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when toggle format what by license comment
Dec 19, 2021 at 20:24 history edited anon.jpg CC BY-SA 4.0
pointing out one of the more obvious errors in my question lmao
Dec 17, 2021 at 5:44 history edited Qmechanic CC BY-SA 4.0
edited tags; edited title
Dec 17, 2021 at 3:22 vote accept anon.jpg
Dec 17, 2021 at 3:21 answer added anon.jpg timeline score: 0
Dec 17, 2021 at 3:00 history tweeted twitter.com/StackPhysics/status/1471676613054062592
Dec 17, 2021 at 2:42 vote accept anon.jpg
Dec 17, 2021 at 3:00
Dec 17, 2021 at 1:42 answer added knzhou timeline score: 4
Dec 17, 2021 at 1:31 comment added anon.jpg @knzhou sowwy it's actually $ 2 L \cdot S$. Here is a ref
Dec 17, 2021 at 1:28 history edited anon.jpg CC BY-SA 4.0
added 3 characters in body
Dec 17, 2021 at 1:27 comment added anon.jpg @ZeroTheHero I got that, but can you elaborate why the multiplication goes the way it does? Because Py and Px for example do not act on the same space; that's why I was saying 2d and 3d and so on; for example Py could be seen as L and Px as S and Pz as... I dunno, the radial coordinate. I assume it has something to do with the trace or something as in the linked question
Dec 17, 2021 at 0:57 comment added knzhou Can you give a reference for $J^2 = L^2 + S^2 + L \cdot S$?
Dec 17, 2021 at 0:28 comment added ZeroTheHero $L$ and $S$ act in different spaces, whereas $p$ and $A$ act on the same space.
Dec 16, 2021 at 23:27 history edited Connor Behan CC BY-SA 4.0
Close bracket
Dec 16, 2021 at 22:56 comment added anon.jpg btw there is a "similar" issue in here . But I don't think this is exactly what I'm looking for; the J situation deals with a space of only two Hilbert spaces since Lx, Ly, Lz are all in the same H space, while in the magnetic-field case we're dealing with a 3 dimensional system where Px , Py, Pz live on separate spaces.
S Dec 16, 2021 at 22:46 review First questions
Dec 16, 2021 at 22:46
S Dec 16, 2021 at 22:46 history asked anon.jpg CC BY-SA 4.0