This problem is designed to help with the transition from 1D QM to dealing with issues that commonly arise in 2- and 3-dimensional QM. This will help you to really understand angular momentum and degeneracy issues when they arise in more complex systems.
As with IP1, the concept and format of this problem, is for you to work on it both individually and collectively, using this site to discuss the problem, what it means, how to approach it, how to solve it, what the solution means, and so on. And then to turn in your own written version of it at the end of the week, which should include a pedagogic description and discussion of your results.
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IP 2.a) Calculate the matrix of the operator Lz in the basis of the 1st 6 energy eigenstates of the symmetric 2D harmonic oscillator. Can you use this matrix to find Lz eigenstates? If so, how? What do you learn from this? Explain. Discuss.****
*****see May 7 comment below
(Why the 1st 6 states?; why not 7 or 5?)
b) Are these Lz eigenstates also energy eigenstates? What energy manifold does each one belong to? Describe the nature and "content" of the degeneracy manifolds of this system. (A nuanced disussion of a is more important than b.)
c) Building on your results from this problem, what other problems can you think of? What are some possible things, related to angular momentum, that one might do next?
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The Lz operator, x p_y - y p_x, is an operator that acts on a continuous function. How do we change this into a matrix from an ordinary operator? Also, the basis spoken of - those are all continous functions as well. Any state written in terms of them seem like they'd just be a number of functions, multiplied by coeffients, added up. How could we put this into the typical eigenvector form, with the seperated values along an orthonormal basis? Why would we do this?
ReplyDeleteThe operator Lz is complex and the eigenfunctions of the harmonic oscillator are real aren't they? How can you write one in terms of the other? Unless this is not the question you are asking in which case perhaps someone can clarify what IS being asked in part a?
ReplyDeleteIs the reason we are using the first 6 eigenstates degeneracy? Since
ReplyDeletePsi_00 has no degeneracy (1 state)
Psi_01 and Psi_10 are degenerate
Psi_11 Psi_20 and Psi_02 are degenerate.
I'm assuming this is what you mean by the first 6, and not the first six non-degenerate states:
Psi_30 Psi_03 Psi_12 Psi_21
Psi_40 Psi_04 Psi_13 Psi_31 Psi_22
Psi_50 Psi_05 Psi_41 Psi_14 Psi_32 Psi_23
If that's what you mean, then those listed as "degenerate" would be energy eigenstates (in our problem we would have 3), but I don't understand how to put any of this in a matrix.
ReplyDeletesame, how to put it in a matrix form??
ReplyDeleteI posted some comments on this problem at
ReplyDeletehttp://mathbin.net/11511
Let me know what y'all think.
Hi Jerome
ReplyDeleteYour math post looks nice. I got the same equation and I see your point, our object has 4 indices's and I believe it would make it a 4th rank tensor or something of that nature. We can always assign to indices a specific value like m=0 and l=1 and generate a matrix of n,k. But i'm stuck as well as you how to order them or pick specific matrix. If you want to work on it together you can email me at rickyfernan@yaoo.com
Jerome,
ReplyDeleteI think your post refers to a different sort of problem. You are both going down a very dangerous path.
I would encourage you to adopt (indeed, embrace) a "flattened" basis. Although you start out with a basis indexed by two integers, |n,m > . I think that if you want to have a “regular” 2 index Hermitian matrix (you do), you must choose an order for your basis. Like today, we started out with:
|0,0> = 1
|0,1> = 2
|1,0> = 3
and then you could say:
|1,1> = 4
and so on.
If you don't do that you will have a “4-dimensional” object (that is, each element is specified by 4 indexes, rather than the usual 2 indexes). That is bad, in the sense that you will get nowhere with that. It sounds cool, but it is not.
Getting out of nested, multi-dimensional tensor spaces like this is so important, that it is the motivation for a entire class of operations in Mathematica called “Flatten” (written by very wise people).
Friends don't let friends operate in 4-dimensional tensor spaces!
With the first 6 basis vectors, which encompass all the eigenvectors of the first 3 degeneracy manifolds, you then get a 6 by 6 matrix. Because there are a lot of zero elements, you will be able to find all the eigenvectors and eigenvalues of this matrix; actually finding eigenvectors and eigenvalues is the goal here.
After you find the eigenvectors and figure out what they mean, (what they are telling you about the form of the Lz eigenstates), you will see, in retrospect, that the order you chose does not matter.
So today we worked out that what defines the matrix is the fact that when the Lz operator is used on an energy eigenstate (say nx=1,ny=0), it switches it around (say to nx=0,ny=1), at which point, if you're to multiply your vector by the matrix, the result will be the appropriate vector, the one that you should have gotten transforming it with the Lz operator. So with above's example, if (nx=1,ny=0)=(0,1,0,0,0,0) and (nx=0,ny=1)=(0,0,1,0,0,0) than your matrix needs to turn the first one into the latter. I THINK this is right, and when I go to figure out the Lz eigenstate, I can get a few, and I know something symetric is going on with the matrix, but I don't know what it is yet. What do you guys think?
ReplyDeleteThis comment has been removed by the author.
ReplyDeleteI get a determinate like:
ReplyDelete0 0 0 0 0 0
0 0 -1 0 0 0
0 1 0 0 0 0
0 0 0 0 sqrt(2) -sqrt(2)
0 0 0 -sqrt(2) 0 0
0 0 0 sqrt(2) 0 0
are other people getting this? My order was Psi_00, 01, 10, 11, 02, 20.
Any thoughts on part b after we find the eigenvectors/eigenvalues for the above 6X6 with mathematica?
You mean "matrix" i think (rather than determinant).
ReplyDeleteRegarding: "Can you use this matrix to find Lz eigenstates? If so, how? What do you learn from this?"
Don't stop when you get the eigenvector of the matrix. What do they mean?
I would like you to figure out how to find a way to express these Lz matrix eigenstates as a function of x and y.
just want to check,
ReplyDeletefor
<01|Lz|11>=0
right?wrong?
alright i guess what i'm really confused about is
ReplyDeletewhat Lz|11>=?
i'm getting |20-02> but then what? or does that even make any sense?
is that where you are getting the sqrt(2) wilson?
Lz|11> is equal to |20> - |02>. depending on what Psi you have on the left, it could be 0 by orthogonality or it would be sqrt(2).
ReplyDeleteremember, that the raising/lowering operators have that sqrt term after they operate:
a+|1> = sqrt(1+1)|2>.
Wilson--my matrix looks identical to yours, and I used the same order, except that my signs are all switched--did you factor out hbar*I or -hbar*I (I factored out just hbar*I).
ReplyDeleteNora--I also got <01|Lz|11>=0, since the 11 either goes to 20 or 02 (20 for ax+*ay and 02 for ax*ay+), and then we use orthogonality.
forgot about that minus....thx
ReplyDelete"does that even make any sense?"
ReplyDeleteNora, What your eqn reflects is that Lz is the sum of two terms, one of which lowers x and raises y, while the other raises the x index and lowers the y index. What could make more sense that that?
okay, i got it, thanks, does make sense, just needed to hear it a different way i guess.
ReplyDeletei've got the same matrix as wilson and mike now.
i'm confused about finding the momentum eigenstates. or are we suppose to find the φ eigenstates? are we suppose to solve for the eigenvalue from Lzφ=aφ where a is the eigenvalue?
ReplyDeletei feel like i'm just missing something really simple here again.
"Can you use this matrix to find Lz eigenstates? If so, how? What do you learn from this? Explain. Discuss."
ReplyDeleteThis problem is all about finding Lz eigenstates. First you find the eigenstates of the matrix, right? Technically those are Lz eigenstates (one representation of them).
****But if you stop there, you won't get nearly to the interesting or most important part. Use those relatively simple vector results as a starting point for further exploration. After you have done this math and found these "6 vectors" (6 of them): what does(do) it (they) mean? What do they look like? what are they trying to tell you (about themselves)?
Imagine Heisenberg has done the 1D H.O. system, with the a+ and a's, and that you are the first one to apply this to look at the 2d H.O. You want to share your result with the world, (at a presentation of "the academy"). You want to present your results with nuance, insight, explaining their meaning in a mathematical and intuitive manner. For many people, intuition works best in x,y space. Can you get back to that? Can you tell a story?
extra credit: Can you make a mixed state that exhibits circular motion? I think the member of the academy might like that.
In response to Zack (may 7 5:01pm)
ReplyDeleteusing the definition of an eigenstate, all i can really think of is that these Lz eigenstates have unchanging angular momentum w.r.t. time; they are "standing waves" of angular momentum. not sure what else to say here, any ideas???
Assuming I did the math right, when the angular momentum eigenstates are written out in the basis of the energy eigenstates they have both imaginary and real components. If these componenets are considered seperately (as in each angular momentum eigenstate is considered the sum of an imaginary vector and a real vector, instead of the vector being the sum of an imaginary and a real component) then the angular momentum eigenstates are degenerate within each energy level.
ReplyDeleteI don't know if this implies something cool or if I have just been staring at this matrix for too long. Does anyone have some insight to offer?
just some things i'm thinking:
ReplyDelete-the Lz matrix is symmetric times -1
-the eigenvectors lie in the xy plane, rotating about z-axis
-wouldn't we see a circle or ellipse?
i haven't calculated the energy eigenstates so i don't really know about the real and imaginary parts.
These Lz eigenvectors must also be energy eigenstates. Yes they are a linear combination but as long as the linear combination only contains eigenstates of a certain ENERGY, then the superposition is too an energy eigenstate. These Lz eigenstates are necessarily superpositions of the energy eigenstates.
ReplyDeleteCheck out this applet of a 2D quantum harmonic oscillator. Whenever you make a combination of states with the same energy, you'll get something that looks like some "angular momentum" but certainly not in the classical sense.
http://www.falstad.com/qm2dosc
This comment has been removed by the author.
ReplyDeletePhysics party tomorrow in the library?
ReplyDeleteWhat time?
ReplyDeleteI don't see how one would get a root 2 for any of the elements.
ReplyDelete<02|Lz|11> = <02|20-02> = <02|20> - <02|02> = -1
when you use the raising and lowering operators remember to include the normalizing coefficients, that is ...
ReplyDeletea+|Psi_n> = sqrt(n+1)|Psi_n+1>
a|Psi_n> = sqrt(n)|Psi_n-1>
Nevermind.
ReplyDeleteP.S. I don't know how to erase comments.
the little trash can at the bottom of your comment lol
ReplyDeleteI can't get much extra meaning out of the eigenvectors. Sure, there seems to be two sets of two and two other randomish vectors, which is cute, but I don't see how this might have enhanced my understanding. Maybe my linear algebra is weak.
ReplyDeleteThe eigenvalues seem pretty cool though, (0,+-hbar,+-2hbar), since we have the first 3 ("nondegenerate") energy states, and if we went up another energy state we'd probably get +-3hbar. Which seems telling.
Maybe I am just imagining things, but what if you try mixing them back together in and focus on the spatial symmetries you can get with re-combinations of them that make real states? That might give you a new perspective on the Lz eigenstates themselves. Also, what about making circular motion? Can you do that?
ReplyDeleteZack, here is an idea. Justen mentioned there there are two sets of Lz eigenstates. I think what he means is that there are two sets of two Lz such that one of the eigenstates is teh complex conjugate of the other. The four Lz eigenstates that form these two sets can be written as a sum of two of the energy eigenstates, one part imaginary and one part real. Therefore if I add together these two states, I get a single energy eigenstates (if I normalize it). What is interesting about this is that these angular momentum eigenstates that form the sets have equal and opposite angular momentum eigenvalues. Perhaps then the energy eigenstate can be thought of as the superposition of two states "spinning" in opposite directions with the same angular momentum. So the net angular momentum expectation value of any of the energy eigenstates will be zero, kind of like momentum. But if we take the expectation value of Lz^2, we will get a finite nonzero answer, also like momentum. This seems exactly analogous to how we said an energy eigenstate can be thought of as two standing waves traveling in opposite directions, only here we have two waves orbiting in opposite directions.
ReplyDeleteIf we think about it this way we might gain some insight into how to make a state that resembles circular orbit (perhaps it is analogous to how we created a state that had a net expectation value for momentum out of the sum of two different states that had zero expectation value for momentum).
Does that seem right of wrong to anyone?