=0.5 nm and 2.5 nm.
\(\psi_2(x) = 0.134 sinh(b_2 x)\), between the two wells (between x=-0.5 nm and +0.5 nm).
\(\psi_2(x) = 0.253 e^{-b_2 (x-2.5 nm)}\), to the right (beyond x= 2.5 nm).
where \(b_2= 2.55 nm^{-1}\).
What do these wave-functions look like? How can we use them to help us with problems 5 & 6? Let's discuss this here.
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videos added on May 1&2: These videos illustrate how to think about and approach this problem.
Does this align with what you did and with how you thought about this problem? Why or why not? Please discuss here how you originally thought about this problem and how you think about it now.
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Midterm 2 solutions
Here are solutions to midterm 2. I see that I did not write it in the solutions, but it is helpful for sp2 type integrals to recall that \(\...
What orders of magnitude should the tunneling time be? Is $10^2$ femtoseconds reasonable?
ReplyDeleteSounds right, using just the first 2 states, I found the half period of oscillation to be 517fs.
DeleteInteresting, I used four states and got about 230 fs. However, I'm not sure if the states for double square well form a complete basis, because I have some discrepancies for the approximated single square well function.
DeleteUsing 2, 3, and 4 states give me almost the same answer, so its probably an overkill to use 4 states. The first two states I have magnitude of coefficient about 0.7, 3rd and 4th states about 0.01.
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DeleteI had the same coefficients, but I don't think they matter much. When you take the magnitude squared you get two time independent terms, then a cosine term, and I just set the argument of the cosine to pi, is that what you did?
DeleteWhat I did is to construct the sums of magnitude squared of 4 complex numbers and find where the sum is 0. Each one is the ith fourier coefficient of the tunneled wave function (centered at x=-1.5nm) minus the ith fourier coefficient of the initial wave (centered at x=1.5nm) times $e^{-i E_i t/\hbar}$.
DeleteI'm not quite sure why you need to set argument to be $\pi$, could you elaborate on that please?
Also, the fourier coefficients I've got for two waves are bit different from each other, on the order of $10^-5$, might be a numerical error? I would expect the coefficients, at least in magnitude, should be the same. What did you get?
Just realized I did not use the correct unit, I now have 560 fs for the first minimum. Sorry about that!
DeleteThe first two coefficients are different, the 1st slightly higher than the 2nd, so that's no error.
DeleteIf we approximate the wavefunction with the first two terms, the only part of the probability density that changes with time has a cosine time dependence, so when the cosine goes through a whole period, we should be back to the original wavefunction centered in the first well, so it should be centered in the 2nd well in half that time. I think our results are mostly agreeing, the final answer can vary significantly depending on when you round things. If I use the energies rounded to 3 significant figures I think I got 540fs, but digoing through my calculator and plugging in the unrounded energies resulted in 517fs.
I meant the first coefficient for the tunneled and original state, not the first two coefficients of one of the states.
DeleteThat's a fantastic idea! I got 570 fs for that.
Another issue is wether the initial state can actually be written as a linear comb. of the 4 bound states. I have differences in wave function up to 0.004 including 4 states, up to 0.015 including first 2 states, it might just be a numerical error problem, or maybe not.
I see, yes I think those coefficients should be the same.
DeleteI can't say much about the completeness of the set, I haven't really got into that part of the book yet.
I think you have the better estimate, I got around 560fs after going back and keeping all the digits throughout the calculation.
DeleteGlad to hear our results agree :)
DeleteA to 3 significant digits is 0.606, and L is 1.52. The rest are good.
ReplyDelete*for 1ES
ReplyDeleteIs there a reason why the cos isn't just shifted by exactly 1.5nm? That would put it exactly in the center of the well.
ReplyDeleteMathematically it just doesn't work out, if you make that assumption and apply the boundary conditions, you will get an equation with no solutions.
DeleteIntuitively, I think the ground state should approach that of a single well as the width of the gap between wells goes to zero, drawing the maximum of the cosines towards the center, though I haven't confirmed this yet.
Since we are showing that an electron in the ground state of a single well will tunnel to another well that appears, it might make sense to think about the well on the left pulling on the probability density on the right, and vise-versa.
That does make sense. Thanks!
DeleteI think your intuition is correct. I tried to use L=1nm instead of 2 to model the single well with L=2nm, with potential shifted by .5001 nm, coefficients are almost the same as single well in the previous homework.
DeleteModeling with L=2nm would not be good , cause there are two additional states
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DeleteI am thinking that, as the space between the wells becomes really large then the "offset" of the cosine function inside the well would become very very small. Does that make sense? I haven't calculated it, but that is what I am thinking. Let's say we made the space between the wells 20 nm wide, so that the right hand well exists between x= 10 and 12 nm and the left-hand well between -10 and -12 nm, then I think the cosine would be centered at 11.00000 nm on the right (and -11.00000 nm on the left). My intuition* is that the offset would be too small to calculate, (and that it would become zero in the limit...). Does that make sense.
Delete