edits:
5:00 PM, Feb 19: [problem 3 clarified. 7 edited. 7 is interactive.]
11:00 AM, Feb 18: [problem 1 note added for clarity.]
Do all these problems seem worth your while? What is your favorite and least favorite? If you were to eliminate one problem, which one would it be?
[Kinetic energy.]
1. Consider a 2 square well system: 2 identical wells of width L spaced a distance S apart. If you put the origin of your coordinate system exactly between the wells, then the wave-function in that center region is of the form: A1cosh(x/a1) for the ground state and A2sinh(x/a2) for the 1st excited state. You are given that: A1=0.2(2L)1/2, A2=(54)A1 and a1=a2=0.2 nm, and the region separating the 2 wells is 0.4 nm wide.
a) Sketch the ground state and 1st excited state (over all regions). Pay particular attention to the center region and make your sketch reasonably consistent with the numbers given for the pre-factors and length scales.
b) For the ground state, calculate the contribution to the kinetic energy integral from just the center region (that is, integrating from x = -0.2 nm to +0.2 nm). (To make your result readable by the graders, and to you, please leave your answer in terms of A1 and a1 and a numerical result for the definite integral over cosh2(x/a1), which will have units of nm. (By the way, those nm will cancel the units of nm−1 from A21 .) There will, of course, be other constants that always are present for a K.E.. )
c) Do the same thing for the 1st-excited state. [Again, please leave your answer in terms of A1 and a1 and a numerical result for the definite integral.]
d) Summarize/discuss your results. What does this suggest about the overall K.E. for these states?
e) (extra credit) Plot the integrand in each case; show the area under it and discuss.
[States of multiple well systems.]
2. Consider a system of 4 equally-spaced identical wells. Suppose that for a single isolated well ψ1 is an energy eigenstate with energy E1.
When there are 4 wells, it turns out you can make 4 linearly independent, orthogonal energy eigenstates from the single well state ψ1. Using a notation where an electron in the left-most well is (1,0,0,0) and an electron in the state ψ1 in the right-most well is (0,0,0,1), one can write the ground state as approximately: 1√4(1,1,1,1)
a) Create 4 orthogonal states using this notation (with nodes that respect the symmetry of the system).
Let's call these states: ψ1,1, ψ1,2, ψ1,3 and ψ1,4.
Let's call their energies: E1,n, n=1 to 4.
b) Order these states according to their energies (lowest to highest).
c) Sketch a plot of each state. While this is just a sketch, be careful to keep the wave-fucntions close to zero in the regions between the wells.
d) How many nodes does each state have? Are they symmetrically located?
3. a) Continuing our exploration of a system of 4 equally-spaced identical wells. Do the same thing as the previous problem, but construct your states from the single-well state ψ2 (instead of from the ground state, ψ1). Does that make sense?
b) What are the lowest and highest energy states in this group of 4?
( Let's call these states: ψ2,1, ψ2,2, ψ2,3 and ψ2,4.
Let's call their energies: E2,n, n=1 to 4.)
b) What are the lowest and highest energy states in this group of 4?
[This problem may be more difficult than it looks.]
[Time dependence.]
4. Consider a 2 square well system: 2 identical wells of width L spaced a distance S apart.
a) Show that you can make a state where the electron wave-function is mostly on the left hand side at t=0 by using a superposition state that combines the ground state and 1st excited state of the double well. Include sketches. What is that superposition state?
b) Guess the expectation value of x for this state. Is it zero? Why or why not?
c) Factoring out an overall phase factor, e−iE1t/ℏ, show that there is a time, not too far from t=0, at which the phase relation between the states is reversed. What time is this? Guess the expectation value of x at this time?
d) As the wells get far apart, how does that effect the relationship between E_1 and E_2 ? How does that effect the time that you got in part c)?
e) How can the electron get from one side of the well to the other when the potential between the wells is higher than the electron energy (in either state). Discuss the relationship between: well proximity, E_1 and E_2, and how long it takes the electron to get from one side to the other.
5. For the 1D harmonic oscillator:
a) Calculate the expectation value of x2 for an electron in the superposition state that is an equal mix of ψ1 and ψ3.
b) Show in a graph how this oscillates as a function of time. What are the minimum and maximum values of ⟨x2⟩ ?
c) What does this imply about the kinetic and potential energy (expectation values) as a function of time for this state??? Discuss and graph.
[Selection rules.]
6. An oscillating electric field in the x direction produces an oscillating potential energy VE+M(→r,t)∝Exxcos(2πft). The strength of a transition between two eigenstates in the presence of this "perturbing potential" is proportional to the integral over all 3D space of the integrand ψ∗i(→r)(Exx)ψf(→r)∝ψ∗i(→r)(x)ψf(→r), where ψi(→r) and ψi(→r) are the initial and final states, respectively. [Note added: There is an x in that integral that comes from the direction of the electric field. Don't overlook the x.]
a) Given this information, which transitions from the ground state to a first-excited state are possible and which are forbidden? For each of the possible (allowed) transitions, what polarization of light do they require? (Polarization refers to the direction of the electric field in this case.)
b) Extra credit: Starting from the initial state ψ21x, which transitions to 2nd excited states of the form ψ30− and ψ31− are possible? (In other words, leaving out d states, i.e. states of the form ψ32−.)
c) extra credit: Starting from the initial state ψ200, which transitions to 2nd excited states of the form ψ30− and ψ31− are possible?
(These are dipole selection rules. They refer to transitions associated with an oscillating electric dipole moment, which are the strongest transitions.)
[Time dependence of a free electron.] This problem involves a superposition state for a free electron expressed as an integral. It is a wave-packet.
This is an interactive question. Some participation is required in order to provide the motivation to finish this complex but interesting problem.
7. In 1D, for an electron in a flat potential, U(x) =0, the energy eigenstates can be written as:{cos(kx), sin(kx)}, k= 0 to infinity.
a) Show that these are energy eigenstates and determine their energies.
b) sin(kx) and cos(kx) have the same energy, and it turns out to be convenient to combine them to form a new basis of states {e^ikx} k = -infinity to infinity. Show that these are momentum and energy eigenstates. How does their momentum and energy depend on their k value?
Although these states are not normalized, one can create a simple normalized state from a superposition of these states, e.g.,
ψ(x)=1(√2πa)1/2e−x2/(4a2)=∫+∞−∞ϕ(k)eikxdx,
where ϕ(k)∝e−k2a2. [In more detail, ϕ(k)=√a(2π)1/4e−k2a2.]
c) Plot ψ(x). Plot ϕ(k). (Two different plots, two different scales and axes.) What does ϕ(k) represent?
d) Important: How could you incorporate time dependence into this superposition state? In other words, without calculating anything, what would you put into the integrand of ψ(x)=∫+∞−∞ϕ(k)eikxdx to get an integral expression for Ψ(x,t)?
Now let's consider a different state: one with a k distribution centered at positive k. This will make a state that moves (to the right) as a function of time. Consider the same expression as above, but with ϕ(k)=√a(2π)1/4e−(k−ko)2a2, where ko is a positive number (corresponding to a positive momentum).
With this ϕ(k) the wave-function now has a new phase factor eikox. That is, ψ(x)=1(√2πa)1/2eikoxe−x2/(4a2). This wave-function (wave-packet) will move to the right as a function of time (at a speed ℏk/m). To show that we need to be able to put time dependence into this state in a systematic and correct manner.
e) What is the integral expression for ψ(x)? How would you put time dependence into the integral expression for ψ(x) to get an integral expression for Ψ(x,t)?
----
Let's stop here for now and we will put the rest of this problem on next week's HW. Does that sound like a good plan?
When we work on this further, we will define a unit-less quantity proportional to time,
T=ℏt2ma2. This makes the exponential time dependence more manageable.
...
Ψ(x,0)==∫+∞−∞ϕ(k)eikxdx,
This integrates to:
where
which can be written as:
The key thing to seeing and understanding the phenomenology of this moving wave-packet is to simplify the notation and parameters involved, e.g., by creating a unit-less time, T, and using variables that make sense to start with: a for length (the width of the packet at t=0 is proportional to a) and ko for the center of the packet...
e) Considering Psi(x,t), where is it centered at T=1? What is its width at T=1?
f) Discuss/describe the phenomenology of this wave packet (as a function of time).
I think i did number one and five correctly but I am not sure about number 3 and 4 and 6. We should go over these problems in class.It would help me out a lot so I may understand the material in this homework.
ReplyDeleteFor part b on number one: are we assuming that L (and therefore A1 and A2) is x dependent, or can we just factor that constant out?
ReplyDeleteYou can just factor it out. It is not x dependent.
DeleteI put some extra commentary into prob 1 to clarify.
DeleteI think we can do that.
ReplyDeleteRegarding problem 1, I put in some extra commentary to clarify that and also to specify a particular form for you to present your result. This is a problem that illustrates an important concept, but with what you are given, the calculation is not hard. It is important to organize and present your answer in the manner indicated.
ReplyDeletefor question 1, what is "the contribution to the kinetic energy integral" ?
ReplyDeleteWe'll cover that in class today and if anyone would like to comment on that here I would appreciate it very much!
DeleteQuestion 2.
ReplyDeleteI don't think the energy states that I wrote down from today's lecture are orthogonal, here is what I wrote (in increasing energy): (1,1,1,1), (-1,-1,1,1), (-1,1,-1,-1) and (-1,1,-1,1) {all with the sqrt(1/4) as well}. I'm dotting them with one another hoping to get zero but that doesn't work out. Perhaps I'm thinking of orthogonality in the wrong light here.
Do you never get zero, or sometimes and not others?
DeleteLet's make the question smaller and more specific. What do you get when you take the inner product of your 1st state with your 2nd state?
PS. Node locations should respect symmetry.
PPS. Sketch the states and look at them.
What you wrote in class is not the most important thing. Think about the process by which you create them.
DeleteDo you have a sense of how to create them? Can you make 4 states (from psi1) that seem reasonable?
I understand how to create them, I believe I see the general method in creating the states while respecting symmetry.
DeleteFollow up from class: We discussed how you can use the (1,0,0,0), (0,1,0,0) ... as a basis for this same space of R4.
DeleteSo the question becomes why do we use this basis of (1,1,1,1)(-1,-1,1,1) ... -> is it because it looks more interesting? Or is it because that's what we actually see in experiments? If those are the types of states we usually see electrons in, when in a 4-well system, is that the case because the (1,1,1,1) state is less energetic than the (1,0,0,0) state? That wouldn't make sense according to our logic here though, because (1,0,0,0) has zero nodes too! Right? It would peak on the first well and then exponentially decay throughout all the other wells.
One thing I didn't understand in Sp2, and Sp3 bases, which seems to have come back here, is this idea of: "the arbitrariness of bases". Can anyone speak to this please?
@ Alex I. That is a profound question. you would be surprised how much that ties into deep questions in physics. when do you use a localized basis, when do you not use a local basis? something to think about for a decade or so.
DeleteIn this particular case, which has lower energy? 1000 or 1111?
DeleteCan we talk about Alex's question in class tomorrow? and maybe talk some more about the orthogonality of these states?
DeleteThat kind of thing sometimes it is difficult to keep a class's attention. It is sort of subtle. People might doze off. Maybe you can discuss it more here?
DeleteInteresting, very interesting! So there isn't really an "arbitrariness" of choosing bases then? It does matter which one you choose.
DeleteI believe (1,0,0,0) has lower energy simply with our intuition that bendier, shorter wavelength states are higher in kinetic energy. I argue again that we can't use the idea from #1, regarding the less negative contribution of the sinh, since we have the same amount of nodes for both of these states (0).
Calculating an < E > =< T >+< U > would not be so unreasonable though, I think, because of the symmetry. I may try to whip that out after I finish 7.
note that 1111 has 3 cosh regions. remember that those tend to lower KE.
DeleteI don't think that 1000 really has any cosh regions, does it?
Oh no, just decaying exponentials on both sides for (1,0,0,0). Oh okay wow, I was a fool!
DeleteSo indeed, by the new found cosh rule, (1,1,1,1) is a lower energy state, and thus by the second law of thermo/entropy (if I'm allowed to say that word in quantum, I am unsure) we get that an electron would rather be in that lower energy/more interesting state across all four wells?
If this is correct, that's nice to see.
i'm not sure i understand. can you come to office hours today?
ReplyDeleteI think a superposition of states will force the expectation value (the one that corresponds to the excited state) to be nonzero. Correct me if I am wrong. In addition, an electron in a superposition of states will have a time dependent expectation value.
ReplyDeleteFor numbers 6 and 7, it seems to me that we only briefly covered the pertinent information in the past 2 lectures, and I am not really sure how to approach these questions. Are we going to talk more about that material on Thursday?
ReplyDeleteI think we have covered what you need for 6. Also, it is in the problem statement. Ask a question about it here for help. (More specific questions are easiest to respond to.)
DeleteRegarding 7, it is an interactive problem. How far can you get with it? Can you do parts a) and b)? How about d)? Let me know where you get stuck.
Problems 6 & 7 are very different. 6 is our last problem related to light-wave--electron-wave interaction. It sort of puts a bow on that subject and connects what we have done with what are called "selection rules". So I don't think we will cover that more in class, though you can ask about it here.
DeleteProblem 7, on the other hand, introduces something new -- a free electron wave packet moving in space. I am hoping to get an idea how much interest and engagement there is for that, because it is a pretty involved calculation and hard to follow and understand unless you are really interested and thinking about it. I think everyone should be able to do parts a) and b) before class tomorrow. If not, post a question here about parts a) or b) of 7. Maybe we should have quiz about that tomorrow?
I agree, my notes seem to be missing a lot of information needed to properly address these problems. I tried to look up the Selection Rules on my own, but they seem quite complicated, having never seen them before.
ReplyDelete6 is all about integrals being zero or not. which ones are zero? those are forbidden. I think that is all you need to know. Your job is to assess which integrals are zero.
DeleteI'm somewhat confused about creating the orthogonal states in number 2. What would qualify two states as orthogonal? Is it as simple as taking the inner products of two states and making sure that they sum to zero, or is there something else to it?
ReplyDeleteYes, it is as simple as that. (That is equivalent to doing the integral more or less.
DeleteDoes it make sense to assume that this applies to number 3 as well?
Deletefor 6a, to figure out which integrals are allowed and not allowed, do we basically compute for psi(1,0,0),psi(2,1,x) and psi(1,0,0)psi(2,1,x) and psi(1,0,0)psi(2,0,0) and figure out which are zero?
ReplyDeletefor 6b, how do we figure out what psi(3,1,x) etc are?
would psi(2,1,x) be allowed with psi(3,0,0) and psi(3,1,x) just like the lower states?
DeleteThis comment has been removed by the author.
DeleteThere is also an x in that integral. Include an x (or y or z, depending on polarization) and then the answer to your question is yes, basically you figure out which integrals are zero. Without the x, all the integrals would be zero.
DeleteI'll make b) extra credit.
anon "I have been working on the homework and i have finally got to number 7. I have no idea where to go on a and b. How would i check to see if it is an eigen state and evaluate its energy?
ReplyDeleteExcellent question. Use the Schrodinger eqn. (WIth U(x) = 0)
Hey Prof,
ReplyDeletethis homework was a bit tamer than last week's problem set.
Many of the questions were either straightforward or easy once I figured out what to do. That being said, I enjoyed the challenge of finding the pattern in numbers 2 and 3. The selection rules in problem 6 were also fun.
The hardest problem to me was number 7, although you cleared up much of my confusion in office hours.
Regarding 6B and 6C, do we need to know what the form of the second excited states are in order to determine which transitions from the first to the second excited states are possible?
ReplyDeleteYes, but you are only asked to do the s and p states. That is, psi300, 31x, 31y and 31z. Those are pretty similar to the s and p 1st excited states. Does that make sense?
Delete( the d states (psi32-) would be harder and extra extra credit.
Ok thank you
DeleteThis homework is very dense for the few notes we have on these problems. I still have not reached problem #7 and it's the day before homework is due. :/ :/ :/ :/ :/ :/ :/ :/ :/ :/ :/ :/ :/ :/ :/ :/ :/ :/ :/ :/ :/ :/ :/ :/ :/ :/ :/ :/ :/ :/ :/ :/ :/ :/ :/ I would like very much to have a good understanding of these problems but I cannot find helpful resources with google.
ReplyDeleteYou still have all of today to do problems... but I kind of agree. I am also taking a while to do problems compared to what was covered in class this past week.
DeleteHere is my take:
Deleteproblem: 1). new and interesting, but not difficult
2) and 3) new and interesting. closely related
4) and 5) same as last weeks problems (time dependence
6) do-able if you read the problem carefully. This type of integral has come up a lot in this class in various contexts, has it not?
7) new and challenging. difficult!
Having a good grasp of time dependence from last week, and integrals of the form x psiA psiB would probably help a lot overall.
I second this. I am having trouble understanding how to make superposition states. How different is making these states in a double well potential compared to the single well?
ReplyDeleteI got to problem 7 and am on part b. I am not sure where to go with this part and am a little confused if I even did part a correctly. Do we just show the 2nd derivatives to show that they are energy eigenstates? If so, what's the different in showing the energy eigenstate versus the momentum eigenstate for part b?
ReplyDeleteyes, that sounds right. One (E) is a 2nd derivative, the other (p) is a first derivative.
DeleteWhat do you mean by what's the difference?
This comment has been removed by the author.
ReplyDeleteyes, that's right.
ReplyDeleteWhat are we supposed to notice in problem 1? All I got out of it was that KE contribution was negative in the center. And we already established that a long time ago.
ReplyDeleteAlso cosh lowers KE more than sinh
Delete@"What are we supposed to notice in problem 1? All I got out of it was that KE contribution was negative in the center."
DeleteWhat are the implications and consequences of that? Hmmm, this might be a good question for a test.
If I'm not mistaken, today in class, we said that the fact that the wave-packet widens as time goes on is a manifestation of Heisenberg's uncertainty principle; I don't see how this is the case for two reasons:
ReplyDelete(1) ∆x∆p >= hbar/2 , ∆E∆t >= hbar/2 --> position is related to momentum, not time. (2) Why would allowing time to go by mean that we know it more precisely, thus meaning we do not know the converse (that is, energy, but apparently maybe position) as precisely.
Or maybe I misunderstood the claim – anyone see the connection here?
I hope I did not say that. In any case, I do not think that the spreading of the wave-packet is a manifestation of an uncertainty principal.
DeleteAh, well I sat in the back thursday – I probably just misheard then.
DeleteI know you said in class yesterday that a solution to the problem similar to #4 was online, but it isn't, and I'm still stuck because I didn't get it correct on last week's hw and there are no corrections on the homework assignment I got back.
ReplyDeleteI see. Yes, you are right, it was not posted. I just added it to the HW5 solution post. My apologies and thank you for persisting.
Delete@Monique
ReplyDeleteBy creating a superposition state we know the expectation value of x will be time dependent. We are asked to have this superposition state start off in the left well (t=0), so we essentially have to choose whether we are going to add or subtract psi2 from ps1. After you sketch it out (kinda depends on what you assume) you should be able to guess the expectation values, I definitely didn't do any calculations. n.B. I don't recommend factoring out the term Zack suggests, might just be me but the time dependence becomes less clear than just using Euler. Hope this helps.