Please let me know of any typos no matter how small. Thanks. Also, question, comments and feedback are greatly appreciated! I may add more, but you can start this anytime.
Regarding units, in this class please use eV (for energy) and nm for length.
Semiconductor physics is based on quantum mechanics and understanding quantum states and energies. These problems are important to establish a foundation for this class.
If you run across any terms or concepts that you don't understand, and I expect that you will, please ask about them here. You get credit for questions. If you don't ask, I'll tend to assume that you understand everything, or that you haven't read the problems, or that you are reluctant for some reason to engage and interact. I feel that education is most effective when you do choose to engage and interact. I am not a fan of a "passive student" approach.
1. a) What width of infinite square well has a ground state energy of 10 eV?
b) For that square well, what is the numerical value of the wave-function in the center of the well?
c) If we center the well at x=0 then the wave function is A cos(kx). What is k? What are its units?
**Please post your thoughts, comments and answers to this question as a comment to this post as soon as you work on it. That is, post right away. Today. Now.
2. Consider a finite square well of the same width as you found for problem 1. Call this L. Let's center this well at x=0. Suppose the bottom of the well is at V(x) = -20 eV, and outside the well V(x)=0 eV.
a) what is the energy of the ground state more or less? (please post this as soon as you work on this. ASAP!)
[Let me suggest some notation here for square well problems including this one. Center the well at x =0. Inside the well write:
\(\psi_1 (x) = A \sqrt{\frac{2}{L}}cos (kx)\) where A and k are to be determined via bc's and wave equation.
outside right side write:
\(\psi_1 (x) = B \sqrt{\frac{2}{L}} e^{-(x-L/2)/b}\). L is the width of the well, right? b is a key parameter to be determined. B is a normalization parameter.
You can figure out b and k, and the state energy, without determining A and B. In this notation, by the way, A and B are unitless, and I think that A is something that can be compared to one to see how much the wave-function amplitude is diminished. Does that make sense?]
b) Sketch the ground state (gs) wave-function.
c) What is the numerical value of the wave-function in the center of the well? Is it smaller or larger than the gs wf of the infinite sq well of problem 1. Why? Discuss briefly. (Post your results and thoughts ASAP.)
d) How far outside the well does the wave-function extend? Comment below if you are not sure what this is asking. How might one quantify this? Speculate.
e).... what else should we explore for this well...? (Post you thoughts or ideas here.)
Note: It might make sense for you to skip problem 3 and go straight to problem 4. It is up to you. At any rate, don't get too bogged down on problem 3. If you have trouble with it you may want to skip it and go on to problem 4, which is a little more important.
3. Consider the limit of an infinitely narrow square well. That is, \( V(x) = -\alpha \delta(x)\) .
a) For what value of alpha is the energy of the ground state -10 eV? (What are the units of alpha?)
b) What is a characteristic length scale associated with this wave function? What is the "size" of the electron in this state? (To what region of space is it localized?)
c) Discuss how this wf relates to the gs wf of a finite square well.
d) What are the electon's KE and PE?
e) what else should we explore for this state?
**I'll make a separate post for question 4 so that we can work on
it collaboratively there via comments on that post. Please go to that post
to comment on problem #4.
4. Consider a well of the width we found for problem 1 of this homework but with a depth of 100 eV:
a) How many bound states does it have and what are their energies.
... (see HW1: part2 post)
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ReplyDeleteI tried the first problem, but got a very thin well...
ReplyDeleteI took the energy of ground state in the infinite square well, set it equal to 10 ev and solved for L:
$$
\frac{\pi^2 \hbar^2}{2mL^2} = E_0 \\
L = (\frac{\pi^2 \hbar^2}{2mE_0})^{1/2}\\
L = (\frac{\pi^2 \hbar^2}{2 \cdot 511 \cdot 10^3 eV/c^2 \cdot 10 ev})^{1/2} = (\frac{\pi^2 \hbar^2 c^2}{2 \cdot 511 \cdot 10^3 eV \cdot 10 ev})^{1/2} = (\frac{\pi^2 (197 ev nm)^2}{2 \cdot 511 \cdot 10^3 eV \cdot 10 ev})^{1/2} = 0.037 nm
$$
Did anyone else get this? - I think it is a very slim well..
slash open-paren x^2 slash close-paren yeilds:
Delete\( x^2\)
Its is latex syntax. you can search on latex to learn about it. It is pretty simple: e^{-x/L} can be put inside slash-parens to become:
Delete\(e^{-x/L}/)
so that second slash is the wrong way so it didn't work.I'll try again:
Delete\(e^{-x/L}\)
Hi Sophie,
DeleteI was looking at the problem and worked it backwards and for a L=.037nm you do not get 10eV, so then I tried to find the error.
All the constants check out....and I found that it was the last step you did not take the square root! Just realized it myself after making the same mistake.
So it should be (.037nm)^(1/2) = .192nm. Hope this helps.
-Valentin Urena
yes, latex syntax
DeleteSounds like you've already foundthe Schr{\"o}dinger equation inside the well! To get A, you need to normalize the equation:
ReplyDelete\begin{align*}
\Psi(x)&=A\cos(kx) \\
1&=\int\limits_{-\infty}^{\infty} \Psi^{*}(x)\Psi(x) dx
\end{align*}
Note that the Psi equation I wrote is only valid inside the well, outside it's just zero.
Here's hoping I didn't mess up the LaTeX for the blog!
Also, I agree that \(k\) should be meters\(^{-1}\).
ReplyDeleteOh! And once you find \(A\), you're good to go. Notice that in the center of the well the cosine term is at its maximum, and go from there.
ReplyDeleteLet's go with nm-1 for k.
ReplyDeleteand \(nm^{-1/2}\) for the units of the the wave-function (and of A).
Also, isn't \( A = \sqrt{2/L}\) ?
Long story short, on problem 1a I got:
ReplyDelete\begin{equation}\label{p1a}
L=\left(\dfrac{\hbar^{2} \pi^{2}}{2m E}\right)^{1/2} = \left(\dfrac{\pi^{2}\hbar^{2}}{2*511\times 10^{3} eV \tfrac{1}{c^{2}}* 10 eV} \right)^{1/2}=\boxed{0.193\text{nm}}
\end{equation}
Which... has already been discussed at length. Also as a sanity check, I thought of the Hydrogen atom. If we a approximated the hydrogen atom as an infinite square well with ground state energy 13.6 eV, Equation \ref{p1a} gives a little bit more than an angstrom. Good!
For Problem 1b, I solved the Schrod equation (finding \(k\) in the process) for the system and normalized the function to find \(A\).
\begin{align}
\Psi(0)&=\sqrt{\dfrac{2}{L}}\cos(0) = \sqrt{\dfrac{2}{L}} \\
k&=\dfrac{\pi n}{L}
\end{align}
Here is what I solved for part 1:
ReplyDelete1a) .192nm same as others have said above
1b) Reference the wave equation for an infinite square well and just plug in the value from part a, it should be = SqRoot(2/L) since Sin(Pi/2)=1.
1c) Just by centering the well at 0 I do not think K changes (nor the energy) so it should still be k = Pi*n/L which has units of 1/m.
For part 2 here is my thinking I have not worked it out all yet:
2a) Reference your old modern physics book and the
E=(K^2*Hbar^2)/(2*m), just plug and solve using n=1.
2b) ground state function has a node in the center and falls off exponentially at the ends (in opposite curvature) and does not have nodes at the ends.
2c)At the center there is a node so the function Y(x)=0. This is smaller than in part 1 because well, there is a node.
2d) It extends infinitely to both ends as e^-x on the right and e^x on the left of the well.
2e) I would like to explore how the wave function will change if we changed the L of the well or the V(x). And the relation between V(x) and L outside the well.
Hope this helps let me know what you guys think. Thanks.
For 1c) I do agree with you that K does not change, but since we are centering the well at X=0 meaning L also equals 0. So shouldn't we use K=sqrt(2mE)/(hbar)? We want to find the actual value of K right?
DeleteI got right around there too, -13.64eV to be specific.
ReplyDeletethat sounds incorrect because -10 eV is close to ground state energy of H and so the length should be not so far from the Bohr length scale.
ReplyDeleteFor 2c, I found the value of the distribution function in the middle of the well to be around \(\boxed{15.77}\). It's a smaller value than what we got for the infinite square well.
ReplyDeleteCompared to the infinite well, a finite well has a more 'smeared' distribution function: there becomes a non-zero chance to find the particle outside the bounds of the well, and a non-certain chance of finding the particle inside the well. We're smearing the probabilities out of the well. \\
If we were to imagine shrinking the depth of the well, the odds of finding the particle outside the well would steadily grow. As the well eventually shrinks into nothing, a uniform particle distribution would form, and the value of the function at \(x=0\) would approach its minimum possible value (assuming an isolated system).
what did you use for A? For the solution at the center of the well I got A•sqrt{2/L} but my answer is coming out to 1.8, which is extremely small right?
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