Timeline for Why are electron orbitals shaped how they are?
Current License: CC BY-SA 4.0
16 events
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| Nov 4, 2023 at 18:35 | comment | added | Jagerber48 | @hft your argument convinces me that isosurface contours are surfaces of minimal volume for a fixed probability enclosed. If you want zero probability enclosed then you can pick the global maximum. If you want a little bit more probability while gaining minimal volume you should add a set of points with the highest possible probability. This will leave you with an isosurface contour. If you want to increase enclosed probability while minimizing volume you will greedily collect points with the highest probability so that you get minimal volume. This will leave you with isosurfaces. | |
| Nov 3, 2023 at 14:50 | history | edited | hft | CC BY-SA 4.0 |
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| Nov 3, 2023 at 14:48 | comment | added | hft | @GiorgioP-DoomsdayClockIsAt-90 Yeah, I'm not sure about that minimum volume condition either. I'm just giving an argument for why it is justified in certain simple cases. I'll update my answer to indicate this is not a general proof. | |
| Nov 3, 2023 at 6:32 | comment | added | GiorgioP-DoomsdayClockIsAt-90 | I am not sure the condition about the minimum volume is correct in general. The question was about orbitals, without further qualification. This means not only atomic but also molecular orbitals (not monotonic around any point) should be considered. | |
| Nov 2, 2023 at 18:27 | vote | accept | Atharv Rao | ||
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| Nov 2, 2023 at 16:03 | history | edited | hft | CC BY-SA 4.0 |
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| Nov 2, 2023 at 15:59 | comment | added | hft | @Jagerber48 It's not trivial, but I think there is a pretty straightforward argument to be made. Definitely easier to see if one just works with a one dimensional probability density rather than 3d. I updated my answer. | |
| Nov 2, 2023 at 15:57 | history | edited | hft | CC BY-SA 4.0 |
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| Nov 2, 2023 at 10:37 | comment | added | Jagerber48 | It's not obvious to me that the shapes we show are "the smallest region that gives us 90% of the probability density". Rather, the shapes we show are "the specific isosurface contour which encloses some chosen percentage of the probability density". Maybe there's a non-trivial (or tirvial?) theorem that the two are equivalent? See my answer. | |
| Nov 1, 2023 at 22:39 | comment | added | hft | @WillO I don't think people are ignoring "the question," per se. The title question is simply: "Why are electron orbitals shaped how they are?" This is likely what most answers (including mine) are focused on. | |
| Nov 1, 2023 at 22:22 | comment | added | WillO | Your update seems to me to be exactly what the question was asking for, while all of the other answers ignore the question entirely. Thank you. | |
| Nov 1, 2023 at 19:56 | history | edited | hft | CC BY-SA 4.0 |
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| Nov 1, 2023 at 19:51 | comment | added | hft | Hmm. Ok. Yeah, that's a strange/worse question, but I can address that... | |
| Nov 1, 2023 at 18:32 | comment | added | WillO | "The choice of "90%" is arbitrary" --- yes, but I think this misses the OP's point. There are multiple (overlapping) regions that satisfy the 90% criterion. The question is, when the standard pictures are drawn, what additional criterion is being used to choose among these? Perhaps your equations answer that question implicitly, but I think the OP is looking for a clear conceptual explanation of how one region is chosen over another. | |
| Nov 1, 2023 at 18:21 | history | edited | hft | CC BY-SA 4.0 |
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| Nov 1, 2023 at 18:08 | history | answered | hft | CC BY-SA 4.0 |