Dear Wilson,. Here is the matrix I promised you. I hope you like it as much as I do. I think it is very beautiful.
. This is the matrix of Lz in the sub-basis: 101, 011, 110, 200, 020, 002; which spans the 6-fold degenerate 2nd-excited state manifold of the 3-dimensional harmonic oscillator. We worked on our choice for basis order, and calculating the matrix, in office hours today. Our guiding aesthetic --that influenced and guided our choice of basis-state order, was connection. We wanted to put states which "connected" next to each other as much as possible. I imagine you are pretty excited right now. I would be too in your shoes.
. If you feel inspired to do so, please solve for the eigenstates and then write a cogent summary of the nature of the states, their organization and meaning. One thing about this matrix is, I believe, that there are two states with (Lz) e.v. of zero? Do you get that too? What does that mean? How can you handle those to get maximum insight? Are some of the eigenstates related to flower states in any way? These are some of the questions I would like to see addressed in your report. I trust that you will also come up with other questions and resolutions.
. I would like to also invite everyone in class to do this, if they like. It provides an excellent opportunity to develop and exhibit your understanding of matrices, degeneracy and angular momentum in QM. Ideally I would suggest you do this by Sunday --you can hand it in at the review section-- that way you will be able to focus on reviewing 1D QM, the hydrogen atom states and degeneracies, and related topics after the Sunday review section.
Best regards,
-Zack
To confirm your calc's Zach, I also get two eigenvectors with eigenvalue = 0.
ReplyDeleteThis comment has been removed by the author.
ReplyDeleteI think these 0 eigenvalues correspond to nodes. I get one eigenvector with a 0 eigenvalue for the first excited state
ReplyDeleteI don't think so.
ReplyDeleteHmmm. Well a matrix multiplied by any eigenvector that churns out zero is considered null space. I was thinking that a zero eigenvalue forced the Matrix/eigenvector product to be zero. If the eigenvector is scaled to zero under this operation, then doesn't this mean mean zero angular momentum about that vector?
ReplyDeleteIf this doesn't describe a node, then perhaps a forbidden eigenvector -----> forbidden eigenstate?
"null space." "forbidden eigenvector?"...
ReplyDeleteI appreciate your creativity and originality of thought, but i am not sure that this comment will be helpful to people trying to understand this problem.
One comment I would like to recommend is the very last comment in the string associated with the MAY 2 post "Interactive Problem #2,Lz eigenstates". People may not have seen that because it was the last one on that post. I think it introduces some key ideas (from John).
So, how does a zero eigenvalue describe angular momentum? I guess instead of r x p where there is strict dependence on position, and particle momentum, there is intrinsic angular momentum with eigenvector scaled to zero.
ReplyDeleteWell this is rotation in classical mechanics. But, in terms of the 2nd excited state of a 3D H.O., where there are two eigenvectors with zero eigenvalue, would this describe spin about those eigenvectors? I'm not well versed in this kind of thinking. I've just been trying think about this visually. Apparently it's not working very well:[
I get in phase flower states for their eigenvectors (eigenvalue = 0). From earlier calculations, it was shown that linear combinations of flower states are out of phase by 45 degrees, suggesting the eigenstates with non-zero eigenvalues are out of phase. To me, these describe the trajectory of ______.
ReplyDeleteDoes it make sense that in phase eigenstates describe the spin of ______?
Here is a link that describes a question we have about the formation of H-atom like states from the degenerate eigenvalues.
ReplyDeletehttp://mathbin.net/16673
Ethan, Jerome and Kelsey
This comment has been removed by the author.
ReplyDeleteThis comment has been removed by the author.
ReplyDeleteEthan, Jerome and Kelsey: Your state is nice, but i think it needs to be tweaked slightly. Can anyone see how?
ReplyDeleteI think where we started in our discussion was by noticing that the two 0 e.v. eigenstates (of the Wilson's matrix) could be combined to make an eigenstate that depending only on r (i.e., independent of theta and phi). So that is step one. Then the next step involves combining the same two eigenvectors to make a state orthogonal to that. Your state is close to that, but not quite. (You have to change the coefficients a little bit.)
Behind all this is the idea that we are moving toward (finding) Y_l,m states. The Y_l,m states are orthogonal* and they have certain symmetry propoerties we can recognize (like l=0 states depend only on r, l=1 states come in bunches of 3, ...)
* I guess this is why we look for a pure radial state and its orthogonal complement...
PS. Here was my first comments that i deleted:
1. Did I recommend that combination? What was i thinking? Looking at it now, in the cool light of day, I don't see why that is so special? Maybe it would be more logical to find a combination that depends only on r - not theta or phi-- and then the complementary state that is orthogonal to that. Does that make any more sense?
2.Oh is see. Maybe that is what your state is trying to be...
We are wondering why Zack recommended the combination of 3D HO states:
ReplyDelete002 - (020 + 200)
which are our degenerate eigenvectors associated with our L_z matrix.
-Ethan, Jerome and Kelsey
--I think Zack made a mistake. What he should have recommended was the state that is: 1) orthognal to 002 + 020 + 200
and, 2) can be made from a linear combination of the two 0 e.v. eigenvectors.
I think those 2 conditions specify a unique state.
I'm a little off about one of the eigenvectors with eigenvalue=0. It could have been many things, but it seems like we choose it to produce an eigenstate with z dependence. Is the purpose of this special choice to fulfill our spatial requirements (a desire for radial symmetry)?
ReplyDeletePsi_(n,l,m)(r,theta,phi) is being described using linear combinations of Harmonic oscillator states (basically a coordinate transformation from spherical to cartesian, but much more telling). If the hydrogen atom's zero eigenvalue describes intrinsic angular momentum, then that just screams radial symmetry.
So, the two e.v.'s with eigenvalue=0 for the 3D harmonic oscillator must combine to produce only r dependence if we are going to make the comparison to the hydrogen atom. Is this a good rationale for that eigenvector choice?
Jon--
ReplyDeleteI think trying to get r-dependence by combining the 2 states with Lz=0 will only yield half the answer. We want one state with L^2 = 0 (that'll be the one we construct which is radially symmetric), but we still have another state to form from these 2 Lz eigenstates. This new state won't have L^2 = 0, because it won't be radially symmetric. In fact, we need to construct it so that it is orthogonal to the radially-symmetric state. What you end up getting (I think) is something like r^2 - 3z^2, which seems related to the flower states (which are of the form a^2-b^2), and so perhaps it has L^2 similar to the 4 Lz eigenstates that have Lz nonzero, which all look like flower states.
I understand now what my issue was with the word "sub-basis" here (i tried to post this just after the review session but it wasn't working for some reason).
ReplyDeleteI thought the problem was saying that this matrix somehow forms a sub-basis *of* the 2nd excited manifold for 3D H.O., when in fact this matrix is the Lz matrix *of* the sub-basis (i.e. the 2nd excited state manifold) of the Hilbert space corresponding to this potential. From the Lz matrix of this space, we can form an "Lz space," whose basis vectors are given by the eigenvectors of the matrix.