Anyone wanting to get extra-credit toward an A, may work on Wilson's problem.
For extra credit toward a B, you could do the same sort of analysis for the matrix:
0, i hbar, o
-i hbar, 0, 0
0, 0, 0
which is the matrix of Lz in the basis: 100, 010, 001 ; which spans the 3-fold degenerate 1st excited state manifold of the 3D H.O.
These should be handed in in Sunday and should be be well-organized, clear and attractive. The key thing is to look deeply at what the eigenvectors are telling you about the nature of the spatial states. The credit begins with that part, and the clear, cogent discussion that accompanies it.
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I'm toying around with the matrix for B credit, and I get pretty much the exact same eigenvalues, eigenvectors, and eigenstates as for the 2D H.O.
ReplyDeleteI can definitely see how the corresponding eigenvectors produce certain mixed states that I believe to be eigenstates of the Lz operator.
For this case, it doesn't look like 001 plays any real role.
Are people getting similar results?
I mean, it's pretty much exactly like the 2D....unless I'm doing something gravely wrong.
Jon, what you are doing seems fine, but just so people are not confused by your comment, the essential thing is to interpret the eigen-vectors --to see what they are telling you about the spatial states. The interpretation can't really be the same as for the 2D H.O. because the basis of states is not exactly the same. One needs to discuss and understand the states. That would include, but not be limited to, understanding the similarities and differences between:
ReplyDeletethe 2D HO and the 3D HO, and also
similarities and differences between the excited states of H and the 3D HO, especially with regard to the angular states. You are not doing anything gravely wrong, but it is essential not to dismiss the most important aspects of the problem.
Here is an extreme example: when we study the H atom we find solutions involving Legendre polynomials. Legendre polynomials may also arise in studying modes of a drum or some E&M problems. But we don't say a hydrogen atom is just like a drum.
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