(Quantum review homework)
*For problems 1.4 and 2.1, let's all use the harmonic oscillator wave-functions posted by Nicolas Blanc in the comments below.
For this homework assignment our main focus is on results and relationships. Integrals can be done via wolfram alpha (see video above); there is no point in doing them by hand or showing work. Please comment and ask questions below. (Extra credit for finding errors, ambiguities, etc..)
Section 1: quantum length scales
1.1 Consider an electron in a 1D quantum potential \(U(x) =\frac{1}{2} k x^2\) where k= 100 \(eV/nm^2\). Suppose that the electron is in a state of the form \(\psi(x) = A e^{-x^2/(2a^2)}\).
a) Find A. Calculate the expectation value of \(x\). Calculate the expectation value of \(x^2\).
b) What is the expectation value of the U ?
c) Calculate the expectation value of the kinetic energy, T . What is the relationship between T and the expectation value of \(p^2\) ?
d) Graph the expectation values of T and U as a function of a. What is each one for a = 1 nm? What about for a = 2 nm?
e) What value of a gives you the lowest value for the total energy, T + U ? How does this value of a depend on each member of the "quantum triumvirate" m, k and \(\hbar\)?
1.2 For the results for problem 1.1, calculate the product of the expectation value of \(p^2\) times the expectation value of \(x^2\). On what does this depend? Discuss.
1.3 a) For an electron in the ground state of a hydrogen atom, express the quantum length scale known as the Bohr radius in terms of m, e and \(\hbar\).
b) For an electron in the ground state of a 1D harmonic oscillator, express the
quantum length scale, "a", in terms of m, k and
\(\hbar\).
c) discuss.
1.3e (see below)
1.4 a) For the 3 lowest energy states of a 1D harmonic oscillator, write the normalized energy eigenstate wave-functions. (Please use the HO quantum length scale to simplify both the normalization factor and the exponential.)
[Please post your answer here. That way we can all converge on the same thing.]*
(see comments!)
b) Plot each of these wave functions as a function of x.
c) Using wolfram alpha to save time, calculate the expectation value of \(x^2\) for each of these states. I am thinking that your answer will be in the form of a number times \(a^2\).
d) Take the square root of that to get something with units of length. Plot that (3 points) as a function of the quantum number. What might this tell you about the nature of these states?
Section 2: (time dependence)
2.1 Consider an electron in a harmonic oscillator potential (the same potential as in problem 1.1). Suppose the electron is in a normalized state that is an equal mix of the ground state and 1st-excited state?
a) Write down a mathematical expression for that state including normalization and time dependence.
[Please post your answer here if you like. That may help us resolve any possible normalization or interpretation issues.]
b) For this state, calculate the expectation value of x.
c) Is the expectation value of x zero or non-zero. Does it depend on time or not?
Section 3: (square wells & multiple well systems) (This part is important for this class.)
3.1 Consider an infinite square well that is 0.1 nm wide.
a) What is the energy of the ground state?
b) What is the energy of the 1st excited state?
c) Is this kinetic or potential energy?
[For energy calculations like the ones in this problem I would suggest multiplying by \(c^2\) and using \(m_e c^2 = .51 \times 10^6 \: eV\) and \(\hbar c = 197 eV*nm\). (are those correct?) Joules and kg don't work well in the quantum realm. eV and nm are much better.] (Also, I think, \(\hbar^2/m_e = 0.076 eV*nm^2\))
3.2 a) sketch the ground state for a finite square well that is 0.1 nm wide (and about 200 eV deep). How is it different from the infinite square well ground state (same width well)?
b) sketch the 1st excited state for a finite square well that is 0.1 nm wide and about 200 eV deep.
3.2 e (see below)
3.3 Consider a potential that involves two square wells each 0.1 nm wide and with a separation between them of about 0.05 nm. (Suppose the potential is zero between the wells and outside them, and that it is about -200 eV within each well.)
a) sketch the 4 lowest energy states for this two-well system.
b) Do the same thing for a system of 4 identical square wells, except in this case sketch the 8 lowest states.
c) think about and discuss the organizing principle for these states. Please feel free to post your thoughts on this below.
d) discuss the organization of the energies of these states. Might there be any clustering? How might that arrange itself? [Post here on clustering.]
3.3e (see below)
Section 4: Displaced functions. (There was an error in this problem. It was corrected on Friday.)
4.1 Suppose a= 0.5 nm.
a) Plot \(e^{-((x)^2/a^2}\)
b) Plot \(e^{-((x-1 nm)^2/a^2}\) as a function of x.
c) Plot the function \(e^{-(x+1 nm)^2/a^2} + e^{-x^2/a^2} + e^{-(x-1 nm)^2/a^2}\) as a function of x.
---------------------------------------------
1.3e (hydrogen atom) For a wave-function of the form \(\psi(r) = (\pi a^3)^{-1/2} e^{-r/a}\) calculate U and T (expectation values) and find the value of a that provides the lowest total energy (T + U). You can post your results here if you like or send them to me by email.
3.2e (finite square well) Using some approximate numerical methods, or any approach you like, find the energies and wave-function of the bound states for a square well 0.1 nm wide and 200 eV deep. Assume the potential is centered at x=0 and that the ground state has the form \(A_1 cos (k_1 x) \) inside the well and \( B_1 e^{-(x-.05 nm)/a_1} \) outside the well on the right. Note that from the Wave Equation (Schrodinger eqn) \(\hbar^2/2ma_1^2 = - E_1\) (which is positive), and \(\hbar^2k_1^2/2m = E_1-V_o\) where \(E_1\) is the ground state energy and \(V_o = - 200 eV\). (Please post results here. It may be useful for the class if someone does this problem because then we can use it as a model atom from which to construct a simple crystalline solid and it will provide us with a particular point of reference.)
3.3e (two square wells) (This is a difficult but not impossible problem that may be pretty interesting and relevant to crystal states.) Calculate the expectation values of the kinetic
energy and potential energy for the ground state and first excited
state of a two well potential. Call the center-to-center distance c and let's try c = 0.15 nm, L= 0.1 nm and let the well depth be -35 eV. [Suggestions: put the origin between the wells and then assume that in the space between the wells the wave function has the form \(C_1 cosh
(x/a_1)\) for the ground state and \(C_2 sinh (x/a_2)\) for the first
excited state. That \(a_n\) parameter will have the same value as the one you use outside the wells. Why? Inside the wells, the form of the wave function is \(\pm A_n cos(k_n(x-c/2)\).
For Problem 1.1(b), do we have to calculation? I have an intuition what the expected value of U to be without doing any calculations.
ReplyDeleteTo other students, please don't worry about asking "stupid questions" in the blog comments. I'm sure you might be clearing up things that your fellow classmates wants to be cleared about but may not have the courage to ask. This isn't Youtube comments after all!
For 1b you can just use the calculation that you did in 1a, so you don't have to make any additional calculation. For 1a the calculation is for the expectation value of x^2 is:
Delete\(\int_{-\infty}^{+\infty} A^2 e^{-x^2/a^2} x^2 dx \)
You have to find A first. Then I think you might get a relationship between the expectation value of x^2 and the wave-function parameter a.
This comment has been removed by the author.
ReplyDeleteFurther LaTeX testing
ReplyDelete$$KE = \hat{p}^{2} / 2m_{e}$$
This comment has been removed by the author.
ReplyDeleteMore testing of LaTeX
ReplyDelete$$KE = \frac{\hat{p}^{2}}{2m_{e}}$$
I like this one!
DeleteI think we just have to look them up and write them for the problem. I am sure there's a way to calculate them but it would be very time consuming.
ReplyDeleteI would like to get the final word from Zack though.
Yes, don't calculate them. Too time consuming and not that interesting. Here is one:
ReplyDelete\(\psi_3 (x) = A_3 (1-\frac{2x^2}/{a^2}) e^{-x^2/2a^2}\)
That is the sort of intuitive form that we want where the quantum length scale is used.
\(\psi_3 (x) = A_3 (1-\frac{2x^2}{a^2}) e^{-x^2/2a^2}\)
ReplyDeleteWhere I believe that
Delete\(A_3 = \frac{1}{\pi^{1/4} (2a)^{1/2}}\)
May I ask what's the source of your wave functions? I'm looking at the Harris book and while I'm assuming they are the same thing, I would like to be consistent with the wave functions being used in the class.
Deleteoops. that's not right. see below.
ReplyDeleteExcellent question. I think that is a good subject for peer discussion. I look forward to seeing that discussion here!
ReplyDeleteI think maybe if you reexamine problem 1.3 that might be a good context in which to think about what quantum length scale means.
ReplyDeleteI think maybe the word "scale" might be confusing things. Maybe we should just call it a quantum length.
ReplyDeleteBy length scale I just sort of mean something that tells us about the characteristic size of something. Like for the earth it might be about 10,000 miles; for the Pacific ocean maybe about 3,000 or 4,000 miles; for the UCSC campus, maybe 2 or 3 miles, and for your apartment about 50 ft.
For 1.3(e), do we assume that y = z = 0?
ReplyDeleteAnd it's been corrected.
DeleteI think you would have to do a 3 dimensional integration for that.
DeleteThanks for pointing that I out. I corrected my mistake in that.
DeleteFor some reason, it seems that I don't need to directly integrate with the angles because of symmetry and the lack of dependance on the angle. I'm thinking that the value for 1.3e is $$ 2\pi^{2} \int_{0}^{\infty} \Psi(r) dr $$
DeleteAm I on the right track or am I just getting too lazy in doing 3D integrals.
I was under the impression that the Jacobian has been taken to account already. I assumed that the function that is given has already been converted from restangular to spherical.
DeleteI got 2\pi^{2} from the d\theta and d\phi integration from 0 to 2\pi and 0 to \pi respectively. Since there isn't a dependence of either angles, we are left with just integrating r.
Repost because I can't LaTeX in the last post.
DeleteI was under the impression that the Jacobian has been taken to account already. I assumed that the function that is given has already been converted from rectangular to spherical.
I got 2\$pi$^{2} from the d$\theta$ and d$\phi$ integration from 0 to 2$\pi$ and 0 to $\pi$ respectively. Since there isn't a dependence of either angles, we are left with just integrating r.
Okay, I realize that LaTeX on the same line doesn't really work well on the comments.
DeleteAfter thinking about your comment, I think I stand corrected. The integral over all space would be given by
Delete$$ \int_{all space} \psi\vec(r) dV $$
where we define $$dV = r^{2} sin^{2}(\phi)$$
We define theta to be from 0 to 2pi and phi to be from 0 to pi.
Doing the substitution and converting it to a triple integral would yield
$$ \int_{0}^{\pi}\int_{0}^{2\pi}\int_{0}^{r} \psi(r, \theta, \phi) r^{2} sin^{2}(\phi) dr d\theta d\phi $$
Which would then yield a mathematically correct result. So then pi would be independent which lets us take the 2pi out. Phi on the other hand would be within the sin factor.
I am under the impression that we have to take angle in consideration since this is for the Hydrogen atom which is a 3-D thing.
As for LaTeX, yeah, the LaTeXing would have to be in a new line of the comments.
Oops, you're right. That was just me being dumb at the late hours.
Delete$$ \int_{0}^{\pi}\int_{0}^{2\pi}\int_{0}^{r} \psi(r, \theta, \phi) r^{2} sin(\phi) dr d\theta d\phi $$
Thanks for catching that. I should really read my post for errors since there isn't an edit button.
DeleteI should post my results for 1.3(e) with my work shown tomorrow.
Right. good point. your way is better.
ReplyDeleteHere is another one.
ReplyDelete\(\hbar^2/m_e = 0.076 eV*nm^2\)
This comment has been removed by the author.
ReplyDeleteI'm getting a negative Kinetic Energy for 1.3e. Not sure if anyone else got that result. The negative is coming from the Kinetic Energy operator and taking the second derivative of the function makes the constant positive.
ReplyDeleteAs discussed during office hours, I will agree with you!
ReplyDeleteThis comment has been removed by the author.
ReplyDeleteThis comment has been removed by the author.
DeleteDid you take the integral from negative infinity to infinity with all positions and from 0 to infinity with respect to time to find the constants that normalized the function and use superposition?
ReplyDeleteI actually should've asked is that do we need to consider the normalization due to the time component of the wave function.
ReplyDeleteI believe it is only for states which satisfy V < E < 0, which is the condition for bound states. I haven't figured out the easiest way to find k yet, though.
ReplyDeleteGood point, I was assuming something similar, anyway thanks. Oh and k can be calculated by graphing the two equations you get from solving S.E. ( well at least that is how I did it)
ReplyDeleteHow many bound states do you think there are?
ReplyDelete