1. Using a variational wave-function (like the one we used in class),
determine the size of an electron in a hydrogen atom, and in an He+
potential (2 protons). How are they different.? Why are they different? Graph U and T in each case and discuss their respective roles in influencing size. [Understanding U and T is the important part of this problem.]
2.
What are the 4 states of 2 spins? On what premise (based on symmetry)
could you divide them into a group of three states and another of just
one state?
3. For 2 electrons in a double well potential,
a)
write a spatial state (using on the states A and B that we discussed in
class Tuesday) that goes with the spin state
\(\frac{1}{\sqrt{2}}(\uparrow \downarrow - \downarrow \uparrow)\) .
[Hint: try starting with:
\(\psi_{A'} =
\frac{1}{\sqrt{1+\delta^2}}\psi_A +
\frac{\delta}{\sqrt{1+\delta^2}}\psi_B \)
and making a 2 electron state
that "respects the symmetry associated with the indistinguishablility of
electrons.]
b) write another spatial state that goes with the spin state \(\frac{1}{\sqrt{2}}(\uparrow \downarrow + \downarrow \uparrow)\).
c) Why do these spatial states turn out to be different?
d) In what way are they different? That is, how does that difference manifest itself?
e) Discuss the consequences of that difference?
4. a) Show that the kinetic energy for an electron in an
infinite-square-well energy-eigenstate has zero uncertainty.
b) Calculate the kinetic energy for an electron in an infinite square
energy eigenstate.
5. Calculate the expectation value of the kinetic energy of a (Gaussian) free electron wave-packet. (Do that at t=0 to make it easier.)
6. Calculate the kinetic energy expectation value for an electron in the ground state of:
a) an infinite square well
b) a harmonic oscillator
7. Calculate the potential energy expectation value for an electron in the ground state of:
a) an infinite square well
b) a harmonic oscillator
8. Do not ignore HW problems related to a finite square well. What is the energy of an electron in a ground state of an infinite well that is 3 nm wide. Approximately what is the energy of an electron in a ground state of an finite well that is 3 nm wide (and has several other bound states).
9. Review the 1st excited states of hydrogen. Do a calculation that shows where the maximum of (\Psi_{21x}\) is located. Review hybridization possibilities for the 1st excited states of H. What are the essential things that make hybridization interesting; how do they work?
Would you consider posting answers (if not solutions), so that we can make sure we are doing these problems right?
ReplyDeleteI haven't done these problems. (I think there may be enough reference material in this blog to give you a pretty good idea.) Please feel free to present some of your results with questions, where you feel the need.
DeleteFor number 5 are we supposed to just apply the K.E. operator to the wave function, or integrate over all space with the operator (the expectation value) even though you don't specify "expectation value," I feel like they would be close if not the same answer.
ReplyDeleteThey are not the same (in this case). Expectation value is the one that makes sense to calculate with this state. (I made that more clear.) Thanks.
DeleteYour question brings up an interesting point. When are those things equivalent and when are they different? I will greatly appreciate whoever answers that here.)
DeleteMy guess is they are only equivalent for the infinite square well?
DeleteWell that is a good guess, and guessing is good, but no, there are many other circumstances in which they may be equivalent.
DeleteInteresting, and off topic, but I noticed there is nothing in this prep about n-p junctions (semi-conductors), is it safe to say these will not be covered on the exam?
DeleteWhich psi function did you use? I tried it with the one we used in class on 2/20, that is psi =
Deletee^ikx e^-((K-K0)^2 a^2) and some constants and the second derivative was huge. Can I just drop the e^ikx factor since I am multiplying by psi star, even though it is inside the derivative? Or is there a simpler psi I can use?
Yeah, I dropped out the e^ikx factor although I didn't get rid of the k^2 you get from it's double derivative, so I was left with the standard answer for K.E. (hbar^2a^2/2m) but also with a k^2 factor, which I hope is unit-less, or if not then it shouldn't be there.
DeleteYour answer seems like it could be right to me, but I don't understand why you can just pull that e^ikx factor out of the double derivative.
DeleteWhat do you think, professor? Can we do the problem like this?
DeleteLike what? Describe step-by-step what you are proposing. Carefully and clearly.
DeleteWe start with the Gaussian that we used in class on 2/20, that is psi = e^ikx e^-((K-K0)^2 a^2) and some constants. We then take the second derivative of this. However this derivative will be quite unmanageable if we take it as is (and I think we will end up with an i in the answer), so we somehow manage to pull the e^ikx factor out of the derivative, since we know it will cancel when we multiply in the psi star later on. This is the part I'm confused about... I know you can't just arbitrarily pull things out of derivatives, but I am unsure of how else to approach this problem. Evan seems to have taken the 2nd derivative of the e^ikx factor separately in his comment above, but I am not sure if that is allowed either.
Deleteare you taking the 2nd derivative with respect to x? (you didn't say, but I assumed that.) Why would that be unmanageable? how would you end up with an i?
DeleteSorry, I made a mistake. The function I gave you was the one we had before we integrated with respect to K. After integrating, we got e^ikx e^-(x^2 / 4a^2) , which you said is a gaussian. This is the function that produces the unmanageable 2nd derivative (when taken with respect to x) that includes an i, and this is what I meant to talk about in my last comment. Is there a better form of a gaussian we can use, or is there some way of dealing with the unruly 2nd derivative?
Deleteit wasn't terribly hard taking the second derivative, its taking the integral of e^(2ikx-x^2/4a^2),and the integral of that times x and x^2. Wolfram alpha comes up with an error function when i try and do that...
DeleteYeah, that's what I mean. Taking the derivative itself isn't that bad, its just that then you're left with this massive function that includes an i to integrate, and I don't know how to do that.
Delete... and then the psi star will get rid of the ikx and it will be a really easy integral.......
Deletei is not a variable, it is a constant it shouldn't effect anything regarding the integral's difficulty
DeleteMaybe I'm doing something wrong, but here is what I get for psi star times the second derivative of psi (with constants left out):
Deletehttp://www.wolframalpha.com/input/?i=%28e^-ikx+e^-%28x^2+%2f+%284a^2%29%29%29%28second+derivative+e^ikx+e^-%28x^2+%2f+%284a^2%29%29%29+
And wolfram will not integrate the result. What is your second derivative?
using the chain rule quite a lot, i get things like -(1/(2a^2))*(x^2)*e^(ikx)*e^(-x^2/4a^2)
Deleteand +(1/(4a^4))*(x^2)*e^(ikx)*e^(-x^2/4a^2).... etc (not exactly correct i am just giving you the general idea). multiply it by psi star and the e^ikx goes away and e^(-x^2/4a^2) becomes e^(-x^2/2a^2) etc. my answer was (hbar ^2 *Knaught^2 / 2ma^2) which sounds correct to me the only real integrals you have to do are e^(-x^2/2a^2) , (x)e^(-x^2/2a^2) (x^2)e^(-x^2/2a^2).
This conversation has progressed a lot since this question was asked, but I'd like to take a crack at the question of when the kinetic energy operator and expectation value are equivalent. The operator will only work correctly if the function's kinetic energy is well-defined; it has to have no uncertainty in that observable (in this case, K.E.). By definition, the wave-function must be an eigenfunction of the operator, with the eigenvalue being the well defined value. If this is the case, the expectation value must be the same, being an "average" value of a single value. In this case, I don't think the kinetic energy is well-defined because the Fourier transform of the wave-packet implies a continuous range of momenta; although each constituent e^ikx is an eigenfunction, i believe the integration over all k values takes away this property for the whole. Particularly at t = 0, where the position is fairly localized, the momentum and kinetic energy cannot be well-defined. As the packet spreads out, it may have a momentum with less uncertainty, but I don't think it will be well defined, although I'm not sure for the limiting case.
Delete@Anonymous 12:41 AM,
DeleteI agree. That seems like a cogent and excellent analysis of both the general concept and the application to this specific case.
@anonymous 12:41 AM Are you Patrick? If not then who are you? (You can send me an email if you want to maintain you anonymity here.)
Deleteyes thats me
DeleteI believe this sounds like number 5, but when working out the problem, does it give us 0?
DeletePatrick you're a boss! Thanks, I have a way better grip on the situation after that argument.
DeleteI'm still not 100% though; I would like to know why you think, "although each constituent e^ikx is an eigenfunction, I believe the integration over all k values takes away this property for the whole." Are you leaning on intuition here? I would have leaned the other way – that because all the k components are eigenfunctions, the sum of them would be too...
And this I argue against: "Particularly at t = 0, where the position is fairly localized, the momentum and kinetic energy cannot be well-defined." It seems here you might be trying to say localized position implies unlocalized momentum implies unlocalized KE – that last bit isn't true! Or did I misunderstand the point? I am pretty sure that we showed on the review day that for the infinite square well, for instance, we have a kinetic energy eigenfunction (which by definition means no uncertainty in KE) but then ∆p = hbar*k = nonzero...which I believe was to show you that while KE is proportional to ∆p^2 it does not necessarily mean it ∆KE is proportional to it.
Okay I am starting to feel VERY iffy about that last bit...any clarifying thoughts on this would be helpful.
After some thinking, I would like to retract the third paragraph of that last post. For the infinite square well, I find psi to be a momentum eigenstate and thus there should be no momentum uncertainty: < p > = -i (hbar) d/dx (sin(kx)) = -i (hbar) *k sin(kx). So the e-value is -i(hbar)k.
DeleteBut then think about this: we know, with certainty, that the electron is in the finite square well. So we can estimate it has a position uncertainty of ∆x = L (width of the well).
What would Heisenberg say here? We have a finite uncertainty in x and 0 uncertainty in p. Something is amiss, and I'm not sure what. It may just be that infinite square wells have their limits as models – a drawback of being so ideal & simple.
Incase anyone is still looking I retract the last thing too! It's not a momentum e state because derivative of sin cos not sin!!! Sorry my mistake!
DeleteDo you think you could post clarification notes of your notes on Problem 1 from class? In the class notes, you said that the expectation value of Kinetic Energy will reduce and give us h(bar)^2 over 2ml^2. However, I, as well as other students I've talked to, do not get this answer when doing the integral using del squared. It seems that in your notes, you only took the 2nd derivative of the variable psi function as opposed to using the del operator.
ReplyDeleteWhat are you getting when you take del^2 of e&{-r/l}?
DeleteAre you using the Jacobian in your integral?
Who are these other students?
What answer do they get? :)
Your question is interesting. Here is why.
DeleteThere is a higher level way of thinking about this. There is only one length scale in this problem (l) as far as I can see. (Can anyone come up with another?) In a quantum problem with a single length scale, since the KE has to have units of energy, I don't think it is possible to get anything other than hbar^2/(m length^2) times some unit-less number. This is a sweeping generalization and I could be wrong, but that is what I think.
For that reason, even without doing the problem, I feel confident about the form of the result. If you said the 2 was wrong, I would be more worried. The "2" could be wrong; but I don't think the form can be different.
Never mind. After some deliberation with the problem I managed to get the correct answer.
DeleteHmmm, well thats strange because I got KE to be hbar^(2)/(2ml^2) using the del squared. Maybe you made a mistake somewhere?
ReplyDeleteFor problem 2, I understand that (up up), (down down), and (up down)+(down up) are part of the spin 1 triplet, and (up down-down up) is spin 0 singlet, but why? What is the difference? Also I understand that the first arrow represents the first electron, so does that mean that a state (up down)+(down up) and a state (up down)+(up down) are different?
ReplyDeleteI'm not sure about spin 1 or spin 0 – but one thing is for sure: the spin triplet is even! So this means to build a fermion (odd) state, you'd want to use one of those triplet spin states with an odd spatial state. The odd singlet, however, needs to be with an even spatial state. The states are on the spin video, which was really helpful for this.
DeleteI'm interested in what people have to say about your second question.
Very clear explanation of the symmetry issue!
DeleteRegarding Sean's 2nd question:
(up down) means electron 1 up and electron 2 down.
(down up) means electron 1 down and electron 2 up.
"Also I understand that the first arrow represents the first electron, so does that mean that a state (up down)+(down up) and a state (up down)+(up down) are different?"
Yes!, they are different.
(Also you probably would not see a state written as: "(up down)+(up down)" since that is 2(up down).
@Alex: your summary of the basis and change of basis in the spin video post is right on and also helps clarify all this.
Deletewhich wave function are you referring to in number 1?
ReplyDeletepropto e^(-r/l), where l is a variable length scale.
DeleteAre you asking for the radius of the "orbit" of the electron, or the classical radius of the electron itself? I'm hoping its the former since I have no idea how to calculate the latter.
ReplyDeletenone of the above. an electron in the H atom ground state has no orbit.
DeleteFor problem #1, I have calculated the length scale L to have the same value as the bohr radius a. How do we go about calculating the size of the electron from this? I assumed that we calculated Δx, and in this I get = L^2, and = 0 by symmetry. sqrt( - ^2) = sqrt(L^2) = L. Is this what you wanted..?
ReplyDeleteFair enough. nothing more to do on that. And for the helium nucleus?
DeleteI should have asked also for you to graph U and T separately and discuss them -their roles is influencing electron size. I'll add that in.
There are two protons (Z = 2) ,the potential would be twice as strong so L will be half as big if my calculations are correct. yes?
DeleteThat's what I got too.
DeleteFor number 8, part 2, when you ask about the energy in a finite potential well, do we need more information than just the length to solve this? Like the height of the well maybe? I have in my notes that the wave functions for the finite well require several constants as "given" information in order to make a valid wave function. The length is certainly necessary, but also in Homework #2 question 5, you give us the whole wave function, and define "a" as .27 nm…It seems to me that we need this kind of information in order to tell what the energy is in the finite well. Am I wrong? If so, is there another way to construct the wave functions from the information that is given (just the length)? Or am I totally missing something else…?
ReplyDeletewhat you say is true.
DeleteWhat the probably says is: "approximately"...
or in other words, what can you infer from what you are given.
is the answer absolutely nothing, or is there something to be inferred?
Does that make sense??
Ohh, yeah that makes sense. It should be a little less than the energy from the infinite well, correct?
DeleteSo to clarify, if you were to ask us to calculate this quantitatively on the final you would have to give us the extra information in order to properly construct the wave functions, yes?
I agree with you on that Christopher: the finite well would require some serious calculations of expectation values, with a given value of a.
DeleteApproximately, we can say one thing for sure (the crown jewel of the midterm): the finite well will have a lower kinetic energy because there's less confinement. Not only that, the total energies are lower as well. That's not as easy to see why though. Killa Whale told me we solved this in a homework once (I'm not sure which one, I have to go back tonight).
What can we make of the fact that this finite square well has "several other bound states"? – this is the question. I think it means that this is not a low potential well – it's high enough to support quite a few higher energy bound states. And if that's the case, we can say it's pretty close to infinitely high --> thus the energy is less than but pretty close to what we got for the infinite square well: .04eV (that's what I got at least).
I hadn't refreshed the page since 6:53, so really I was talking to the former Chris Powell!
DeleteAwesome, thanks for the response! That was pretty much my reasoning, so I'm glad I had the right idea. But you make a good point about the height of the well- the more states it supports then the higher (or deeper…) the well must be, and thus the energy of the infinite well will be a closer approximation. Thanks for adding some dialogue Alex.
DeleteO.K. If anybody is up studying or Professor if you see this before the final... for number 4. showing that the uncertainty in the K.E. is zero. Since the potential energy in the the infinite well is zero and we're assuming we're in an energy eigenstate which means there is a quantized energy and all the energy is kinetic, I think it makes sense that the uncertainty would be zero. But how do we go about showing this. Is it sqrt { - }? We would be taking something like a 4th derivative then for ke^2. Is it just manipulating the Schroedinger equation? Anybody have any ideas?
ReplyDeleteWell, just because I was curious, I took 4 derivatives of the ground state wave function for the infinite square well and then calculated the expectation value for {KE^2} (I'm using curly brackets here since blogspot won't allow the angle brackets). Then I calculated {KE}. Finally, I said that the KE uncertainty = sqrt( {KE^2} - {KE}^2) and I did get 0..... I don't know if the KE uncertainty is defined that way.
DeleteI think we are supose to draw the well then specifically draw the energy and lable it. then write E = KE + U then U = 0 in the well. Now E=KE+0 this shows that KE is well defined because total energy is labled.
DeleteIn TA office hours there was a discussion about how one could do this out mathematically and one could conceptually argue it (or both). You can also use the expectation of momentum and momentum squared to solve it, since it's related to KE.
DeleteWhoops that should be sqrt( ke^2 - ke) using the expectation values. Guess it doesn't like angle brackets.
ReplyDeleteI know this isn't really helpful at all, but what you said about a quantized energy Eigenstate is all that I could come up with myself. Quantitatively it makes sense to me to go from there to the assumption that the uncertainty would be zero…but I don't actually know how to show it numerically. Once again sorry I didn't help, I just had the same issue. Hopefully someone else has a better idea.
ReplyDeletesorry this is late, but for # 1. in class we said that the derivative of dE(L)/dL is zero. Could anyone explain to me why this is?
ReplyDeletealso, how is energy a function of the length scale? in my notes it says to put our calculated value of L into the energy function to get the energy scale.
We set the derivative equal to zero because we are trying to find the minimum of the Energy as a function of L. In other words, we are solving for the L that will make the energy minimal.
DeleteEnergy is a function of the length scale because the only variable that is in the solved Schrodinger equation is L. All the other values are constants of some sort. Thus, it is E(l). Substitute L for "a" in Psi_100 and go from there.
Does that help?