1. Graph the (spatial) wave-functions Psi1r and Psi2r as a function of r. (Note that r cannot be negative.) Where is "a" in your graph?
2. Graph Psi1r and Psi2r as a function of x along the line y=z=0 (i.e., the x axis). Indicate where "a" is in your graph.
3. Graph Psi2x as a function of x along the line y=z=0 (i.e., the x axis). Why is it pointless to graph Psi2x along the y or z axis?
4. Illustrate the nature of Psi2x using either a polar plot or contour plot of the x-y plane.
(This is an important problem!)
5. Calculate the expectation value of x for:
a) an electron in the state Psi2r,
b) an electron in the state Psi2x,
c) an electron in the state (sqrt(1/3)*Psi2r + sqrt(2/3)*Psi2x)
6. a) Calculate the expectation values of x and y, respectively, (2 different calculations) for an electron in the state (sqrt(1/3)*Psi2r - sqrt(1/6)*Psi2x+ sqrt(1/2)*Psi2y).
**Please note that there was a critical typo in this problem. See discussion below to get an idea where the error was. (This error propagated into 8 and 10.)
7. For which of the states in problems 5 and 6 would the expectation values (of x or y) be time dependent? Are you sure? Explain.
8. Calculate the expectation value of y for an electron in the state (sqrt(1/3)*Psi2r + sqrt(2/3)*Psi2x) and then, on a graph where the axes are the x-y plane indicate, as a dot:
i) the location of the (x,y) expectation values for that state (sqrt(1/3)*Psi2r + sqrt(2/3)*Psi2x),
ii) the location of the (x,y) expectation values for the state (sqrt(1/3)*Psi2r - sqrt(1/6)*Psi2x+ sqrt(1/2)*Psi2y).
(This is a fairly important problem to do carefully and understand.)
9. never mind.
10. Advanced. Are the states:
(sqrt(1/3)*Psi2r + sqrt(2/3)*Psi2x) and
(sqrt(1/3)*Psi2r - sqrt(1/6)*Psi2x+ sqrt(1/2)*Psi2y)
orthogonal (in the sense of the inner product we discussed involving integrals over 3D space)? (For that determination you may assume that Psi2r, Psi2x and Psi2y are mutually orthogonal.)
11. Can you find another one or two states that have an energy E2 (1st excited states) and that are orthogonal to the 2 states in problem 10 (and to each other)?
12. Optional graph, thought and discussion problem. Please post any thoughts, questions or results for this problem here:
Suppose you do not like to do grungy calculations and integrals, but you want to get an idea of whether the expectation value of x is positive, negative or zero for the state sqrt(1/2)[Psi2r+Psi2x]. Can you perhaps graph that along just the x-axis (where y and z are zero)? (Why is that a pretty easy graph?) Can you then guess as to whether the expectation value of x is positive, negative or zero from looking at that graph?
Solutions follow:
Professor,
ReplyDeleteFor problems 1-3 are you asking us to graph the wavefunction vs. r or the probability densities vs. r?
The wavefunction.
Deletein questions 6,8, and 10, we have the state {sqrt(1/3)*Psi2r-sqrt(1/6)*Psi2x+sqrt(1/2)*Psi2x}
ReplyDeletedid you mean to mix together two Psi2x's or was one of them supposed to be something else like Psi2y?
Kevin: very good point. These 3 problems do all require the same state, and the state is not correct as presented now.
DeleteI would like to ask each member of the class: what is it that provides evidence that this is really not a reasonable problem statement or likely to be what was intended? (there are several things I imagine).
In addressing the question: "what should this state really be?", it might be helpful to realize that it is part of a triplet of states that can be constructed with Psi2r, Psi2x and Psi2y. (and that the state from problem 5 is part of that triplet as well.)
I am looking forward to seeing some discussion of what the state for problem 6 (and the related parts of 8 and 10) should actually be.
Good call. Yes, one if them must be Psi2y. Can you decide which one?
ReplyDeleteyes how about the last one with the 1/sqrt(2)
ReplyDeleteSounds right. I guess I will edit the problems to correct that. Thanks.
DeleteRegarding questions 1-3, since we're graphing the wavefunction, i.e. Psi2r=A*(1-r/2a)*e^-(r/2a) vs. r (correct me if I'm mistaken), is all we're supposed to see in #1 a simple decaying exponential (e^-r)? What is that telling us exactly?
ReplyDeleteIn #2, does Psi2r depend on x? Is the graph constant for Psi2r then different for Psi2x?
In 1: Psi1r is a simple decaying exponential. Psi1r is not a simple decaying exponential, there is more to it (because of the (1/r/2a) factor that you correctly mention).
Delete"What is that telling us exactly?"
Well it is partly given us a sense of what these two functions look like, and partly it is a warm-up problem for the later, more difficult problems. Also, very importantly, it is helping you understand how to plot a function of r vs x for y=z=0.
"In #2, does Psi2r depend on x?"
DeletePsi2r very much does definitely depend on x. Think of it this way: for y=z=z, that is, along the x axis, r= sqrt(x^2+0+0), right? So r depends on x, and thus Psi2r depends on x... So Psi2r is not constant (vs x). Does that make sense?
Oops, I meant: (because of the (1-r/2a) factor that you correctly mention).
DeleteThree posts up when you said "plot a function of r vs. x" you didn't mean use r and x as your axes did you? In the problems calling for plotting along a specific axis I plotted the value of the wave function against that axis. That's what you meant isn't it?
Deletei think you did it correctly.
DeleteI am going to sign off for tonight, but I would encourage you to post questions anyway and to answer each other's questions.
ReplyDeleteHmmm, no questions. That's disappointing. If we do get some questions, and they are hard to answer just in writing, then I can make a video reply showing you how to do some things people may be stuck on.
ReplyDeleteWhat does it mean for Psi(2r) to be negative wrt r? Is this some kind of phase shift? It's strange to think of this Psi(2r) form as a wave that is stationary, and has a phase shift at r = 2a, where Psi = 0 then becomes negative...and a phase shift of probabilities?
ReplyDelete"What does it mean for Psi(2r) to be negative?
DeleteIs this some kind of phase shift?"
yes, definitely I would call that a phase shift. On one side of the node it has one phase (positive), and on the other side it has another phase, negative.
"It's strange to think of this Psi(2r) form as a wave that is stationary, and has a phase shift at r = 2a, where Psi = 0 then becomes negative."
Well, true, but this is a first excited state, and, just like the first excited state of an infinite square well, it must have one node. All first excited states that i know of have one node. In the sq. well, the node is at L/2 (by symmetry); in this case the node is at r=2a. Actually there is a sphere of nodes, right?, with a radius of 2a (centered at zero). So the wave function is positive everywhere inside that sphere and negative outside it.
It is a stationary wave, but we have examples of other stationary waves with nodes (and every node is a phase shift boundary). For example, the 1st harmonic oscillation of a wave on a string between to fixed posts (guitar string), has a node in the center, and it is a stationary wave with a classic stationary wave oscillation.
And, also, the excited states of the infinite sq well, as mentioned before.
In higher dimensions there is also the classical example of the modes of drumheads, which have lines of nodes.
"…and a phase shift of probabilities?"
i don't think so…
Coming back to your original question (""What does it mean for Psi(2r) to be negative? Is this some kind of phase shift?" To answer what it means is a huge question that will take years to answer satisfactorily. Phase is a critical aspect of quantum mechanics.
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ReplyDelete*Edited from previous post* (No edit button?)
ReplyDeleteWhat is the significance of the coefficients preceding each of the states in problems 6 and 8? i.e sqrt(1/3)*Psi2r
Those are selected to create states that are are relevant to sp2 bonding. It is true that they are very specific and may seem arbitrary. In a way it would probably be preferable to allow people to make up their own states and explore their nature, but that sort of less structured approach is more difficult and tends to take more time than we generally have in a quarter system class.
DeleteOne way in which they are constrained is that their squares have to add up to 1. (for example, 1/3 + 2/3 =1)
I think they will start to make more sense after we finish and analyze this assignment.
Higher up in the comments, before the coefficients for #6 were corrected, it was posited that the term with the 1/sqrt(2) coefficient should be Psi2y. How would the problem change if the state had been 1/sqrt(3)Psi2r - 1/sqrt(6)2y + 1/sqrt(2)Psi2x ? The squares of the coefficients still add up to 1. Would it just change the expectation values of x and y?
Deleteyes, good point. it is another possible state, but its orientation would be different (and the exp values of x and y would be sort of reversed...)
DeleteThat makes a lot more sense after lecture today. Thanks!
DeleteI'm having trouble with number 7. I just don't understand where any information regarding time dependence or independence is found. Perhaps a hint?
ReplyDeleteConsider Psi(x,t) = Psi(x)e^-(iwt) where w=E/hbar. If you use this when calculating expectation values, taking the complex conjugate will lead to some nice cancellations. And remember, all the states we are dealing with in problems 5 and 6 are 1st excited states, aka have the same energy. Hope that helps!
DeleteWhen you calculate for one of the mixed state (i.e 5c),you should get something like (1/3)|Psi_2r|^2 + (2/3)|Psi_2x|^2 + cross term. Adding the full wave equation (Psi * e^(iwt)) the cross term will include the time dependent part. However, if the cross term of two wave function are in the same excited states, their energy will be the same, and since w is related to E, the cross term will be zero. So, if the two wave function are not in the same state, there will be a time dependent part due to different in energy.
Deletewhen you calculate [x_bar] (did not come out for some reason)
DeleteThese replies are really good i think. Maybe I would say "if the two wave functions were not of the same energy there would be time dependence".
DeleteSince you asked, and you have been thinking about it, I will give you a very direct answer. There is no time dependence in any of the expectation values for 5 and 6!
But it is very important for you to think about why and how to explain that (which requires a real understanding of where time dependence comes from in QM).
My thoughts are that the oscillation of (the expectation value of r=sqrt[(x^2)+(y^2)+(z^2)]) due to the time dependence associated with a mixed state would be important for an electron to make the actual physical transition between ground and excited states. Actually, this brings up a question. Do electrons transition into higher energy states with the same orientation about the origin, as in Psi2x to Psi3x? I recall in lecture talking about the polarization of the electromagnetic field of the incoming photon. Is that what determines the higher energy state that the electron will transition to?
DeleteIn number 8, when we plot the expectation values (x,y) for each of the two combinations of the second energy state, are we to interpenetrate this as a bonding angle? I know above it was mentioned that these combinations are crucial for pi bonding.
ReplyDeleteYes, but actually these combinations are crucial for sp2 sigma bonding. It is the leftover state that does the pi bonding.
DeleteIs anyone having trouble with the integral for expectation value on 5 (c)? I've arrived at the integral of (1-r/2a)*x^2*e^(-r/a)dx and can't solve it. Any suggestions?
ReplyDeleteRight now your integral is in kind of a mix of cartesian and polar coordinates. If you switch to polar coordinates (don't forget the Jacobian), you should be able to finish the integration. Does that help?
Deleteyou describe that very well. spot on.
DeleteIsn't r a spherical coordinate in this case (since it is radially dependent on 3 variables)? in which case you would need the jacobian to be r²cosθ
Deleter²sinθ, with our co-ordinate system choice (theta from zero to pi...
DeletePS. how did you make that theta and the superscript? I just copied yours.
the superscript is done by typing "²" followed by ; the theta is "θ" with ;. I had to look up the character's hex value to make it work so it's kind of inefficient.
DeleteAnyway, in the problem, transforming the integrand (substitute x = r cosϕ sinθ and multiply by the jacobian r²sinθ) gives (1-r/2a)e^(-r/a) rcos²ϕsin²θ r²sinθ gives an odd powered sinusoid term, which is 0 when integrated about its period and is multiplied by the whole integral, meaning the expectation of the hybridized wavefn is 0. This result seems wrong by inspection but I cant really understand where I went wrong mathematically.
Ah nevermind I figured it out, the theta is evaluated about a half period and therefore the odd powered term does not equal 0.
Delete@krobers: i had exactly the same concern when i first saw that sin^3 term. anyway, you figured t out. good.
DeleteAny chance on extra credit if we can find the mutually orthogonal set of wave functions that correspond to a tetrahedral bonding structure?
ReplyDeleteSounds good to me. Though I should say there are several natural ways to do that. (I think maybe one of them has all sqrt(1/4)'s with different signs, while others have some zero coefficients...(one state oriented along a principle axis, like we did for sp2).
Delete