Timeline for Is there a good mathematical explanation for why orbital lengths in the periodic table are perfect squares doubled?
Current License: CC BY-SA 4.0
20 events
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| Mar 29, 2022 at 17:44 | comment | added | Dan Romik | For anyone who's interested, I posted a follow-up question (following a suggestion from Carlo Beenakker) covering aspects of this question that I feel were not covered by the answers to the current question. | |
| Mar 24, 2022 at 1:16 | vote | accept | Eugene Stern | ||
| Mar 22, 2022 at 13:31 | answer | added | Aaron Bergman | timeline score: 11 | |
| Mar 22, 2022 at 7:51 | history | edited | Daniele Tampieri | CC BY-SA 4.0 |
Minor grammar improvement
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| Mar 22, 2022 at 0:02 | comment | added | Dan Romik | A good explanation should also explain why the initial “2” appears only once and the subsequent numbers (8, 18, 32, …) each appear twice. | |
| Mar 21, 2022 at 19:31 | history | edited | YCor | CC BY-SA 4.0 |
formatting, added tag
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| Mar 21, 2022 at 19:30 | comment | added | Jojo | Here is a paper that I found doing this based on a quick search. (I remember the one I was looking at a while back was more readable than this one, so could be worth some more searching) researchgate.net/publication/… | |
| Mar 21, 2022 at 19:21 | comment | added | Jojo | Yeah it should be something like so(4) = so(3)×so(3) (at least, at the level of the algebra and maybe with su(2) instead of so(3)). Then one set of $|l,m\rangle$ spherical harmonics is for the rotational motion as in any radial potential, and there's another one for the radial motion in this case due to the augmented symmetry group? Although this should be something like $|n,q\rangle$, I'm not sure what the $q$ quantum number is about. I'm pretty sure I saw a paper doing this a while back | |
| Mar 21, 2022 at 19:06 | comment | added | Aaron Bergman | Right, and the answer to the question is the existence of an $n^2$ dimensional representation of so(4) as I recall. | |
| Mar 21, 2022 at 16:04 | comment | added | Michael Engelhardt | ... and it's also there quantum mechanically (one has to define the Runge-Lenz operator in a symmetrized fashion). | |
| Mar 21, 2022 at 14:40 | comment | added | Aaron Bergman | The $SO(4)$ symmetry is there classically. The extra conserved quantity is the Runge-Lenz vector. @johnbaez has written a bit about this, say, math.ucr.edu/home/baez/gravitational.html | |
| Mar 21, 2022 at 13:06 | history | became hot network question | |||
| Mar 21, 2022 at 12:14 | answer | added | Carlo Beenakker | timeline score: 29 | |
| Mar 20, 2022 at 23:52 | comment | added | Will Sawin | My understanding (from something I read elsewhere on MO) is that, when the Hamiltonian operator for the hydrogen atom is expressed in the Fourier basis, it is invariant under $SO(4)$ (viewing $\mathbb R^3$ as $S^3$ minus a point by spherical projection) and not just $ SO(3)$. These numbers should be the dimensions of irreps of $SO(3)$ (the spherical harmonics, specifically. | |
| Mar 20, 2022 at 23:40 | comment | added | Eugene Stern | The rows correspond to atomic numbers 1-2 (length=2), 3-10 (length=8), 11-18 (length=8), 19-36 (length=18), 37-54 (length=18), 55-86 (length=32), 87-118 (length=32). The two sections of length 14 that you're talking about are subsections of the two rows of length 32 (look at the atomic numbers and see). | |
| Mar 20, 2022 at 23:18 | comment | added | markvs | Section of which row? There are two rows of length $14$. | |
| Mar 20, 2022 at 23:02 | comment | added | Eugene Stern | The 14 elements are just a section of that row, the entire row is $18 + 14 = 32 = 16 \times 2$ elements. Each section of $14 = 7 \times 2$ elements corresponds to $V_6 \otimes W$, and the $7$ gets added on to $1 + 3 + 5$ to give you your next perfect square. | |
| Mar 20, 2022 at 22:27 | comment | added | markvs | What is $...$? How many rows are in the periodic table? The last two rows have $14$ elements each. $14$ is not $2$ times a square. | |
| S Mar 20, 2022 at 22:11 | review | First questions | |||
| Mar 20, 2022 at 22:13 | |||||
| S Mar 20, 2022 at 22:11 | history | asked | Eugene Stern | CC BY-SA 4.0 |